TS Inter 1st Year

Telangana TSBIE TS Inter 1st Year Accountancy Study Material 9th Lesson Final Accounts of Sole Trading Concerns Textbook Questions and Answers.

TS Inter 1st Year Accountancy Study Material 9th Lesson Final Accounts of Sole Trading Concerns

Essay Questions:

Question 1.
Define final accounts and explain various stages in their preparation.
Answer:
Final Accounts :

1. To know the business results and financial position at the end of the period, the business prepares various statement which are called as “Final Accounts”.
2. These final accounts of a sole trader are prepared based on the nature of the business establishments.
3. If the business organization is trading concern it has to prepare Trading Account, profit & loss account and Balance sheet, if the concern is manufacturing entities they prepare manufacturing account, trading account, profit and loss account and Balance sheet.

Various stages in Financial Accounts :
1. Trading account:

• Preparation of Trading Account is the first stage in final accounts. It is prepared to find the “Gross Profit” or “Gross Loss”, which is transferred to profit and loss account.
• Trading account is a “nominal account” in nature. All the trading expenses should be debited and trading incomes should be credited to this trading account.

2. Profit & Loss account:

• The second stage of the preparation of final account is the preparation of profit & loss account. Profit & loss account is prepared to find out the Net Profit or Net Loss of the business. This is a nominal account.
• The balance of profit and loss account is transfered to capital account in Balance sheet.

3. Balance Sheet:
Balance sheet is the statement prepared to find out financial position i.e., assets and liabilities of a concern on a given date. In preparing Balance Sheet, liabilities and capital are shown on the left hand side and the assets are shown on the right hand side.

Question 2.
Describe various types of expenses and incomes with suitable examples.
Answer:
In the preparation of final accounts, clear distinction is made between capital and revenue items. The allocation of expenditure and incomes between capital and revenue plays a vital role in the preparation of correct and true financial statements of the business concerns.
Business expenditure of a concern can be broadly classified into :

1. Capital expenditure,
2. Revenue expenditure,
3. Deferred revenue expenditure.

1. Capital Expenditure :
Capital expenditure is the expenditure which is normally incurred for acquiring fixed assets which increases the earning capacity of the business. Benefits of capital expenditure extends over number of years. Examples of capital expenditure are purchase of fixed assets like plant, building, installation of machinery and their improvement. These are shown as assets in Balance Sheet.

2. Revenue Expenditure :
Revenue expenditure is the expenditure which is incurred in normal course of the business activities. The benefit of revenue expenditure is limited to only one accounting year. Examples of revenue expenditure are payment of salaries, rent, carriage, advertisement etc. These are debited to profit and loss account.

3. Deferred Revenue Expenditure :
Any amount of revenue expenditure spent in huge sums and its benefits spread over more than one year is called deferred revenue expen-diture. It includes the expenses like preliminary expenses, discount on issue of shares and debentures, heavy amount spent of advertisement, shifting of business premises etc.

Incomes are divided into capital receipts and revenue receipts :
1. Capital Receipts / Capital Income :
Amount received as investment by the owners, raised by way of loans and sale proceeds of fixed assets is called capital receipts (or) capital income. Ex. Capital, sale of machinery. All the capital receipts are recorded as liability in the Balance Sheet.

2. Revenue Receipts / Revenue Income :
Amount received in the normal course of business is called revenue receipts and includes sale of goods, interest, discount, commission etc. These incomes are credited to profit and loss a/c.

3. Deferred Income :
This consists of income which spreads over several years is called deferred income. Ex. Rent or interest received for more than one year.

Question 3.
Write the differences between trading account and profit & loss account.
Solution:

Question 4.
Present the main features, advantages and limitations of final accounts.
Answer:
Features :

1. Final accounts consider monetary transactions only.
2. Final accounts considers only those transactions which are of historical nature, i.e., the transactions which are already taken place.
3. Preparation of final accounts is a legal requirement.
4. Final accounts are used by internal and external users for their decision making.

Advantages :

1. Business profit or loss can be known to the trader through the trading account and profit and loss account (Income statements).
2. Financial position can be revealed by the preparation of Balance Sheet.
3. Final accounts are important source of finance information and this help the trader or management to plan the financial activities of the business concern for any period or time.
4. Financial statement help the trader to take business decisions by comparing current year results with the results of the previous year statements.
5. As the profit & loss account discloses either profit or loss, based on which a trader prepares himself to pay the taxes correctly.
6. Tax authorities also needs financial statements to determine the account of tax exactly.
7. As financial statements reveals the solvency position of the organization, the banks and other lending organizations may con v r for extending the loan facility.

Limitations :

1. Do not reflect the current prices as they are based on the historical costs. *
2. Do not consider qualitative data, such as, qualLy, efficiency of workforce, employee and employer relationship, motivation level of employees, value of human resources etc.
3. Do not reveal the accurate picture of the business, as certain values of assets and some expenses and income items are based on the judgement of the management, who may have prejudice.

Short Answer Questions:

Question 1.
Give the objectives of final accounts.
Answer:
The main objectives of preparation of final accounts are :

1. To ascertain the profit or loss of the business for a particular period. (By preparing Trading and Profit & Loss Account.)
2. To find out the financial position of the business concern on a specified date or period. (By preparing the Balance Sheet).

Question 2.
Write the features of Trading Account.
Answer:
Features :

1. It is the first stage of final accounts.
2. Trading account is a Nominal account.
3. It helps to find the Gross Profit or Gross Loss.
4. It records only the net sales and direct cost of goods sold.
5. The balance of trading account is transferred to profit and loss account.

Question 3.
Explain the meaning and main characteristics of profit & loss account
Answer:
After preparing the trading account, profit and loss account is prepared to find out the net profit or net loss of the business. This is also a nominal account. So, all the expenses and losses should be debited and all the incomes and gains to be credited to profit and loss account.

The balance of profit and loss account is either net profit or net loss and the same is added or deducted from the capital account in the balance sheet.

Characteristics of Profit and Loss Account :

1. Profit & Loss account is a Nominal account.
2. It is prepared at end of the financial year.
3. It records all current year indirect expenses and incomes.
4. It reveals net profit or net loss.
5. Net profit ratio can be calculated.
6. Current year administration expenses and other expenses can be compared with the previous years expenses.
7. It facilitates for the preparation of Balance Sheet.

Question 4.
Briefly explain various types of assets.
Answer:
Assets :
Property of any description belonging to a person or business organization can be named as an asset. Assets are the ‘ownings’ of a business and they may be classified as follows :

1. Fixed Assets :
The assets permanent in nature and not intended for the re-sale, but are used to carry the business continuously are called fixed assets. They earn profit or gain to the business.
Ex : Land and buildings, machinery, furniture etc. These assets are recorded in the balance sheet after deducting depreciation if any.

Fixed Assets can be further classified as :
a) Tangible Assets – which can be seen and touched, e.g., Furniture, Machinery etc.
b) Intangible assets – which can neither be seen nor touched, e.g., Patents, Goodwill etc.

2. Current Assets :
These assets are held for resale or can be converted into cash on a later date. These are also known as floating or circulating assets. Cash, Stock, Debtors, Bank balance etc., are the examples of these assets.

Question 5.
Define Manufacturing Account and give it contents.
Answer:
Manufacturing Account:
A trader or a business concern who undertakes the job of manufacturing or production of goods will prepare this account.

As trading account it is also a nominal account. Hence, all the expenses incurred on goods and factory will be considered and rec orded at debit side of it and in progress and sale of scrap etc., are recorded at creuit side and the difference debit and credit will be termed as ‘Cost of Production’, wnieh will be transferred to the debit side of the trading account.

This accovnt may be prepared either on horizontal form or on vertical form.

Horizontal form Proforma:

Vertical form Proforma:

Manufacturing Account of XYZ Co., for the year ending XX, XX, XXX

Very Short Answer Questions:

Question 1.
Define deferred expenditure.
Answer:
The Deferred Expenditure consists of revenue and capital items. Benefits from these expenses are spread over several years. In other words, the expenses which are incurred in one accounting year but benefits of the same accrue in the future years are called as Deferred Expenses.
Ex : amount of expenditure on advertisement in the initial years, expenditure incurred on research and development, innovation etc.

Question 2.
Define current assets & current liabilities.
Answer:
a. Current Assets:
Current assets are those assets which are for resale or can be converted into in short period, usually within one year. These are also known as floating or circulating assets.
Ex.: Cash in hand, Cash at bank, Sundry debtors, Stock-in-trade etc.

b. Current Liabilities :
These liabilities are payable by the business organisation within one accounting period. These are short term liabilities as they are repayable in not more than 12 months from the date of acquiring them.
Ex.: Bills payable, Sundry creditors, Bank overdraft etc.

Question 3.
Give examples of intangible assets.
Answer:
Intangible assets are those which cannot be seen and touched.
Ex.: Patents, Goodwill etc.

Question 4.
Write accounting equation.
Answer:

1. The relationship of assets with liabilities and owners equity in the equation form is known as “Accounting Equation”.
2. Accounting Equation: Assets = Liabilities + Capital
3. The complete picture of the balance sheet in manner of accounting equation.

Question 5.
Write a short note on Opening Stock & Closing stock.
Answer:
a) Opening Stock :
The opening stock is the first item to be shown on the debit side of trading account. The closing stock of the preceeding year is the opening stock of the current year.

b) Closing Stock :
It is the unsold stock left with the organization on the last day of the accounting year. The closing stock of the current year becomes the opening stock of the next year.

If the closing stock is given in the adjustments, record it in the trading account on credit side and also in Balance Sheet on assets side.

Additional Questions:

Question 1.
Capital expenditure.
Answer:

1. Capital expenditure is the expenditure which is normally incurred for acquiring fixed assets or assets which increase the earning capacity of the business. Benefits of capital expenditure are extended over a number of years.
2. Ex. : Purchase of fixed assets like machinery, building, furniture etc.

Question 2.
Revenue expenditure
Answer:

1. Revenue expenditure is the expenditure incurred in the normal course of the business activities. The benefit of revenue expenditure is restricted to only one accounting year.
2. Ex: Office expenses such as rent, salaries, selling expenses like carriage outwards, advertisement expenses.

Question 3.
Capital income.
Answer:

1. Any amount received as investment by the owners, raised by way of loans and income received on sale of fixed assets is called capital income.
2. Ex.: Capital, sale of machinery.

Problems:

Question 1.
Prepare Trading Account of Srikanth Traders for the year ended 31.12.2015.

Solution:

Question 2.
Prepare Trading Account from the following particulars for the year ended 31.03.2017 :

Solution:

Question 3.
Prepare Trading Account:

Solution:

Question 4.
Prepare Trading Account of Hyderabad Traders as on 31,12.2017 :

Solution:

Note :
Manufacturing expenses given twice so, 2nd one is not taken into problem.

Question 5.
From the following particulars, prepare Profit & Loss a/c as on 31.03.2019 :

Solution:

Note :
Stationary is given twice so we take 1st one and ignore 2nd one.

Question 6.
Prepare Profit & Loss a/c from the following particulars :

Solution:

Question 7.
From the following particulars, prepare Trading and Profit & Loss a/c as on 31.12.2018.

Solution:

Question 8.
Prepare Trading a/c & Profit & Loss a/c of Suresh Traders for the year ending 31.12.2017.

Solution:

Question 9.
From the following Trial Balance, prepare Trading a/c, and Profit & Loss a/c of Ms. Veena Reddy for the year ended 31.12.2018.

Solution:

Question 10.
Prepare Trading Account and Profit & Loss a/c (amounts in rupees).

Solution:

Question 11.
Prepare Balance Sheet from the following details of Yadagiri as on 31.12.2016.

Solution:
Balance Sheet of Yadagiri as on 31.12.16

Question 12.
Prepare Balance Sheet of Karnakar Reddy Traders from the following as on 31.12.2017.

Solution:
Balance Sheet of Karnakar Reddy as on 31.12.2017

Question 13.
Prepare Balance Sheet of Madhavi Traders for the year ended on 31.12.2016.

Solution:
Balance Sheet of Madhavi Traders as on 31.12.16

Question 14.
From the following Trial Balance, prepare Trading, P & L Account and Balance Sheet.

Solution:

Balance Sheet

Question 15.
From the following trial balances prepare final accounts for the year ending 31.03.2015.

Solution:

Balance Sheet

Textual Questions:

Question 1.
From the following information, prepare Trading Account of Shylaja Traders for the year ending 31.12.2018:

Solution:

Question 2.
From the following information, prepare Trading Account of Swamy as on 31.12.2018 :

Solution:

Question 3.
Prepare Trading Account of Ayyappa Traders for the year ending 31.12.2018.

Solution:

Question 4.
From the following particulars, prepare Profit & Loss a/c of Sathwlka for the year ending 31.12.2018 :

Solution:

Question 5.
Prepare Profit & Loss a/c of Naveen Kumar Traders as on 31.03.2018.

Solution:

Question 6.
From the following balances, prepare Profit & Loss a/C of Harini for the year ended 31.12.2018.

Solution:

Question 7.
Prepare leading Account and Profit & Loss a/c from the following particulars.

Solution:

Question 8.
Prepare Balance Sheet of Srinivas from the following particulars as on 31.12.2017 :

Solution:
Balance Sheet of Srinivas as on 31.12.2018

Telangana TSBIE TS Inter 1st Year Accountancy Study Material 10th Lesson Preparation of Final Accounts Textbook Questions and Answers.

TS Inter 1st Year Accountancy Study Material 10th Lesson Preparation of Final Accounts

Essay Questions:

Question 1.
Describe the various types of adjustments with suitable examples.
Ans.
The following are the various adjustments :
1. Outstanding expenses:
Outstanding expenses are the expenses relating to the current year but unpaid during the year and are to be paid in the next year is known as outstanding expenses. E.g. Salary or rent for the month of March is due but not paid. These expenses are added to the concerned expenditure either in trading or profit and loss account debit side. These expenses are again shown as liability in balance sheet.

2. Prepaid expenses :
Prepaid expenses are the expenses relating to the next year but paid during the year. E.g. Insurance or taxes are paid for the next year. These expenses are deducted from the concerned expenditure either in trading or profit and loss account debit side. These expenses are shown as assets in the balance sheet.

3. Income to be received :
It is also known as accured income. It is the income relating to the current year which is not received but to be received in the next year. These incomes are added to the concerned income in profit and loss account credit side. These incomes are shown as assets in the balance sheet.

4. Income received in advance:
Income relating to the next year but received in advance during the current year. These incomes are deducted from the concerned income in profit and loss account credit side. These incomes are shown as liability on the Balance Sheet.

5. Depreciation on fixed assets:
The value of fixed assets like machinery, plant, buildings will decrease year after year due to reasons like wear and tear, obsolescence etc. Such decline is called as depreciation. It is considered as an expense and generally calculated as a percentage on the value of asset. It is debited to profit and loss account and again deducted from the asset on the balance sheet.

6. Interest on capital :
Interest paid on the owners capital is treated as expenditure is debited profit and loss account. It is added to the capital on the liabilities side of the balance sheet.

7. Interest on drawings :
The amount of cash or goods taken for personal use are known as drawings. Interest on such drawings is credited to profit and loss a/c and again it should be deducted from capital in balance sheet.

8. Closing stock :
If the closing stock is given as adjustment, it is credited to trading a/c and shown as an asset in the balance sheet.

9. Bad debts :
The debts which cannot be recovered are called as bad debts.
i) If the bad debts are given in the trial balance. It is debited to profit and loss account only.
ii) If the bad debts are given as adjustment, they should be debited to profit and loss account and deducted from the sundry debtors on the asset side of the balance sheet.

10. Provision for bad debts :
Some of the debts of a year may or may not be recovered in the next year. These are known as doubtful debts. So, the trader create some amount on current year to meet the doubtful debts of the next year which is called provision for bad and doubtful debts when the provision is given as an adjustment, the amount so calculated on debtors is debited to profit and loss a/c. It is again deducted from the debtors on the asset side of the balance sheet.

11. When provision for doubtful debts is given in the trial balance and also in adjustments. The provision given in the trial balance is created in the last year, is the old reserve. If the new provision is more than the old provision the difference is debited to profit and loss account, and the new provision is deducted from the debtors in the balance sheet.

On the other hand, if the new provision is less than old provision. The difference is credited to profit and loss account. New provision only deducted form sundry debtors in the balance sheet.

Short Answer Questions:

Question 1.
What do you mean by Adjustments ? State the importance of Adjustments.
Answer:
On the date of preparing the final accounts, all the expenses whether paid or not and all incomes whether received or not should be taken, similarly, expenses and incomes relating to next year should not taken. All these items are to be adjustment in final accounts by way of adjustments entries. Addition or subtraction of the amounts from revenue items is called adjustment.

Importance of adjustments :

1. Expenses or incomes relating to the accounting period can be ascertained accurately.
2. Profit or loss can be calculated accurately.
3. True value of assets and liabilities can be ascertained easily.
4. Unrecorded transactions can be brought into the books of accounts.

Question 2.
Write the following :
a) Interest on capital.
b) Interest on Drawings.
Answer:
a) Interest on capital :
It is the amount of interest payable on owner’s capital by the business firm. It is an expenditure.

Adjustment entry :
Interest on capital a/c Dr
To Capital a/c (Being the interest payable capital)

Calculate interest on capital at a given percentage and is debited to profit and loss account. Interest on capital is added to capital on the liabilities side of balance sheet.

b) Interest on drawings:
The amount of cash or goods taken by the proprietor for personal use is called drawings. Interest should be calculated at a given rate and it is credited to profit and loss account as it is income to business. The interest on drawings is deducted from the capital on the liabilities side of balance sheet.

Interest on Drawings entry :
Capital / Drawings a/c Dr
To Interest on Drawings a/c (Being interest receivable on drawings).

Question 3.
Briefly explain about the following.
a) Depreciation
b) Loss of stock by Fire.
Answer:
a) Depreciation :
The value of fixed assets such as Buildings, Machinery, Furniture, Loose tools etc., will decrease year after year due to various reasons, such as wear and tear, obsolescence etc. Such decline in the value of fixed asses is called “Depreciation”.

Depreciation may be defined as “the gradual decrease in the value of an asset”. Depreciation should be considered as a business expense and is to be charged to the Profit and Loss Account. Generally depreciation is calculated as a percentage on the value of the asset given in Trial Balance.

Adjustment Entry :
Depreciation a/c Dr XX
To Asset a/c (individually) XX
(Being the depreciation provided on asset)

b) Loss of stock by Fire Accident:
Sometimes business stock may be lost due to fire accident. A provision has to be made for such loss. Stock lost by fire accident is an abnormal loss and hence it should be treated properly in the final accounts. Firstly, the total loss of stock by fire should be shown on credit side of the Trading Account. Again, it should be treated according to the following situations :
a) When Stock is fully insured.
b) When Stock is partly insured.
c) When Stock is not insured at all.

a) When Stock is fully insured and full claim is admitted by the Insurance Company : Adjustment Entries
i) Loss of Stock by Fire Accident a/c …………. Dr XX
To Trading a/c XX
(Being loss of stock by fire accident)

ii) Insurance company a/c …………. Dr XX
To Loss of stock by Fire Accident a/c XX
(Being claim admitted by the Insurance Company)

b) When stock is partly insured and Insurance company admits a part of the claim :
i) Loss of Stock by Fire Accident a/c …………. Dr XX
To Trading a/c XX
(Being loss of stock by fire accident)

ii) Insurance company a/c …………. Dr XX
Profit and Loss a/c …………. Dr XX
To Loss of stock by fire accident a/c XX
(Being part of the claim admitted by the Insurance Company and remaining loss transferred to P & L a/c)

c) When Stock is not insured at all:
i) Loss of Stock by Fire Accident a/c …………. Dr XX
To Trading a/c XX
(Being loss of stock by Fire Accident)

ii) Profit and Loss a/c …………. Dr XX
To Loss of Stock by Fire Accident a/c XX
(Being loss of stock transferred to P & L a/c).

Question 4.
Explain the treatment of Adjustment on Debtors in Final accounts.
Answer:
In final accounts, Bad Debts, Provision for Bad and Doubtful Debts and Provision for Discount on Debtors may be given as adjustments relating to Debtors.

I. Bad Debts :
When goods are sold on credit basis, some cf the customers may not pay die amount. The debts which are not collected and irreco\s ole are known as Bad Debts which are a loss.

After preparing the Trial Balance, the trader may notice some amounts are irrecoverable debts. These bad debts are called Further Bad Debts or Additional Bad Debts.

Adjustment Entry :
Bad Debts a/c Dr XX
To Sundry Debtors a/c XX
(Being bad debts written off)

Accounting Treatment in Final Accounts :

a) When bad debts are given only in the adjustments :
Accounting Treatment :

1. Bad Debts are to be shown on debit side of the Profit and Loss Account.
2. Bad Debts should be deducted from the debtors on the assets side of the Balance Sheet.

If the Bad debts are given in Trial Balance only, it should be shown on debit side of the Profit & Loss Account as a loss and need not be shown in Balance sheet.

b) When Bad Debts are given in both Trial Balance and Adjustments :
Accounting Treatment in Final Accounts :

1. Both the bad debts (Bad debts given in Trial Balance and given in adjustment) are ~ to be shown on debit side of the Profit and Loss Account.
2. Bad debts given only in the adjustments are to be deducted from debtors on the assets side of the Balance Sheet.

II. Provision for Bad and Doubtful Debts :
According to the principle of Conservatism, anticipated losses must be provided. Therefore, the trader creates some provision on the basis of some percentage which is fixed on the basis of past experience. Such provision is called as Provision for Bad and Doubtful Debts. It is also termed as Reserve for Bad and Doubtful Debts.

a) When provision is given as an adjustment:
Adjustment entry:
Profit and Loss a/c Dr XX
To provision for Bad and Doubtful Debts a/c XX
(Being provision for bad and doubtful debts created on debtors)

Accounting Treatment in Final Accounts :

1. Provision for Bad and Doubtful Debts should be shown on debit side of the Profit and Loss Account.
2. It will be deducted from the Debtors on the assets side of the Balance Sheet.

When Provision for Bad and Doubtful Debts is give in the Trial Balance, it should be deducted from the Debtors on the assets side of the Balance Sheet and need not be shown in the Profit and Loss Account.

b) When provision for doubtful debts is given in Trial Balance and also in adjustments :
The provision for doubtful debts given in Trial Balance is created as created in the last year and is known as the old provision or existing provision.

The provision for doubtful debts given in the adjustments is treated as to be created in the current year and is known as the new provision or provision required.

Accounting Treatment in Final Accounts :
1. Compare the old provision (given in trial balance) with new provision (given in the adjustments), if the new provision for doubtful debts is more than the old provision for doubtful debts, the difference amount (New provision – old provision) should be debited to the Profit and Loss Account.

On the other hand, new provision for doubtful debts is less than the old provision for doubtful debts, the difference amount (old provision – new provision) should be recorded on the credit side of the Profit & Loss Account.

2. Deduct only the amount of new provision for bad and doubtful debts from sundry debtors on the assets side of the Balance Sheet.

c) If tne bad debts are given both in trial balance and in adjustments, and also provision for bad and doubtful debts given in adjustments:

Accounting Treatment in Final Accounts :
1. Firstly, calculate the provision for Bad and Doubtful Debts as per the percentage given in the adjustments after deducting the further bad debts given in the adjustments from the debtors. Then both the Bad Debts (given in Trial Balance and in adjustments) and provision for Bad and Doubtful Debts should be shown on debit side of the Profit and Loss Account.

2. Further, Bad Debts given in the adjustments and Provision for Bad and Doubtful Debts should be deducted from the Debtors on the assets side of the Balance Sheet.

III. Provision for Discount on Debtors :
When a customer (debtor) pays the amount of sales within certain period, trader normally allows certain percentage of discount. The provision created towards the discount on debtors is called as “Provision for Discount on Debtor”.

Adjustment Entry :
Profit and Loss a/c Dr XX
To provision for Discount on Debtor a/c X X
(Being provision for discount on debtors created)

Accounting Treatment in Final Accounts :

1. Provision for Discount on Debtors is shown on debit side of the Profit and Loss Account.
2. It will be again shown by way of deduction from the Debtors on the assets side of the Balance Sheet.

However, if Provision for Discount on Debtors is given in the Trial Balance, it should be deducted directly from the Debtors on the assets side of the Balance Sheet.

Question 5.
Describe the following.
a) Goods Distributed as Free Samples.
b) Accrued Income.
Answer:
a) Goods Distributed as Free Samples :
With a view to promote the sales, some of the goods are distributed among customers at free of cost. These free samples are treated as advertisement expenses.

Adjustment Entry:
Advertisement a/c …………. Dr XX
To Purchases a/c XX
(Being goods distributed as free samples)

Accounting Treatment in Final Accounts :
1. Goods distributed as Free Samples should be deducted from the purchases on debit side of the Trading Account.
2. These Free Sample should be treated as Advertisement Expense and is to be shown on debit side of the Profit and Loss a/c. No need to be shown in the Balance Sheet.

b) Accrued Income :
The income which is earned but not received during the accounting year is called Accrued Income or Income Receivable.
Adjustment Entry :
Accrued Income a/c Dr X X
To Income a/c XX
(Being the income receivable)

Accounting Treatment in Final Accounts :

1. Accrued Income should be added to the concerned income on credit side of the Profit and Loss Account.
2. It will be shown on the assets side of the Balance Sheet.
   If Accrued income is given in the Trial Balance, the amount should be shown only once on the assets side of the Balance Sheet, i.e., need not be shown in the Profit & Loss a/c.

Very Short Answer Questions:

Question 1.
What is the meaning of Adjustment ?
Answer:
On the date of preparing the final accounts, all the expenses whether paid or not and all incomes whether received or not should be taken, similarly, expenses and incomes relating to next year should not taken. All these items are to be adjustment in final accounts by way of adjustments entries. Addition or substraction of the amounts from revenue items is called adjustment.

Question 2.
Explain the importance of Adjustment.
Answer:Importance of adjustments :

1. Expenses or incomes relating to the accounting period can be ascertained accurately.
2. Profit or loss can be calculated accurately.
3. True value of assets and liabilities can be ascertained easily.
4. Unrecorded transactions can be brought into the books of accounts.

Question 3.
Give the meaning of bad debts.
Answer:

1. The trader sells the goods on credit to some of the customers. The customer who has taken the credit may not pay the amount. These debts which are not collected and irrecoverable are known as bad debts.
2. Bad debts is a loss to the business. It shown in debit side of the Profit & Loss a/c.

Question 4.
Describe provision for Discount on Creditors.
Answer:
When the payment is made to creditors within the scheduled period, the discount is likely to be earned. Such discount on creditors is an anticipated profit. The reserve created on creditors is known as Reserve for Discount on Creditors.

Adjustment Entry :
Reserve for Discount on Creditors a/c …………. Dr XX
To Profit and Loss a/c XX
(Being reserve for discount on creditor created)

Question 5.
Explain the accounting treatment of Manager’s Commission.
Answer:
Manager’s Commission: In order to encourage the manager to work more and to increase the profits, manager may be given commission on net profits of the business. It can be given at a certain percentage on the net profits in any of the following two ways :

Manager’s Commission paid on :
a) Net Profit before charging such commission.
b) Net Profit after charging such commission.

Adjustment Entry :
Profit and Loss a/c ………… Dr XX
To Commission Payable a/c XX
(Being commission payable to manager).

Problems:

Question 1.
From the following Trial Balance, prepare final accounts of SathishTraders as on 31-12-2013.

Adjustments :
1) Closing Stock : ₹ 2,100
2) Outstanding Stationery : ₹ 600
3) Depreciationof Machinery : 10%
4) Bad Debts : ₹ 500
5) Prepaid Wages : ₹ 500
Solution:

Question 2.
From the following particulars, prepare final accounts of Godrej Traders for the year ending 31.12.2018:

Solution:

Question 3.
From the following Trial Balance, prepare final accounts of Sachin Trader’s for the year ended 31-3-2019.

Adjustments :
1) Outstanding Salaries : ₹ 500
2) Closing Stock : ₹ 4,500
3) Prepaid Insurance : ₹ 400
4) Outstanding Wages : ₹ 300
5) Depreciation on Machinery : 10%.
Solution:

Question 4.
From the following Trial Balance of M/s Manoj & Son’s Trader’s prepare Trading, Profit & Loss a/c and Balance Sheet for the year ended 31-3-2018.

1) Closing Stock : ₹ 9,000
2) Outstanding Salaries : ₹ 1,000
3) Prepaid insurance : ₹ 100
4) Create 5% Provision for Bad Debts
5) Depreciation on Furniture : 200 and on Machinery : 600
6) Outstanding Wages : ₹ 1,200.
Solution:

Question 5.
From the following particulars of Anasuya Traders, prepare final accounts for the year ended 31-12-2018.

Adjustments :
1) Closing Stock : ₹ 15,500
2) Interest on Capital: 6%
3) Write off ₹ 2,000 as Bad Debt and provide 5% Reserve for Doubtful Debts
4) Outstanding Wages : ₹ 1,000
5) Depreciation on Furniture : 10%.
Solution:

Question 6.
Prepare Final Accounts from the Trial Balance of Brindavan Business for the year ended 31-3-2019.

Adjustments :
1) Outstanding Wages : ₹ 2,000.
2) Outstanding Salaries : ₹ 1,000.
3) Prepaid Insurance: ₹ 50.
4) Create 5% Reserve for Bad Debts on Debtors.
5) Depreciation on Furniture ₹ 150 and on Machinery ₹ 500.
6) Closing Stock ₹ 11,000.
Solution:

Question 7.
Prepare Final Accounts of Pawan Enterprises for the ending 31-3-2018.

Adjustments :
1) Closing Stock : ₹ 6,800.
2) Depreciation of Motor Van : 5%.
3) Reserve for Bad & Doubtful Debts : 6%.
4) Outstanding Rent: ₹ 500
5) Prepaid Insurance ₹ 300
6) Create Reserve for Discount on Creditors : 3%.
Solution:

Working Note :
Accounting for Reserve for Bad & Doubtful Debts :
Old Reserve for Bad & Doubtful Debts 200
New Reserve for Bad & Doubtful Debts [10,000 × \(\frac{6}{100}\)] = 600
New RBD is > Old RBD
⇒ 600 – 200 = 400

Accounting Treatment :
1. Debit the difference (New Reserve – Old Reserve) to the Profit & Loss a/c
2. Deduct new provision for bad debts from debtors on assets side of Balance Sheet.

Question 8.
Prepare Final Accounts from the following Trial Balance of Mamaji Trader’s for the year ended 31-12-2018.

Adjustments :
1) Closing Stock (31-12-2018) : ₹ 2,500.
2) Outstanding Stationary : ₹ 600
3) Depreciation on Buildings : 10%.
4) Bad Debts : ₹ 500 and Provision for Bad and Doubtful Debts : 5%
5) Prepaid Wages : ₹ 500
6) Interest on Drawings : 6%.
Solution:

Working Note :
i) Accounting Related with Debtors :
Calculation of Provision for Bad and Doubtful Debts = (Debtors – New Bad Debts) (Rate of PBD / 100)
= (12,500 – 500) \(\frac{5}{100}\)
= 12,000 × \(\frac{5}{100}\) = 600
∴ PB & DD = 600.

Question 9.
From the following Trial Balance of Vinayaka Enterprises, prepare Final Accounts for the year ending 31-12-2018.

Adjustments :
1) Closing Stock : ₹ 10,500.
2) Bad Debts : ₹ 1,500 and Provide Reserve for Doubtful Debts : 5%.
3) Outstanding Factory Rent: ₹ 400.
4) Depreciation of Furniture : 10%.
5) Interest received in advance : ₹ 500.
6) Stock worth of: 10,000 destroyed in fire and Insurance Co. admitted a claim of ? 7,500.
Solution:

Working Note:
1) Renne for Doubtfùl Debts = (Debtors – New Bad debts) (RDD Rate / 100)
= (13,500 – 1,500) × \(\frac{5}{100}\)
= 12000 × \(\frac{5}{100}\)
∴ RDD = 600.

2) Calculation of actual loss on stock destroyed by fire :
Value of stock destroyed = 10,000
(-) Insurance claim = (-) 7,500

Actual loss = 2,500 (Transfer to P & L a/c).

Question 10.
Prepare Final Accounts of Raghavendra Trader’s for the year ended 31-12-2019.

Adjustments :
1) Closing Stock Value : ₹ 7,500.
2) Depreciation on Machinery : 10%
3) Commission Received in Advance: ₹ 1,000.
4) Outstanding Salaries : ₹ 1,500.
5) Further Bad Debts : ₹ 400 and Provision for Bad & Doubtful Debts : 5%
6) Good worth of ₹ 5,000 withdrawn by the owner for his personal use.
Solution:

Working Note :
Accounting treatment for provision for Bad & Doubtful debts (PB & DD) :
i) Calculation of new PB & DD = (Debtors – Bad debts) (Rate of PB & DD / 100)
= (40,000 – 400) × \(\frac{5}{100}\)
= 39,600 × \(\frac{5}{100}\) = 1,980

ii) Accounting : Old PB & DD (2,500) > New PB & DD (1,980)
Difference is record on Cr. side of P & L a/c [i.e., 2,500 – 1,980 = 520]
Deduct new PB & DD from Debtors in Balance Sheet [i.e., 1,980].

Question 11.
Prepare final accounts of Yadadri Trader’s as on 31-12-2018.

Adjustments :
1) Closing Stock : ₹ 6,500.
2) Prepaid Salaries & Wages : ₹ 400
3) Create 6% for Provision for Bad & Doubtful Debts.
4) Depreciation on Machinery : 10%.
5) Interest on Capital: 6%
6) MD’s Commission on Net Profit is 10% before charging such commission.
Solution:

Working Note :
Calculation of MD’s Commission :
M.D’s commission on Net profit before charging commission
= Net profit before commission × (Rate of commission / 100)
= 5,900 × \(\frac{10}{100}\)
∴ M.D’s Commission = 590.

Question 12.
Prepare Veena Enterprises Final Accounts for the year ended 31-12-2018.

Adjustments :
1) Closing Stock Value : ₹ 8,300.
2) Reserve for Bad Debts : 5%
3) Depreciation on Patents 10%
4) Outstanding Rent: ₹ 600
5) Commission Receivable : ₹ 400
6) Interest on Capital: 6%.
Solution:

Question 13.
Prepare Final Accounts of Srivasthava Traders as on 31-12-2018.

Adjustments :
1) Closing Stock Value : ₹ 6,000
2) Outstanding Salary : ₹ 450
3) Interest on Drawings : 10%
4) Provide Reserve for Doubtful Debts : 5% and Create Provision for Discount on Debtors 2%.
5) Depreciation on Buildings : 10%
6) Goods worth of ₹ 2,000 distributed as free samples.
Solution:

Working Note :
Treatment of Reserve for Doubtful Debts :
Old RDD = 2,000
NewRDD = 15,000 = 750
Old RDD > New RDD

Accounting Treatment :
Difference of Old & New RDD record on Cr. side in profit and loss a/c New RDD deduct from Debtors in Balance Sheet.

Question 14.
From the following Trial Balance, prepare Final Accounts for the year ended 31-03-2019.

Adjustments :
1) Closing Stock Value : ₹ 17,800
2) Outstanding Wages : ₹ 400
3) Prepaid Rent & Taxes : ₹ 380
4) Provide Reserve for Bad & Doubtful Debts on Sundry Debtors 5%.
5) Depreciation on Furniture : 10%
6) Reserve for Discount on Creditors : 5%
Solution:

Working Note :
Calculation of New Reserve for Bad & Doubtful Debts (RB & DD) :
New RB & DD = Debtors × RB & DD Rate / 100
= 10,400 × \(\frac{5}{100}\) = 520.

Accounting Treatment :
Old RB & DD <800) > New RB & DD (520)
∴ → Credit the difference (i.e., 800 – 520 = 280) in Profit & Loss a/c.
→ Deduct New RB & DD from Debtors in Balance Sheet.

Question 15.
From the following Trial Balance of Sudheer Trader’s prepare Final Accounts for the year ended 31-03-2018.

Adjustments :
1) Closing Stock Value : ₹ 16,400.
2) Depreciation on Furniture : 5%, on Machinery : 10%
3) Outstanding Wages : ₹ 500.
4) Write off ₹ 600 as Bad Debts and create Provision for Bad & Doubtful Debts : 6%
5) Interest on Drawings : 5%
6) Stock worth of ₹ 12,000 destroyed by fire and Insurance Company admitted a cliam of ₹ 8,500.
Solution:

Working Note :
Calculation of Actual loss of goods by fire :
Value of goods/stock destroyed by fire = 12,000
(-) Insurance claim = (-) 8,500
Actual loss = 3,500

Accounting Treatment :
→ Deduct total loss from purchases in trading a/c (i.e., 12,000)
→ Debit actual loss (ie 3,500) in Profit & Loss a/c.

Question 16.
Prepare Final Accounts for the year ended 31-03-2018.

Adjustments :
1) Closing Stock Value : ₹ 53,000.
2) Outstanding Salaries : ₹ 5,000.
3) Bad Debts : ₹ 2,000, Create Reserve for Bad Debts : 5% and Provision for Discount on Debtors : 2%.
4) Depreciation on Machinery : 5%
5) Lighting : Factory ₹ 4,000 and Office ₹ 2,000.
6) Manager’s Commission is 10% on Net Profit after charging such commission.
Solution:

Working Note :
1. Calculation of Manager’s commission on net profit after charging such commission:
Manager s commission = Net profit before commission × (Rate of commission / (100 + Rate of commission))
= 1,54,330 × \(\frac{10}{100+10}\)
= l,54,330 × \(\frac{10}{110}\) = 14,030.

2. Accounting treatment for RDD :
Old RDD (1000) < New RDD (4500)
→ Debit the difference (3,500) on Profit & Loss a/c.
→ Deduct New RDD (4,500) from Debtors in Balance Sheet.

Question 17.
From the following Trial Balance and additional information of Lalitha, prepare Trading and Profit and Loss Account for the year ended 31st Dec. 2018 and Balance Sheet as on that date.

Adjustments :
1) Closing Stock : ₹ 28,800.
2) Depreciate 10% on Machinery and 20% on Patents.
3) Outstanding Salaries : ₹ 3,600.
4) Unexpired Insurance : ₹ 230.
5) Create 5% Provision for Bad Debts on Debtors.
6) Create 3% Resere for Discount on Creditors.
Solution:
Trading and Profit & Loss a/c of Lalitha as on 31.12.2018.

Question 18.
From the following Trial Balance of Mallikaijuna Trader’s prepare the Trading, Profit and Loss Account and Balance Sheet for the year ended 31-12-2018.

Adjustments :
1) Closing Stock : ₹ 32,500.
2) Outstanding Salaries : ₹ 5,300.
3) Depreciate Plant and Machinery by 5%.
4) Prepaid Insurance ₹ 1,800.
5) 5% Provision is to be made for Bad and Doubtful Debts.
6) Goods worth of ₹ 3,000 used by Owner for his personal use.
Solution:
Trading and Profit & Loss a/c of Mallikaijuna Trader’s as on 31.12.2018

Question 19.
From the following Trial Balance, prepare final accounts of Sri Raja Rajeshwara Traders as on 31-3-2019.

Adjustments :
1) Closing Stock : ₹ 7,500.
2) Outstanding Wages : ₹ 400.
3) Accrued Interest: ₹ 600
4) Interest on Capital: 6%.
5) Write off ₹ 1,000 as Bad Debts and create 6% Provision for Bad & Doubtful Debts
6) Depreciation on Furniture : 10%.
Solution:
Trading and Profit & Loss a/c of Sri Raja Rajeswara Trader’s as on 31.3.2019

Working Note :
Calculation of New provision for Bad & Doubtful Debts (PB&DD)
New PB & DD = (Debtors – Bad debts) × (Rate of PB&DD / 100)
= (10,000 – 1,000) \(\frac{6}{100}\)
= 9000 × \(\frac{4}{4}\) = 540.

Accounting Treatment:
Old PB & DD (300) < New PB & DD (540)
→ Debit the difference (540 – 300 = 240) in profit & loss a/c.
→ Deduct New PB & DD (540) from Debtors in Balance Sheet.

Question 20.
From the Trial Balance of M/s Sathyam Trader’s, prepare Trading and Profit & Loss Account and Balance Sheet for the year ended 31-3-2018.

Adjustments :
1) Closing Stock : ₹ 34,500.
2) Outstanding Salaries : ₹ 5,500.
3) Depreciation on Machinery : 5%.
4) Prepaid Insurance : ₹ 1,500.
5) Provide for Bad Debts : 5%.
Solution:
Trading and Profit & Loss a/c of M/s Sathyam Trader’s as on 31.3.2018

Working Note :
1. Calculation of Depreciation on Machinery = 80,000 × \(\frac{5}{100}\) = 4,000.
2. Calculation of Bad Debts = 50,000 × \(\frac{5}{100}\) = 2,500.

Textual Examples:

Question 1.
From the following trial balance of Mahindra Traders, prepare final accounts for the year ended 31.12.2018.

Adjustments :
(1) Value of Closing Stock ₹ 2,500
(2) Prepaid Insurance ₹ 250
(3) Outstanding Salaries ₹ 300
Solution:

Question 2.
Prepare Final Accounts of Telangana Traders for the year ended 31.12.2018.

Solution:

Question 3.
From the following trial balance, prepare Mahesh Trader’s final accounts for the year ended 31.03.2018 :

Adjustments :
1) Values of Closing Stock : ₹ 5,400
2) Prepaid Wages : ₹ 300
3) Outstanding Rent: ₹ 400
4) Depreciation on Machinery : 5%, Depreciation on Furniture : 10%.
Solution:

Question 4.
From the following Trial Balance of Warangal Trader’s, prepare final accounts for the year ended 31.12.2018.

Adjustments :
1) Closing Stock ₹ 15,000.
2) Outstanding Wages ₹ 600; Accrued Interest ₹ 1,000.
3) Unexpired Insurance ₹ 200.
4) Depreciation on Plant & Machinery is 10%.
5) Bad Debts to written off ₹ 1,000 and provide 5% for Bad and Doubtful Debts.
6) Create 2% Provision for Discount on Debtors and on Creditors.
Solution:

Question 5.
From the following trial balance, and adjustments, prepare final accounts of Revanth Traders as on 31.03.2019.

Adjustments :
1) Closing Stock : 15,000.
2) Reserve for Bad Debts : 5%.
3) Goods worth of ₹ 1,000 withdrawn by owner for his personal use.
4) Stock destroyed by fire ₹ 6,000 and Insurance company admitted a claim to the extent of ₹ 4,500.
5) Manager’s Commission is 5% on net profit after charging such commission.
6) Goods worth of ₹ 300 distributed as free samples.
Solution:

Question 6.
Prepare final accounts of Mahathi Trader for the year ending 31.03.2018.

Adjustments :
1) Closing Stock : ₹ 6,000
2) Reserve for Bad and Doubtful Debts : 5%
3) Interest on Capital: 5%
4) Outstanding Wages : ₹ 300
5) Depreciation on Machinery : 5%.
Solution:

Question 7.
From the following trial balance, prepare final accounts of Shankari Traders for the year ending 31.12.2018.

Adjustments :
1) Closing Stock : ₹ 6,500
2) Prepaid Salaries : ₹ 600
3) Accrued Interest: ₹ 200
4) Write off Bad Debts : ₹ 1,000, Provision for Bad and Doubtful Debts : 5% and Provision for Discount on Debtors : 2%.
5) Reserve for Discount of Creditors : 2%.
6) Stock of worth ₹ 3,000 destroyed in Fire and Insurance Company admitted a claim of ₹ 1,500
Solution:

Question 8.
From the following trial balance of Mahishmathi Traders, prepare final accounts for the year ended 31.03.2019.
Trial Balance as on 31.03.2019

Adjustments :
1) Closing stock : ₹ 12,000
2) Interest on capital : 5%
3) Bad Debts : 1000 and Reserve for Bad Debts : 5%
4) Depreciation on Machinery : 10%
5) Manager’s Commission is 5% on net profit before charging such commission.
6) Reserve for Discount on Creditors : 3%
Solution:

Students must practice these TS Inter 1st Year Maths 1B Study Material Chapter 9 Differentiation Ex 9(a) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Differentiation 9(a)

I.
Question 1.
Find the derivatives of the following functions f(x). (V.S.A.Q.)
(i) √x + 2x^3/4 + 3x^5/6 (x > 0)
Answer:

(ii) \(\sqrt{2 x-3}+\sqrt{7-3 x}\)
Answer:

(iii) (x² – 3) (4x³ + 1)
Answer:
y = (x² – 3) (4x³ + 1)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (x² – 3) \(\frac{\mathrm{d}}{\mathrm{dx}}\) (4x³ + 1) + (4x³ + 1) \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² – 3)
= (x² – 3) (12x²) + (4x³ + 1) (2x)
= 12x⁴ – 36x² + 8x⁴ + 2x = 20x⁴ – 36x² + 2x

(iv) (√x – 3x) (x + \(\frac{1}{x}\))
Answer:

(v) (√x + 1) (x² – 4x + 2) (x > 0)
Answer:
y = (√x + 1) (x² – 4x + 2)
\(\frac{d y}{d x}\) = (√x + 1) \(\frac{d}{d x}\) (x² – 4x + 2) + (x² – 4x + 2) \(\frac{d}{d x}\) (√x + 1)
= (√x + 1) (2x – 4) + (x² – 4x + 2) \(\left(\frac{1}{2 \sqrt{x}}\right)\)
= (√x + 1) (2x – 4) + \(\frac{x^2-4 x+2}{2 \sqrt{x}}\)

(vi) (ax + b)ⁿ (cx + d)^m
Answer:
Let y = (ax + b)ⁿ (cx + d)^m
\(\frac{d y}{d x}\) = (ax + b)ⁿ . m(cx + d)^{m – 1}(c) + (cx + d)^m n (ax + b)^{n – 1} (a)
= (ax + b)ⁿ (cm (cx + d)^{m – 1}] + (cx + d)^m [an(ax + b)^{n – 1}]
= (ax + b)^{n – 1} (cx + d)^{m – 1} [cm (ax + b) + an(cx + d)]
= (ax + b)ⁿ (cx + d)^m \(\left[\frac{a n}{a x+b}+\frac{c m}{c x+d}\right]\)

(vii) 5 sin x + e^x log x
Answer:
y = 5 sin x + e^x log x

(viii) 5^x + log x + x³ e^x
Answer:
y = 5^x + log x + x³ e^x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 5^x log 5 + \(\frac{1}{x}\) + [x³e^x + 3x²e^x]

(ix) e^x + sin x cos x
Answer:
y = e^x + sin x cos x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x + sin x (- sin x) + cos x (cos x) dx
= e^x + cos² x – sin² x = e^x + cos 2x

(x) \(\frac{p x^2+q x+r}{a x+b}\) (|a| + |b| ≠ 0)
Answer:

(xi) log₇(log x) (x > 0)
Answer:
y = log₇(log x)

(xii) \(\frac{1}{a x^2+b x+c}\) (|a| + |b| + |c| ≠ 0)
Answer:

(xiii) e^2x log (3x + 4) (x > – 4/3)
Answer:
y = e^2x log (3x + 4) (x > – 4/3)

(xiv) (4 + x²) e^2x
Answer:
y = (4 + x²) e^2x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (4 + x²) \(\frac{\mathrm{d}}{\mathrm{dx}}\) (e^2x) + e^2x . \(\frac{\mathrm{d}}{\mathrm{dx}}\) (4 + x²)
= (4 + x²) (2e^2x) + e^2x (2x)
= 2e^2x (x² + x + 4)

(xv) \(\frac{a x+b}{c x+d}\) (|c| + |d| ≠ 0) (May 2022)
Answer:
y = \(\frac{a x+b}{c x+d}\)

(xvi) a^x.e^x2 (Board New Model Paper)
Answer:
y = a^x.e^x2
Then
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a^x . \(\frac{\mathrm{d}}{\mathrm{dx}}\) (e^x2) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (a^x)
= a^x (2x) e^x2 + e^x2 . a^x log_ea
= a^x e^x2 (2x + log_e^a)
= y (2x + log_e^a)

Question 2.
If f(x) = (1 + x + x² + …… + x¹⁰⁰) then find f ‘(1). (V.S.A.Q.)
Answer:
f(x) = (1 + x + x² + ……. + x¹⁰⁰)
Then f(x) = 1 + 2x+ 3x² + …………….. + 100x⁹⁹
and f’(1) = 1 + 2 + 3 + ………………. + 100
= \(\frac{100(100+1)}{2}\) = 101 × 50 = 5050
(∵ Σn = \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\))

Question 3.
If f(x) = 2x + 3x – 5 then prove that f ‘(0) + 3f ‘( -1) = 0. (V.S.A.Q.)
Answer:
Given f(x) = 2x² + 3x – 5
Then f'(x) = 4x + 3
∴ f(0) + 3f (- 1) = 3 + 3 [4 (- 1) + 3] = 3 – 3 = 0

II.
Question 1.
Find the derivatives of the following functions from the first principle. (S.A.Q.)
(i) x³
Answer:
Let f(x) = x³ then

(ii) x⁴ + 4
Answer:
let f(x) = x⁴ + 4

(iii) ax² + bx + c
Answer:
Let f(x) = ax² + bx + c

(iv) √x + 1
Answer:
Let f(x) = √x + 1

(v) sin 2x (Board New Model Paper)
Answer:
Let f(x) = sin 2x

(vi) cos ax (May 2014, March 2013, 2011)
Answer:
Let f(x) = cos ax

(vii) tan 2x (March 2014, May 2011)
Answer:
Let f(x) = tan 2x

(viii) cot x
Answer:
let f(x) = cot x
then

(ix) sec 3x
Answer:
Let f(x) = sec 3x
then

(x) x sin x
Answer:
Let f(x) = x sin x
Then f(x + h) – f(x) = (x + h) sin (x + h) – x sin x
= x [sin(x + h) – sin x] + h sin (x + h)

= x cos x × 1 + sin x
= x cos x + sin x

(xi) cos²x
Answer:
Let f(x) = cos²x
Then f (x + h) – f(x) = cos² (x + h) – cos²x
= [1 – sin²(x + h)] – (1 – sin²x)
= sin²x – sin²(x + h)
= sin (x + x + h) sin (x – x – h)
= sin (2x + h) sin (- h)

= – sin 2x × 1 = – sin 2x

Question 2.
Find the derivatives of the following functions. (S.A.Q.)
(i) \(\frac{1-x \sqrt{x}}{1+x \sqrt{x}}\) (x > 0)
Answer:

(ii) xⁿ n^x log (nx), (x > 0, n ∈ N)
Answer:
Let y = xⁿ n^x log (nx)
Then

(iii) ax²ⁿ log x + bxⁿ e^-x
Answer:
Let y = ax²ⁿ log x + bxⁿ e^-x

= a [x^{2n – 1} + 2nx^{2n – 1} log x] + b [- xⁿ e^-x + ne^-x x^{n – 1}]
= ax^{2n – 1} + 2anx^{2n – 1} log x – bxⁿe^-x + bne^-x x^{n – 1}

(iv) \(\left(\frac{1}{x}-x\right)^3\) e^x
Answer:

Question 3.
Show that the function f(x) = | x | + | x – 1| x ∈ R is differentiable for all real numbers except for 0 and 1. (S.A.Q.)
Answer:
f(x) = |x| + |x – 1| ∀ x ∈ R
then f(x) = x + x – 1 = 2x – 1, x ≥ 1
= x – (x – 1) = x – x + 1 = 1, 0 < x < 1
= – x – (x – 1)
= – x – x + 1 = 1 – 2x; x ≤ 0
∴ f(x) = 2x – 1, x ≥ 1
= 1, 0 < x < 1 = 1 – 2x; x ≤ 0 If x > 1, then f(x) = 2x – 1 which is a polynomial differentiable for all x > 1.
If 0 < x ≤ 1, then f(x) = 1 = Constant which is differentiable.

Case (i): Differentiability at x = 0:

∵ Lf'(0) ≠ Rf'(0), the function f is not differentiable at x = 0.

Case (i): Differentiability at x = 1:

∵ Lf'(1) ≠ Rf'(1) we say that f(x) is not differentiable at x = 1.
∴ f(x) is differentiable for all real x except zero and one.

Question 4.
Verify whether the following function is differentiable at 1 and 3. (S.A.Q.)

Answer:
Case (i):

Case (ii):

Since Lf'(3) ≠ Rf'(3), f(x) is not differentiable at x = 3.

Question 5.
Is the following function f differentiable at 2 ? Justify. (S.A.Q.)

Answer:

Since Lf'(2) ≠ Rf'(2), we say f(x) is not differentiable at x = 2.

Students must practice these TS Intermediate Maths 1A Solutions Chapter 3 Matrices Ex 3(h) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(h)

Question 1.
Solve the following systems of equations.
(i) By using Cramer’s rule and Matrix inversion method, when the coefficient matrix is non-singular.
(ii) By using Gauss Jordan Method. Also deter-mine whether the system has a unique solution or infinite number of solutions or no solution and find the solutions if exist.
I) 5x – 6y + 4z = 15
7x + 4y-3z = 17
2x + y + 6z = 46
Answer:
i) Cramer’s rule
Δ = \(\left|\begin{array}{rrr}
5 & -6 & 4 \\
7 & 4 & -3 \\
2 & 1 & 6
\end{array}\right|\)
= 5(24 + 3) + 6(42 + 6) + 4(7 – 8)
= 135 + 288-4 = 419 ≠ 0
Hence Cramer’s rule is applicable.

Δ₁ = \(\left|\begin{array}{rrr}
15 & -6 & 4 \\
19 & 4 & -3 \\
46 & 1 & 6
\end{array}\right|\)
= 15(24 + 3) + 6(114 + 138) + 4(19 -184)
= 405 + 1512 – 660
= 1917 – 660
= 1257

Δ₂ = \(\left|\begin{array}{rrr}
5 & 15 & 4 \\
7 & 19 & -3 \\
2 & 46 & 6
\end{array}\right|\)
= = 5(114 + 138) -15(42 + 6) + 4(322 – 38)
= 1260 – 720 + 1136
= 1676

Δ₃ = \(\left|\begin{array}{rrr}
5 & -6 & 15 \\
7 & 4 & 19 \\
2 & 1 & 46
\end{array}\right|\)
= 5(184 – 19) + 6 (322 – 38) + 15 (7 – 8)
= 825 + 1704 – 15
= 2529 – 15
= 2514

∴ x = \(\frac{\Delta_1}{\Delta}=\frac{1257}{419}\) = 3
y = \(\frac{\Delta_2}{\Delta}=\frac{1676}{419}\) = 4
z = \(\frac{\Delta_3}{\Delta}=\frac{2514}{419}\) = 6
∴ Solution is x = 3, y = 4, and z = 6

ii) Matrix Inversion method:
Use the formula A^-1 = \(\frac{{Adj} A}{{det} A}\)

iii) Gauss Jordan Method:
Augmented matrix of the system is

The given system is consistent and has a unique solution given by x = 3, y = 4, z = 6.

Question 2.
x + y + z = 1
2x + 2y + 3z = 6
x + 4y + 9z = 3
Answer:
i) Cramer’s rule :
Δ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 3 \\
1 & 4 & 9
\end{array}\right|\)
= 1(18 – 12) – 1(18 – 3) + 1(8 – 2)
= 6- 15 + 6
= -3

Δ₁ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
6 & 2 & 3 \\
3 & 4 & 9
\end{array}\right|\)
= 1(18 – 12) – 1(54 – 9) + 1(24 – 6)
= 6 – 45 + 18
= -21

Δ₂ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 6 & 3 \\
1 & 3 & 9
\end{array}\right|\)
= 1(54 – 9) – 1(18 – 3) + 1(6 – 6)
= 45 – 15
= 30

∴ Solution is x = 7, y = -10, and z = 4.

ii) Matrix Inversion Method:

∴ Solution is x = 7, y = -10 and z = 4

iii) Gauss Jordan Method:
Augmented matrix of the system

The given system is consistent and has a unique solution given by x = 7,y = -10,z = 4. [∵ ρ(A) = ρ(AB)]

Question 3.
x – y + 3z = 5 (March 2015-T.S)
4x + 2y – z = 0
-x + 3y + z = 5
Answer:
i) Cramer’s rule
Δ = \(\left|\begin{array}{rrr}
1 & -1 & 3 \\
4 & 2 & -1 \\
-1 & 3 & 1
\end{array}\right|\)
= 1(2 + 3) + 1(4-1) + 3(12 + 2)
= 5 + 3 + 42
= 50 ≠ 0
Cramer’s rule is applicable.

Δ₁ = \(\left|\begin{array}{rrr}
5 & -1 & 3 \\
0 & 2 & -1 \\
5 & 3 & 1
\end{array}\right|\)
= 5(2 + 3) + 1(0+ 5) + 3(0 -10)
= 25 + 5 – 30
= 0

Δ₂ = \(\left|\begin{array}{rrr}
1 & 5 & 3 \\
4 & 0 & -1 \\
-1 & 5 & 1
\end{array}\right|\)
= 1(0+ 5)-5(4-1)+ 3(20)
= 5 – 15 + 60 = 0

Δ₃ = \(\left|\begin{array}{rrr}
1 & -1 & 5 \\
4 & 2 & 0 \\
-1 & 3 & 5
\end{array}\right|\)
= 1(10 – 0) + 1(20 – 0)+ 5(12 + 2)
= 10 + 20 + 70
= 100

∴ x = \(\frac{\Delta_1}{\Delta}\) = 0
y = \(\frac{\Delta_2}{\Delta}=\frac{50}{50}\) = 1
z = \(\frac{\Delta_3}{\Delta}=\frac{100}{50}\) = 2
∴ x = 0, y = 1 and z = 2.

ii) Matrix Inversion Method:

iii) Gauss Jordan Method:

The given system of equations is consistent since ρ(A) = ρ(AB) = 3 and the system has a unique solution. x = 0, y = 1, z = 2.

Question 4.
2x + 6y + 11 = 0
6x + 20y – 6z + 3 = 0
6y – 18z + 1 = 0
Answer:
Δ = \(\left|\begin{array}{rrr}
2 & 6 & 0 \\
6 & 20 & -6 \\
0 & 6 & -18
\end{array}\right|\)
= 2(- 360 + 36) – 6(-108-0) + 0(36)
= -648 + 648 = 0
Cramer’s and matrix inversion methods are not applicable.

Gauss Jordan Process:
The augmented matrix of the given system is

Since ρ(A) = 2 and ρ(AB) = 3, the given system is not consistent and has no solution.

Question 5.
2x – y + 3z = 9
x + y + z = 6
x – y + z = 2 (May 2014, Mar. ’14, ’05, ’02)
Answer:
i) Cramer’s rule :

ii) Martix Inversion Method:

det A = 2(2) + 1(0) + 3(- 2)
= 4 -6 = -2 ≠ 0
∴ A^-1 exists and A^-1 = \(\frac{{Adj} \mathrm{A}}{{det} \mathrm{A}}\)

∴ x = 1, y = 2, z = 3 is the solution.

iii) Gauss Jordan Method:
Augmented matrix of the system

Since ρ(A) = ρ(AB) = 3; the system is consistent and has a unique solution given by x = 1, y = 2, z = 3.

Question 6.
2x- y + 8z = 13
3x + 4y + 5z = 18
5x – 2y + 7z = 20 (March 2004, 03, ’01) (Board New Model Paper)
Answer:
i) Cramer’s rule :

ii) Matrix Inversion Method:

∴ Solution is x = 3, y = 1 and z = 1

iii) Gauss Jordan Method:
Augmented matrix of the system

Since ρ(A) = ρ(AB) = 3, the system is consistent and has a unique solution given by x = 3, y = 1 and z = 1.
Then the cofactors of elements in the matrix A are

Question 7.
2x – y + 3z = 8
-x + 2y + z = 4
3x + y – 4z = 0
Answer:
i) Cramer’s rule :
Δ = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
-1 & 2 & 1 \\
3 & 1 & -4
\end{array}\right|\)
= 2(- 8 – 1) + 1(4 – 3) + 3(- 1-6)
= -18 + 1 – 21 = -38 ≠ 0
Cramer’s rule is applicable.

Δ₁ = \(\left|\begin{array}{ccc}
8 & -1 & 3 \\
4 & 2 & 1 \\
0 & 1 & -4
\end{array}\right|\)
= 8(- 8 – 1) + 1(- 16 – 0) + 3(4 – 0)
= -72 – 16 + 12
= -76

Δ₂ = \(\left|\begin{array}{ccc}
2 & 8 & 3 \\
-1 & 4 & 1 \\
3 & 0 & -4
\end{array}\right|\)
= 2(- 16 – 0) – 8(4 – 3) + 3(0 – 12)
= – 32 – 8 – 36 = – 76

Δ₃ = \(\left|\begin{array}{ccc}
2 & -1 & 8 \\
-1 & 2 & 4 \\
3 & 1 & 0
\end{array}\right|\)
= 2(0 – 4) + 1(0 – 12) + 8(-1 – 6)
= -8 – 12 – 56
= -76
∴ x = \(\frac{\Delta_1}{\Delta}=\frac{-76}{-38}\) = 2
y = \(\frac{\Delta_2}{\Delta}=\frac{-76}{-38}\) = 2
z = \(\frac{\Delta_3}{\Delta}=\frac{-76}{-38}\) = 2
∴ Solution is x = 2, y =2, z = 2.

Matrix Inversion Method:

∴ Solution is x = 2, y = 2 and z = 2

iii) Gauss Jordan Method:
The augmented matrix of the system is

∴ ρ(A) = ρ(AB) = 3 ; the system is consistent and has a unique solution, x = 2, y = 2 and z = 2.

Question 8.
x + y + z = 9
2x + 5y + 7z = 52
2x + y-z = 0 (May 2011)
Answer:
i) Cramer’s rule :
Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 5 & 7 \\
2 & 1 & -1
\end{array}\right|\)
= 1(-5 – 7) – 1(-2 – 14) + 1(2 – 10)
= – 12 + 16-8 = -4 ≠ 0
∴ The Cramer’s method is applicable.

Δ₁ = \(\left|\begin{array}{ccc}
9 & 1 & 1 \\
52 & 5 & 7 \\
0 & 1 & -1
\end{array}\right|\)
= 9(-5 – 7) -1(-52) + 1(52)
= -108 + 52 + 52 =-4

Δ₂ = \(\left|\begin{array}{ccc}
1 & 9 & 1 \\
2 & 52 & 7 \\
2 & 0 & -1
\end{array}\right|\)
= 1(- 52 – 0) – 9(- 2 – 14) + 1(0 – 104)
= -52 + 144 – 104
= -12

Δ₃ = \(\left|\begin{array}{ccc}
1 & 1 & 9 \\
2 & 5 & 52 \\
2 & 1 & 0
\end{array}\right|\)
= 1(0 – 52) -1(0 – 104) + 9(2 – 10)
= -52 + 104 – 72
= -20
∴ x = \(\frac{\Delta_1}{\Delta}=\frac{-4}{-4}\) = 1
y = \(\frac{\Delta_2}{\Delta}=\frac{-12}{-4}\) = 3
z = \(\frac{\Delta_3}{\Delta}=\frac{-20}{-4}\) = 5
x = 1, y = 3, z = 5 is a solution.

ii) Matrix Inversion Method:

∴ x = 1, y = 3, z = 5 is the solution.

iii) Gauss Jordan Method:
Augmented matrix

ρ(A) = ρ(AB) = 3 and the system is consistent.
The system has a unique solution given by
x = 1, y = 3 and z = 5.

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

TS Inter 1st Year Maths 1B Mean Value Theorems Important Questions

Question 1.
Verify Rolle’s theorem for the function y = f(x) = x² + 4 in [-3, 3]. [Mar. ’19 (TS); Mar. ’17 (AP), ’15 (TS); B.P.; May ’15 (TS)]
Solution:
f is a polynomial function that is continuous and differentiable in [-3, 3]
Given f(x) = x² + 4
∴ f(-3) = (-3)² + 4 = 13
and f(3) = 3² + 4 = 13
∴ f(-3) = f(3).
Hence the function satisfies the conditions of Rolle’s theorem and f'(x) = 2x
∴ f'(c) = 2c = 0
c = 0 ∈ (-3, 3)
Thus Roll’s theorem is verified.

Question 2.
Verify Rolle’s theorem for the following function x² – 1 on [-1, 1]. [Mar. ’14; May ’13]
Solution:
Let f(x) = x² – 1 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 satisfies 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.

Question 3.
Let f(x) = (x – 1) (x – 2) (x – 3). Prove that there is more than one ‘c’ in (1, 3) such that f'(c) = 0. [Mar. ’13]
Solution:
f is a polynomial in x which is continuous over [1, 3] and derivable over (1, 3).
Given f(x) = (x – 1) (x – 2) (x – 3)
We have f'(x) = (x – 1) (x – 2) + (x – 2) (x – 3) + (x – 1) (x – 3)
= x² – 3x + 2 + x² – 5x + 6 + x² – 4x + 3
= 3x² – 12x + 11
f'(c) = 0
3c² – 12c + 11 = 0
c = \(\frac{12 \pm \sqrt{144-132}}{6}\) = 2 ± \(\frac{1}{\sqrt{3}}\)
2 + \(\frac{1}{\sqrt{3}}\) ∈ (1, 3) and 2 – \(\frac{1}{\sqrt{3}}\) ∈ (1, 3)
∴ There exists more than one ‘c’ in (1, 3) such that f'(c) = 0.

Question 4.
Verify Rolle’s theorem for the function f(x) = x(x + 3) \(\mathrm{e}^{-\mathrm{x} / 2}\) in [-3, 0]. [Mar. ’18 (AP)]
Solution:
f is continuous in [-3, 0] and differentiable in (-3, 0).
Also f(-3) = (-3) (-3 + 3) e^3/2 = 0
f(0) = 0 (0 . 3) e^3/2 = 0
∴ f(-3) = f (0)
So f satisfies the conditions of Rolle’s theorem.
Given f(x) = x(x + 3) e^3/2
Then f'(x) = x(x + 3) \(\left(\frac{-1}{2}\right) \mathrm{e}^{-\mathrm{x} / 2}\) + \(x\left(e^{\frac{-x}{2}}\right)+(x+3) e^{\frac{-x}{2}}\)

f'(e) = 0
-e² + e + 6 = 0
e = -2 or 3
Now -2 ∈ (- 3, 0) and c = -2
∴ Rolle’s theorem was verified.

Question 5.
Verify Lagrange’s mean value theorem for the function f(x) = x² on [2, 4].
Solution:
Being a polynomial function
(i) f(x) = x² is continuous on [2, 4] and
(ii) differentiable in (2, 4)
By Lagrange’s theorem there should exist c ∈ (2, 4) such

Hence Lagrange’s theorem is verified.

Question 6.
On the curve y = x², find a point at which the tangent is parallel to the chord joining (0, 0) and (1, 1).
Solution:
The slope of the chord joining (0, 0) and (1, 1) is
m₁ = \(\frac{1-0}{1-0}\) = 1
Given curve is y = x²
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2x
∴ The slope of the tangent at any point on the curve is m₂ = 2x
If the tangent is parallel to the chord, then m₁ = m₂
2x = 1
x = \(\frac{1}{2}\)
Now, y = x² = \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
∴ The required point on the curve is \(\left(\frac{1}{2}, \frac{1}{4}\right)\)

Question 7.
Verify Rolle’s Theorem for the following function sin x – sin 2x on [0, π].
Solution:
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 satisfies the 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.

Question 8.
Show that there is no real number k, for which the equation x² – 3x + k = 0 has two distinct roots in [0, 1].
Solution:
Let f(x) = x² – 3x + k and suppose there exist 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 [α, β] and differentiable in (α, β).
Also f(α) = α² – 3α + k = 0 and
f(β) = β² – 3β + k = 0
∴ f(α) = f(β) = 0
∴ f satisfies 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 do not exist roots α, β in (0, 1).
So, there is no real number k for which the equation x² – 3x + k = 0 has two distinct roots in [0, 1].

Question 9.
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).
Solution:
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
\(\frac{d y}{d x}\) = 2(x – 3)
Slope of the chord 2(x – 3) = 1
2x = 7
x = \(\frac{7}{2}\)
∴ y = (x – 3)²
= \(\left(\frac{7}{2}-3\right)^2=\frac{1}{4}\)
∴ The point on the curve = \(\left(\frac{7}{2},\frac{1}{4}\right)\)

Question 10.
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).
Solution:
Let the points be A(1, 1) and B(3, 27)

Question 11.
Find ‘c’ so that f'(c) = \(\frac{\mathbf{f}(\mathbf{b})-\mathbf{f}(\mathbf{a})}{b-\mathbf{a}}\) where f(x) = e^x; a = 0, b = 1.
Solution:

Question 12.
Verify Rolle’s theorem for the function (x² – 1) (x – 2) on [-1, 2]. Find a point in the interval where the derivative vanishes.
Solution:
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 satisfies all the conditions of Rolle’s theorem.
f(x) = 3x² – 4x – 1 and f'(c) = 0
3c² – 4c – 1 = 0

∴ Rolle’s theorem is verified.

Question 13.
Verify the conditions of Lagrange’s mean value theorem for the function x² – 1 on [2, 3]. [Mar. ’18 (TS); Mar. ’16 (AP)]
Solution:
Let f(x) = x² – 1 defined over [2, 3]. This is a polynomial in x, continuous on [2, 3] and differentiable 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, f(2) = 3
∴ 2c = \(\frac{8-3}{1}\) = 5
c = \(\frac{5}{2}\) ∈ (2, 3)

Question 14.
Verify the conditions of Lagrange’s mean value theorem for the function sin x – sin 2x on [0, π].
Solution:
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 Lagrange’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 cos²c – 1) = 0
4cos²c – 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)

Question 15.
Verify the conditions of Lagrange’s mean value theorem for the function log x on [1, 2].
Solution:
Let f(x) = log x defined over [1, 2].
This is continuous over [1, 2] and differentiable over (1, 2) since
f'(x)= \(\frac{1}{x}\) ∀ x ∈ (1, 2)

Hence Lagrange’s mean value theorem is satisfied.

Question 16.
Verify Rolle’s theorem for the function f : [-3, 8] → R be defined by f(x) = x² – 5x + 6. [Mar. ’17 (TS)]
Solution:
Given f(x) = x² – 5x + 6 defined on [-3, 8].
Since the function f(x) is a polynomial,
it is continuous on [-3, 8] and differentiable on [-3, 8)
Also f(-3) = (-3)² – 5(-3) + 6
= 9 + 15 + 6
= 30
f(8) = (8)² – 5(8) + 6
= 64 – 40 + 6
= 30
∴ f(-3) = f(8)
Hence f satisfies all conditions of Rolle’s theorem.
Now we have to find a point C ∈ (-3, 8) such that f'(C) = 0
f'(x) = 2x – 5 and f'(c) = 0
2C – 5 = 0
C = \(\frac{5}{2}\) ∈ (-3, 8)
Hence Rolle’s theorem is verified.

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

TS Inter 1st Year Maths 1B Rate Measure Important Questions

Question 1.
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. [Mar. ’19 (AP); Mar. ’12; May ’96]
Solution:
Given, s = t³ – 9t² + 24t – 18
Velocity, v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
= 3t² – 9(2t) + 24(1) – 0
= 3t² – 18t + 24
Given, the particle comes to rest then the velocity = 0
3t² – 18t + 24 = 0
t² – 6t + 8 = 0
t² – 4t – 2t + 8 = 0
t(t – 4) – 2(t – 4) = 0
(t – 4) (t – 2) = 0
t – 4 = 0 (or) t – 2 = 0
t = 4 (or) t = 2
t = 2, 4
The velocity is zero when t = 2 sec and 4 sec.
Case 1:
If t = 2 sec, then s = t³ – 9t² + 24t – 18
= (2)³ – 9(2)² + 24(2) – 18
= 8 – 36 + 48 – 18
= 2 units
Case 2:
If t = 4 sec, then s = (4)³ – 9(4)² + 24(4) – 18
= 64 – 144 + 96 – 18
= 160 – 162
= -2 units
∴ The particles are at a distance of two units from the starting point ‘0’ on either side.

Question 2.
The displacement s of a particle travelling in a straight line in ‘t’ seconds is given by s = 45t + 11t² – t³, finding the time when the particle comes to rest. [May ’13 (Old), ’06; Mar. ’07]
Solution:
Given, s = 45t + 11t² – t³
Velocity, v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
= 45(1) + 11(2t) – 3t²
= 45 + 22t – 3t²
The particle comes to rest then velocity, v = 0
45 + 22t – 3t² = 0
3t² – 22t – 45 = 0
3t² – 27t + 5t – 45 = 0
3t(t – 9) + 5(t – 9) = 0
(t – 9) (3t + 5) = 0
t – 9 = 0 (or) 3t + 5 = 0
t = 9 (or) t = \(\frac{-5}{3}\)
t = 9 Since t is positive then t = 9
∴ The particle comes to rest when t = 9 sec.

Question 3.
A particle is moving in a straight line so that after t seconds its distance is s (in cms) from a fixed point on the line given by s = f(t) = 8t + t³. Find
(i) the velocity at time t = 2 sec [Mar. ’17, ’15 (AP)]
(ii) the initial velocity
(iii) acceleration at t = 2 sec [May ’12; B.P.]
Solution:
The distance s and time t are connected by the relation s = f(t) = 8t + t³
Velocity, v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
= 8(1) + 3t²
= 8 + 3t²
(i) Velocity at t = 2 sec, then,
v = \(\left(\frac{\mathrm{ds}}{\mathrm{dt}}\right)_{\mathrm{t}=2}\)
= 8 + 3(2)²
= 8 + 12
= 20 cm/sec
acceleration, a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\)
= 0 + 3(2t)
= 6t
(ii) Initial velocity (t = 0) is v = \(\left(\frac{\mathrm{ds}}{\mathrm{dt}}\right)_{\mathrm{t}=0}\)
= 8 + 3(0)²
= 8 cm/sec
(iii) The acceleration at t = 2 sec, then
a = \(\left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)_{\mathrm{t}=2}\)
= 6(2)
= 12 cm/sec²

Question 4.
A container is in the shape of an inverted cone and has a height of 8 cm and a radius of 6 m at the top. If it is filled with water at the rate of 2 m³/minute, how fast height of the water change when the level is 4m? [Mar. ’19 (TS): May ’13. ’11, ’09; Mar. ’08, ’06]
Solution:
Let h be the height, r be the radius of the water level and V be the volume of the water in the container at the time ‘t’.
Given that, \(\frac{d v}{d t}\) = 2m3/min.


∴ The height of the water changes at a rate of \(\frac{2}{9 \pi}\) m/min.

Question 5.
The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of the edge is 10 centimeters? [Mar. ’17 (TS), ’16 (AP), ’13; May ’15 (TS)]
Solution:
Let x be the length of the edge of the cube.
V is its volume and S is its surface area.
Given that rate of change of volume
\(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 9 cm³/sec
x = 10 cm
The volume of a cube, V = x³
Differentiating v with respect to ‘t’, we get

Question 6.
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 the y-coordinate is increasing when the point is at (2, 8). [Mar. ’16 (TS), ’13; May ’08]
Solution:
Let P(x, y) be any point on the curve y = 2x²
Given, \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 4 units/sec
Given point (x, y) = (2, 8)
Now, y = 2x²
Differentiating on both sides with respect to ‘t’
\(\frac{\mathrm{dy}}{\mathrm{dt}}=2(2 \mathrm{x}) \cdot \frac{\mathrm{dx}}{\mathrm{dt}}\)
= 4(2) . 4
= 32 units/sec
∴ The y-coordinate is increasing at the rate of 32 units/sec.

Some More Maths 1B Rate Measure Important Questions

Question 7.
A container in the shape of an inverted cone has a height of 12 cm and a radius of 6 cm at the top. If it is filled with water at the rate of 12 cm³/sec. What is the rate of change in the height of the water level when the tank is filled 8 cm?
Solution:
Let h be the height, r be the radius of the water level and v be the volume of water in the container at a time ‘t’.

Given, \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 12 cm³/sec
height, h = 8 cm
In triangle OAB, OCD are similar, then


Hence the rate of change of water level is \(\frac{3}{4 \pi}\) cm/sec when the water level of the tank is 8 cm.

Question 8.
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?
Solution:
Let x be the length of the edge of the cube, V be its volume and S be its surface area.
Given that, the rate of change in volume \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 8 cm³/sec
x = 12 cm
The volume of a cube V = x³
Differentiating V on with respect to ‘t’, we get

Question 9.
A particle moving along straight has the relation s = t³ + 2t + 3, connecting the distance s described by the particle in time t. Find the velocity and acceleration of the particle at t = 4 seconds.
Solution:
Given, s = t³ + 2t + 3
velocity, v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
= 3t² + 2 + 0
= 3t² + 2
velocity at t = 4 sec, then, v = \(\left(\frac{\mathrm{ds}}{\mathrm{dt}}\right)_{t=4}\)
= 3(4)² + 2
= 50 units/sec
acceleration, a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\)
= 3(2t) + 0
= 6t
acceleration at t = 4, then
a = \(\left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)_{\mathrm{t}=4}\)
= 6(4)
= 24 units/sec²

Question 10.
A particle is moving along a line according to s = f(t) = 4t³ – 3t² + 5t – 1 where s is measured in metres and t is measured in seconds. Find the velocity and acceleration at time t. At what time the acceleration is zero? [Mar. ’18 (AP); Mar. ’18, ’15 (TS)]
Solution:
Given, s = f(t) = 4t³ – 3t² + 5t – 1
velocity, v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
= 4(3t²) – 3(2t) + 5(1) – 0
= 12t² – 6t + 5
acceleration, a = \(\frac{d v}{d t}\)
= 12(2t) – 6(1) + 0
= 24t – 6
∴ Acceleration is 0 if 24t – 6 = 0
24t = 6
t = \(\frac{1}{4}\) sec
∴ Acceleration of particle is 0 at t = \(\frac{1}{4}\) sec

Question 11.
A stone is dropped into a quiet lake and ripples move in circles at speed of 5 cm/sec. At the instant when the radius of the circular ripple is 8 cm, how fast is the enclosed area increased?
Solution:
Let A be the area of the circle with radius ‘r’.
Given, radius, r = 8 cm
\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 5 cm/sec
Area of the circle, A = πr²
Differentiating on A with respect to ‘t’
\(\frac{\mathrm{dA}}{\mathrm{dt}}=\pi .2 \mathrm{r}, \frac{\mathrm{dr}}{\mathrm{dt}}\)
= 2π (8) (5)
= 80π cm2/sec

Question 12.
The radius of a circle is increasing at the rate of 0.7 cm/sec. What is the rate of increase in its circumference?
Solution:
Let r be the radius and P be the circumference of a circle.
Given, \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 0.7 cm/sec
Circumference of a circle, P = 2πr
Differentiating P on both sides with respect to ‘t’
\(\frac{\mathrm{dp}}{\mathrm{dt}}=2 \pi \cdot \frac{\mathrm{dr}}{\mathrm{dt}}\)
= 2π(0.7)
= 1.4π cm/sec
∴ Rate of increase of circumference of circle = 1.4π cm/sec.

Question 13.
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic cm of gas per sec. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Solution:
Let r be the radius and V be the volume of the spherical balloon.
Given, \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 900 cu.cm/sec
radius, r = 15 cm
Volume of the spherical balloon, V = \(\frac{4}{3} \pi r^3\)
Differentiating on both sides with respect to ‘t’

∴ Rate of increase of radius of the balloon \(\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{\pi}\) cm/sec

Question 14.
The radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cubic. cm/sec. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution:
Let r be the radius and V be the volume of an air bubble.
Given, \(\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{2}\) cm/sec
radius, r = 1 cm
Volume of a air bubble, V = \(\frac{4}{3} \pi r^3\)
Differentiating on both sides with respect to ‘t’

∴ The volume of the air bubble increasing at r = 1 cm is 2π cu. cm/sec.

Question 15.
Let a kind of bacteria grow in such a way that at time t sec. There are \(\mathbf{t}^{3 / 2}\) bacteria. Find the rate of growth at time t = 4 hours.
Solution:
Let g be the amount of growth of bacteria at t sec.
Let g(t) = \(\mathbf{t}^{3 / 2}\)

Question 16.
Let a kind of bacteria grow by t³ (t in sec). At what time the rate of growth of the bacteria is 300 bacteria per sec?
Solution:
Let g be the amount of growth of bacteria at t sec.
Then g(t) = t³
Now, the growth rate at time ‘t’ is g'(t) = 3t²
Given, the rate of growth of the bacteria is 300 per sec.
3t² = 300
t² = 100
t = 10 sec.
∴ After t = 10 sec, the growth rate of bacteria should be 300 bacteria/sec.

Question 17.
Suppose we have a rectangular aquarium with dimensions of length 8m, width 4m, and height 3m. Suppose we are filling the tank with water at a rate of 0.4 m³/sec. How fast is the height of water changing when the water level is 2.5 m?
Solution:
Let h be the height of the water level in the aquarium at the time t.
Given that length, l = 8 m
breadth (or) width, b = 4 cm
height, h = 3 m
Volume of the aquarium, V = lbh
V = 8 . 4 . 3 = 96 cu.m.
Given, \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 0.4 cu.m/sec
height, h = 2.5 m
volume of aquarium, V = lbh
log V = log(lbh)
log V = log l + log b + log h
Differentiating with respect to ‘t’


(Or)
Let h be the height of the water level in the aquarium at the volume of the aquarium,
V = lbh = 8 . 4 . h = 32h …….(1)
Given that, \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 0.4 cu.cm/sec
height, h = 2.5 m
Differentiating on both sides with respect to ‘t’

The weight of water changes at the rate of \(\frac{1}{80}\) m.sec.

Question 18.
The total cost c(x) in rupees associated with the production of x units of an item is given by c(x) = 0.007x³ – 0.003x² + 15x + 4000. Find the marginal cost when 17 units are produced.
Solution:
Let M represent the marginal cost.
Given, c(x) = 0.007x³ – 0.003x² + 15x + 4000
The rate of change of total cost is M = \(\frac{\mathrm{dc}}{\mathrm{dx}}\)
= (0.007)(3x²) – 0.003(2x) + 15(1) + (0)
= 0.021x² – 0.006x + 15
∴ The marginal cost at x = 17 is \(\left(\frac{d c}{d x}\right)_{x=17}\)
= (0.021)(17)² – 0.006(17) + 15
= (0.021)(289) – 0.102 + 15
= 20.967

Question 19.
The total cost c(x) in rupees associated with the production of x units of an item is given by c(x) = 0.005x³ – 0.02x² + 30x + 500. Find the marginal cost when 3 units are produced (marginal cost is the rate of change of total cost). [May ’15 (AP)]
Solution:
Given relation c(x) = 0.005x³ – 0.02x² + 30x + 500
\(\frac{\mathrm{dc}}{\mathrm{dx}}\) = (0.005)(3x²) – 0.02(2x) + 30(1)
= (0.015)x² – 0.04x + 30
∴ Marginal cost = \(\left(\frac{\mathrm{dc}}{\mathrm{dx}}\right)_{\mathrm{x}=3}\)
= (0.015) (3)² – 0.04(3) + 30
= (0.015) (9) – 0.04 (3) + 30
= 0.135 – 0.12 + 30
= 30.015

Students must practice these TS Intermediate Maths 1A Solutions Chapter 3 Matrices Ex 3(g) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(g)

I.
Examine whether the following systems of equations are consistent or inconsistent and if consistent find the complete solutions,
Question 1.
x + y + z = 4
2x + 5y – 2z = 3
x + 7y – 7z = 5
Answer:
Augmented matrix of the above system is

Rank of the matrix ρ(A) = 2 and ρ(AB) = 3.
Since ρ(A) ≠ ρ(AB), the given system of equa¬tions are inconsistent.

Question 2.
x + y + z = 6
x – y + z = 2
2x – y + 3z = 9
Answer:
Augmented matrix [AB] = \(\left[\begin{array}{rrrr}
1 & 1 & 1 & 6 \\
1 & -1 & 1 & 2 \\
2 & -1 & 3 & 9
\end{array}\right]\)
Apply operations R₂ – R₁, R₃ – 2R₁, we get

Here ρ(A) = 3 and ρ(AB) = 3
Since ρ(A) = ρ(AB), the given system is consistent and has a unique solution.

Question 3.
x + y + z = 1
2x + y + z = 2
x + 2y + 2z = 1 (March 2015-T.S)
Answer:
Augmented matrix of the system is

ρ(AB) = 2 and ρ(A) = 2 and ρ(A) = ρ(AB) < 3
The given system of equations is consis¬tent and has infinitely many solutions.
The given system is equivalent to x + y + z = 1 and y + z = 0.
Solution set is

Question 4.
x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0
Answer:
Augmented matrix of the system

Here ρ(A) = ρ(AB) = 3; and the system of given equations is consistent; and has a unique solution.
Also x = 1, y = 3, z = 5 form the solution.

Question 5.
x + y + z = 6
x + 2y + 3z = 10
x + 2y + 4z = 1
Answer:
Augmented matrix of the system is

Question 6.
x – 3y – 8z = – 10
3x + y – 4z = 0
2x + 5y + 6z = 13
Answer:
The augmented matrix of the above system is

Since ρ(A) = 2 = ρ(AB) < 3, given system of equations is consistent with infinitely many solutions.
The given system is equivalent to
x – 3y – 8z = – 10,
y + 2z = 3
Put z = t then y = 3 – 2t
∴ x = – 10 + 3(3 – 2t) + 8t
= -10 + 9 – 6t + 8t
= 2t – 1
Hence the solutions are given by
x = 2t – 1, y = 3 – 2t and z = t
Where t is any scalar.

Question 7.
2x + 3y + z = 9
x + 2y + 3z = 6
3x + y + 2z = 8
Answer:
Augmented matrix of the above system is

ρ(A) = ρ(AB) = 3; system is consistent and has a unique solution given by
x = \(\frac{35}{18}\), y = \(\frac{29}{18}\), z = \(\frac{5}{18}\)

Question 8.
x + y + 4z = 6
3x + 2y – 2z = 9
5x + y + 2z = 13
Answer:
Augmented matrix of the system

ρ(A) = ρ(AB) = 3;
Hence the system is consistent and has a unique solution given by
x = 2, y = 2, z = \(\frac{1}{2}\).

Students must practice these TS Inter 1st Year Maths 1B Study Material Chapter 8 Limits and Continuity Ex 8(e) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Limits and Continuity 8(e)

Question 1.
Is the function f, defined by

continuous on R? (V.S.A.Q.) (May 2011)
Answer:

The function is continuous at ‘1’ hence at continuous on R.

Question 2.
Is f defined by

continuous at ‘0’ (V.S.A.Q.) (May 2012)
Answer:
Given f(0) = 1

Hence f is continuous at ‘0’.

Question 3.
Show that the function
f(x) = [cos(x¹⁰ + 1)]^{\(\frac{1}{3}\)}; x ∈ ℛ is a continuous function. (V.S.A.Q.)
Answer:
Since cos x is continuous for every x ∈ R, we have f(x) = [cos (x¹⁰ + 1)]^1/3 is also continuous over R.

II.
Question 1.
Check the continuity of the following function at 2. (S.A.Q.)

Answer:
Given f(2) = 0

Hence f(x) is not continuous at x = 2.

Question 2.
Check the continuity of f given by (S.A.Q.) (Mar. ’14, ’13)

Answer:
Given f(3) = 1.5 and

Since \(\lim _{x \rightarrow 3}\) f(x) = f(3) we have f is continuous at the point ‘3’.

Question 3.
Show that f given by f(x) = \(\frac{\mathbf{x}-|\mathbf{x}|}{\mathbf{x}}\) (x ≠ 0) is continuous on R – {0}. (S.A.Q.)
Answer:

In both the cases \(\lim _{x \rightarrow a}\) f(x) = f(a)
But if a = 0 then f(0) is not defined f(x) is not continuous at ‘0’.
Hence f(x) is continuous on R – {0}

Question 4.
If f is a function defined by (S.A.Q.)

then discuss the continuity of f.
Answer:

Question 5.
If f is given by

is a continuous function on R, then find the values of ‘k’. (S.A.Q.)
Answer:

Given that f is continuous on R, then it is continuous at ‘1’.
∴ k² – k = 2 ⇒ k² – k – 2 = 0
⇒ (k – 2) (k + 1) = 0
⇒ k = 2 or k = – 1

Question 6.
Prove that the function ‘sin x’ and ‘cos x’ are continuous on R. (S.A.Q.)
Answer:
(i) Let f(x) = sin x
If f is continuous at a point ‘a’ then
\(\lim _{x \rightarrow \mathrm{a}}\) f(x) = f(a)
∴ \(\lim _{x \rightarrow \mathrm{a}}\) (sin x) = sin a = f(a)
(or) by definition we have for ∈ > 0, ∃ a δ > 0
such that |sin x – sin a| < E for |x – a| < δ

Hence f(x) = sin x is continuous at ‘a’
Hence f is continuous over R.

(ii) f(x) = cos x
Let a ∈ R,
∴ \(\lim _{x \rightarrow a}\) f(x) = \(\lim _{x \rightarrow a}\) (cos x) = cos a = f(a)
∴ f is continuous at ‘a’
Hence f is continuous over R.

III.
Question 1.
Check the continuity of f given by

at the point 0, 1 and 2.
Answer:

Question 2.
Find real constants a, b so that the function f given by

is continuous on R
Answer:
Given f is continuous over R ⇒ f is continuous at 0, 1 and 3.
If f is continuous at ‘0’ then

∴ In all cases if f is continuous at 0, 1, 3 and continuous over R.
The values are a = 0,
b = – 2.

Question 3.
Show that

where a and b are real constants, is continuous at ‘0’. (Board New Model Paper) (E.Q.)
Answer:

Students must practice these TS Inter 1st Year Maths 1B Study Material Chapter 8 Limits and Continuity Ex 8(d) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Limits and Continuity 8(d)

Compute the following limits.

Question 1.
\(\lim _{x \rightarrow 3}\left(\frac{x^2+3 x+2}{x^2-6 x+9}\right)\) (V.S.A.Q.)
Answer:

Question 2.
\(\lim _{x \rightarrow 1}\left(\frac{1+5 x^3}{1-x^2}\right)\) (V.S.A.Q.)
Answer:

Question 3.
\(\lim _{x \rightarrow \infty}\left(\frac{3 x^2+4 x+5}{2 x^3+3 x-7}\right)\) (V.S.A.Q.)
Answer:

Question 4.
\(\lim _{x \rightarrow \infty}\left(\frac{6 x^2-x+7}{x+3}\right)\) (V.S.A.Q.)
Answer:

Question 5.
\(\lim _{x \rightarrow \infty}\) (e^-x2 (V.S.A.Q.)
Answer:

Question 6.
\(\lim _{x \rightarrow \infty}\left(\frac{\sqrt{x^2+6}}{2 x^2-1}\right)\) (V.S.A.Q.)
Answer:

II.
Question 1.
\(\lim _{x \rightarrow \infty} \frac{8|x|+3 x}{3|x|-2 x}\) (S.A.Q.) [March 2012]
Answer:

Question 2.
\(\lim _{x \rightarrow \infty}\left(\frac{x^2+5 x+2}{2 x^2-5 x+1}\right)\) (May 2014) (S.A.Q.)
Answer:

Question 3.
\(\lim _{x \rightarrow-\infty}\left(\frac{2 x^2-x+3}{x^2-2 x+5}\right)\) (S.A.Q.)
Answer:

Question 4.
\(\lim _{x \rightarrow \infty}\left(\frac{11 x^3-3 x+4}{13 x^3-5 x^2-7}\right)\) (V.S.A.Q.)
Answer:

Question 5.
\(\lim _{x \rightarrow 2}\left(\frac{1}{x-2}-\frac{4}{x^2-4}\right)\) (S.A.Q.)
Answer:

Question 6.
\(\lim _{x \rightarrow-\infty} \frac{5 x^3+4}{\sqrt{2 x^4+1}}\) (S.A.Q.)
Answer:

Question 7.
\(\lim _{x \rightarrow \infty}\) (√x + 1 – √x) (S.A.Q.)
Answer:

Question 8.
\(\lim _{x \rightarrow \infty}\) (\(\sqrt{x^2+x}\) – x) (S.A.Q.) [March 2011]
Answer:

III.
Question 1.
\(\lim _{x \rightarrow-\infty}\left(\frac{2 x+3}{\sqrt{x^2-1}}\right)\) (S.A.Q.)
Answer:

Question 2.
\(\lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right)\) (E.Q.)
Answer:
\(\lim _{x \rightarrow \infty} \frac{2+\sin x}{x^2+3}\)
We have – 1 ≤ sin x ≤ 1
⇒ 2 – 1 ≤ 2 + sin x ≤ 2 + 1
⇒ 1 ≤ 2 + sin x ≤ 3

Question 3.
\(\lim _{x \rightarrow \infty}\left(\frac{2+\cos ^2 x}{x+2007}\right)\) (E.Q.)
Answer:

Question 4.
\(\lim _{x \rightarrow \infty}\left(\frac{6 x^2-\cos 3 x}{x^2+5}\right)\) (V.S.A.Q.)
Answer:

Question 5.
\(\lim _{x \rightarrow \infty}\left(\frac{\cos x+\sin ^2 x}{x+1}\right)\) (E.Q.)
Answer:
We know that – 1 ≤ cos x ≤ 1 and 0 ≤ sin² x ≤ 1
∴ – 1 + 0 ≤ cos x + sin² x ≤ 2

Here students can locate TS Inter 1st Year Economics Notes Chapter 8 Theories of Employment and Public Finance to prepare for their exam.

TS Inter 1st Year Economics Notes Chapter 8 Theories of Employment and Public Finance

→ Aggregate Supply Price: Aggregate supply is the total supply of goods and services in the economy. Aggregate supply price refers to the minimum total income that the entrepreneurs in the economy expect to receive from the sale of the total output.

→ Aggregate Demand Price: Aggregate demand is the total demand for the goods and services in the economy. Aggregate demand price refers to the amount that is expected to be spent on the total output. This is also called the total expenditure (Aggregate Demand = Consumption expenditure + Investment).

→ Effective Demand: Effective demand is that aggregate demand which becomes equal to the aggregate supply. This refers to the aggregate demand at equilibrium.

→ Wage-Cut Policy: This is the policy advocated by A.C. Pigou (classical economist) to reduce unemployment in the economy. It suggests that the wages of the labour should be reduced so that the unemployed can be employed.

→ Budget: Budget is the annual statement of the estimated receipts (income) and the estimated expenditure of the government for the ensuing financial year (April 1 to March 31).

→ Vote on Account: Vote – on the account is an interim budget presented for a few months pending the presentation of the regular budget.

→ Surplus Budget: Surplus budget refers to the budget in which the total receipts exceed the total expenditure.

→ Deficit Budget: Deficit budget refers to the budget in which the total expenditure exceeds total receipts.

→ Balanced Budget: Balanced budget is a budget in which the total receipts and total expenditures are equal.

→ Budget Deficit: A budget deficit is the difference between the total receipts and the total expenditure in the budget.

→ Revenue Deficit: Revenue deficit is the difference between the revenue receipts and the revenue expenditure.

→ Fiscal Deficit: It is the budget deficit plus the market borrowings.

→ Primary Deficit: It is the fiscal deficit minus the interest payments.

→ Redemption of Debt: It means repayment of public debt.

→ Public Finance: It deals with the income and expenditure of the public authorities (central, state and local governments).

→ Capital levy: It is a once – for – all tax imposed on capital assets and estates.

→ Federal Finance: It means divisions and coordinations of different heads of income and expenditure between central, state, and local governments.

→ Finance Commission: This was established in the year 1951 by the president of India under Article 280 of the Indian constitution with a duration of five years. It recommends the distribution of financial resources between the center and the state. So far, fifteen Finance Commissions have been constituted in India.

TS Inter 1st Year Economics Notes Chapter 8 ఉద్యోగిత సిద్ధాంతాలు మరియు ప్రభుత్వ విత్తం

→ సాంప్రదాయ ఆర్థికవేత్తలయిన ఆడమస్మిత్, రికార్డో, మాల్డస్, J. B. సే మొదలగు ఆర్థికవేత్తలు రూపొందించిన ఉద్యోగితా సిద్ధాంతాన్నే సాంప్రదాయ సిద్ధాంతం అంటారు.

→ ఈ సిద్ధాంతంలో ప్రభుత్వ జోక్యం ఉండదు. సంపూర్ణ పోటీ ఉద్యోగిత, వేతనాల సరళత్వం మొదలగు ప్రమేయాలపై ఆధారపడింది.

→ సంపూర్ణ ఉద్యోగితా స్థాయిని పొందక ముందే సమిష్టి సప్లై, సమిష్టి డిమాండ్ సమానమైన సమతౌల్యాన్ని సాధించడం జరుగుతుందని కీన్స్ అభిప్రాయం. ఆర్థిక వ్యవస్థలో కొంత మంది నిరుద్యోగులు ఉంటారు. ఈ స్థితినే అల్ప ఉద్యోగితా సమతౌల్యం అని అంటారు.

→ ఆర్థిక వ్యవస్థలో మొత్తం వస్తురాశికుండే డిమాండ్ సమిష్టి డిమాండ్.

→ ఆర్థిక వ్యవస్థలో మొత్తం వస్తురాశి సప్లైని తెలియజేయునది సమిష్టి సప్లై, సమిష్టి డిమాండ్. సప్లై సమానమైన స్థితిని సార్థక డిమాండ్ అంటారు.

→ ప్రభుత్వం ఆదాయ, వ్యయాలను గురించి వీటి మధ్య సర్దుబాట్లకు సంబంధించిన విషయాలను అధ్యయనం చేసేదే ‘ప్రభుత్వ విత్తశాస్త్రం.

→ ప్రభుత్వ ఆదాయ వ్యయాల పట్టికను బడ్జెట్ అంటారు. కొన్ని సందర్భాలలో ప్రభుత్వాలు పూర్తి బడ్జెట్ను సమర్పించే వీలు లేనప్పుడు ‘ఓట్ ఆన్ అకౌంట్’ బడ్జెట్ను తాత్కాలికంగా కొన్ని నెలల కోసం సమర్పిస్తుంది.

Here students can locate TS Inter 1st Year Economics Notes Chapter 7 National Income Analysis to prepare for their exam.

TS Inter 1st Year Economics Notes Chapter 7 National Income Analysis

→ National Income: National income is the market value of goods and services produced annually in a country.

→ Gross National Product (GNP): Gross national product is the current monetary value of goods and services produced in a year in a country and to this net income from abroad has to be added.

→ Depreciation: Depreciation is the wear and tear of machines or replacement cost.

→ Net National Product (NNP): Depreciation has to be deducted from the gross national product to arrive at NNR.

→ Gross Domestic Product (GDP): Gross domestic, the product is the current monetary value of goods and services produced in a year in a country by their own factors of production. Net income from abroad is not included in gross domestic product.

→ Transfer Payments: Payments made in the form of pensions, unemployment allowances etc., which need not be returned.

→ Subsidies: When a producer sells his product at the price less than its cost he incurs loss and this will be paid by the government to the producer in the form of subsidies.

→ Per Capita Income: Per capita income is the average income of people in a country in a particular year. It can be arrived at by dividing national income with population.

→ Disposable Income: Disposable income is the part of personal income which is left with the individual after payment of direct taxes.

TS Inter 1st Year Economics Notes Chapter 7 జాతీయాదాయ విశ్లేషణ

→ ఒక దేశంలో ఒక సంవత్సర కాలంలో ఉత్పత్తి అయిన వస్తు సేవల మొత్తం నికర విలువ.

→ ఫిషర్ నిర్వచనం: వినియోగదారులు మానవ వనరుల నుండి లేదా భౌతిక వనరుల నుండి పొందే వస్తు సేవల సముదాయమే జాతీయాదాయం.

→ జాతీయాదాయాన్ని ప్రకృతి వనరులు, శ్రమ, మూలధనం, వ్యవస్థాపన, ఇతర కారకాలు నిర్ణయిస్తాయి.

→ స్థూల జాతీయోత్పత్తి, స్థూల దేశీయోత్పత్తి, నికర జాతీయోత్పత్తి, ఉత్పత్తి కారకాల ఖరీదు దృష్ట్యా జాతీయాదాయం, వ్యష్టి ఆదాయం, వ్యయార్హ ఆదాయం, తలసరి ఆదాయం మొదలైనవి జాతీయాదా యానికి సంబంధించిన వివిధ భావనలు.

→ వినియోగం, స్థూల దేశీయ పెట్టుబడి, ప్రభుత్వ వ్యయం, నికర దేశీయ పెట్టుబడి మొదలగునవి జాతీయ ఆదాయంలోని వివిధ భాగాలు.

→ జాతీయాదాయ మదింపు మూడు పద్ధతుల ద్వారా లెక్కించవచ్చు.

1. ఉత్పత్తి మదింపు పద్ధతి
2. వ్యయాల మదింపు పద్ధతి
3. ఆదాయాల మదింపు పద్ధతి.

→ ద్రవ్య రూపంలో కొన్ని వస్తువుల విలువను చెప్పలేకపోవటం, కొన్ని సంస్థలలో లెక్కలు సరిగా లేకపోవటం, ప్రభుత్వ ఖర్చులు, పన్నులు మొదలైన వాటి లెక్కల సమగ్ర సేకరణ జరగకపోవటం, మనదేశంలో వృత్తుల ప్రత్యేకీకరణ లేకపోవటం మొదలైన అనేక సమస్యలు ఉన్నాయి.

→ ఉత్పత్తి కారకాల మధ్య పంపిణీ సంవత్సర కాలంలో ఉత్పత్తియైన వస్తురాశిని అంచనా వేయటం, జీవన ప్రమాణ స్థాయిని తెలుసుకోవటం, వివిధ దేశాల అభివృద్ధిని గుర్తించటం, ఆర్థిక సమస్యలను గుర్తించటం, ఆర్థిక ప్రణాళికల రూపకల్పన, వినియోగ వ్యయం, పొదుపు మొదలగు అంశాల అంచనాకు జాతీయాదాయ లెక్కలు తోడ్పడతాయి.