TS Inter 1st Year

Telangana TSBIE TS Inter 1st Year Accountancy Study Material 3rd Lesson Subsidiary Books Textbook Questions and Answers.

TS Inter 1st Year Accountancy Study Materia 3rd Lesson Subsidiary Books

Short Answer Questions:

Question 1.
Explain various types of Subsidiary Books.
Answer:
All day – to – day transactions are first recorded in the journal. In a small business unit, it is possible to record all business transactions in one book. But in a big organisation where the transactions are large in number, it is c ficult to record all business transactions in one place.

Hence, in big organisations, there is need to sub divide journal into special journals. Each journal is meant for recording the transactions of separate category and of repetitive nature. These special journals are called ‘Subsidiary Books’ or Day Books.

Types of Subsidiary Books:
Keep in view the nature and need of the business transactions, the subsidiary books are classified as follows.

1.Purchases Book :
Only credit purchases of goods are recorded in this book. Cash purchases and purchase of assets are not recorded in this book. The recording of entries in this book is based on the invoices and bills received by the firm from the suppliers.

2. Sales Book :
The credit sale of the goods are recorded in sales book. Cash sales and sale of assets are not recorded in this book. The recording of entries in this book is based on the outward invoices or bills prepared by the trader on the customer.

3. Purchase Returns Book :
Sometimes, goods purchased are returned to the suppliers for various reasons such as the goods are not of the required quality or defective, not matching to the specifications, price difference etc., such return of goods is recorded in the purchase returns book. For every return, a debit note is prepared and send it to the supplier.

4. Sales Return Book :
This book is used to record the return of the goods by the customers. Sometimes, customers return a part of the goods purchased from the trader. On receipt of goods from the customer, credit note is prepared.

5. Cash Book :
The cash book is maintained to record all cash transactions. All cash receipts are debited and all cash payments are credited in this book.

6. Bills Receivable Book :
The bill on which the amount is yet to be received and promissory note drawn by the seller or creditor are recorded in bills receivable book.

7. Bills Payable Book:
All bills and promissory note accepted by the buyer or debtor are recorded in the bills payable book.

8. Journal Proper:
This book is used to record only those transactions which cannot be recorded in any of the above mentioned subsidiary books.

Question 2.
State advantages of subsidiary books.
Answer:
The following are the advantages of subsidiary books.

1. No need of writing journal entries. The transactions are entered directly into their respective journals.
2. Ledger accounts can be prepared on the basis of subsidiary books.
3. Recording of transactions is very fast and easy. .
4. By entrusting different subsidiary books to different persons, division of work principle can be implemented.
5. Accounting work will be done efficiently by allotting work to different experts who prepare the special books.
6. Labour involved and time can be saved.
7. Since separate books are maintained to record a particular set of transactions, errors can be easily detected.
8. Information relating to similar types of transactions will be available at one place.
   E.g. Relating to sales, purchases, cash etc.

Question 3.
Explain about purchases and sales book draw the formats.
Answer:
Purchases book :
In this book, all credit purchases of goods are recorded i.e. cash purchases of goods and purchase of assets are not recorded.
The recording of entries in this book is based on the invoices or bills received by the firm from the suppliers.

Proforma of purchases :

Sales Book:
The credit sale of the goods are recorded in sales book. Cash sales and sale of assets are not recorded in this book. The recording of entries in the book is based on the outward invoices or bills prepared by the trader on the customer.

Question 4.
What is Journal Proper ? Write any five items which are recorded in Journal Proper.
Answer:
All business transactions are recorded in different subsidiary books according to their nature. But there are certain transactions which occassionally happen and these are not recorded in any of the subsidiary books.

However, they are recorded in a special book known as ‘Journal Proper’. That is, the transactions which are not recorded is the respective subsidiary books are recorded in the journal proper.

The following transactions are recorded in the journal proper.

1. Opening entries :
Opening entries are used to record at the beginning of the financial year to open the books by recording assets, liabilities and capital appearing in the Balance Sheet.

2. Closing entries :
Closing entries are used at the end of the accounting year for closing all accounts relating to expenses and revenues. These accounts are closed by transfer-ring their balances to Trading account and Profit and Loss account.

3. Purchase and sale of assets on credit:
Every organisation purchase the assets or sell the assets either on cash basis or on credit basis. When they are purchased or sold on credit basis, they are recorded in journal proper.

4. Rectification entries :
While recording the transactions in the books of accounts, some errors may arise due to negligence of the clerk or due to ignorance of fundamental principle of accounts. In such cases profit or loss cannot be determined correctly. So, necessary rectification entries are passed in journal proper.

5. Adjustment entries :
There are certain incomes and expenses are not recorded even at the date of closing the accounts. If they are not taken into account, correct profit or loss cannot be ascertained. So, necessary adjustment entries with regarding to those expenses and incomes are recorded in journal proper.

6. Transfer entries :
These entries in the journal proper for transferring an item from one account to another account.
Ex : Transfer of profit to reserve fund, transfer of drawings a/c to capital a/c.

7. In addition to the above, some transactions like loss of goods due to fire, Dishonour of cheques or bills of exchange, Goods sent on consignment are recorded in journal proper.

Very Short Answer Questions:

Question 1.
What is Debit Note ?
Answer:

1. It is a document sent to the supplier while returning the goods purchased on credit from him.
2. It intimating that his account is debited to the extent of goods returned and the reasons for returning them. It is prepared by the purchaser.

Question 2.
What is Credit Note ?
Answer:

1. It is a note (document)prepared and sent to the customer to inform that his account is credited with the amount of the goods returned by him.
2. It is a common to make it in red ink. It contain name and address of customs, value of goods returned and reasons for return of goods.

Question 3.
What is Invoice ?
Answer:

1. It is a statement sent by the supplier along with the goods or in advance, to the trader, stating that the goods are supplied along with the price, discount offered and other terms and conditions.
2. The document is called ‘Inward Invoice”, as it is received by the trader from the supplier.

Question 4.
Trade Discount.
Answer:

1. The rebate offered by the supplier on the catalogue price is known as trade discount.
2. If trade discount is given, it is to be deducted from the purchase price and only net amount is to be recorded in the books.

Question 5.
Journal Proper.
Answer:

1. Journal proper is also known as “Journal Residual” or “General Journal”.
2. This book is used to record only those transactions which cannot recorded in purchase book or sales book or cash book, purchase returns or sales return book, Bills Receivable or Bills Payable book.
3. E.g. When a firm purchased machinery on credit.

Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B The Plane Important Questions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1B The Plane Important Questions

Question 1.
Reduce the equation x + 2y – 3z – 6 = 0 of the plane to the normal form. [Mar. ’19 (TS); Mar. ’16 (AP & TS), ’14, ’01; May ’13]
Solution:
Given the equation of the plane is,
x + 2y – 3z – 6 = 0
x + 2y – 3z = 6
On dividing both sides by

∴ which is the normal form.

Question 2.
Find the equation of the plane whose intercepts on X, Y, Z axes are 1, 2, 4 respectively. [May ’14; Mar ’10]
Solution:
Given that,
x-intercept, a = 1
y-intercept, b = 2
z-intercept, c = 4
∴ The equation of the plane is

4x + 2y + z = 4

Question 3.
Find the directions of the normal to the plane x + 2y + 2z – 4 = 0. [May ’15 (TS); Mar. ’13; May ’12]
Solution:
Given, equation of the plane is x + 2y + 2z – 4 = 0
Directions ratio’s of the normal to the plane are (a, b, c) = (1, 2, 2).
Directions cosines of the normal

Question 4.
Write the equation of the plane 4x – 4y + 2z + 5 = 0 in the intercept form. [Mar. ’19 (AP); Mar. ’12]
Solution:
Given, equation of the plane is 4x – 4y + 2z + 5 = 0
4x – 4y + 2z = -5
\(\frac{4 x}{-5}-\frac{4 y}{-5}+\frac{2 z}{-5}=1\)
\(\frac{x}{\frac{-5}{4}}-\frac{y}{\frac{-5}{4}}+\frac{z}{\frac{-5}{2}}=1\)
which is the intercept form.

Question 5.
Find the angle between the planes x + 2y + 2z – 5 = 0 and 3x + 3y + 2z – 8 = 0. [Mar. ’17, ’15 (TS), ’09 ; May ’15 (AP)]
Solution:
Given equations of the planes are
x + 2y + 2z – 5 = 0 ……….(1)
3x + 3y + 2z – 8 = 0 ……..(2)
Comparing (1) with a₁x + b₁y + c₁z + d₁ = 0, we get
a₁ = 1, b₁ = 2, c₁ = 2, d₁ = -5
Comparing (2) with a₂x + b₂y + c₂z + d₂ = 0, we get
a₂ = 3, b₂ = 3, c₂ = 2, d₂ = -8
If ‘θ’ is the angle between the planes (1) & (2) then

Question 6.
Find the equation of the plane passing through (1, 1, 1) and parallel to the plane x + 2y + 3z – 7 = 0. [May ’13 (Old), ’11, ’10, ’09]
Solution:
Given, the equation of the plane is
x + 2y + 3z – 7 = 0 ……….(1)
Let the given point P = (1, 1, 1)
The equation of any plane parallel to x + 2y + 3z – 7 = 0 is x + 2y + 3z + k = 0
If equation (1) passes through the point p(1, 1, 1), then
1 + 2(1) + 3(1) + k = 0
1 + 2 + 3 + k = 0
6 + k = 0
k = -6
Substituting the value of ‘k’ in equation (1)
∴ The equation of the required plane is x + 2y + 3z – 6 = 0

Question 7.
Find the equation of the plane bisecting the line segment joining (2, 0, 6) and (-6, 2, 4) and perpendicular to it.
Solution:
Let A(2, 0, 6), B(-6, 2, 4) are the given points

Directions of \(\overline{\mathrm{AB}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (-6 – 2, 2 – 0, 4 – 6)
= (-8, 2, -2)
= (a, b, c)
Since the plane is ⊥ to \(\overline{\mathrm{AB}}\) then the line \(\overline{\mathrm{AB}}\) is normal to the plane.
Also, ‘C'(-2, 1, 5) lies in the plane.
∴ The equation of the plane passing through C(-2, 1, 5) & having direction ratios of the normal to the plane is (-8, 2, -2) is a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
⇒ -8(x + 2) + 2(y – 1) – 2(z – 5) = 0
⇒ -8x – 16 + 2y – 2 – 2z + 10 = 0
⇒ -8x + 2y – 2x – 8 = 0
⇒ 4x – y + z + 4 = 0

Question 8.
Find the equation of the plane through the points (2, 2, -1), (3, 4, 2), (7, 0, 6). [May ’00]
Solution:
Let the given points are
A(x₁, y₁, z₁) = (2, 2, -1)
B(x₂, y₂, z₂) = (3, 4, 2)
C(x₃, y₃, z₃) = (7, 0, 6)
The equation of the plane passing through the points A(2, 2, -1), B(3, 4, 2), C(7, 0, 6) is

(x – 2)(14 + 6) – (y – 2)(7 – 15) + (z + 1)(-2 – 10) = 0
(x – 2)(20) – (y – 2)(-8) + (z + 1)(-12) = 0
(x – 2)(5) – (y – 2)(-2) + (z + 1)(-3) = 0
5x – 10 + 2y – 4 – 3z – 3 = 0
5x + 2y – 3z – 17 = 0

Question 9.
A plane meets the coordinate axes in A, B, C. If the centroid of ∆ABC is (a, b, c) show that the equation to the plane is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=3\). [Mar. ’01]
Solution:
Let the equation of the plane be
\(\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1\) ……..(1)

Plane (1) cuts the X-axis, Y-axis & Z-axis at A, B, C respectively.
Then, A = (p, 0, 0), B = (0, q, 0), C = (0, 0, r)
The centroid of triangle ABC is

Given that, the centroid of triangle ABC is G = (a, b, c)
∴ (a, b, c) = \(\left(\frac{\mathrm{p}}{3}, \frac{\mathrm{q}}{3}, \frac{\mathrm{r}}{3}\right)\)
a = \(\frac{p}{3}\), b = \(\frac{q}{3}\), c = \(\frac{r}{3}\)
p = 3a, q = 3b, r = 3c
∴ The equation to the required plane is from (1)

Some More Maths 1B The Plane Important Questions

Question 10.
Find the angle between the planes, 2x – y + z = 6 and x + y + 2z = 7. [Mar. ’11; B.P.]
Solution:
Given equations of the plane are
2x – y + z = 6 ……..(1)
x + y + 2z = 7 ……..(2)
Comparing (1) with a₁x + b₁y + c₁z + d₁ = 0, we get
a₁ = 2, b₁ = -1, c₁ = 1, d₁ = -6
Comparing (2) with a₂x + b₂y + c₂z + d₂ = 0, we get
a₂ = 1, b₂ = 1, c₂ = 2, d₂ = -7
If ‘θ’ is the angle between the planes (1) & (2) then

Question 11.
Find the equation of the plane through the point (α, β, γ) and parallel to the plane ax + by + cz = 0.
Solution:
Given the equation of the plane is
ax + by + cz = 0 ………(1)
Let the given point p = (α, β, γ)
∴ The equation of any plane parallel to ax + by + cz = 0 is ax + by + cz + k = 0
If equation (1) passes through the point p(α, β, γ) then
a(α) + b(β) + c(γ) + k = 0
⇒ k = -aα – bβ – cγ
Substituting the value of ‘k’ in equation (1)
∴ The equation of the required plane is ax + by + cz – aα – bβ – cγ = 0
a(x – α) + b(y – β) + c(z – γ) = 0

Question 12.
Show that the plane through (1, 1, 1), (1, -1, 1), and (-7, -3, -5) is parallel to Y-axis.
Solution:
Let the given points are
A(x₁, y₁, z₁) = (1, 1, 1)
B(x₂, y₂, z₂) = (1, -1, 1)
C(x₃, y₃, z₃) = (-7, -3, -5)
The equation of the plane passing through the points A(1, 1, 1), B(1, -1, 1), C(-7, -3, -5) is

(x – 1)(12 – 0) – (y – 1)(0 – 0) + (z – 1)(0 – 16) = 0
(x – 1)(12) – 0 + (z – 1)(-16) = 0
12x – 12 – 16z + 16 = 0
12x – 16z + 4 = 0
3x – 4z + 1 = 0
Since the coefficient of y is zero, the plane is perpendicular to xz-plane and hence the plane is parallel to the y-axis.

Question 13.
Find the equation to the plane through the points (0, -1, -1), (4, 5, 1) and (3, 9, 4).
Solution:
Let the given points are
A(x₁, y₁, z₁) = (0, -1, -1)
B(x₂, y₂, z₂) = (4, 5, 1)
C(x₃, y₃, z₃) = (3, 9, 4)
The equation of the plane passing through the points A(0, -1, -1), B(4, 5, 1), C(3, 9, 4) is


x(30 – 20) – (y + 1)(20 – 6) + (z + 1)(40 – 18) = 0
x(10) – (y + 1)(14) + (z + 1)(22) = 0
10x – 14y – 14 + 22z + 22 = 0
10x – 14y + 22z + 8 = 0
5x – 7y + 11z + 4 = 0

Question 14.
Find the equation of the plane if the foot of the perpendicular from the origin to the plane is (1, 3, -5).
Solution:
Let O(0, 0, 0) be the origin.
P(1, 3, -5) is the foot of the perpendicular.
The plane passing through P and is perpendicular to the line segment \(\overline{\mathrm{OP}}\) is normal to the plane.
Directions of \(\overline{\mathrm{OP}}\) = (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (1 – 0, 3 – 0, -5 – 0)
= (1, 3, -5)
∴ The equation of the plane passing through P(1, 3, -5) having direction ratios of the normal to the plane are (1, 3, -5) is a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
⇒ 1(x – 1) + 3(y – 3) – 5(z + 5) = 0
⇒ x – 1 + 3y – 9 – 5z – 25 = 0
⇒ x + 3y – 5z – 35 = 0
⇒ x + 3y – 5z = 35

Question 15.
Find the equation of the plane if the foot of the perpendicular from the origin to the plane is (2, 3, -5).
Solution:

Let O(0, 0, 0) be the origin.
P(2, 3, -5) is the foot of the perpendicular.
The plane passing through P is perpendicular to \(\overline{\mathrm{OP}}\) the line segment \(\overline{\mathrm{OP}}\) is normal to the plane.
Directions of \(\overline{\mathrm{OP}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (2 – 0, 3 – 0, -5 – 0)
= (2, 3, -5)
The equation of the plane passing through P(2, 3, -5) having direction ratios of the normal to the plane are (2, 3, -5) is a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
2(x – 2) + 3(y – 3) + (-5)(z + 5) = 0
2x – 4 + 3y – 9 – 10 – 5z = 0
2x + 3y – 5z = 0

Question 16.
Find the intercepts of the plane 4x + 3y – 2z + 2 = 0 on the coordinate axes. [Mar. ’18 (AP & TS)]
Solution:
Given, the equation of the plane is
4x + 3y – 2z + 2 = 0
4x + 3y – 2z = -2

x-intercept, a = \(\frac{-1}{2}\)
y-intercept, b = \(\frac{-2}{3}\)
z-intercept, c = 1

Question 17.
Find the equation of the plane passing through (-2, 1, 3) and having (3, -5, 4) as directions of its normal.
Solution:
Let the given point (x₁, y₁, z₁) = (-2, 1, 3)
Direction ratio’s of the normal to the plane are (a, b, c) = (3, -5, 4)
∴ The equation of the plane passing through the point (-2, 1, 3) and having directions of (3, -5, 4) is a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
3(x + 2) + (-5)(y – 1) + 4(z – 3) = 0
3x + 6 – 5y + 5 + 4z – 12 = 0
3x – 5y + 4z – 1 = 0

Question 18.
Find the equation of the plane passing through (2, 3, 4) and perpendicular to the X-axis.
Solution:
If the plane is perpendicular to X-axis then the plane is parallel to yz-plane.
∴ The equation of a plane parallel to yz axis plane is x = k ……….(1)
If plane (1) passes through the point (2, 3, 4) then
2 = k
∴ k = 2
Substituting the value of k in equation (1)
∴ The equation of the required plane is x = 2.

Question 19.
Find the equation to the plane parallel to the zx-plane and passing through (0, 4, 4).
Solution:
The equation of the plane parallel to zx plane is y = k ……..(1)
If plane (1) passes through point (0, 4, 4) then
4 = k
∴ k = 4
Substituting ‘k’ in equation (1), y = 4
The equation of the required plane is y = 4.

Question 20.
Find the equation of the plane through (-1, 6, 2) and perpendicular to the join of (1, 2, 3) and (-2, 3, 4).
Solution:
Let A(1, 2, 3), B(-2, 3, 4) are the given points
Directions of \(\overline{\mathrm{AB}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (-2 – 1, 3 – 2, 4 – 3)
= (-3, 1, 1)
Directions of normal to the plane are (a, b, c) = (-3, 1, 1)
The equation of the plane passing through (-1, 6, 2) and having directions of normal to the plane are (-3, 1, 1) is a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
-3(x + 1) + 1(y – 6) + 1(z – 2) = 0
-3x – 3 + y – 6 + z – 2 = 0
-3x + y + z – 11 = 0
3x – y – z + 11 = 0

Students must practice these TS Intermediate Maths 1B Solutions Chapter 3 Straight Lines Ex 3(a) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Straight Lines Solutions Exercise 3(a)

I.
Question 1.
Find the slopes of the lines x + y = 0 and x – y = 0. (V.S.A.Q.)
Answer:
Slope of x + y = 0 is – \(\frac{a}{b}\) = – 1
Slope of x – y = 0 is 1

Question 2.
Find the equation of the line containing the points (2, -3 ) and (0, – 3 ). (V.S.A.Q.)
Answer:
Equation of the line is \(\frac{y-y_1}{y_1-y_2}=\frac{x-x_1}{x_1-x_2}\)
⇒ \(\frac{y+3}{-3+3}=\frac{x-2}{2-0}\) ⇒ \(\frac{y+3}{0}=\frac{x-2}{2}\)
⇒ 2y + 6 = 0 ⇒ y + 3 = 0

Question 3.
Find the equation of the line containing the points (1, 2) and (1, – 2). (V.S.A.Q.)
Answer:
Equation of the line is \(\frac{y-y_1}{y_1-y_2}=\frac{x-x_1}{x_1-x_2}\)
⇒ \(\frac{y-2}{2+2}=\frac{x-1}{1-1}\)
⇒ (y – 2) (0) = (x – 1) 4
⇒ x – 1 = 0

Question 4.
Find the angle which the straight line y = √3x – 4 makes with the Y-axis. (V.S.A.Q.)
Answer:
Equation of the line is y = √3x – 4
Slope m = √3 = tan\(\frac{\pi}{3}\)
Angle made with X- axis = \(\frac{\pi}{3}\)
∴ Angle made with Y – axis = \(\frac{\pi}{2}-\frac{\pi}{3}\) = \(\frac{\pi}{6}\)

Question 5.
Write the equation of reflection of the line x = 1 in the Y-axis. (V.S.A.Q.)
Answer:
Equation of AB is x = 1. Reflection about Y – axis is x = – 1 i.e., x + 1 = 0

Question 6.
Find the condition for the points ( a, 0 ), (h, k ) and ( 0, b ) where ab * 0, to be collinear. (V.S.A.Q.)
Answer:
Let A (a, 0), B (h, k) and C (0, b) are collinear
Slope of AB = \(\frac{\mathrm{k}-0}{\mathrm{~h}-\mathrm{a}}\)
Slope of AC = \(\frac{0-b}{a-0}\) = \(\frac{-b}{a}\)
Since A,B,C are collinear
slope of AB = slope of AC
⇒ \(\frac{k}{h-a}=\frac{-b}{a}\)
⇒ ak = – bh + ab
⇒ ak + bh = ab
⇒ \(\frac{h}{a}+\frac{k}{b}\) = 1

Question 7.
Write the equations of the straight lines parallel to X-axis and
(i) at a distance of 3 units above the X-axis and
(ii) at a distance of 4 units below the X-axis. (V.S.A.Q.)
Answer:
(i) Equation of straight line parallel to X-axis which is at a distance of 3 units above the X – axis is y = 3 ⇒ y – 3 = 0
(ii) Equation of straight line parallel to X-axis which is at a distance of 4 units below the X- axis is y = -4 ⇒ y + 4 = 0

Question 8.
Write the equations of the straight lines parallel to Y-axis and
(i) at a distance of 2 units from the Y-axis to the right of it
(ii) at a distance of 5 units from the Y-axis to the left of it. (V.S.A.Q.)
Answer:
Equation of straight line parallel to Y – axis and
(i) At a distance of 2 units from the Y – axis to the right of it is x = 2 or x – 2 = 0

(ii) Equation of the straight line parallel to Y- axis and at a distance of 5 units from the Y-axis to the left of it is x = – 5 or

II.
Question 1.
Find the slopes of the straight lines passing through the following pairs of points
(i) (-3, 8), (10, 5)
(ii) (3, 4), (7, -6)
(iii) (8, 1), (-1, 7)
(iv) ( – P, q ), ( q, – P), ( pq ≠ 0 ) (S.A.Q.)
Answer:
(i) (- 3, 8) ( 10, 5)
Slope = \(\frac{y_1-y_2}{x_1-x_2}=\frac{8-5}{-3-10}=\frac{3}{-13}\)

(ii) (3, 4), (7, – 6)
Slope = \(\frac{4+6}{3-7}=\frac{10}{-4}\)

(ii) (8 ,1), (-1, 7)
Slope = \(\frac{1-7}{8+1}=\frac{-6}{9}=\frac{-2}{3}\)

(iv) ( -P, q ), ( q, – P )
Slope = \(\frac{q+p}{-p-q}=\frac{q+p}{-(p+q)}\) = – 1

Question 2.
Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2. (S.A.Q.)
Answer:
Slope = \(\frac{y_1-y_2}{x_1-x_2}=\frac{5-3}{2-x}\)
Given that slope is ‘2’
∴ \(\frac{2}{2-x}\) = 2 ⇒ 2 – x = 1 ⇒ x = 2 – 1 = 1 ⇒ x = 1

Question 3.
Find the value of y, if the line joining the points ( 3, y ) and ( 2, 7 ) is parallel to the line joining the points (-1, 4) and (0, 6). (Mar. 14) (S.A.Q.)
Answer:
Let A = ( 3, y ), B = ( 2, 7 ), C = ( -1, 4 ) and D = (0, 6) are given points.
m₁ = Slope of AB = \(\frac{\mathrm{y}-7}{3-2}\) = y – 7
m₂ = Slope of CD = \(\frac{4-6}{-1-0}=\frac{-2}{-1}\) = 2
AB and CD are parallel ⇒ Slopes are equal
∴ m₁ = m₂ ⇒ y – 7 = 2
⇒ y = 9

Question 4.
Find the slopes of the lines (i) parallel to and (IQ perpendicular to the line passing through (6, 3) and ( -4, 5 ). (S.A.Q.)
Answer:
Let A = (6, 3), B = (- 4, 5) be the given points.
Slope of AB = \(\) = m
(i) Slope of the line parallel to AB = \(\frac{3-5}{6+4}=\frac{-2}{10}=-\frac{1}{5}\)
(ii) Slope of the line perpendicular to AB = – \(\frac{1}{m}\) = 5

Question 5.
Find the equations of the straight lines which make the following angles with the positive X – axis in the positive direction and which passes through the points given below.
(i) \(\frac{\pi}{4}\) and ( 0, 0 ).
(ii) \(\frac{\pi}{3}\) and (1, 2)
(iii) 135° and (3, -2)
(iv) 150° and (-2, -1)
Answer:
(i) \(\frac{\pi}{4}\) and (0, 0).
m = Slope = tan 45° = 1
Equation of the line is y – y₁ = m(x – x₁)
∴ Equation of the line passing through
(0, 0) with slope ‘1’ is
y – 0 = 1 (x – 0) ⇒ x – y = 0

(ii) \(\frac{\pi}{3}\) and (1, 2 )
m = tan \(\frac{\pi}{3}\) = √3 and equation of line passing through (1, 2) with slope √3 is
y – 2 = √3 (x – 1) ⇒ √3x – y + (2 – √3) = 0

(iii) 135° and (3,-2)
m = tan 135° = tan (180° – 45°) = – tan 45° = – 1
∴ Equation of line passing through (3, – 2) with -1 as slope is
y + 2 = -1 (x – 3)
⇒ y + 2 = – x + 3
⇒ x + y – 1 = 0

(iv) 150° and (-2, -1)
m = tan 150° = tan(180° – 30°)
= – tan 30° = – \(\frac{1}{\sqrt{3}}\)
∴ Equation of line passing through (- 2, – 1) with slope – \(\frac{1}{\sqrt{3}}\) is
y + 1 = – \(\frac{1}{\sqrt{3}}\) (x + 2)
⇒ √3y + √3 = – x – 2
⇒ x + √3y + (2+ √3) = 0

Question 6.
Find the equations of the straight lines passing through the origin and making equal angles with the coordinate axes. (S.A.Q.)
Answer:

Case (i): AA¹ makes an angle 45° with the positive X – axis m = tan 45° = 1
AA¹ passes through O (0, 0)
Equation of AA¹ is y – 0 = 1 (x – 0) ⇒ x – y = 0

Case (ii): BB¹ makes an angle 135° with positive X – axis
m = tan 135° = tan (180° – 45°) = – tan 45°
Equation of BB¹ is y – 0 = – 1 (x – 0) = – 1 ⇒ x + y = 0

Question 7.
The angle made by a straight line with the positive X – axis in the positive direction and the Y – intercept cut off by it are given below. Find the equation of the straight line. (S.A.Q.)
(i) 60°, 3
(ii) 150°, 2
(iii) 45°, -2
(iv) tan^-1 \(\left(\frac{2}{3}\right)\), 3
Answer:
(i) 60°, 3
We have the equation of a line with Y- intercept ‘c’ and slope m is y = mx + c
m = tan 60° = √3 , c = 3
∴ Equation of the line is y = √3 x + 3
⇒ √3x – y + 3 = 0

(ii) 150°, 2
m = tan 150° = tan (180° – 30°) = – tan 30° = – \(\frac{1}{\sqrt{3}}\), c = 2
∴ Equation of the line is y = – \(\frac{1}{\sqrt{3}}\)x + 2
⇒ √y = – x +2√3
⇒ √3 y + x = 2√3

(iii) 45°, -2
m = tan 45° = 1, c = – 2
∴ Equation of the line is y = x – 2
⇒ x – y – 2 = 0

(iv) tan^-1\(\left(\frac{2}{3}\right)\), 3
⇒ θ = tan^-1 ⇒ m = tan θ = \(\frac{2}{3}\), c = 3
∴ Equation of the line is y = \(\frac{2}{3}\)x + 3
⇒ 3y = 2x + 9
⇒ 2x – 3y + 9 = 0

Question 8.
Find the equation of the straight line passing through (- 4, 5) and cutting off equal and non-zero intercepts on the coordinate axes. (S.A.Q.)
Answer:
Equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1
Given equal and non – zero intercept s ⇒ a = b
Equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 1 ⇒ x + y = a
This line passes through P (- 4, 5) ⇒ 4 + 5 = a ⇒ a = 1
∴ Equation of the required line is x + y – 1 = 0

Question 9.
Find the equation of the straight line passing through (- 2, 4 ) and making non-zero intercepts whose sum is zero. (S.A.Q.)
Answer:
Equation of the line in the intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1
a b
Given a + b = 0 ⇒ b = -a
∴ Equation of the line is x – y = a
This line passes through P (- 2, 4)
∴ – 2 – 4 = a ⇒ a = – 6
∴ Equation of the required line is x – y + 6 = 0

III.
Question 1.
Find the equation of the straight line passing through the point ( 3, -4 ) and making X and Y – intercepts which are in the ratio 2 : 3. (E.Q.)
Answer:
Equation of the line in the intercept form is
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}\) = 1
Given a : b = 2 : 3
⇒ \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{2}{3}\) ⇒ b = \(\frac{3 a}{2}\)
Equation of the line is \(\frac{x}{a}+\frac{2 y}{3 a}\) = 1
⇒ 3x + 2y = 3a
The line passes through the point p(3, – 4) then 9 – 8 = 3a ⇒ 3a = 1 ⇒ a = \(\frac{1}{3}\)
∴ Equation of the required line is 3x + 2y = 1
⇒ 3x + 2y – 1 = 0

Question 2.
Find the equation of the straight line passing through the point (4, -3) and perpendicular to the line passing through the points (1, 1) and (2, 3 ). (E.Q.)
Answer:
Let A = (1, 1) and B = (2, 3) be the given points.
m = Slope of AB = \(\frac{1-3}{1-2}\) = 2 = m ( suppose)
∴ Slope of the perpendicular line is \(-\left(\frac{1}{m}\right)=-\frac{1}{2}\)
∴ Equation of the line passing through (4, -3) with slope – \(\frac{1}{2}\) is y – y₁ = m (x – x₁)
⇒ y + 3 = – \(\frac{1}{2}\) (x – 4)
⇒ 2y + 6 = – x + 4
⇒ x + 2y + 2 = 0

Question 3.
Show that the following sets of points are collinear and find the equation of the line L containing them,
(i) (- 5, 1), (5, 5), (10, 7)
(ii) (1, 3), (-2, -6), (2, 6)
(iii) (a, b + c), (b, c + a), (c, a + b) (E.Q.)
Answer:
(i) Let A (-5, 1), B (5, 5), C (10, 7) be the given points.
Equation of AB is \(\frac{y-y_1}{y_1-y_2}=\frac{x-x_1}{x_1-x_2}\)
⇒ \(\frac{y-1}{1-5}=\frac{x+5}{-5-5}\)
⇒ \(\frac{y-1}{-4}=\frac{x+5}{-10}\)
⇒ – 4x – 20 = – 10y + 10
⇒ 4x – 10y + 30 = 0
⇒ 2x – 5y + 15 = 0
If C (10, 7) lies on the above line then 2(10) – 5 (7) + 15 = 20 – 35 + 15 = 0
Hence A, B, C are collinear and equation of line containing the points is 2x – 5y + 15 = 0

(ii) Let A (1, 3), B (-2, -6), C (2, 6) be the given points. Then equation of AB is
(y – 3 ) (1 + 2) = (x – 1) (3 + 6)
⇒ 3(y – 3) = 9(x – 1)
⇒ y – 3 = 3(x – 1)
⇒ 3x – y = 0
If C (2, 6) lies on the above line then 3(2) – 6 = 6 – 6 = 0
Hence A, B, C are collinear and equation of line containing these points is 3x – y = 0

(iii) Let A ( a, b + c), B ( b, c + a) ; C ( c, a + b) be the given points.
Equation of AB is
[y – (b + c)l (a – b) = (x – a) ( b + c – c – a)
⇒ (y – b – c) (a – b) = – (a – b) ( x – a)
⇒ y – b – c = – x + a ⇒ x + y – (a + b + c) = 0
If the point (c, a + b) lies on the lines then c + a + b – a – b – c = 0
∴ A, B, C are collinear and the equation of the line containing the points is
x + y = a + b + c

Question 4.
A (10, 4), B (- 4, 9) and C (- 2, -1) are the vertices of a triangle. Find the equations of
(i) \(\overleftrightarrow{\mathbf{A B}}\)
(ii) the median through A
(iii) the altitude through B
(iv) the perpendicular bisector of the side \(\overleftrightarrow{\mathbf{A B}}\). (E.Q.)
Answer:
Given A ( 10, 4 ), B ( – 4, 9 ) and C =( – 2, – 1)

(i) The equation of AB is
(y – 4) ( 10 + 4) = ( x – 10) ( 4 – 9)
⇒ 14y – 56 = – 5x + 50
⇒ 5x + 14y – 106 = 0

(ii) D is the midpoint of BC and AD is the Median through A coordinates of D are
\(\left(\frac{-4-2}{2}, \frac{9-1}{2}\right)\) = (- 3, 4)
∴ Equator of Median AD is
(y – 4) ( 10 + 3) = ( x + 3) ( 4 – 4)
⇒ 13 (y – 4) = 0
⇒ y – 4 = 0 ⇒ y = 4

(iii) BE is the altitude through B, perpendicular to AC
Slope of AC = \(\frac{4+1}{10+2}=\frac{5}{12}\)
∴ Slope of BE = – \(\frac{12}{5}\)
∴ Equation of altitude BC is
y – 9 = – \(\frac{12}{5}\) ( x + 4)
⇒ 5y – 45 = – 12x – 48
⇒ 12x + 5y + 3 = 0

(iv) Mid point of AB = \(\left(\frac{10-4}{2}, \frac{4+9}{2}\right)\)
∴ F = \(\left(3, \frac{13}{2}\right)\)
Slope of AB = \(\frac{4-9}{10+4}=-\frac{5}{14}\)
∴ Slope of perpendicular bisector SF is = \(\overline{\mathrm{AB}}\)
∴ Equation of the perpendicular bisector of the side AB is
y – \(\frac{13}{2}\) = \(\frac{14}{5}\) (x – 3)
⇒ \(\frac{2 y-13}{2}\) = \(\frac{14}{5}\) (x – 3)
⇒ 10y – 65 = 28 ( x – 3)
⇒ 28x – 10y – 19 = 0

Telangana TSBIE TS Inter 1st Year Accountancy Study Material 2nd Lesson Business Transactions Textbook Questions and Answers.

TS Inter 1st Year Accountancy Study Materia 2nd Lesson Recording of Business Transactions

Short Answer Questions:

Question 1.
How vouchers are prepared?
Answer:
A ‘voucher’ is a source document for recording the business transactions. It is a piece of printed paper that is used for some purpose. The document provide proof / evidence of a business transaction and is known as ‘voucher’. The source documents include invoice, pay in slip, cash memo, cheque, debit note, credit note, wage sheets, salary slips etc.

Preparation of Vouchers :
a) Cash Memo :
When goods are purchased or sold for cash, the firm gives or receives Cash Memo. It provides details regarding cash transactions.

b) Purchase Invoice:
When goods are purchased on credit, firm receives purchase invoice from the supplier of goods. It contains date, description of goods, quantity, value of goods etc.

c) Sales Invoice :
When goods are sold on credit, sales invoice is prepared. It contains date, name of customer, description of goods, quantity, value of goods etc. The original copy of invoice is sent to customer and duplicate copy is preserved by the firm.

d) Receipt:
When a firm receives cash from customers, it issues a receipt as a proof of cash receipt. It is prepared in duplicate. The original copy is handed over to the parties making payment and the duplicate copy is kept as record for reference.

e) Pay in Slip :
Pay in slip is available when firm deposits cash or deposits a cheque in the bank account of party to whom it is paid. The counterfoil of pay in slip issued by banker serves as evidence.

f) Cheque :
A cheque is a direction to the banker to pay certain sum of money to a person stated in the cheque by the depositor (firm). When cheques are issued, the details are recorded in the counterfoil of the cheque book giving the information as to party name, amount and date of issue etc.

g) Debit Note :
It is a written note sent by the customer to the supplier when the goods are returned (Purchase returns). The note received gives information of goods returned, their quantity and value. It states the amount debited to suppliers account. This note received is filed for future reference.

h) Credit Note :
It is a written note sent by the supplier of the goods to the customer when the goods are returned by the customer. This note issued to customer stating that the party (customer) account is credited with the value of goods returned. Original copy is sent to the customer and the second copy is retained for reference.

Question 2.
Explain Accounting equation with an example.
Answer:
Accounting equation is based on dual aspect concept of accounting principle. The accounting equation shows the relationship between the economic resources of a business and claims created against those resources. This is presented in the form of an accounting equation as under.

Assets (Resources) = Equities (Claims)
(Or)
Assets = Capital + Liabilities
Capital = Assets – Liabilities
Liabilities = Assets – Capital

The following example explains effect of each transaction on accounting equation :
1. Mr. Ganesh commenced business with cash of ₹ 50, 000. When we fit it into equation.
Cash 50000 = Capital ₹ 50,000 + Liabilities ₹ 0

2. Purchased furniture for cash for ₹ 10,000, then the equation stands as :
Cash 40,000 + Furniture ₹ 10,000 – Capital ₹ 50,000 + Liabilities ₹ 0

3. Purchased goods on credit from Mr. Sankar for ₹ 15,000

Assets = Capital + Liabilities
Cash + Furniture + Stock = Capital + Creditors
40,000 + 10,000 + 15,000 = 50,000 + 15,000
65,000 = 65,000.

Question 3.
Explain briefly systems of recording transation.
Answer:
Book-keeping is the science of recording business transactions of financial nature in regular and systematic manner. Such recording of transactions may be done in any one of the following ways :

1. Single Entry System : .

1. It is treated as an incomplete system of book – keeping which is usually followed by small business units. According to Koehler, it is a system of book – keeping in which, as a rule, only records of cash and personal accounts are maintained.
2. Hence, it is not treated as a reliable system of book – keeping. In fact, this is not accepted as a system. Instead, it is often called as accounts from incomplete records.

2. Double Entry System :

1. It is a system in which both the aspects of all the business transactions are recorded. Every transaction has two aspects, giving and receiving.
2. For every transaction, two accounts are affected at the same time and with the same amount – one account about giving the benefit and the other account about receiving the benefit. One is called Debit and the other is called Credit. Double entry is a system which recognises and records both the aspects (Debit and Credit) of a transaction.

Question 4.
Explain classification of accounts with examples.
Answer:
The accounts are broadly divided into two types.

• Personal Accounts
• Impersonal accounts.

1. Personal Accounts :
These accounts which relate to the persons, group of persons or institutions are called personal accounts.
Ex: Rama’s a/c, Andhra Bank a/c., LIC a/c, Infosys Ltd. The rule in personal accounts is “Debit the receiver and credit the giver”. According to this, benefit receivers account is debited and benefit givers account is credited.

2. Impersonal Accounts:
Impersonal accounts are those accounts which are not personal accounts. They are again divided into

1. Real Accounts
2. Nominal Accounts.

i) Real Accounts :
These accounts related to the assets and properties of the business firm. Ex.: Building, machinery, stock, goodwill etc. The rule in real accounts is “Debit what comes in and credit what goes out”. When an asset is received the asset account is debited and when the asset goes out of the business, the asset is credited.

ii) Nominal Accounts:
These accounts relate to expenses, incomes or gains or losses. Ex.: Salary a/c, Rent a/c, Commission received a/c etc. The rule in nominal accounts is “Debit all expenses and losses and credit all incomes and gains”.

Question 5.
State the rules of debit and credit with examples.
Answer:
Every transaction has two aspects, one is called debit aspect and the other is credit aspect. Under double entry Book keeping system, to identify which aspect is to be debited or credited, rules of Debit and Credit are framed. They are :
Rule:
1. Personal Accounts : Debit the Receiver
(Natural, Artificial, Representative persons) Credit the Giver

A person who receives benefit from the business, his account is to be debited, similarly, a person who gives benefit to the business, his account is to be credited. For example, goods sold on credit to Mr. Ramesh for ₹ 5,000. In this case, Ramesh is the benefit receiver, hence, his account is to be debited. In the same way, credit purchase of goods from Mahesh ₹ 2,000, Mahesh account is to be credited, as he is benefit giver.

Rule :
2. Real Accounts : Debit What comes in
(Assets) Credit What goes out.
As per this rule the assets coming into the business are to be debited and assets going out of the business are to be credited.

3. Rule:
Nominal Accounts Debit all expenses and losses
(Expenses, Losses, Incomes and Gains) Credit all incomes and gains.
According to this rule, all expenses incurred and losses suffered by business are to be debited. All incomes and gains of the business are to be credited.
The following table explains application of Debit and Credit rules and the account to which each aspect of a transaction is related.

Some transactions are presented as under :

Question 6.
Explain briefly the systems of accounting.
Answer:
There are two types of accounting systems which are given below :

1. Cash System of Accounting :

1. Under this system, entries made in the records only for transactions which involve receipt and payment of cash. Outstanding aspects such as, expenses payable, incomes receivable or accrued have no place in cash system of accounting. Usually, Government accounts are prepared on cash system.
2. They record only the actual income received but, while recording the expenses, they take into account the expenses actually paid and outstanding as well. To some extent, they follow the principle of conservatism. In such case, their income statement is shown as receipts and expenditure account.

2. Mercantile or Accrual System of Accounting :

1. Under the Mercantile or accrual system, the whole effect of business transactions are recorded, i.e., the amount received and receivable, the expenses paid and outstanding are recorded.
2. In other words, this system, takes into account, while preparing financial statements, all expenses whether paid or due, all incomes earned whether received in cash or accrued / receivable if they are pertaining the financial year for which final accounts are prepared, eg., salaries due, rent receivable etc.
3. Thus, under accrual system of accounting all expenses incurred and all incomes accrued for a given financial year are accounted in the same financial year irrespective of their actual receipt or payment, by passing necessary journal entries in the books of accounts.

Question 7.
What do you mean by posting ? Explain the rules relating to posting.
Answer:
The process of transferring the entries recorded in the journal or subsidiary books to the respective accounts in the ledger is called posting. So, posting means the process of grouping of all the transactions relating a particular account at one place.

It is necessary to post all the journal entries in the ledger because posting keeps to know the net effect of various transactions during a period of time on a particular account.

Rules Relating to Posting :

1. Each transaction affects minimum two accounts for which separate accounts are to be opened in the ledger.
2. If an account is debited in the journal, posting will be made on the debit side of the account in the ledger. Similarly, if an account is credited, posting will be made on the credit side of the account.
3. While writing the debit side, commerce with the word To’ and write the name of the account which is credited in the journal. Write the word ‘By’ on the credit side before writing the name of the account that is debited in the journal.
4. The difference between debit and credit totals of an account is the net position of the account known as balance of account.

Very Short Answer Questions:

Question 1.
What is Voucher ?
Answer:

1. Voucher is a source document. It forms the basis for recording transactions in the books of accounts.
2. A voucher may be in the form of cash memo, invoice, bill, debit note, credit note etc. Vouchers are to be preserved for verification of books of accounts.

Question 2.
State accounting equation.
Answer:
1. Accounting equation is based on the duel aspect concept (i.e., debit and credit). The accounting equation shows the relationship between the economic resources of the business and claims against these resources.
Economic resources = claims.

2. Another term of economic resources is assets. The claims consists of liabilities and owners claims or equity. The accounting equation is as follows :
Assets = equities or capital + liabilities.

Question 3.
What is cash system of accounting ?
Answer:
Cash System of Accounting :

1. Under this system, entries made in the records only for transactions which involve receipt and payment of cash. Outstanding aspects such as, expenses payable, incomes receivable or accrued have no place in cash system of accounting. Usually, Government accounts are prepared on cash system.
2. They record only the actual income received but, while recording the expenses, they take into account the expenses actually paid and outstanding as well. To some extent, they follow the principle of conservation. In such case, their income statement is shown as receipts and expenditure account.

Question 4.
What is mercantile system of accounting ?
Answer:
Mercantile or Accrual System of Accounting :

1. Under the Mercantile or accrual system, the whole effect of business transactions are recorded, i.e., the amount received and receivable, the expenses paid and outstanding are recorded.
2.  In other words, this system, takes into account, while preparing financial statements, all expenses whether paid or due, all incomes earned whether received in cash or accrued / receivable if they are pertaining the financial year for which final accounts are prepared, eg. salaries due, rent receivable etc.
3. Thus, under accrual system of accounting all expenses incurred and all incomes accrued for a given financial year are accounted in the same financial year irrespective of their actual receipt or payment, by passing necessary journal entries in the books of accounts.

Question 5.
What is an account ?
Answer:
Every transaction has two aspects and each aspect has an account. An account is a summary of relevant transactions at one place relating to a particular head. An account has three parts.
i) A title describes the name of the account.
ii) A left side or debit side.
iii) A right side or credit side.
The form of an account is called T account because of the similarity to the letter T.

Question 6.
State the types of accounts.
Answer:
1) Accounts are classified into two types : (1) Personal Accounts and (2) Impersonal Accounts.
2) Personal accounts may relate to natural, artificial and representative persons.
3) Impersonal accounts are further divided into two types.
They are :

• Real Accounts and
• Nominal Accounts.

Question 7.
What are personal accounts ?
Answer:
Personal Accounts :

1. These accounts which relate to the persons, group of persons or institutions are called personal accounts.
2. Ex.: Rama’s a/c, Andhra Bank a/c., LIC a/c, Infosys Ltd.
3. The rule in personal accounts is “Debit the receiver” “Credit the giver”.
4. According to this, benefit receivers account is debited and benefit givers account is credited.

Question 8.
What are nominal accounts.
Answer:
Nominal Accounts : Nominal accounts are related to expenses, incomes or gains or losses.
Ex.: Salary a/c, rent a/c, commission received a/c etc.
The rule in nominal accounts is : “Debit all expenses and losses

Question 9.
State different types of personal accounts.
Answer:
Personal Accounts :
Personal accounts may relate to natural persons, artificial persons and representative persons. They are explained as under with examples :

a) Natural Persons :
Accounts which relate to individual human beings. For example, Ram, Ramesh, Suresh, Robert, Akbar, Laxmi etc. They are natural persons.

b) Artificial Persons:
They relate to a group of persons, firms or institutions. For example, Infosys Ltd., Andhra Bank, Life Insurance Corporation of India, Lions Club, L & T Ltd., Wipro Ltd., etc.

c) Representative Persons :
These accounts are also personal in nature, e.g., salaries payable (to employees) account, rent receivable (from tenant) account, insurance premium paid in advance (to an insurance company) account. These accounts represents or refers to a particular person or group of persons.

Question 10.
What is Journal ?
Answer:

1. The book in which the business transactions are recorded in chronological order, after analysing them and classifying the benefits according to the principles of debit and credit is called ‘Journal’.
2. Journal is also called “original entry” or “prime entry”.

Question 11.
What is Journalising ?
Answer:

1. The process of recording a transaction in the Journal is called Journalising.
2. For journalising transactions, it is essential to identify the two accounts involved in the transaction, then apply the rule of debit and credit and make the entry in Journal in their respective columns.

Question 12.
What is Journal Entry ?
Answer:

1. The process of recording business transactions in a chronological order in Journal after analysing, classifying and identifying as debit and credit is called Entry.
2. All the transactions recorded in the Journal are in the form of entries. Hence, they are called as Journal Entries.

Question 13.
What is ledger ?
Answer:

1. Ledger is also called as “Book of Secondary Entry”.
2. Ledger is a book which facilitates recording of all type of transactions related to personal, real and nominal accounts separately.
3. According to Cropper, “The book which contains a classified and permanent record of all the transactions of a business is called ledger”.

Question 14.
What is posting ?
Answer:

1. The process of transferring the entries recorded in the journal or subsidiary books to the respective accounts in the ledger is called posting.
2. So, posting means the process of grouping of all the transactions relating to a particular account at one place.

Question 15.
What do you mean by balancing of an account ?
Answer:

1. Balancing is the process of finding the difference between the total debits and total credits of an account.
2. When posting in done, many accounts may have entries on the debit side as well as on – the credit side. The net result of such debits and credits in an account is the balance.

Question 16.
What is debit balance ?
Answer:
The excess of debit total over the credit total is called the debit balance.

Question 17.
What is credit balance ?
Answer:
The excess of credit total over the debit total is called the credit balance.

Problems:

Question 1.
Mr. Anil started business with cash ₹ 75,000 on 1st January 2018. The details of business transactions for the month of January are as follows. Prepare Journal.
2018 Jan
Jan 02 Cash Sales – ₹ 10,000
Jan 05 Goods Purchased for Cash – ₹ 12,000
Jan 07 Goods sold on credit to Rahim – ₹ 20,000
Jan 08 Sale of Goods – ₹ 20,000
Jan 09 Wages Paid – ₹ 5,000
Jan 10 Cash Deposited into bank – ₹ 8,000
Jan 13 Cash paid to Mahesh – ₹ 6,000
Jan 15 Machinery purchased for cash – ₹ 12,000
Jan 18 Purchased goods from Anitha – ₹ 5,000
Jan 20 Cash received from Mrs. Ramya – ₹ 3,000
Jan 22 Paid commission – ₹ 1,500
Jan 24 Paid for Postage and Stationery – ₹ 500
Jan 27 Cash drawn from Bank for personal use – ₹ 7,000
Jan 30 Received rent – ₹ 1,200
Solution:
Journal Entries in the book of Mr. Anil

Question 2.
Pass journal entries in the books of Mr. Ram
2019 March
March 01 Ram Commenced business with cash – ₹ 1,00,000
March 02 Cash deposited into Canara Bank – ₹ 60,000
March 04 Purchased from Rama – ₹ 4,000
March 06 Purchased computer and paid by cheque – ₹ 15,000
March 09 Ram withdrew cash for his personal expense – ₹ 5,000
March 13 Furniture purchased – ₹ 10,000
March 15 Returned goods to Rama – ₹ 500
March 18 Amir returned goods – ₹ 1,000
March 20 AdvertIsement Expenses paid – ₹ 1,000
March 23 Deposited cash into Canara Bank – ₹ 3,000
March 25 Goods withdrawn for personal work – ₹ 2,000
March 27 Cash paid to Ramesh Rs. 3,900 and discount received – ₹ 100
March 28 Cash received from Ramu Rs. 2,800 and discount allowed – ₹ 200
March 29 Cash withdrawn from bank for office use – ₹ 6,000
March 31 Salaries paid by cheque – ₹ 8,000
Solution:
Journal Entries in the book of Mr. Anil

Question 3.
Journalize the following transactions in the books of Akbar.
2019 March
1st Akbar Started business with cash – ₹ 50,000
March 02 Cash Sales – ₹ 30,000
March 04 Cash Purchases – ₹ 40,000
March 06 Sold goods to Mahesh – ₹ 35,000
March 09 Bought goods from Radhika – ₹ 25,000
March 11 Sold goods to Swathi for cash – ₹ 10,000
March 15 Mahesh returned goods – ₹ 5,000
March 18 Commission Received – ₹ 1,000
March 19 Office Expenses paid – ₹ 500
March 20 Cash paid to Pramod – ₹ 6,000
March 22 Returned goods to Radhika – ₹ 2,000
March 25 Goods withdrawn for domestic use – ₹ 5,000
March 27 Cash received from Anand Rs. 3800 and discount allowed – ₹ 200
March 28 Interest received – ₹ 500
March 29 Cash paid to Raman Rs. 4,900 discount allowed by him – ₹ 100
March 31 Commission paid – ₹ 300
Solution:
Journal Entries in the books of Mr. Akbar

Question 4.
Journalize the following transactions in the books of Bhagat.
2019 January
January01 Commenced business with cash 40,000 and furniture worth 10,000
January02 Sold goods to Suchitra – ₹ 20,000
January03 Purchased Machinery – ₹ 30,000
January05 Paid rent – ₹ 5,000
January09 Paid Electricity bill – ₹ 1,000
January12 Sold goods for cash – ₹ 6,000
January15 Bought goods on credit from Nikhil – ₹ 10,000
January18 Paid wages – ₹ 5,000
January21 Interest received through cheque – ₹ 5,000
January25 Advertisement Expenses paid – ₹ 3,000
Solution:
Journal entries in the books of Mr. Bhagat

Question 5.
Journalize the following transactions in the books of Dinesh.
2019 April
April 01 Dinesh Started Business with cash – ₹ 50,000
April 02 Sale of goods for cash – ₹ 10,000
April 04 Purchase of goods for cash – ₹ 15,000
April 06 Sold Plant and Machinery for cash – ₹ 5,000
April 10 Cash paid to Rahim – ₹ 3,000
April 14 Salaries paid – ₹ 8,000
Solution:
Journal entries in the books of Mr. Dinesh

Question 6.
Journahze the following transactions in the books of Atma Ram.
2019 January
January 01 Started business with cash – ₹ 25,000
January 03 Purchased computer – ₹ 5,000
January 06 Purchased goods for cash – ₹ 6,000
January 07 Purchased goods on credit from Mahesh – ₹ 8,000
January 10 Purchased goods for and paid by cheques – ₹ 7,000
January 12 Sold goods for cash – ₹ 10,000
January 15 Goods sold for Rs. 15,000 of which Rs. 8,000 received in cash and by a cheque. – ₹ 7,000
Solution:
Journal entries in the books of Atma Ram

Question 7.
Journahze the following transactions
March 2019
March 01 MrAnthony Commenced business with cash ₹ 13,000 and stock ₹ 7,000
March 02 Bought plant – ₹ 5,000
March 03 Paid for postage – ₹ 500
March 05 Paid for Sundry Expenses – ₹ 500
March 06 Paid into bank – ₹ 10,000
March 07 Paid salaries – ₹ 5,000
March 09 Cash withdrawn from bank for office use – ₹ 1,200
Solution:
Journal entries

Question 8.
Write journal entries in the books of Sudha.
2019 April
April 01 Started business with a capital of – ₹ 90,000
April 01 Purchased goods from Tharun on credit – ₹ 20,000
April 02 Sold goods to Sonu – ₹ 30,000
April 03 Purchased goods from Ragav for cash – ₹ 25,000
April 04 Sold goods to Toni for cash – ₹ 16,000
April 05 Goods retuned to Tharun – ₹ 5,000
April 06 Bought furniture for cash – ₹ 15,000
April 18 Sold goods to Sudip – ₹ 12,500
April 19 Goods returned by Sudip – ₹ 2,000
April 25 Cash received from Sudip – ₹ 5,500
April 28 Goods taken by Sudha for domestic use – ₹ 3,000
Solution:
Journal entries in the books of Sudha

Question 9.
Journalise the following transactions of Mr. Prahlad and post them in the ledger and balance the same
2019 February
February 1 Prahlad invested ₹ 50,000 cash in the business
February 3 Paid into Bank – ₹ 6,000
February 5 Purchased Furniture for – ₹ 2,000
February 7 Purchased goods for – ₹ 5,000
February 10 Sold goods for – ₹ 10,000
February 15 Withdrew cash from bank – ₹ 2,000
February 25 Paid interest – ₹ 800
February 28 Paid Salary – ₹ 8,000
[Answer: Bank Balance ₹ 4,000 Dr. Cash balance ₹ 40, 200 Dr.]
Solution:
Journal entries in the books of Mr.Prahlad

Question 10.
Prepare Pavan account from the following :
2018 March
March 1 Goods purchased from Pavan – ₹ 38,000
March 6 Cash paid to Pavan – ₹ 5,000
March 10 Goods returned to Pavan – ₹ 1,500
March 14 Paid to Pavan by Cheque – ₹ 6,800
March 20 Discount allowed by Pavan – ₹ 500
March 26 Goods purchased from Pavan for Cash – ₹ 2,500
March 28 Furniture purchased from Pavan – ₹ 8,000
Solution:

Question 11.
Prepare Sudha account from the following :
2018 February
February 1 Amount due from Sudha – ₹ 8,000
February 4 Goods sold to Sudha – ₹ 11,000
February 12 Goods returned by Sudha – ₹ 4,000
February 16 Cash received from Sudha – ₹ 3,000
February 22 Received cheque from Sudha – ₹ 6,000
February 28 Sudha account settled with 10% discount
Solution:

Note:
Discount = 6000 × \(\frac{10}{100}\) = 600.

Question 12.
Prepare Swami’s account from the following :
2018 January
January 2 Amount due to Swami – ₹ 12,000
January 8 Goods purchased from Swami – ₹ 16,000
January 15 Goods returned to Swami – ₹ 5,000
January 20 Cash paid to Swami – ₹ 16,000
January 24 Goods purchased from Swami – ₹ 9,000
January 30 Swami’s account is settled by cheque with 10% discount
[Ans: Paid₹ 14,400; Discount received ₹ 1,600]
Solution:

Note :
Discount = 16,000 × \(\frac{10}{100}\) = 1600

Question 13.
Prepare Machinery account from the following :
2018 March
March 1 Purchased machinery from Vikram & co – ₹ 42,000
March 6 Machinery purchased from Virat – ₹ 16,000
March 12 Machinery costing ? 8,000 sold for – ₹ 5,000
March 16 Depreciation provided on Machinery – ₹ 3,000
March 22 Goods purchased from Swami – ₹ 9,000
March 30 Purchase of machinery for cash – ₹ 8,000
Solution:

Question 14.
Prepare the Ledger accounts from the following particulars in the Books of Rani
2018 June
June 1 Received cash from Shiva – ₹ 75,000
June 4 Bought goods for cash – ₹ 40,000
June 6 Sold to Suresh – ₹ 40,000
June 12 Bought goods from Praveen – ₹ 50,000
June 16 Sold goods to Ganesh – ₹ 35,000
June 20 Withdrew cash for personal use – ₹ 20,000
June 26 Received Commission – ₹ 2,000
June 30 Paid rent – ₹ 5,000
June 30 Paid salary – ₹ 10,000
Solution:
Ledger Accounts in this books of Rani:

Question 15.
From the following information prepare Praveen’s Account as on 31-03-2014
2018 March
March 7 Balance due from Praveen – ₹ 3,500
March 7 Sold goods to Praveen – ₹ 1,500
March 10 Purchased goods from Praveen – ₹ 1,000
March 15 Paid cash to Praveen – ₹ 800
March 23 Received cash from Praveen – ₹ 500
March 25 Returned goods to Praveen – ₹ 200
Praveen settled account with 10% discount.
Solution:

Note:
Discount = 4500 × \(\frac{10}{100}\) = 450.

Question 16.
Prepare Vamsls Account from the following
2018 August
August 1 Balance due to Vamsi – ₹ 4,400
August 5 Purchased goods from Vamsi – ₹ 1,500
August 10 SoldgoodstoVamsi – ₹ 1,200
August 13 Received cheque from Vamsi – ₹ 1,000
August 17 Paid cash to Vamsi – ₹ 100
August 23 Vamsi returned goods – ₹ 200
August 29 Purchased goods from Vainsi – ₹ 500
Vamsi account settled with 5% discount.
Solution:

Note:
Discount = 4500 × \(\frac{5}{100}\) = 315.

Question 17.
Prepare Anurudh’s Account from the following 2018
2018 December
December 1 Balance due from Anurudh – ₹ 1,900
December 9 Sold goods to Anurudh – ₹ 1,000
December 12 Purchased goods from Anurudh – ₹ 700
December 15 Returned goods to Anurudh – ₹ 200
December 20 Anurudh returned goods – ₹ 100
December 25 Received cheque from Anurudh – ₹ 400
December 28 Paid cash to Anurudh – ₹ 600
Solution:

Textual Examples:

Question 1.
Journalise the following transactions.
Date 2019 March
March 11 Paid for printing expenses – ₹ 900
March 13 Goods returned to Sankar on account – ₹ 600
March 14 Cash sales – ₹ 18,000
March 15 Wages paid – ₹ 3,000
March 17 Mahesh returned goods – ₹ 2,000
March 18 Paid to Sankar – ₹ 3,400 on account
March 20 Received from Mahesh on account – ₹ 7,000
March 23 Rent paid – ₹ 1,500
March 25 Commission received – ₹ 1,200
March 28 Paid salaries – ₹ 5,000
March 30 Ganesh (proprietor) taken – ₹ 1,000 for personal expenses
March 31 Goods taken for personal use – ₹ 800
Solution:
Journal Entries in the book of Mr. Ganesh

Question 2.
Journalise the following transactions.
Date 2019 January
January 01 Mr.Ram Commenced business with ₹ 98,000
January 02 Cash deposited into State Bank of India ₹ 50,000
January 04 Purchased office furniture for ₹ 10,000 paid through bank
January 05 Purchased goods from Amar ₹ 12,000
January 07 Purchased goods for cash ₹ 5,000 from Ramesh
January 08 Cash sales ₹ 11,000
January 10 Goods sold to Akbar for ₹ 10,000
January 12 Paid rent by cheque ₹ 4,000
January 14 Paid to Amar ₹ 6,000 on account
January 15 Goods returned by Akbar ₹ 1,000
January 16 Goods returned to Amar ₹ 1,500
January 18 Paid for advertising ₹ 1,200
January 19 Received from Akbar by cheque ₹ 3,000
January 21 Loan taken / borrowed from Raju ₹ 9,000
January 25 Goods purchased for ₹ 15,000 paid by cheque
January 28 Drawings from Bank by Ram ₹ 1,500
January 31 Salaries paid by cheque ₹ 12,000.
Solution:
Journal of Mr. Ram

Question 3.
Journalise the following transactions.
Date 2019 January
January 05 Received cash from Ramesh ₹ 2,800, discount allowed ₹ 200.
January 10 Paid to Mohan ₹ 8,500 infull satisfaction of his account ₹ 9,000.
January 18 Rahim is a customer from whom ₹ 5,000 due, he became insolvent, only ₹ 3,000 received as final divident from his estate.
January 24 Interest received on investment by cheque (our Banker’s is Canara Ban ₹ 1,200).
January 28 Commission paid ₹ 1,200.
January 30 Interest paid on Loan ₹ 3,000.
Solution:
Journal entries

Question 4.
Journalise the following transactions and post them into their respective accounts in the ledger.
Date 2019 January
January 05 Ramesh commenced business with ₹ 25,000
January 10 Purchased office furniture for cash ₹ 10,000
January 12 Goods purchased on credit from Rao ₹ 8,000
January 25 Sold goods on credit to Ram ₹ 12,000
Solution:
Journal entries of Ramesh

Question 5.
Journalise the following transactions and post them into ledger and balance the accounts.
Date 2019 January
January 1st Mr.Ganesh commenced business with – ₹ 40,000
January 2nd Purchased a Computer for office use – ₹ 5,000
January 4th Purchased Furniture from Godrej Co. for – ₹ 15,000
January 5th Goods purchased from Srinivas – ₹ 6,000
January 7th Goods Sold for Cash – ₹ 8,000
January 9th Purchased goods for Cash – ₹ 2,000
January 10th Paid to Godrej Co. – ₹ 15,000
January 12th Paid for stationery – ₹ 500
January 13th Wages paid – ₹ 800
January 15th Goods purchased from Naresh – ₹ 10,000
January 16th Paid to Srinivas on account – ₹ 4,000
January 17th Goods sold to Ramesh – ₹ 12,000
January 18th Rent received – ₹ 1,800
January 19th Ganesh (owner) taken cash for personal expenses – ₹ 600
January 20th Goods returned to Srinivas – ₹ 700
January 23rd Received cash from Ramesh on account – ₹ 8,000
January 25th Paid for printing – ₹ 900
January 27th Goods returned by Ramesh – ₹ 1,000
January 28th Paid Salaries – ₹ 3,500
January 29th Purchased goods from Sunder – ₹ 4,000
January 30th Sold goods to Rav – ₹ 5,000
Solution:
Journal Entries

Telangana TSBIE TS Inter 1st Year Hindi Study Material Grammar वर्तनी, शब्द विचार, उपसर्ग, प्रत्यय Questions and Answers.

TS Inter 1st Year Hindi Grammar वर्तनी, शब्द विचार, उपसर्ग, प्रत्यय

హిందీ మన దేశపు జాతీయ భాష కాక అధికార భాష కూడా. ఇప్పుడు ఈ భాష మనదేశంలోనే కాక ప్రపంచంలో సుమారు 170 విశ్యవిద్యాలయాల్లో భోదించబడుతుంది పరిపాలన, బ్యాకింగు, వాణిజ్య వ్యాపారాలు, పత్రికారంగం, ప్రసార మాధ్యమాలు మొదలైన వాటన్నింటిలో హిందీ వ్యాప్తి క్రమేపి పెరుగుతూ వస్తోంది. ఇలా ప్రాచుర్యం పెంచుకుంటున్న హిందీ భాష యొక్క శుద్ద రూపాన్ని (correct form) నేర్చుకోవటం. మాట్లాడటం, వ్రాయడం చాలా అవసరం. భాష యొక్క శుద్ధ రూపాన్ని मानक भाषा [standard Language] అంటారు. విద్య, పరిపాలన తదితర రంగాల్లో हिन्दी యొక్క मानक रूप మాత్రమే ఉపయోగపడుతుంది.

हिन्दी भाषा యొక్క मानक रूप ని నేర్చుకోవటం కోసం క్రమం తప్పకుండా చదవటం, వ్రాయటం చాలా అవసరం. పదాల ఉచ్చారణ – వ్రాత (लेखन) పట్ల దృష్టి కేంద్రీకరించాలి. दोष (Mistakes) లేకుండా ఉండేందుకు ఎల్లప్పుడూ జాగరూకులై ఉండాలి. वर्तनी दोष [Spelling] విషయంలో నిర్లక్ష్యం ఎంత మాత్రం తగదు. క్రమం తప్పని అభ్యాసంతో షెగే ని నివారించవచ్చు.

हिन्दी యొక్క ప్రస్తుత రూపం खड़ीबोली, దీని లిపిని देवनागरी అంటారు. लिपि యొక్క సరియైన రూపాన్ని బాగా అర్థం చేసుకుంటే वर्तनी दोष రాకుండా జాగ్రత్త పడవచ్చు. ఉచ్చారణే వ్రాతకు ఆధారం కనుక స్పష్టమైన ఉచ్చారణ అభ్యసించటం చాలా అవసరం.

अ) वर्णमाला :
स्वर : अ आ इ ई उ ऊ ऋ
ए ऐ ओ औ अं अः

व्यंजन : क ख ग घ ङ
च छ ज झ ञ
ट ठ ड ढ ण
त थ द ध न
प फ ब भ म
य र ल व
श ष स ह
क्ष त्र ज्ञ

वर्ण (परिभाषा) : ‘लिखित भाषा की उस छोटी से छोटी मूल ध्वनि को वर्ण कहते हैं, जिसके टुकड़े नही किए जा सकते’ । मूल रूप में वर्ण वे चिह्न हाते हैं, जो हमारे मुख से निकली हुई ध्वनियों के लिखित रूप होते हैं ।

आ) मात्राएँ :

इ) हस्व और दिर्घ : ఉచ్చారణ వ్రాతల్లో ‘हस्व’, ‘दीर्घ’, యొక్క తేడాను గమనించాలి.
a) जब उच्चरण अल्प होता है, तब उस ध्वनि को ह्रस्व कहा जाता है ।
b) जब उच्चरण दीर्घ होता है, तब उस ध्वनी का दीर्घ कहा जाता है ।
(इ, उ मात्राएँ)
नियम, विजय, सितार, किसान
मिठाई, विपुल, गिलास, बिल्ली
बुराई, पुस्तक, लुहार, बुढ़ापा
सुनार, कुम्हार, गुड, भुलाना

(ई, ऊ मात्राएँ)
कीमत, गीत, खीर, पीर
वीर, क्षीर, भाई, नीरज
पूर्व, भूत, मूर्ख, दूध
झूला, दूसरा, पेटू, चूहा

ई) अल्प प्राण और महा प्राण : హల్లులు (व्यंजन) లోని ప్రతి వర్గపు మొదటి అక్షరాన్ని ‘अल्प-प्राण’ అని, రెండవ అక్షరాన్ని ‘महा-प्राण’ అని వ్యవహరిస్తారు. ఒకటి ఉచ్చరించ / వ్రాయవలసిన చోట మరొకటి ఉపయోగించటం తప్పు.
उदा : अल्प प्राण – क, च, ट, त, प
महा-प्राण – ख, छ, ठ, थ, फ

अशुद्ध (✗) – शुद्ध (✓)
बोजन – भोजन
आँक – आँख
तोडा – थोड़ा
रोठी – रोटी
चाया – छाया
पूल – फूल
चतरी – छतरी

उ) अनुनासिकता और अनुस्वार : హిందీలో అనునాసిక శబ్దాల సంఖ్య చాలా హెచ్చు చాలా శబ్దాలు अनुस्वार జోడించడం ద్వారా बहुवचन గా మారతాయి. अनुस्वार चंद्र बिन्दा (ँ) లేక (बिन्दी) (.) చేర్చటం ద్వారా వ్యక్తమౌతాయి. వీటిని ఉచ్చారణ వ్రాతల్లో విధిగా పాటించాలి.

उदाहरण :
गाय – गायें
आँख – आँखें
काँटा – काँटे
बहन – बहनें
माता – माताएँ
बहूएँ – बहू
रानी – रानियाँ
दवा – दवाएँ
चिडिया – चिड़ियाँ
बुढिया – बुढ़ियाँ
परीक्षा – परीक्षाएँ
लड़की – लड़कियाँ
मछलियाँ – मछली
महिला – महिलाएँ
स्त्री – स्त्रीयाँ

ऊ) पर भाषा शब्द : హిందీలో చాలా శబ్దాల अरबी, फारसी మరియు భాషల నుండి వచ్చినవి. తారసపడతాయి. వాటిని సరిగా ఉచ్చరించటం కోసం అక్షరం క్రింద ఉంచబడుతుంది.
उदाहरण : अंदाज, कागज, काजल, कमीज, कफन, फायदा, जागीर, एतराज़, इल्जाम, जरा, आजाद, पाजटिव, जीरों, डज़न, जेबरा, आदि ।

ॠ) ‘ऋ’ मात्रा का सही प्रयोग : ‘ऋ’ मात्रा (c) వ్రాయవలసిన చోట (‘र’) मात्रा दुई తో వ్రాయటం తప్పు.
उदाहरण :
अशुद्ध (✗) – शुद्ध (✓)
प्रुथ्वी – पृथ्वी
क्रुष्ण – कृष्ण
स्रुटि – सृष्टि
क्रुपा – कृपा

ए) ‘र’ का संयुक्ताक्षर : హిందీలో (‘र’) ధ్వని మూడు రకాలుగా సంయుక్తాక్షరంగా వ్రాయబడుతుంది.
प्र – क्रम, ग्रंथ, व्रत, प्रचार, भ्रमर, द्रविड, आदि ।
र्प – धर्म, कर्म, मर्म, गन्धर्व, आर्य, पर्व आदि ।
ट्र – राष्ट्र, दंष्ट्रा उष्ट्र, पौण्ड्रवर्धन, आदि ।

ऐ) ‘घ’ और ‘ध्य’ : ‘द्य’ వ్రాయవలసిన చోట ‘ध्य’ వ్రాయటం, ఉచ్చరించుట తప్పుగా పరిగణింపబడుతుంది. ‘द’ ని ఉచ్చరించవలసిన చోట ‘ध्य’ ప్రయోగించాలి.
उदाहरण :
अशुद्ध (✗) – शुद्ध (✓)
खद्य – खाद्य
विद्या – विद्या
मद्यपान – मद्यपान
विध्यार्थी – विद्यार्थी
विध्यलय – विद्यालय
पध्य – पद्य

ओ) ‘ध्य’ केल००’ ‘द्य’ వ్రాయటం ఉచ్చరించుటం తప్ప .
उदाहरण :
अशुद्ध (✗) – शुद्ध (✓)
अद्याय – अध्याय
उपाधाय – उपाध्याय
अद्यापक – अध्यापक

औ) ‘ए’ और ‘ये’ का प्रयोग : క్రియారూపంలో ‘ए’ మరియు ‘ये’ ప్రయోగించటం విషయంలో జాగ్రత్త వహించాలి. క్రియ పూర్ణకాలికమైతే ये, క్రియ, विधि స్వరూపమైతే ఉపయోగించాలి.
उदाहरण :
अशुद्ध (✗) – शुद्ध (✓)
गाया – गाये
दिया – दिये
सोया – साये
आया – आये
रोया – रोये
दीजिये – दीजिए
बैठिये – बैठिए
लाइये – लाइए
खाइये – खाइए

शुद्ध वर्तनी केलिए नियम

1. भाषा में उच्चारण की स्पष्टता लानी होगी, जब उच्चारण ठीक हो, तो वर्तनी भी ठीक होगा । क्योंकि देवनागरी लिपि ऐसी वैज्ञानिक लिपि है कि जो पढा जाता है, वही लिखा जाता है ।
2. निरंतर पढने – लिखने के अभ्यास से वर्तनी दोषों पर रोक लगायी जा सकती है ।
3. स्वर – व्यंजन, हलंत – लिंग, वचन, प्रत्यय, संधि, समास आदि के प्रयोग में निरंतर अभ्याय की जरूरत है ।

హిందీలో వ్రాతను మెరుగుపరచుకోవాలంటే ముందు ఉచ్చారణ స్పష్టంగా ఉండేలా చూసుకోవాలి. ఉచ్చారణ సరిగ్గా ఉంటేనే అక్షర దోషాలు రాకుండా ఉంటాయి. దేవనాగరి లిపిలో ఉచ్చారణ ఎలా ఉంటుందో వ్రాత కూడా అలాగే ఉంటుంది. కాబట్టి ముందుగా ఉచ్చారణ సరి చేసుకుంటే చాలా వరకు వ్రాతలోని అక్షర దోషాలను నివారించవచ్చు. దీనికి ఒకటే మార్గం. నిరంతరం చదవడం, వ్రాయడం చదువుతూ ఉంటే మన కళ్ళతో అక్షరాలకు ఒక రిలేషన్ లాంటిది ఏర్పడాలి. దానినే ‘చక్షురక్షర సంయోగం’ అంటారు. ఇది ఒక్కసారి కుదిరిందంటే ఇక వ్రాసేటప్పుడు అక్షర దోషాలు రావు.

हिन्दी में वर्तनी दोषों के असंख्य उदाहरण बन सकते हैं जिनका उल्लेख करना यहाँ संभव नही है । अतः उदाहरण के लिए जिन शब्दों के साथ अधिक वर्तनी – दोष पाये गये हैं उनका यहाँ प्रस्तुत किया जा रहा है ।

హిందీ వ్రాసేటప్పుడు అసంఖ్యాకమైన అక్షర దోషాలు విద్యార్థులలో కనబడతాయి. అయితే వాటన్నింటినీ ఇక్కడ ప్రస్తావించడం కష్టం. కాబట్టి ఉదాహరణ కోసం కొన్ని అక్షర దోషాలను ఇక్కడ ఇవ్వడం జరుగుతున్నది.

अशुद्ध रूप – शुद्ध रूप
टंडा – ठंडा
अहार – आहार
आदमि – आदमी
आरामान – आसमान
अन्यधा – अन्यथा
अत्याधिक – अत्यधिक
आँक – आँख
ईक – ईख
इमलि – इमली
उन्नति – उन्नति
उपनयास – उपन्यास
उध्योग – उद्योग
ऊचा – ऊँचा
चांद – चाँद
घंठा – घंटा
काना – खाना
बादा – बांधा
धोका – धोखा
पौदा – पौधा
ग्यान – ज्ञान
दीर्घ – दीर्घ
भूक – भूख
बिक्षुक – भिक्षुक
पंढित – पंडित
पुरानि – पुरानी
वस्तू – वस्तु
आँसु – आँसू
निष्टा – निष्टठा
पूजारी – पुजारी
हमेसा – हमेशा
लीखिए – लिखिए
दीजीए – दीजिए
प्रतिक्षा – प्रतीक्षा
उत्तीण – उत्तीर्ण
प्रवेस – प्रवेश
परकाश – प्रकाश
यात्र – छाञ
राष्ट्र – राष्ट्र
शरन – शरण
बारत – भारत

अभ्यास

(✗) – (✓)

पुरसकार – पुरस्कार
प्रर्दसन – प्रदर्शन
सममेलन – सम्मेलन
मचली – मछली
बोजन – भोजन
कृष्ण – कृष्ण
विध्यार्थी – विद्यार्थी
स्मरन – स्मरण
पूस्तक – पुस्तक
हीन्दी – हिन्दी
कताब – किताब
गूण – गुण
धरमात्मा – धर्मात्मा
गोर – घोर
भादा – बाधा
कारन – कारण
प्रान – प्राण
कन – कण
उदहरण – उदाहरण
इस्पष्ट – स्पष्ट
सीक – सीख
पच्चिस – पच्चीस
कर्तब्य – कर्तव्य

शब्द विचार

जिसने शब्दों के भेद ( उनके प्रयोग) रूपांतर और व्युत्पति का निरूहण किया जाता है, उसे शब्द साधान (या) शब्द विचार कहते है ।

ధ్వనుల కలయికతో ఏర్పడే అర్థపూరిత అక్షర సమూహాన్ని ‘शब्द’ అంటారు. ప్రతి ‘शब्द’ కి దాని అర్థం ఉంటుంది. ఏ ‘शब्द’ ల అర్థం స్పష్టంగా వ్యక్తమౌతుందో వాటిని ‘सार्थक शब्द’ అంటారు. ఏ భాషలోనైతే ఒక शब्द కి అర్థం ఉండదో దానిని ‘निरर्थक शब्द’ గా పరిగణిస్తారు. హిందీ భాషలో कल’ सार्थक शब्द’ అని, దీనికి సందర్భాన్ని బట్టి ‘నిన్న’, ‘రేపు’, ‘యంత్రము’ అనే అర్థాలున్నాయి. కాని लक ఈ భాషలో नरर्थक शब्द, అలాగే नरर्थक शब्द’, ‘जल’, सार्थक, ‘लज’, निरर्थक అని చెప్పవచ్చును.

परिभाषा : ఒకటి లేక అంతకంటే ఎక్కువ అక్షరాలు అర్థపూరితమైన కలయికతో ఏర్పడే దాన్ని ‘शब्द’ అంటారు. ‘जा’, ‘खा’, ‘गा’ ‘आ’ మొదలైనవి. ఏకాక్షర శబ్దాలు ‘चिड़िया’, ‘पहाड़’, ‘आसमान’ మొదలైనవి ఏకాధిక అక్షరాలతో ఏర్పడిన శబ్దాలు.

शब्दों के प्रकाश : కొన్ని ‘शब्द’ ఒకే అర్థానిచ్చేవిగా ఉంటాయి. వాటిని అని అంటారు. ఉదా : स्वागत, यश, कृपा, पुस्तक వీటికి ఏ సందర్భంలోనైనా ఒకే అర్థం గోచరిస్తుంది. కొన్ని ‘शब्द’ బహు అర్థాలు ఇచ్చేవిగా ‘अनेकार्थी शब्द’ అంటారు. వేర్వేరు సందర్భాలలో ఒక్క. ‘शब्द’ కే వేర్వేరు అర్ధాలు ఉంటాయి.

उदा : अर्थ = कारण, मतलब, लिए, धन
पानी = जल, सम्मान, प्रतिष्ठा, चमक
कर = क्रिया, हाथ, किरण, लगना, सूँड ।

కొన్ని शब्द ‘समानार्थक शब्द’ గా ఉంటాయి.
उदा : बंधु और मित्र
दक्ष, कुशल और निपुण
शोक और दुख
कष्ट और पीड़ा
आज्ञा और आदेश

शब्द కు ‘विपरीतार्थक शब्द’ కూడా ఉంటాయి.
उदा : मान – अपमान
अग्रज – अनुज
उत्तम – अधम
अल्पायु – दीर्घायु
उदय – अस्त
उच्च – निम्न
आय – व्यय
आजादी – गुलामी

शब्दों का एक प्रकार : ధ్వని సంబంధమైన అనుకరణనం బట్టి కూడా ఉంటుంది. పశుపక్ష్యాదుల అరుపులు, ధ్వనులు ప్రతి భాషలో తనశైలిలో అనుకరించబడతాయి. హిందీలో ఇటువంటి ధ్వనుల అనుకరణ క్రింది విధంగా ఉంటుంది.

चिड़ियाँ खनकती हैं ।
घड़ी टिक-टिक चलती है ।
शेर दहाड़ता है ।
बिल्ली म्याऊँ म्याऊँ करती है ।
कुत्ता भौकता है ।
कौआ कॉव-कॉव करता है ।
गधा रेंकता है ।
मुर्गा कूंकडूं कूँ करता है ।
दिल धक धक करता हैं ।
बादल गरजते हैं ।
बिजली कड़कती है ।

ఇతర భాషలవలె हिन्दी లో కూడా मुहावरे और कहावतें ఉంటాయి. వీటికి సామాన్య అర్థంకాక విశిష్టమైన అర్థం వ్యక్తమౌతుంది.
ఉదా : अँगूठा दिखाना जरूरत होने पर धोखा देना
आम के आम गुठलियों के दाम = बहुत फायदा होना

शब्द स्त्रोत భాష దృష్ట్యా హిందీ – సంస్కృతం యొక్క అనేక ‘मूल शब्द’ యధాతథంగా ఒకేలా ఉంటాయి. కాని అనేక సంస్కృత భాషా శబ్దాలు పాళి, ప్రాకృత, అపభ్రంశ భాషల్లోకి వెళ్ళి మార్పుచెంది హిందీలోకి ప్రవేశించాయి. మరికొన్ని శబ్దాలు హిందీ భాషా ప్రాంతాల్లోని వివిధ ఉపభాషల నుండి हिन्दी లోనికి వచ్చాయి. ఇంకొన్ని విదేశి భాషల శబ్దాల్ని हिन्दी भाषा గ్రహించింది. ఈ రకంగా हिन्दी లో నాలుగు రకాలైన शब्द ఉంటాయి. तत्सम तद्भव, देशज तथा विदेशी ।

1. तत्सम : హిందీ భాషలో చెలామణిలో ఉన్ संस्कृत मूल शब्द ను ‘तत्सम शब्द’ అంటారు अग्नि, पुष्प, चतुर्थ, मयूर, क्षीर మొదలైన సంస్కృత శబ్దాలు హిందీలో యధాతదంగా ప్రయోగింపబడతాయి. ఇటువంటి శబ్దాల సంఖ్య హిందీలో వేలాదిగా ఉంటుంది.

2. तद्भव : సంస్కృత భాష నుండి ప్రాకృతంలోనికి వెళ్ళి మార్పుచెంది ‘హిందీ భాషలోనికి వచ్చిన శబ్దాల్ని ‘तद्रभव शब्द’ అంటారు.
संस्कृत – हिन्दी
मया = मैं
उष्ट्र = ऊँट
आम = आम
मयूर = मोर
मध्य = में
दन्त = दांत
तिक्त = तीता
शत = सौ
चंचु = चोंच
पुष्प = फूल
घोटक = घोड़ा
चतुर्थ = चौथा
क्षीर = खीर
त्वरित = तुरंत
वन्स = बच्चा
भक्त = भात
नव = ना
सूचि = सुई

3. देशज : ఏ శబ్దాలైతే हिन्दी भाषा ప్రాంతాల్లో వ్యాప్తిలో ఉన్న ఉప భాషల నుండి हिन्दी లోకి వచ్చాయో, వాటిని ‘देशज शब्द ‘ అంటారు.
उदा : कटरा, पाडा, तेन्दुआ, खिडकी, कटोरा, चिड़िया, ठुमरी, जूता, खिचडी, लोटा, डोल, डाब, डोंगी, पगडी, कलाई, ठेठ, डिबिया మొదలైనవి ఈ కోవకు చెందిన శబ్దాలే.

4. विदेशी : ఏ శబ్దాలైతే విదేశీ భాషల నుండి हिन्दी లోనికి వచ్చాయో, . వాటిని ‘विदेशी शब्द’ అంటారు. अरबी, फारसी, तुर्की, पुर्तगाली, మొదలైన భాషల్లోని వేలాది శబ్దాలు హిందీలోనికి వచ్చాయి. వాటిని యథాతదంగాగాని, లేక హిందీభాషా స్వభావాన్ననుసరించి కొద్దిపాటి మార్పుచేసి గాని, हिन्दी లో ప్రయోగించటం జరుగుతుంది.
उदा :
अ) अरबी के शब्द : फेसला, हैजा, कीमत, अल्लाह, ईमान, कायदा, नशा, कसरत, नहर, तरफ, मौसम, वापस, जवाहर, तकिया, दुनिया, अमीर, अदा, अजब, अजीब, अक्ल, असर, अदावत, अजायब, आखिर, आदमी, आदत, इज्जत, इनाम, इस्तीफा, इमारत, इलाज, उम्र, औरत, ओसत, ओलाद, एहसान, कदम, कब्र, कसूर, कसर, कमाल, कर्ज, किस्त, किस्मत, किला, किस्सा, कसम, किताब, कुर्सी, कातिल, खबर, खत्म, खिदमत, खत, खयाल, खराब, गरीब, जलजा, जिस्म, जनाब, जाहिल, जवाब जहाज, जिक्र, जालिम, तकदीर, तमाम, तमाशा, तारीख, तरक्की, तय, तजुर्बा, दवा, दावा, दफ्तर, दावत, दिमाग, दाखिल, दुआ, दगा, दुकान, दौलत, दान, नतीजा, नकल, फकीर, फायदा, बहस, बाकी, मदद, मरजी, मुहावरा, माल, मजबूर, मामूली, मतलब, मालूम, मुकदमा, मुल्लाह, मुक्त, मौका, मुसाफिर, मशहूर, राय, लिहाज, लफ्त, लायक, शराब, वकील, हिम्मत, हरामी, हिसाब, हम, हक, हुक्म, हाल, हाजिर, हमला, हाकिम, हौसला, हजाम, हवालात आदि ।

आ) फारसी शब्द : अफसोस, आतिशबाजी, आबरू, आराम, आवारा, आमदनी, आफत, आईना, आवाज, उम्मीद, कबूतर, कुश्ती, कमरबन्द, किशमिश, किनारा, खाल, खुद, खुश, खरगोश, खामोश, खूब, गुम, गलाह, गरम, गिरफ्तार, गुलाब, गोश्त, चश्मा, चर्खा, चिराग, चादर, चेहरा, चूँकि, चाशनी, जंग, जहर, दिंदगी, जादू, जान, जरमाना, जिगर, जोश, तेज, तनख्वाह, दीवार, देहान्त, दिल, नाव, पलंग, पैदावर, पलक, रेशा, पुल, पारा, पैमाना, बहरा, मजा, मुर्दा, मुफ्त, मुर्गा, मरहम, रंग, राह, लेकिन, लगाम, शादी, शोर, सितार, सरदार, सरकार, सौदागर, हजार, हफ्ता आदि ।

इ) तुर्किशब्द : चम्मच, चुगल, आका, उजबक, उर्दू, कैंची, काबू, कालीन कुली, चिक, चरमक, चेचक, तमगा, तलाश, तीप, बहादूर, बेगम, लफंगा, मुगल, लाश, सुराग, सौगात आदि ।

ई) पुर्तगाली भाषा के शब्द : अनन्नास, आलपीन, अलकता, आलमारी, आया, इस्तरी, इस्पात, कमीज, कमरा, किरानी, काजू, तम्बाकू, फोता, बाल्टी, तौलिया, गोभी, गोदाम, गमला, नीलाम, पादरी, हिस्तौल, बटन, मेज, लबादा आदि ।

उ) अंग्रेजी के शब्द : डांक्टर, इंजीनियर, स्कूल, कॉलेज, प्रोफेसर, बस, स्टेशन, टिकट, स्लेट, अफसर, पुलिस, जज, मिनिस्टर, ऑईर, इंच, एजेंसी, कम्पनी, कमीशन, कमिशनर, क्लास, क्रिकेट, गार्ड, जेल, चेयरमैन, ट्यूशन, डायरी, पेंसिल, पेपर, पेन, नम्बर, नोटिस, नर्स, थर्मामीटर, पार्सल, पेट्रोल, मीटिंग, कोर्ट, कोट, कॉलर, प्रेस, फेम, फोटो, कार, स्कूटर, रिक्शा, साइकिल, मोटर, इंजन, बल्ब, रेडियो, टेलीविजन, कम्यूटर, ट्रेन, क्लर्क कमिटी, वोर्ड, रोड, कैमरा, गेट, टाई, सिनेमा, फिल्म हीरो आदि ।

व्युत्पति: शब्दों का निर्माण वर्णा के मेल से होता है । वर्णों के मिलने से शब्द – खण्ड और शब्द बनाये जाते हैं। शब्द बनाने की यह प्रक्रिया ‘व्युत्पत्ति’ कहलाती है ।

उदाहरण : ‘घर’ शब्द ‘घ’ और ‘र’ दो वर्णों से बना है ।
शब्द – खण्डों के मेल से भी नये शब्द बनते हैं ।

उदा : ‘राज’ और ‘भवन इन दो शब्दों के मेल से ‘राजभवन’ शब्द बना है । वास्तव में ये शब्द खण्ड स्वतंत्र शब्द भी हैं !

संधि था समास की प्रक्रिया के द्वारा शब्द खण्ड (या. स्वतंत्र शब्द ) आपस में मिलकर थोडा बहुत परिवर्तित भी हो जाते हैं। तथा अलग अर्थ देते है ।

उदा ‘सूर्य’ और ‘उदय’ – इन दो शब्दों के मेल से ‘सुर्योदय’ शब्द बना है ।

व्युत्पति की दृष्टि से हिन्दी में शब्दो के तीन प्रकार होते हैं – रूढ़, यौगिक और योगरूढ़ ।

‘रूढ़’ शब्द से आशय वैसे शब्दों से है, जो निरर्थक वर्णों के मेल से बने हैं। जैसे- ‘ठं + डा ठंडा,’ ‘ग + र + म = गरम,’ ‘मी + ठा = मीठा’, ‘ती + ता = तीता’, ‘का + म = काम’, अलग – अलग ये सभी वर्ण निरर्थक हैं, लेकिन ये वर्ण मिलकर अर्थ देंने लगते हैं । ऐसा क्यों है ? ऐसा रूढि [ जन्म से अथवा प्रथा या रिवाज] के चलते हैं । ‘यौगिक’ शब्द वे हैं जो शब्द खण्डों अथवा शब्दों के जुड़ने से बने है । अर्थात इनके खण्ड सार्थक होते हैं ।

उदा : शिव + आलय = शिवालय, विद्य + आलय = विद्यालय, महा + सागर, महासागर, जय + माला = जयमाला, जन + जागरण = जनजागरण, महा + उत्सव महोत्सव आदी ! यहाँ हम देखते हैं कि सभी शब्दांश शार्थक हैं।

‘थोगरूढ़’ उन शब्दों को कहते हैं, जिनके शब्द – खण्ड सार्थक तो होते हैं, लेकिन उनके योग. होने पर शब्द – खण्ड अपने सावान्य अर्थ खोकर एक विशेष अर्थ देते हैं । यह विशेष अर्थ परम्परा से चला आया है ।

उदा : गण (समुह ) + ईश (स्वामी) – गणेश (शिव के पुत्र)
हिम (बर्फ) + आलय (घर) – हिमालय उत्तर में स्थित.
विशाल पर्वत श्रृंखला
जल (पानी) + ज ( जन्मा) – जलज – कमल
चक्र ( पहिया ) + प्राणि (हाथ) – चक्रपाणि (विष्णु)
पीत (पीला) + अम्बर (वस्त्र) + धारी ( धारण करने वाला) – पीतांबरधारी (विष्णु)
धन (बादल) + श्याम (काला) = धनश्याम (कृष्ण)

रचना के आधार पर शब्दों के दो भेद होते है. ।

1. विकारी
2. अविकारी

विकारी : विकारी शब्द वे हैं, जिनके रूपों में लिंग, वचन तथा. कारक के अनुसार विकार या परिवर्तन आ जाता है ।
लड़का – लड़का, लड़की, लड़के, लड़कों ।
देखना – देख, देखना, देखो, देखी, देखेँ, देखिए ।
वह – वे वहा, तू, तुम, आप

विकारी शब्द है :

1. संज्ञा,
2. सर्वनाम,
3. विशेषण,
4. क्रिया

अविकारी : अविकारी वे शब्द हैं जिनके रूप नही बदलते । इनका रूप सभी लिंग, विभक्तियों तथा वचनों में हमेशा एक सा रहता है । परिर्तिन नही होता हे । इसलिए इन्हे अव्यय भी कहते है ।
1. क्रिया विशेषण
2. सम्बन्ध – बोधक
3. समुच्चय बोधक
4. विस्मयादि बोधक

शब्द रचना

हम भाषा का व्यवहार अपने भावों व विचारों के संप्रेषण हेतु करते हैं. ओर कम परिश्रम करके अधिक अर्थ लाना चाहते हैं । अतः शब्दों के आरंभ में अथवा अंत में नये शब्दांश जोड़कर अथवा स्वर या व्यंजनों का परिवर्तन करके हम नया अर्थ लाते है। शब्दों के अरंभ में जुड़नेवाले शब्दांशों को ‘उपसर्ग’ तथा अंत में जुड़नेवाले शब्दांशों को ‘प्रत्यय’ कहा जाता है । अतः इन शब्दांशों से परिचित होना ही इस पाठ का लक्ष्य है ।

उपसर्ग

उपसर्ग में ‘उप’ का अर्थ है- ‘समीप’ तथा ‘सर्ग’ का अर्थ ‘सृष्टि करना’ है । अर्थात ‘उपसर्ग’ उस शब्दांश को कहते हैं, जो शब्द के समीप आकर नया शब्द बनाये । इनके जुड़ने के कारण मूल शब्द के अर्थ में या तो विशेषगुण का प्रतिपादन होता है । अथवा अर्थ बदल जाता है । वस्तुतः अव्यय होने के कारण थे अविकारी होते हैं । अर्थात इनमें लिंग, वचन, कारक आदि के कारण कोई परिवर्तन नही आयेगा । उपसर्गों का स्वतंत्र रूप में उपयोग नही होता । उपसर्गों का विवरण इस प्रकार है –
क) संस्कृत के उपसर्ग :

ख) हिन्दी के उपसर्ग :

ग) अरबी – फारसी और उर्दू के उपसर्ग :

प्रत्यय

शब्द के पीछे जुडनेवाले शब्दांश को ‘प्रत्यय’ कहा जाता है । प्रत्यय भी उपसर्ग की भांति अव्यय होते हैं । प्रत्यय शब्द में ‘प्रति’ का अर्थ है ‘साथ में’ । फिर अय का अर्थ बनता है. ‘चलनेवाला’ इस प्रकार प्रत्यय का अर्थ वह शब्दांश है जो बाद में जुडता हो । प्रत्यय के दो प्रकार हैं – कृदन्त और तध्दित । जो प्रत्यय क्रिया था धातु के अंत में प्रयुक्त होते हैं, उनको कृत प्रत्यय कहा जाता है । कृत प्रत्यय से बने शब्दों को कृदन्त कहा जाता है । जो धातु को छोड़कर अन्य शब्दों अर्थात् संज्ञा, सर्वनाम व विशेषण से जुड़ते हैं उनको तध्दित प्रत्यय कहा जाता है ।

कृदन्त प्रत्यय :

पर्यायवाची या समानार्थी शब्द

एक ही अर्थ को प्रकट करनेवाले भिन्न- भिन्न शब्दों को पर्यायवाची था समानार्थी शब्द कहा जाता है। जो जितने पर्यायवाची जानता है, वह अपने भावों व विचारों का संप्रेषण उतना ही सफलतापूर्वक कर सकता है । संमानार्थी शब्दों के जानने से शब्द भण्डार में वृध्दि होती है और शब्द भण्डार सफल अभिव्यक्ति का साधन होता है ।

शब्द – पर्यायवाची शब्द
1. आग – अग्नि, अनल, पावक, हुताशन, जातवेद, वैश्वानर
2. घोड़ा – अश्व, हय, वाजि, घोटक, तुरंग, तुरंगम
3. आकाश – नभ, अंबर, आसमान, गगन, व्योम, अंतरिक्ष
4. इच्छा – चाह, अभिलाषा, कामना, वांछा, आकांक्षा, मनोरथ
5. कमल – पंकज, नीरज, राजीव, सरोज, पदम, अरविंद
6. घर – भवन, सदन, मकान, गृह, आवास, निलय
7. नदी – सरिता, तटिनी, तरंगीणी, निर्झरिणी, सलिला
8. पक्षी – खग, विहंग, पखेरू, विहग, परिंदा, अंडज
9. पृथ्वी – भूमि, धरा, धरती, धरणी, वसुधा, धरित्री
10. समुद्र – सागर, जलधि, स्नाकर, पयोधि, जलनिधि, सिंधु
11. सूर्य – दिनकर, भास्कर, रवि, प्रभाकर, आदित्य, मित्र
12. अपमान – अनादर, बेइज्जती, अवमानना, निरादर, तिरस्कार
13. अनुपम – अनुप, अपूर्व, अतुल, अनोखा, अदभूत, अनन्य
14. अनुपम – तम, तिमिर, अंधेरा, अंधियारा, तमस, तमिस
15. जल – पानी, नीर, सलील, वारि, अंबु
16. वायु – पवन, समीर, हवा, मारुत, अनील
17. पेड़ – वृक्ष, तरु, झाड़, विटप
18. नारी – औरत, महिला, स्त्री, सबला
19. मनुष्य – मानव, नर, इंसान, आदमी
20. चाँद – चंद्र, राशि, रजनीश, सोम, कलानिधि
21. दिन – दिवस, दिवा, वार, याम
22. शिक्षक – गुरु; अध्यापक, आचार्य
23. धन – दौलत, संपति, संपदा, किन्त, अर्थ
24. फूल – पुष्प, सुमन, कुसुम
25. माता – माँ, जननी, अम्मा, अंबा
26. अहंकार – दंभ, गर्व, अभिमान, दर्प, मद, घमण्ड
27. अमृत – सुधा, अमिय, पीयुष
28. असुर – दैत्य, दानव, राक्षस, निशाचर
29. अतिथि – अभ्यागत, मेहमान, पाहूना
30. आँख – लोचन, नयन, नेत्र, चक्षु, दूग, विलोचन, अक्षि
31. ईश्वर – प्रभु, परमात्मा, ईश, जगदीश, भगवान, परमेश्वर
32. ओंठ – ओष्ठ, होंठ, अधर
33. क्रोध – रोष, गुस्सा, अमर्ष, नाराज़गी
34. किरण – रश्मी, मरीचि, मयूख, अंशु
35. अलि – भ्रमर, भँवरा, भोरा, षट्पद्, मधुकर, मिलिंद
36. उद्योग – परिश्रम, श्रम, मेहनत, प्रयास, उद्यम
37. पर्वत – पहाड़, गिरि, नग, शैल
38. राजा – नृप, नृपति, सम्राट, नरेश

विलोम शब्द

किसी शब्द के उल्टे या विपरीत अर्थ देनेवाले शब्द को उसका विलोम शब्द कहा जाता है । इन्हें भिन्नार्थक शब्द भी कहा जाता है। इन शब्दों का निर्माण अ, अन, दू, नि आदि उपसर्ग (या) हीन, पूर्ण आदि प्रत्ययों से अथवा कभी – कभी स्वतंत्र शब्द के रूप में भी होता है।
अंत × आदि
अति × अल्प
अपना × पराया
अपराजित × पराजित
अर्वाचीन × प्राचीन
अकाल × सुकाल
अनन्त × अन्त, सान्त
अज्ञ × विज्ञ
अपराधी × निरपराधी
अमृत × विष
अतिथि × आतिथेय
गर्मी × सर्दी
अंकाश × पाताल
घृणा × प्रेम
आचार × अनाचार
जल × थल
इच्छा × अनिच्छा
दया × क्रूरता
इत्थान × पतन
दक्षिण × उत्तर
उपस्थित × अनुपस्थित
दिन × रात
उत्कृष्ट × निकृष्ट
दुर्गन्ध × सुगन्ध
उपकार × अपकार
दुरुपयोग × सजुपयोग

उदघाटन × समापन
दुर × पास
उत्पति × विनाश
धूप × छाँव
उपचार × अपचार
नया × पुराना
उषा × संध्या
नरवर × शाश्वत
उत्तीर्ण × अनुत्तीर्ण
निर्बल × सबल
उल्लास × विषाद
निन्दा × स्तुति
उधार × नकद
निरक्षर × साक्षर
एक × अनेक
प्रश्न × उत्तर
कायर × वीर
बलवान × कमजोर
कडुवा × मीठा
कठिन × सरल
कमजोर × बलवान
शुभ × अशुभ
क्रोध × शान्त
जन्म × मृत्यु
कीर्ति × अपकीर्ति
ज्ञान × अज्ञान
ग्राम × नगर
स्वदेश × विदेश
हिंसा × अहिंसा
लाथक × नालायक
न्याय × अन्याय
कठिनाई × सरलता
उपस्थित × अनुपस्थित
राजा × रंक

आशा × निराशा
गुण × अनगुण / दोष
गलत × सही
स्वीकृत × तिरस्कृत
स्वर्ग × नरक
कृत्रिम × स्वाभाविक
कनिष्ठ × ज्येष्ठ, वरिष्ठ
घीटा × फ़ायदा
जड़ × चेतन
चतुर × मूर्ख
प्रसन्न × अप्रसन्न
भूरि × अल्प

Students must practice these TS Intermediate Maths 1B Solutions Chapter 3 Straight Lines Ex 3(b) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Straight Lines Solutions Exercise 3(b)

I.
Question 1.
Find the sum of squares of the intercepts of the line 4x – 3y = 12 on the axes of coordinates. (VS.A.Q.)
Answer:
Equation of the given line is 4x – 3y = 12
Writing in the intercepts form \(\frac{x}{a}+\frac{y}{b}\) = 1

∴ a = 3, b = – 4
∴ Sum of the squares = a² + b² = 9 + 16 – 25

Question 2.
If the portion of a straight line intercepted between the axes of coordinates is bisected at (2p, 2q), write the equation of the straight (V.S.A.Q.)
Answer:

Coordinates of A = (a, 0) and B = (0, b)
M is the midpoint of AB
∴ M = \(\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) = (2p, 2q)
∴ \(\frac{a}{2}\) = 2p and \(\frac{b}{2}\) = 2q ⇒ a = 4p, b = 4q

Question 3.
If the linear equations ax + by + c = 0 (abc ≠ 0) and lx + my + n = 0 represents the same line and r = \(\frac{l}{a}\) = \(\frac{n}{c}\), write the value of r in terms of m and b. (V.S.A.Q.)
Answer:
ax + by + c = 0 and lx + my + n = 0 represents the same line.
Hence coefficients are proportional
∴ \(\frac{l}{a}=\frac{m}{b}=\frac{n}{c}\) = r
r = \(\frac{\mathrm{m}}{\mathrm{b}}\) is the value of ’r’ interms of m and b.

Question 4.
Find the angle made by the straight line y = – √3 x + 3 with the positive direction of the X – axis measured in the counter clock-wise direction. (V.S.A.Q.)
Answer:
Equation of the given line is y = – √3x + 3 which is of the form y = mx + c where
m = tan α = √3 = tan \(\frac{2 \pi}{3}\) (angle made by the line in counter clock-wise direction)
∴ α = \(\frac{2 \pi}{3}\)

Question 5.
The intercepts of a straight line on the axes of coordinates are a and b. If p is the length of the perpendicular drawn from the origin to this line, write the value of p in terms of a and b. (V.S.A.Q.)
Answer:
Equation of the line in the intercepts form is
\(\frac{x}{a}+\frac{y}{b}\) – 1 = 0
p = length of the perpendicular drawn from origin to the line then

II.
Question 1.
In what follows, p denotes the distance of the straight line from the origin and a denotes the angle made by the normal ray drawn from the origin to the straight line with \(\overleftrightarrow{O x}\) measured in the anti-clockwise sense. Find the equations of the straight lines with the following values of p and α. (V.S.A.Q.)
(i) p = 5, α = 60°
(ii) p = 6, α = 150°
(iii) p = 1, α = \(\frac{7 \pi}{4}\)
(iv) p = 4, α = 90°
(v) p = 0, α = 0
vi) P = 2√2, α = \(\frac{5 \pi}{4}\)
Answer:
Equation of line in the normal form is x cos α + y sin α = p
(i) p = 5, α = 60°
Equation of line is x cos 60°+ y sin 60°= 5
⇒ \(\frac{x}{2}\) + y\(\frac{\sqrt{3}}{2}\) = 5
⇒ x + √3y = 10

(ii) p = 6, α = 150°
Equation of line is x cos 150° + y sin 150°= 6
⇒ x cos ( 180° – 30°) + y sin (180° -30°) = 6
⇒ – x cos 30° + y sin 30° = 6
⇒ – x \(\frac{\sqrt{3}}{2}\) + y \(\frac{1}{2}\) = 6
⇒ √3x + y = 12
⇒ √3x – y + 12 = 0

(iii) P = 1, α = \(\frac{7 \pi}{4}\)
Equation of the line is x cos \(\frac{7 \pi}{4}\) + y sin \(\frac{7 \pi}{4}\) = 1
⇒ x cos 315° + y sin 315° = 1
⇒ x cos (360° – 45°) + y sin (360° – 45°) = 1
⇒ x cos 45° – y sin 45° = 1
⇒ \(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}\) = 1
⇒ x – y – √2 = 0

(iv) p = 4, α = 90°
Equation of the line is
x cos 90° + y sin 90° = 4
⇒ x(0) + y(1) = 4
⇒ y – 4 = 0

(v) p = 0, α = 0
Equation of the line is
x cos 0 + y sin 0 = 0
⇒ x = 0

(vi) p = 2√2 , α = \(\frac{5 \pi}{4}\)
Equation of the line is
x cos \(\frac{5 \pi}{4}\) + y sin \(\frac{5 \pi}{4}\) = 2√2
⇒ x cos 225° + y sin 225° = 2√2
⇒ x cos (180° + 45°) + y sin( 180° + 45°) = 2√2
⇒ – x cos 45° – y sin 45° = 2√2
⇒ – \(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}\) = 2√2
⇒ x + y + 4 = 0

Question 2.
Find the equations of the straight lines in the symmetric form, given the slope and a point on the line in each part of the question. (S.A.Q.)
(i) √3, (2, 3)
(ii) – \(\frac{1}{\sqrt{3}}\), (- 2, 0)
(iii) – 1, (1, 1)
Answer:
(i) Equation of the line in symmetric form is
\(\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}\) = r
(x₁, y₁) = (2, 3), slope m = tan θ = √3 ⇒ θ = 60°
∴ Equation of the line is
\(\frac{x-2}{\cos \theta}=\frac{y-3}{\sin \theta}\) = √3
⇒ \(\frac{x-2}{\cos 60^{\circ}}=\frac{y-3}{\sin 60^{\circ}}\)

(ii) (x₁, y₁) = (-2, 0)
Answer:
Slope m = tan θ = – \(\frac{1}{\sqrt{3}}\)
⇒ θ = 180° – 30° = 150°
∴ Equation of the line is \(\frac{x+2}{\cos 150^{\circ}}=\frac{y}{\sin 150^{\circ}}\)

(iii) tan α = – 1, α = 180 – 45°
(x₁, y₁) = (1, 1)
∴ Equation of the line is
\(\frac{x-1}{\cos \left(\frac{3 \pi}{4}\right)}\) = \(\frac{y-1}{\sin \left(\frac{3 \pi}{4}\right)}\)

Question 3.
Transform the following equations into
(a) slope – intercept form (b) intercept form and (c) normal form. (S.A.Q.)
(i) 3x + 4y = 5
(ii) 4x – 3y + 12 = 0,
(iii) √3x + y = 4
(iv) x + y + 2 = 0
(v) x + y – 2 = 0
(vi) √3x + y + 10 = 0
Answer:
(i) 3x + 4y = 5
(a) Slope – intercept form : 4y = 5 – 3x
⇒ y = \(\frac{5}{4}-\frac{3}{4}\)
Slope m = – \(\frac{3}{4}\) and Y- intercept = \(\frac{5}{4}\)

(b) Intercept form : 3x + 4y = 5
⇒ \(\frac{x}{\left(\frac{5}{3}\right)}+\frac{y}{\left(\frac{5}{4}\right)}\) = 1
This is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1

(c) Normal form: 3x + 4y = 5,
Dividing by \(\sqrt{9+16}\) = 5 on both sides

(ii) 4x – 3y + 12 = 0,
(a) Slope – intercept form: 3y = 4x + 12
⇒ y = \(\left(\frac{4}{3}\right)\)x + 4
Slope = \(\frac{4}{3}\) and y-intercept = 4

(b) Intercept form: 4x – 3y = – 12
⇒ \(\frac{4}{-12} x+\frac{3}{12} y\) = 1
⇒ \(\frac{x}{(-3)}+\frac{y}{(4)}\) = 1

(c) Normal form : 4x – 3y + 12 = 0
⇒ – 4x + 3y = 12
Dividing by \(\sqrt{16+9}\) = 5 on both sides

(iii) √3x + y = 4
Answer:
(a) Slope – intercept form : y = -√3x + 4
Slope = – √3 , Y-intercept = 4

(b) Intercept form: √3 x + y = 4

(c) Normal form: √3x + y = 4
Dividing by √3 + 1 = 2 on both sides

(iv) x + y + 2 = 0
(a) Slope – intercept form: y = – x – 2
Slope = – 1, Y – intercept = – 2,

(b) Intercept form: x + y = – 2
⇒ \(\frac{x}{-2}+\frac{y}{-2}\) = 1

(c) Normal form : x + y + 2 = 0
⇒ – x – y = 2
Dividing by \(\sqrt{1+1}\) = √2 on both sides

(v) x + y – 2 = 0
(a) Slope – intercept form: y = – x + 2
Slope = -1, Y – intercept = 2

(b) Intercept form: x + y -2 = 0
⇒ x + y = 2
⇒ \(\frac{x}{2}+\frac{y}{2}\) = 1

(c) Normal form: x + y -2 = 0
⇒ x + y = 2
Dividing by \(\sqrt{1+1}\) = √2 on both sides

(vi) √3x + y + 10 = 0
Answer:
(a) Slope – intercept form : y = -√3x – 10
Slope = – √3 , Y – intercept = – 10

(b) Intercept form: √3 x + y = – 10

(c) Normal form: √3 x + y = – 10
Dividing by \(\sqrt{3+1}\) = 2 both sides

Question 4.
If the product of the intercepts made by the straight line x tan α + y sec α = 1 (0 ≤ α ≤ \(\frac{\pi}{2}\))on the coordinate axes is equal to sin α, find α. (S.A.Q.)
Answer:
Equation of the lines x tan α + y sec α = 1
⇒ \(\frac{x}{\cot \alpha}+\frac{y}{\cos \alpha}\) = 1
This is in intercepts form and
a = cot α, b = cos α
given ab = product of intercepts on the axes = sin α
∴ ab = sin α
⇒ cot α . cos α = sin α
⇒ \(\frac{\cos \alpha}{\sin \alpha}\) .cos α = sin α
⇒ cos² α = sin² α
⇒ tan² α = 1 ⇒ tan α = ± 1
(∵ 0 ≤ α ≤ \(\frac{\pi}{2}\)) we take α = \(\frac{\pi}{4}\) = 45°

Question 5.
If the sum of the reciprocals of the intercepts made by a variable straight line on the axes of coordinates is a constant, then prove that the line always passes through a fixed point. (S.A.Q.)
Answer:
Equation of the line in the intercept form is
\(\frac{x}{a}+\frac{x}{b}+\frac{x}{c}\) = 1 ……………. (1)
Sum of the reciprocals of the intercepts

Question 6.
Line L has intercepts a and b on the axes of the coordinates. When the axes are rotated through a given angle, keeping the origin fixed, the same line L has intercepts p and q on the transformed axes. Prove that \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}\). (S.A.Q.)
Answer:
Equation of the line in the original system in the intercepts form is \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ \(\frac{x}{a}+\frac{y}{b}\) – 1 = 0
Length of the perpendicular from origin
= \(\frac{|0+0-1|}{\sqrt{\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}}}\) ………………. (1)
Equation of the line in the second system in the intercepts form is

Question 7.
Transform the equation \(\frac{x}{a}+\frac{y}{b}\) = 1 into the normal form when a > 0 and b > 0. If the perpendicular distance of the straight line from the origin is ‘p’, deduce that \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\). (S.A.Q)
Answer:

III.
Question 1.
A straight line passing through A (- 2, 1) makes an angle of 30° with \(\overline{\mathrm{ox}}\) in the positive direction. Find the points on the straight line whose distance from A is 4 units. (S.A.Q.)
Answer:
We have symmetric form of the line as
\(\frac{x-x_1}{\cos \alpha}=\frac{y-y_1}{\sin \alpha}\) = r
∴ Coordinates of any point on the given line are (x₁ + r cos α , y₁ + r sin α)
Given α = 30°
⇒ cos α = cos 30° = \(\frac{\sqrt{3}}{2}\)
⇒ sin α = sin 30° = \(\frac{1}{2}\)
∴ (x₁, y₁) = (- 2, 1)
⇒ x₁ = – 2, y₁ = 1
Taking r = 4, coordinates of P
= [- 2 + 4\(\left(\frac{\sqrt{3}}{2}\right)\), 1 + 4\(\left(\frac{1}{2}\right)\)]
= [- 2 + 2√3, 3]
Taking r = – 4 coordinates of P’ are
= [- 2 + 4\(\left(\frac{\sqrt{3}}{2}\right)\), 1 – 4\(\left(\frac{1}{2}\right)\)]
= (- 2 + 2√3, – 1)

Question 2.
Find the points on the line 3x – 4y – 1 = 0 which are at a distance of 5 units from the point ( 3, 2 ). (S.A.Q.)
Answer:
Equation of the line in symmetric form is
\(\frac{x-3}{\cos \alpha}=\frac{y-2}{\sin \alpha}\) = r
Coordinates of the point P are
= ( 3 + r cos α, 2 + r sin α)
= ( 3 + 5 cos α, 2 + 5 sin α)
P is a point on the line 3x – 4y – 1 = 0
⇒ 3 (3 + 5 cos α) – 4(2 + 5 sin α) – 1 = 0
⇒ 15 cos α – 20 sin α = 0
⇒ 15 cos α = 20 sin α
⇒ tan α = \(\frac{15}{20}\) = \(\frac{3}{4}\)

Case (i): When α is in the first quadrant
cos α = \(\frac{-4}{5}\), sin α = – \(\frac{3}{5}\)

Case (ii): When α is in the third quadrant then
cos α = \(\frac{-4}{5}\), sin α = – \(\frac{3}{5}\)
From case (i) coordinates of P are
[3 + 5\(\left(\frac{4}{5}\right)\), 2 + 5\(\left(\frac{3}{5}\right)\)]
Case (iii): Coordinates of P are
[3 – 5\(\left(\frac{4}{5}\right)\), 2 – 5\(\left(\frac{3}{5}\right)\)] = [- 1, 1]

Question 3.
A straight line whose inclination with the positive direction of the X-axis measured in the anti clockwise sense is \(\frac{\pi}{3}\) makes positive intercept on the Y – axis. If the straight line is at a distance of 4 from the origin, find its equation. (S.A.Q.)
Answer:
Given α = \(\frac{\pi}{3}\), p = 4
We have m = tan α = tan 60° = √3
∴ Equation of the line in slope – intercept form is y = √3 x + c
⇒ √3x – y + c = 0
Distance from the origin = 4
∴ \(\frac{|0-0+c|}{\sqrt{3+1}}\) = 4
⇒ |c| = 2 × 4 = 8
⇒ c = ±8
When c > 0 we have c = 8
Hence the equation of the line is
√3x – y + 8 = 0

Question 4.
A straight line L is drawn through the point A (2, 1) such that its point of intersection with the straight line x + y = 9 is at a distance
of 3√2 from A. Find the angle which the line L makes with the positive direction of the X-axis. (S.A.Q.)
Answer:
Suppose a is the angle made by L with the positive X-axis.
Any point on the line is (x₁ + r cos α, y₁ + r sin α)
= (2 + 3√2 cos α, 1 + 3√2 sin α)
This is a point over the line x + y = 9
∴ (2 + 3√2 cos α) + ( 1 + 3√2 sin α) = 9
⇒ 3 √2 ( cos α + sin α) = 6
⇒ cos α + sin α = \(\frac{6}{3 \sqrt{2}}\) = √2
⇒ \(\frac{1}{\sqrt{2}}\) cos α + \(\frac{1}{\sqrt{2}}\) sin α = 1
⇒ cos α . cos 45° + sin α . sin 45° = 1
⇒ cos (α – 45°) = cos 0°
⇒ α – 45° = 0 ⇒ α = 45° = \(\frac{\pi}{4}\)

Question 5.
A straight line L with negative slope passes through the point ( 8, 2 ) and cuts positive coordinate axes at the points P and Q. Find the minimum value of OP + OQ as L varies, where O is the origin. (S.A.Q.)
Answer:
Equation of the line passing through A (8, 2) with negative slope – m is
y – 2 = – m(x – 8)
⇒ mx + y – (2 + 8m) = 0
⇒ mx + y = 2 + 8m


∴ Minimum value of OP + OQ as L varies where O is the origin is 18.

Students must practice this TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(f) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f)

Question 1.
Verify Rolle’s theorem for the following functions. (V.S.A.Q.)
(i) x² – 1 on [-1, 1]
Answer:
Let f(x) = x² defined on [-1, 1].
Since the function f(x) is a polynomial, it is continuous on [-1, 1] and differentiable on (-1, 1)
Also f(1) = 1 – 1 = 0, f(- 1) = (- 1)² – 1 = 0
∴ f (-1) = f (1). Hence f satisfy all conditions of Rolle’s theorem. Now we have to find a point c ∈ (-1,1) such that f'(c) = 0, f'(x) = 2x and f'(c) = 0 ⇒ 2c = 0
⇒ c = 0 ∈ (-1, 1)
Hence Rolle’s theorem is verified.

(ii) sin x – sin 2x on [0, π]
Answer:
Let f(x) = sin x – sin 2x defined over [0, π] f is continuous over [0, π] and differentiable over (0, π).
Also f(0) = 0 and f(π) = sin π – sin 2π = 0
∴ f(0) = f(π) = 0.
Hence f satisfy conditions of Rolle’s theorem’.
Also f'(x) = cos x – 2 cos 2x and f’ (c) = 0
⇒ cos c – 2 cos 2c = 0
⇒ cos c = 2 (2 cos²c – 1)
⇒ 4 cos²c – cos c – 2 = 0

Hence conditions of Rolle’s theorem are verified.

(iii) log (x² + 2) – log 3; over [-1, 1]
Answer:
f(x) = log (x² + 2) – log 3
This function is continuous over [-1,1] and differentiable over (-1,1)
f (-1) = log (1 + 2) – log 3 = 0
f(1) = log 3 – log 3 = 0
f (-1) = f (1)
Hence f satisfy all the conditions of Rolle’s theorem. So we have to find c ∈ (-1,1)
such that f’ (c) = 0 and f’ (c) = 0
⇒ c = 0 ∈ (-1, 1)
So Rolle’s theorem is verified.

Question 2.
It is given that Rolle’s theorem holds for the function f(x) = x³ + bx² + ax on [1, 3] with c = 2 + \(\frac{1}{\sqrt{3}}\). Find the values of a and b.
Answer:
Given that f(x) = x³ + bx² + ax defined over [1, 3] satisfy all the conditions of Rolle’s theorem.
∴ f ’(x) = 3x² + 2bx + a

b = -6, and b² – 3a = 3
⇒ 36 – 3a = 3 ⇒ a = 11
Hence a = 11 and b = – 6

Question 3.
Show that there is no real number k, for which the equation x² – 3x + k = 0 has two distinct roots in [0, 1]. (S.A.Q.)
Answer:
Let f(x) = x² – 3x + k and suppose there exists two different roots α, β (α < β). Since f is a polynomial in x, it is continuous over [0, 1] and differentiable in (0, 1). f is continuous over [a, 3] and differentiable in (a, β).
Also f(α) = α² – 3α + k = 0 and
f(β) = β² – 3β + k = 0
f(α) = f(β) = o :
f satisfy the conditions of Rolle’s theorem.
Also f'(c) = 0 ⇒ 3c² – 3 = 0
⇒ c² = 1
⇒ c = ± 1
This is a contradiction since
0 < α < c < β < 1
Hence there does not exists roots α, β in (0, 1).
So, there is no real number k for which the equation x² – 3x + k = 0 has distinct roots in[0, 1]

Question 4.
Find a point on the graph of the curve y = (x – 3)² where the tangent is parallel to the chord joining (3, 0) and (4, 1). (S.A.Q.)
Answer:
Let the points be A (3, 0) and B (4, 1)
∴ Slope of chord AB = \(\frac{1-0}{4-3}\) = 1
Given y = (x – 3)²
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2(x – 3)
Slope of the chord 2 (x – 3) = 1
⇒ 2x = 7 ⇒ x = \(\frac{7}{2}\)
∴ y = (x – 3)² = (\(\frac{7}{2}\) – 3) = \(\frac{1}{4}\)
The point on the curve = \(\left(\frac{7}{2}, \frac{1}{4}\right)\)

Question 5.
Find a point on the graph of the curve y = x³ where the tangent is parallel to the chord joining (1, 1) and (3, 27). (S.A.Q.)
Answer:
Let the points be A (1, 1) and B (3, 27)
Slope of chord AB = \(\frac{27-1}{3-1}=\frac{26}{2}\) = 13

Question 6.
Find ‘c’ so that f'(c) = \(\frac{f(b)-f(a)}{b-a}\) in the following cases. (S.A.Q)
(i) f(x) = x² – 3x – 1; a = \(\frac{-11}{7}\), b = \(\frac{13}{7}\)
Answer:

Question 7.
Verify the Rolle’s theorem for the function (x² – 1) (x – 2) on [- 1, 2]. Find a point in the interval where the derivative vanishes. (S.A.Q.)
Answer:
Let f(x) = (x² – 1) (x – 2)
= x³ – 2x² – x + 2 defined over [-1, 2] f being a polynomial in ‘x’, it is continuous over [-1, 2] and differentiable over (- 1, 2) f (- 1) = 0, f(2) = 0 f(-1) = f(2)
∴ f satisfy all the conditions of Rolle’s theorem.
f'(x) = 3x² – 4x – 1
and f'(c) = 0 ⇒ 3c² – 4c – 1 = 0

= 1.549 ∈ (-1, 2)
Rolle’s theorem is verified.

Question 8.
Verify the conditions of the Lagrange’s mean value theorem for the following functions. In each case find a point ‘c’ in the interval as stated by the theorem. (S.A.Q.)
(i) x² – 1 on [2, 3]
Answer:
Let f(x) = x² – 1 defined over [2, 3]. This being a polynomial in x, continuous on [2,3] and djfferentiable over (2,3). So by Lagrange’s mean value theorem there exists a point ‘c’ ∈ (2, 3) such that
f'(c) = \(\frac{\mathrm{f}(3)-\mathrm{f}(2)}{3-2}\)
f'(x) = 2x ⇒ f'(c) = 2c f(3) = 8, 1(2) = 3
∴ 2c = \(\frac{8-3}{1}\) = 5 ⇒ c = \(\frac{5}{2}\) ∈ (2, 3)

(ii) sin x – sin 2x on [0, π]
Answer:
Let f(x) = sin x – sin 2x defined over [0, π]. This is continuous on [0, π] and differentiable over (0, π) since
f’ (x) = cos x – 2 cos 2x exists for all x ∈ (0, π). So by Lrgrange s mean value theorem
f'(c) = \(\frac{\mathrm{f}(\pi)-\mathrm{f}(0)}{\pi-0}\), f(π) = 0, f(0) = 0
∴ cos c – 2 cos 2c = \(\frac{0}{\pi}\) =0
⇒ cos c – 2 cos 2c = 0
⇒ cos c – 2 (2 cos² c – 1) = 0
⇒ 4cos² c – cos c – 2 = 0
⇒ cos c = \(\frac{1 \pm \sqrt{1+32}}{8}=\frac{1 \pm \sqrt{33}}{8}\)
∴ c ∈ cos^-1\(\left(\frac{1+\sqrt{33}}{8}\right)\) ∈ (0, π)

(ii) log x on [1, 2]
Let f(x) = log x defined over [1,2]
This is continuous over [1, 2] and differentiable over (1, 2) since

Hence Lagrange’s mean value theorem is satisfied.

Students must practice this TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(e) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(e)

I.
Question 1.
At time ‘t’ the distance ‘x’ of a particle moving in a straight line is given by s = – 4t² + 2t. Find the average velocity between t = 2 and t = 8 sec. (V.S.A.Q.)
Answer:
s = – 4t² + 2t
∴ v = \(\frac{\mathrm{d} s}{\mathrm{dt}}\) = -8t + 2
Velocity at t = 2 is -16 + 2 = -14 units/sec.
Velocity at t = 8 is – 64 + 2 = – 62 units/sec.
Average velocity between t = 2 and t = 8 sec
= \(\frac{-62-14}{2}\) = – 38 units/sec.

Question 2.
If y = x⁴ then find the average rate of change of y between x = 2 and x = 4. (V.S.A.Q.)
Answer:

Question 3.
A particle moving along a straight line has the relation s = t³ + 2t + 3 connecting the distance s described by a particle in time t. Find the velocity and acceleration of the particle at t = 4 sec. (V.S.A.Q.)
Answer:
s = t³ + 2t + 3

Question 4.
The distance – time formula for the motion of a particle along a straight line is s = t³ – 9t² + 24t – 18. Find when and Where the velocity is zero. (S.A.Q.)
Answer:
Given s = t³ – 9t² + 24t – 18
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 3t² – 18t + 24
v = 0 ⇒ \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 0 ⇒ 3 (t² – 6t + 8) = 0
⇒ t = 2 or 4
The velocity is zero after 2 and 4 seconds

Case – (i) :
When t = 2, we have s = t3 – 9t2 + 24t – 18
= 8 – 36 + 48 – 13
= 56 – 54
= 2

Case – (ii) :
When t = 4, we have
s = t3 – 9t2 + 24t – 18
= 64 – 144 + 96- 18
= 160 – 162
= – 2
The particle is at a distance of 2 units from the starting point 0 on either side.

Question 5.
The displacement s of a particle travelling in a straight line in t seconds is given by s = 45t + lit2 – t3. Find the time when the particle comes to rest. (S.A.Q.)
Answer:
s = 45t + 11t² – t³
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 45 + 22t – 3t²
If the particle comes to rest then v = 0
⇒ 45 + 22t – 3t² = 0
⇒ 3t² – 22t – 45 = 0
⇒ 3t² – 27t + 5t – 45 = 0
⇒ (3t + 5) (t – 9) = 0
⇒ t = \(\frac{-5}{3}\) or t = 9
∴ t = 9 and the particle comes to rest after t = 9 seconds.

II.
Question 1.
The volume of a cube is increasing at the rate of 8 cm3/sec. How fast is the surface area increasing when the length of an edge is 12 cm ? (Mar. ’14) (S.A.Q.)
Answer:
Suppose ‘a’ is the edge of the cube and V be the volume of the cube.
V = a³ ………….(1)

Question 2.
A stone is dropped into a quiet lake and ripples move in circles at the speed of 5 cm/sec. At the instant when the radius of circular ripple is 8 cm, how fast is the enclosed area increases ? (S.A.Q.)
Answer:
Suppose r is the radius of the outer ripple and A be its area.
Then the area A = πr2
dA „ dr dr _
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = 2πr \(\frac{\mathrm{dr}}{\mathrm{dt}}\); Given r = 8, \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 5
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = 2π(8) (5) = 80 π cm² / sec

Question 3.
The radius of a circle is increasing at the rate of 0.7 cm/sec. What is the rate of Increase of its circumference ? (S.A.Q.)
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 0.7 cm/sec at
Circumference C = 2πr
\(\frac{\mathrm{dC}}{\mathrm{dt}}\) = 2π\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 2π(0.7) = 1.4π cm/sec

Question 4.
A balloon which always remains spherical on inflation is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of balloon increases when the radius is 15 cm. (S.A.Q.)
Answer:
Given \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 900 c.c/sec and r = 15 cm
Volume of the sphere V = \(\frac{4}{3}\)πr³

Question 5.
The radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cm/sec. At what rate is the volume of the bubble is increasing when the radius is 1 cm ? (S.A.Q.)
Answer:
\(\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{2}\) cm/sec end radius r = 1 cm
Given volume of sphere V = \(\frac{4}{3}\)πr³
\(\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3}\)π3r² = \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 4πr²\(\frac{\mathrm{dr}}{\mathrm{dt}}\)
= 4π × 1 × \(\frac{1}{2}\) = 2π cm³/sec

Question 6.
Assume that an object is launched upward at 980 m/sec. Its position would be given by s = – 4.9t2 + 9801. Find the maximum height attained by the object. (S.A.Q.)
Answer:
Given relation between s and t as s = – 4.9t² + 980 t
= – 9.8t + 980
For maximum height \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 0
s = – 4.9 (100)² + 980 (100)
= – 49000 + 98000 = 49000 units
The maximum height attained by the object is 49000 units.

Question 7.
Let a kind of bacteria grow in such a way that at time t sec, there are t3^2 bacteria. Find the rate of growth at time t = 4 hours. (S.A.Q.)
Answer:
Let ‘s’ be the amount of growth of bacteria at time’t’. Then s(t) = t^3/2
The growth rate at time’t’ is given by
\(\frac{\mathrm{ds}}{\mathrm{dt}}=\frac{3}{2}\)
When t = 4, \(\frac{3}{2}\) (4 × 60 × 60)
= \(\frac{3}{2}\) (2 × 60) = 180 units.

Question 8.
Suppose we have a rectangular aquarium with dimensions of length 8 m, width 4 in and height 3 m. Suppose we are filling the tank with water at the rate of 0.4 m3/sec. How feist is the height of water changing when the water level is 3 m ? (S.A.Q.)
Answer:
Length of aquarium l = 8 m
Width of aquarium b = 4 m
Height of aquarium h = 3 m
\(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 0.4 m/sec dt
V = lbh = 8 (4) (3) = 96
and log V = log l + log b + log h
\(\frac{1}{V} \frac{d V}{d t}=\frac{1}{h} \frac{d h}{d t}\) (∵l, b are constants)
⇒ \(\frac{0.4}{96}=\frac{1}{3} \frac{\mathrm{dh}}{\mathrm{dt}}\)
⇒ \(\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{3 \times 0.4}{96}=\frac{1}{80}\)

Question 9.
A container is in the shape of an inverted cone has height 8 m and radius 6 m at the top. If it is filled with water at the rate of 2 m³/minuite how fast is the height of the water changing when the level is 4 m ? (S.A.Q.)
Answer:
Given h = 8 m = OA
r = 6 m = AB

The above measurements are at instant of time’t’
In ΔOAB, ΔOCD we have

Question 10.
The total cost c(x) in rupees associated with the production of x units of an item is given by c(x) = 0. 007 x³ – 0.003 x² + 15x + 4000. Find the marginal cost when 17 units are produced. (S.A.Q.)
Answer:
Let m represents the marginal cost then m = \(\frac{\mathrm{dc}}{\mathrm{dx}}\)
Hence
m = \(\frac{d}{d x}\)(0.007x³ – 0.003x² + 15x + 4000)
= 0.007 (3x²) – 0.003 (2x) + 15
The marginal cost at x = 17 is
(m)_x=17 = 17 = (0.007) (3 × 172) – 0.003 (34) + 15
= (0.007) (3 × 289) – (0.003) (34) + 15
= (0.007) 867 – (0.003) (34) + 15
= 20.967

Question 11.
The total revenue in rupees received from the sale of x units of a produce is given by R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7. (S.A.Q.)
Answer:
Let m denotes the margin revenue. Then
m = \(\frac{\mathrm{d}}{\mathrm{dx}}\)
Given R(x) = 13x² + 26x + 15 m = 26x + 26
The marginal revenue when x = 7 is
(m)_x=7 = 7= 26 (7) + 26 = 208

Question 12.
A point P is moving on the curve y = 2x². The x coordinate of P is increasing at the rate of 4 units per second. Find the rate at which y coordinate is increasing when the point is at (2, 8). (S.A.Q.)
Given equation of the curve is y = 2x² ………….. (1)
Let P (x₁, y₁) be any point on the curve (1)
Let y₁ = 2x₁
x₁ = 2 and y₁ = 8
Differentiating (2) w.r.t ‘t’
\(\frac{d y_1}{d t}\) = 4x₁ \(\frac{d x_1}{d t}\)
= 4 (2) (4)
= 32 units/sec.
∴ y co-ordinate is increasing at the rate of 32 units/sec.

Students must practice this TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(d) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(d)

I. Find the angle between the curves given below. (March 2014) (E.Q.)

Question 1.
x + y + 2 = 0, x² + y² – 10y = 0
Answer:
x + y + 2 = 0 ⇒ x = – (y + 2)
x² + y² – 10y = 0
⇒ (y + 2)² + y² – 10y = 0
⇒ y² + 4y + 4 + y² – 10y = 0
⇒ 2y² – 6y + 4 = 0
⇒ y² – 3y + 2 = 0
⇒ (y – 2) (y – 1) = 0
⇒ y = 1 (or) y = 2
x = – (y + 2)
y = 1 ⇒ x = -(1 + 2) = -3
y = 2 ⇒ x = -(2 + 2) = -4
The points of intersection at P (- 3, 1) and Q (- 4, 2)
Equation of curve is x + y – 10y = 0
Differentiating w.r.to ‘x’

Question 2.
y² = 4x, x² + y² = 5. (E.Q.) (May 2007)
Answer:
The given equations of curves are y² = 4x and x² + y² = 5
Eliminating y, we get x² + 4x = 5
⇒ x² + 4x – 5 = 0
⇒ (x – 1) (x + 5) = 0
⇒ x = 1 or – 5
From y² = 4x and x = 1
⇒ y² = 4 ⇒ y = ±2
But when x = -5, y is imaginary and not real
Points of Intersection are
P(1, 2) and Q(1, -2)
From the equation y² = 4x
⇒ 2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 4
⇒ \(\frac{d y}{d x}=\frac{z}{y}\) …….(1)
At P(1, 2), slope of first curve from (1)
m₁ = \(\frac{2}{2}\) = 1
and m₂ = slope of second curve from (2)
= \(\frac{-1}{2}\)

Question 3.
x² + 3y = 3, x² – y² + 25 = 0 (E.Q.)
Answer:
x² + 3y = 3 ……….(1),
x² – y² + 25 = 0 ………….(2)
are the equations of given curves.
From (1) x² = 3 – 3y
From (2), 3 – 3y – y² + 25 = 0
⇒ – y² – 3y + 28 = 0
⇒ y² + 3y – 28 = 0
⇒ (y + 7)(y – 4) = 0
⇒ y = 4 or y = -7
If y = 4 then x² = 3 – 12 = -9
Values of x are not real.
If y = – 7 then x² = 3 + 21
⇒ x² = 24 = ± 2√6
∴ Points of intersection are
P (2√6 , -7) and Q (-2√6 , -7)
From (1), 2x + 3 \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=\frac{-2 x}{3}\)
Slope of the tangent at P (2√6 , -7) to the curve (1) is m₁ = \(\frac{-2(2 \sqrt{6})}{3}=\frac{-4 \sqrt{6}}{3}\)

Question 6.
y² = 8x, 4x² + y² = 32
Answer:
y = 8x ….(1)
4x² + y² = 32
⇒ 4x² + 8x – 32 = 0
⇒ x² + 2x – 8 = 0
⇒ (x + 4) (x – 2) = 0
⇒ x = 2 or x = – 4
When x = 2, y² = 16 ⇒ y = ± 4
∴ Points of intersection are P (2, 4) and Q (2, – 4)
From (1), 2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 8 ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{\mathrm{y}}\)

Question 7.
x²y = 4, y(x² + 4) = 8.
Answer:
x²y = 4 ………………(1)
y(x² + 4) = 8 …………(2)
From (1), x² = \(\frac{4}{y}\)

Substitute x² value in equation (2)
∴ y(\(\frac{4}{y}\)+4) = 8 ⇒ y\(\left(\frac{4+4 y}{y}\right)\) = 8
⇒ 4 + 4y = 8 ⇒ y = 1
∴ x² = \(\frac{4}{y}\)
⇒ x² = 4
⇒ x = ±2
∴ Points of intersection are P(2, 1), Q(-2, 1)
From (1)

Question 8.
Show that the curves 6x² – 5x + 2y = 0 and 4x² + 8y² = 3 touch each other at \(\left(\frac{1}{2}, \frac{1}{2}\right)\). (E.Q)
Answer:
The given curves are 6x² – 5x + 2y = 0 ………..(1)
and 4x² + 8y² = 3 ………..(2)

Slopes of two curves at \(\left(\frac{1}{2}, \frac{1}{2}\right)\) are equal and hence the given curves touch each other at \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

Students must practice this TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(c) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c)

I.
Question 1.
Find the length of subtangent and sub¬normal at a point of the curve y = b sin\(\frac{x}{a}\).
Answer:
Equation of the curve is y = b sin\(\left(\frac{x}{a}\right)\)
\(\frac{d y}{d x}\) = b cos\(\left(\frac{x}{a}\right) \cdot \frac{1}{a}=\frac{b}{a}\) cos\(\left(\frac{x}{a}\right)\)

Question 2.
Show that the length of the subnormal at any point in the curve xy = a² varies as the cube of the ordinate of the point. (V.S.A.Q.)
Answer:
Equation of the curve is xy = a²

∴ Length of the subnormal αy₁³ = cube of the ordinate.

Question 3.
Show that at any point (x, y) on the curve y = be^x/a, the length of the subtangent is a constant and the length of the subnormal is \(\frac{y^2}{a}\). (V.S.A.Q)
Answer:
Equation of the given curve is y = be^x/a
∴ \(\frac{d y}{d x}=\frac{b}{a}\)e^x/a = \(\frac{\mathrm{y}}{\mathrm{a}}\)
∴ Slope at any point P(x, y) = \(\frac{y}{a}\)
Length of the subtangent = |y₁/f'(x₁)|
= |y/\(\frac{y}{a}\)| = a = constant
Length of the subnormal = |y₁/f'(x₁)|
= |y.\(\frac{y}{a}\)| = \(\left|\frac{y^2}{a}\right|\)

II.
Question 1.
Find the value of k so that the length of the subnormal at any point on the curve xy^k = a^k+1 is a constant. (S.A.Q.)
Answer:
Equation of the curve is xy^k = a^k+1
Let P(xj, yO be any point on the curve then
x₁y₁^k = a^k+1 …………….(1)
Differentiating w.r. to x,

Length of the subnormal is constant at any point on the curve is independent of x₁ and y₁.
\(\frac{\mathrm{y}_1^{\mathrm{k}+2}}{\mathrm{k} \cdot \mathrm{a}^{\mathrm{k}+1}}\) is independent of x₁ y₁.
∴ k + 2 = 0
⇒ k = – 2

Question 2.
At any point t on the curve x = a (t + sin t), y = a (1 – cos t), find the lengths of tangent, normal, subtangent and subnormal. (S.A.Q.) (June 2004, Board Model Paper)
Answer:
Equation of the curve is
x = a (t + sin t), y = a (1 – cos t)

Question 3.
Find the length of normal and subnormal at a point on the curve y = \(\frac{a}{2}\left(e^{\frac{x}{a}}+e^{\frac{-x}{a}}\right)\) (S.AQ.) (March 2013)
Answer:
Equation of the curve is y = \(\frac{a}{2}\left(e^{\frac{x}{a}}+e^{\frac{-x}{a}}\right)\)
= a cos h\(\left(\frac{x}{a}\right)\)

Question 4.
Find the lengths of subtangent, subnormal at a point’t’ on the curve x = a (cos t + t sin t), y = a (sin t – t cos t) (May 2014) (S.A.Q.)
Answer:
Equation of the curve is
x = a (cos t + t sin t)
y = a (sin t – t cos t)
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = a(-sint + tcost + sint) = at cos t
and \(\frac{\mathrm{dy}}{\mathrm{dt}}\)= a (cost + tsint – cost) = at sin t dt
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} / \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{at} \sin \mathrm{t}}{\mathrm{at} \cos \mathrm{t}}\) = tan t
= \(\left|\frac{\mathrm{a}(\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t})}{\tan \mathrm{t}}\right|\)
= |a cot t(sin t – t cos t)|
Length of the subnormal = |y₁. f'(x₁)|
= |a(sin t – t cos t) tan t|
= |a tan t(sin t – t cos t)|

Students must practice this TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(b) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(b)

I.
Question 1.
Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 4. (V.S.A.Q.)
Answer:
Equation of the given curve is y = 3x⁴ – 4x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 12x³ – 4
Slope of the tangent at x = 4 is 12(4)³ – 4
= 12(64) – 4
= 764

Question 2.
Find the slope of the tangent to the curve y = \(\frac{x-1}{x-2}\) x ≠ 2 at x = 10. (V.S.A.Q.)
Answer:
Equation of the curve is y = \(\frac{x-1}{x-2}\)
= 1 + \(\frac{1}{x-2}\)
∴ \(\frac{d y}{d x}=-\frac{1}{(x-2)^2}\)
and slope of the tangent at x = 10 is
= \(-\frac{1}{(10-2)^2}=-\frac{1}{64}\)

Question 3.
Find the slope of the tangent to the curve y = x³ – x + 1 at the point whose x coordinate is ‘2’. (V.S.A.Q.)
Answer:
Equation of the curve is y = x³ – x + 1
∴ \(\frac{d y}{d x}\) = 3x² – 1
Slope of the tangent at x = 2 is 3(2)² – 1 = 11

Question 4.
Find the slope of the tangent to the curve y = x³ – 3x + 2 at the point whose x co-ordinate is ‘3’. (V.S.A.Q.)
Answer:
Equation of the curve is y = x³ – 3x + 2
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3x² – 3
Slope of the tangent at x = 3 is 3 (3)² – 3 = 24

Question 5.
Find the slope of the normal to the curve
x = a cos³θ, y = a sin³θ at θ = \(\frac{\pi}{4}\). (V.S.A.Q.)
Answer:
x = a cos³θ ⇒ \(\frac{d x}{d \theta}\) = 3a cos²θ (-sin θ) dθ
and y = a sin³θ ⇒ \(\frac{d y}{d \theta}\) = 3a sin²θ (cos θ)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d} \theta} / \frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{3 \mathrm{a} \sin ^2 \theta \cos \theta}{-3 \mathrm{a} \cos ^2 \theta \sin \theta}\) = – tan θ
∴ Slope of the tangent at θ = \(\frac{\pi}{4}\) is – tan \(\frac{\pi}{4}\) = -1
and hence the slope of the normal = 1.

Question 6.
Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos²θ at θ = \(\frac{\pi}{2}\). (V.S.A.Q.)
Answer:

Question 7.
Find the points at which the tangent to the curve y = x³ – 3x² – 9x + 7 Is parallel to the X-axis. (V.SA.Q.)
Answer:
Equation of the curve is y = x³ – 3x² – 9x + 7
∴ \(\frac{d y}{d x}\) = 3x² – 6x – 9

If the tangent is parallel to X – axis then
\(\frac{d y}{d x}\) = 0, ⇒ 3x² – 6x – 9 = 0
⇒ x² – 2x – 3 = 0
x = 3 or – 1
When x = 3, then
y = 3³ – 3(3)² – 9(3) + 7 = – 20
When x = – 1, then y = (- 1)³ – 3(- 1)² – 9 = (-1) + 7 = 12
∴ The required points are (3, – 20), (-1, 12)

Question 8.
Find a point on the curve y = ( x – 2)² at which the tangent is parallel to the chord joining the points (2,0) and (4, 4). (V.S.A.Q.)
Answer:
Equation of the curve is y = (x – 2)²
∴ \(\frac{d y}{d x}\) = 2(x – 2)
Slope of the chord joining points (2, 0) and (4, 4) is = \(\frac{4}{2}\) = 2
The tangent is parallel to the chord 2 (x – 2) = 2
⇒ 2x = 6
⇒ x = 3
y = (x – 2)² = (3 – 2)² = 1
The required point = (3, 1)

Question 9.
Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11. (V.S.A.Q.)
Answer:
Equation of the curve isy = x³ – 11x + 5
∴ \(\frac{d y}{d x}\) = 3x² – 11
The tangent is y = x – 11
Slope of the tangent is 3x² -11 = 1
⇒ 3x² = 12
⇒ x² = 4
⇒ x = ± 2
y = x – 11, if x = 2 ⇒ y = – 9
The point on the curve is (2, – 9)

Question 10.
Find the equations of all lines having slope zero which are tangents to the curve y = \(\frac{1}{x^2-2 x+3}\) (V.S.A.Q)
Answer:
Equation of the given curve is y = \(\frac{1}{x^2-2 x+3}\)
∴ \(\frac{d y}{d x}=\frac{-1(2 x-2)}{\left(x^2-2 x+3\right)^2}\)
Given slope of the tangent is ‘O’ .
\(\frac{-2(x-1)}{\left(x^2-2 x+3\right)^2}\) = 0 ⇒ x = 1
At x = 1 ⇒ y = \(\frac{1}{1-2+3}=\frac{1}{2}\)
The point is (1, \(\frac{1}{2}\))
Slope of the tangent is ‘0’
∴ Equation of the tangent is
y – \(\frac{1}{2}\) = 0 (x – 1) ⇒ 2y – 1 = 0

II.

Question 1.
Find the equations of tangent and normal to the following curves at the points indicated against. (S.A.Q.)
(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)
Answer:
\(\frac{d y}{d x}\) = 4x³ – 18x² + 26x – 10
At x = 0, slope of the tangent = – 10
∴ Equation of tangent is
y – 5 = – 10 (x – 0) = – 10x
⇒ 10x + y – 5 = 0
Slope of the normal = \(\frac{1}{10}\)
∴ Equation of normal is
y – 5 = \(\frac{1}{10}\)(x – 0) ⇒ 10y – 50 = x
⇒ x – 10y + 50 = 0

(ii) y = x³ at (1, 1)
Answer:
\(\frac{d y}{d x}\) = 3x²
At (1, 1), slope of the tangent = 3 (1)2 = 3
Equation of the tangent at P (1, 1) is
y – 1 = 3 (x – 1)
⇒ 3x – y – 2 = 0
Slope of the normal = –\(\frac{1}{3}\)

∴ Equation of the normal is
y – 1 = –\(\frac{1}{3}\)(x – 1)
⇒ 3y – 3 = -x + 1
⇒ x + 3y – 4 = 0

(iii) y = x² at (0, 0)
Equation of the curve is y = x²
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2x
At (0, 0) the slope of the tangent is ‘O’ Equation of the tangent is y – 0 = 0 (x – 0)
⇒ y = o
Let equation of normal which is perpendicular to the tangent is of the form x = k
Since the normal passes through (0, 0) we have k = 0
∴ Equation of the normal is x = 0

(iv) x = cos t, y = sin t at t = \(\frac{\pi}{4}\)
Answer:

(v) y = x² – 4x + 2 at (4, 2)
Answer:
Equation of the curve is y = x² – 4x + 2
∴ \(\frac{d y}{d x}\) = 2x – 4
Slope of tangent at (4, 2) is 2(4) – 4 = 4
Equation of the tangent is y – 2 = 4(x – 4)
⇒ 4x – y – 14 = 0
Slope of the normal = \(\frac{-1}{4}\)
∴ Equation of the normal is
y – 2= \(\frac{-1}{4}\)(x – 4)
⇒ 4y – 8 = -x + 4
⇒ x + 4y – 12 = 0

(vi) y = \(\frac{1}{1+x^2}\) at (0, 1)
Answer:
Given equation of the curve is y = \(\frac{1}{1+x^2}\)
∴ \(\frac{d y}{d x}=\frac{-2 x}{\left(1+x^2\right)^2}\)
At (0, 1), slope of the tangent = 0

Equation of the tangent is y – 1 = 0 (x – 0)
⇒ y = 1
Slope of the normal = ∞

∴ Equation of the normal is
y – 1 = \(\frac{1}{0}\) (x – 0)
⇒ x = 0

Question 2.
Find the equations of tangent and normal to the curve xy = 10 at (2, 5). (S.A.Q.)
Answer:
Given equation of the curve is xy = 10
y = \(\frac{10}{x} \Rightarrow \frac{d y}{d x}=\frac{-10}{x^2}\)

Slope of the tangent at the point (2, 5) is \(-\frac{10}{4}=-\frac{5}{2}\)
Equation of the tangent is
y – 5 = \(-\frac{5}{2}\) (x-2)
2y – 10 = -5x + 10
⇒ 5x + 2y – 20 = 0

Slope of the normal is \(\frac{2}{5}\)
∴ Equation of normal at (2, 5) is
y – 5 = \(\frac{2}{5}\)(x – 2)
⇒ 5y – 25 = 2x – 4
⇒ 2x – 5y + 21 = 0

Question 3.
Find the equations of tangent and normal to the curve y = x³ + 4x² at (-1, 3). (S.A.Q.) (May 2014)
Answer:
Equation of the curve is y = x³ + 4x²
∴ \(\frac{d y}{d x}\) = 3x² + 8x
Slope of the tangent at (- 1, 3) is
= 3 (- 1)2 + 8 (- 1) = – 5
Equation of tangent at (- 1, 3) is
y – 3 = – 5 (x + 1)
⇒ 5x + y + 2 = 0

Slope of the normal at (- 1, 3) is \(\frac{1}{5}\)
∴ Equation of normal at (- 1, 3) is
y – 3 = \(\frac{1}{5}\)(x + 1)
⇒ 5y – 15 = x + 1
⇒ x – 5y + 16 = 0

Question 4.
If the slope of the tangent to the curve x² – 2xy + 4y = 0 at a point on it is \(\frac{-3}{2}\), then find the equations of tangent and normal at that point. (S.A.Q.)
Answer:
Equation of the curve is
x² – 2xy + 4y = 0
Differentiating with respect to x

⇒ 2x – 2y = – 3x + 6
⇒ 5x – 2y – 6 = 0 (2)
⇒ 2y = 5x – 6
P (x, y) is a point on the curve and x² – x (5x – 6) + 2 (5x – 6) = 0
⇒ x² – 5x² + 6x + 10x -12 = 0
⇒ -4x² + 16x- 12 = 0
⇒ – 4 (x² – 4x + 3) = 0
⇒ x² – 4x + 3 = 0
⇒ (x – 1) (x – 3) = 0

Case – (i): x = 1, then form (1)
1 – 2y + 4y = 0 ⇒ 2y + 1 = 0 ⇒ y = –\(\frac{1}{2}\)
∴ Required point is P(1, –\(\frac{1}{2}\))

Equation of tangent is
y + \(\frac{1}{2}=\frac{-3}{2}\)(x – 1)
⇒ 2y + 1 = -3x + 3
⇒ 3x + 2y – 2 = 0

Equation of normal is
y + \(\frac{1}{2}=\frac{2}{3}\)(x – 1)
⇒ \(\frac{2 y+1}{2}=\frac{2}{3}\)(x – 1)
⇒ 6y + 3 = 4x – 4
⇒ 4x – 6y – 7 = 0

Case – (ii): When x = 3
∴ From (1); 9 – 6y + 4y = 0
⇒ y = \(\frac{9}{2}\)

Equation of the tangent is
y – \(\frac{9}{2}=\frac{-3}{2}\)(x – 33)
⇒ 2y – 9 = -3x + 9
⇒ 3x + 2y – 18 = 0

Equation of the normal is
y – \(\frac{9}{2}=\frac{2}{3}\)(x – 3)
⇒ \(\frac{2 y-9}{2}=\frac{2}{3}\)(x – 3)
⇒ 6y- 27 = 4x- 12
⇒ 4x – 6y + 15 = 0

Question 5.
If the slope of the tangent to the curve y = x log x at a point on it is \(\frac{3}{2}\), then find the equations of tangent and normal at that point. (S.A.Q.)
Answer:
Equation of the curve is y = x log x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x . \(\frac{1}{x}\) + log x = 1 + log x

Question 6.
Find the tangent and normal to the curve y = 2e^-x/3 at the point where the curve meets the Y-axis.
Answer:
Equation of the curve is y = 2e^-x/3
Equation of Y – axis is x = 0 ⇒ y = 2
Required point = (0, 2)
Equation of tangent at P (0, 2) is
y – 2 = \(\frac{-2}{3}\)(x – 0)
⇒ 3y – 6 = – 2x
⇒ 2x + 3y – 6 = 0
Equation of the normal is
y – 2 = y(x – 0)
⇒ 2y – 4 = 3x
⇒ 3x – 2y + 4 = 0

III.

Question 1.
Show that the tangent at P(x₁, y₁) on the curve curved √x + √y = √a is yy₁^{\(\frac{-1}{2}\)} – 2 +xx₁^{\(\frac{-1}{2}\)} = a^{\(\frac{1}{2}\)} (E.Q.) [June 2004]
Answer:
Equation of the curve is √x + √y = √a
Differentiating with respect to x,

Question 2.
At what points on the curve x2 – y2 = 2, the slopes of the tangents are equal to 2 ?(E.Q.)
Answer:
Given equation of the curve
x² – y² = 2 ……….(1)
Differentiating with respect to ‘x’
2x – 2y\(\frac{d y}{d x}\) = 0
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{\mathrm{y}}\)

∴ Slope of the tangent at P(x₁, y₁) = 2
Also \(\frac{x_1}{y_1}\) = 2 ⇒ x₁ = 2y₁

Since P(x,, y,) is a point on the curve

Question 3.
Show that the curves x² + y² = 2 and 3x² + y² = 4x have a common tangent at the point (1, 1). (E.Q.)
Answer:
Equations of the curves are
x² + y² = 2 ……….(1)
3x² + y² = 4x ……(2)
From (1) we have 2x + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-x}{y}\)
Slope of the tangent to the curve (1) at (1, 1) is m₁ = \(\frac{-1}{1}\) = -1

Also from (2), 6x + 2y = 4 dy
⇒ 2y\(\frac{d y}{d x}\) = 4 – 6x
⇒ \(\frac{d y}{d x}=\frac{4-6 x}{2 y}\)

Slope of the tangent to the curve (2) at (1, 1) is m₂ = \(\frac{4-6}{2}=-\frac{2}{2}\) = -1
The slope of the tangents to both the curves at P(1, 1) are same and pass through the same point (1,1).
∴ The given curves have a common tangent at P(1, 1).

Question 4.
At a point (x₁, y₁) on the curve x³ + y³ = 3axy show that the tangent is (x₁² – ay₁)x + (y₁² – ax₁) y = ax₁y₁ (E.Q.)
Answer:
Equation of the curve is x³ + y³ = 3axy
Differentiating w.r. to ‘x’

⇒ y (y₁² – ax₁) – y₁(y₁² – ax₁) = -x(x₁² – ay₁) – x₁(x₁² – ay₁)
⇒ x(x₁² – ay₁) + y(y₁² – ax₁) = -x(x₁² – ay₁) + x₁(x₁² – ay₁)
⇒ x(x₁² – ay₁) +y₁(y₁² – ax₁)
= x₁³ – ax₁y₁ + y₁³ – ax₁y₁
= x₁³ – ax₁y₁ + y₁³ – ax₁y₁
= x₁³ + y₁³ – 2ax₁y₁
= 3ax₁y₁ – 2ax₁y₁
= ax₁y₁
(∵ (x₁, y₁) is a point on the given curve and x₁³ + y₁³ = 3ax₁y₁)

Question 5.
Show that the tangent at the point P(2, – 2) on the curve y(1 – x) = x makes intercepts of equal length on the coordinate axes and the normal at P passes through the origin. (E.Q.)
Answer:
Given equation of the curve is y(1 – x) = x
⇒ y = \(\frac{x}{1-x}\)
∴ Differentiating w.r.t to ‘x’
\(\frac{d y}{d x}=\frac{(1-x)-x(-1)}{(1-x)^2}=\frac{1}{(1-x)^2}\)
At P(2, -2), slope of the tangent = \(\frac{1}{(1-2)^2}\) = 1

Equation of tangent at P(2, – 2) is
y + 2 = 1 (x – 2)
⇒ x – y = 4
⇒ \(\frac{x}{4}+\frac{y}{(-4)}\) = 1 (Intercepts form}
The tangent makes equal intercepts on the co-ordinate axes but they make intercepts with opposite sign.
Equation of normal at P is
y – y₁ = – 1 (x – x₁)
⇒ y + 2 = – 1(x – 2)
⇒ x + y = 0
This passes through the origin.

Question 6.
If the tangpnt at any point on the curve x^2/3+ y^2/3 – a^2/3 intersects the co-ordinate axes in A and B, then show that the length AB is aconstant (E.Q.) [Mar.’14,T3, ’08, ’07, ’05]
Answer:

Equation of the curve is
x^2/3+ y^2/3 = a^2/3 ……………..(1)
Differentiating w.r. to ‘x’

Question 7.
If the tangent at any point P on the curve xm y” = am+“ (mn 5* 0) meets the co-ordinate axes in A and B then show that AP : PB is a constant. (E.Q.)
Answer:
Given curve is x^myⁿ = a^m+n, (mn ≠ 0)
Differentiating with respect to ’x’
x^mny^n-1.\(\frac{d y}{d x}\) + yⁿ. mx^m-1 = 0
⇒ \(\frac{d y}{d x}=-\frac{y^n m x^{m-1}}{x^m n y^{n-1}}=-\frac{m}{n} \cdot \frac{y}{x}\)
Slope of the tangent at (x₁, y₁) = \(\frac{-\mathrm{my}_1}{\mathrm{nx}_1}\)

∴ Equation of the tangent at (x₁, y₁) is
y – y₁ = nx₁ (x – x₁)
⇒ nx₁y – nx₁y₁ = \(\frac{-m y_1}{n x_1}\)mxy₁ + mx₁y₁
⇒ mxy₁ + nx₁y = (m + n) x₁y₁

If P divides AB in the ratio m₁ : m₂ then co-ordinate of P are

∴ Point P divides AB in the ratio n : m and AP : PB = n : m = a constant.