TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces

Telangana SCERT 10th Class Physics Study Material Telangana 1st Lesson Reflection of Light at Curved Surfaces Textbook Questions and Answers.

TS 10th Class Physical Science 1st Lesson Questions and Answers Reflection of Light at Curved Surfaces

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I. Reflections on concepts

Question 1.
Where will the image be formed when we place an object, on the principal axis of a concave mirror at a point between focus and centre of curvature?
Answer:
1. When we place an object on the principal axis of concave mirror at a point between focus and centre of curvature, the image is formed beyond centre of curvature.
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 1
2. The image so formed is real, inverted and magnified.

Question 2.
State the differences between convex and concave mirrors.
Answer:

Convex mirror Concave mirror
1. A parallel beam of light falling on this mirror appears to diverge from a point after reflection. 1. A parallel beam of light falling on this mirror converges at a point after reflection.
2. The reflecting surface of convex mirror is bulged out. 2. The reflecting surface of a concave mirror curve inward.
3. Radius of curvature and focal length are negative. 3. Radius of curvature and focal length are positive.
4. It’s magnification has positive only. 4. It’s magnification has both positive and negative.
5. Magnification of convex mirror is in between zero and one. 5. Magnification value of concave mirror having all values except zero to one.
6. The image formed by convex mirror always diminished. 6. The image formed by concave mirror may be magnified or diminished.
7. This mirror produces only virtual image. 7. This mirror produces both real and virtual images depending upon position of object.

Question 3.
Distinguish between real and virtual images.
Answer:

Real image Virtual image
1. Real image is formed if light after 1. Virtual image is formed when rays reflection or refraction converges after reflection appear to be comto a point. ing from a point.
2. Here the rays actually meet at the from the image point. 2. Here the rays appear to diverge image point.
3. It can be captured on screen. 3. It cannot be captured on screen.
4. It is always inverted. 4. It is always erect.

 

TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces

Question 4.
How do you get a virtual image using a concave mirror?
Answer:
A. When an object is kept between pole and focus of a concave mirror virtual image is formed behind the mirror.
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 2

Question 5.
What do you know about the terms given below related to spherical mirrors?
(a) pole
(b) centre of curvature
(c) focus
(d) Radius of curvature
(e) Focal length
(f) principal axis
(g) object distance
(h) image distance
(i) Magnification.
Answer:
(a) Pole: The point on the principal axis of spherical mirror with respect to which all the measurements are made. Usually, it is the mid point of the curvature of mirror.

(b) Centre of curvature: The centre of the sphere of which the curved surface of the mirror is a part.

(c) Focus: The light rays coming from a source parallel to the principal axis converge at a point after reflection. This point is called focus or focal point.

(d) Radius of curvature: The radius of the sphere of which the curved surface is a part is called radius of curvature.

(e) Focal length: The distance between the pole of the mirror and focus is called focal length of mirror.

(f) Principal axis: The straight line passing through the centre of curvature and pole of curved mirror is called principal axis.

(g) Object distance: The distance between the pole of the mirror and object position of object is known as object distance.

(h) Image distance: The distance between the pole of the mirror and position of image is called image distance.

(i) Magnification: Magnification of a spherical mirror is the ratio between size (height) of image to the size (height) of object. Also, m= v/u.

Question 6.
What do you infer from the experiment which you did to measure the object distance and image distance?
Answer:
Inference from the experiment which I did with concave mirror :
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 3
Inference:

  1. As the object moves away from the mirror the image approaches the mirror.
  2. As the object moves away from mirror the size of the image becomes smaller.

Question 7.
Write the rules for sign convention.
Answer:
Rules for sign convension:

  1. AN distances should be measured from the pole.
  2. The distances measured in the direction of incident light, are taken positive and the opposite direction of incident light are taken negative.
  3. Height of object (H0) and height of image (H1) are positive if measured upward
    from the axis and negative if measured downward.
  4. For a concave mirror ‘f’ and ‘R’ are negative and for a convex mirror these are positive.

II. Application of concepts

Question 1.
Find the distance of the image, when an object is placed on the principal axis, at a distance of 10 cm in front of a concave mirror whose radius of curvature is 8 cm.
Answer:
Object distance u = -10 cm
Radius of curvature (r) = -8cm
Image distance v=?
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 4
The image distance (v) = 6.7 cm.
i.e., Real image is formed at same side of the mirror.

TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces

Question 2.
The magnification product by a plane mirror is +1. What does it mean?
Answer:

  1. Magnification = \(\frac{\text { height of the image }}{\text { height of the object }}=\frac{\text { distance of the image }}{\text { distance of the object }}\)
  2. The magnification produced by a plane mirror is +1 means then the size of the image is equal to the size of the object.
  3. + sign indicates that the image is erect. Magnification ‘+1’ indicates the image is erect and size of the image is equal to size of the object.

Question 3.
If the spherical mirrors were not known to human beings, guess the consequences.
Answer:
If spherical mirrors are not known to human beings

  1. Many optical instruments would not have been invented.
  2. We cannot increase the size of images of the objects.
  3. The problem of lateral inversion of images will not be solved.
  4. Now a days spherical mirrors are used as shaving mirrors, head mirrors for ENT specialists, in headlights of motor vehicles, in solar furnaces and as rearview mirror. If spherical mirrors are not known all these are not possible.

Question 4.
Draw suitable rays by which we can guess the position of the imge formed by a concave mirror.
Answer:
The following rays are used to guess the position of the image formed by a concave mirror.
(i) A ray parallel to the principal axis passes through principal focus (F) after reflection from a concave mirror.
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 5
(ii) A ray passing through ‘F’ becomes parallel to principal axis after reflection from a concave mirror.
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 6
(iii) A ray passing through ‘C’ is reflected back along the same path after reflection from a concave mirror.
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 7
(iv) A ray incident obliquely to the principal axis towards the pole P, on the concave mirror is reflected obliquely, following the laws of reflection.
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 8

Question 5.
Show the formation of image with a ray diagram, when an object is placed on the principal axis of a concave mirror away from the centre of curvature.
Answer:
When an object is placed on the principal axis of a concave mirror and beyond (away from) its centre of curvature ‘C’, the image is formed between the focus (F) and the centre of curvature (C). The ray which is parallel to principle axis will pass through focus after reflection and the ray which passes through focus will travels parallel to principal axis after reflection. These two rays will converge between F and C of the mirror.
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 9
The image is real, inverted and diminished in size.

Question 6.
Why do we prefer a convex mirror as a rear-view mirror in the vehicles?
Answer:
We use convex mirror as a rear-view mirror in the vehicles because

  1. Convex mirror always forms virtual, erect, and diminished images irrespective of distance of the object.
  2. A convex mirror enables a driver to view large area of the traffic behind him.
  3. Convex mirror forms very small image than the object. Due to this reason convex mirrors are used as rear-view mirrors in vehicles.

TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces

III. Higher Order Thinking Questions

Question 1.
A convex mirror with a radius of curvature of 3 m is used as a rearview mirror for a vehicle. If a bus is located at 5m from this mirror, find the position, nature, and size of the image.
Answer:
According to the sign convention :
Radius of curvature = R = + 3m
Object distance = u = -5 m (negative sign)
Image distance = v =?
Focal length, f = \(\frac{R}{2}=\frac{3}{2} m\) = 1.5 m
Formula : \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 10
The image is formed at a distance of 1.15 m at the back of the mirror.
Magnification, m = \(\frac{h_i}{h_o}=-\frac{v}{u}=\frac{-1.15 m}{-5 m}=\frac{1.15}{5}\) = 0.23
The image is virtual, erect, and diminished to 0.23 times of the size of the object.

Question 2.
To form the image on the object itself, how should we place the object in front of a concave mirror? Explain with a ray diagram.
Answer:
To form the image on the object itself, the object should be kept at center of curvature of a concave mirror.

  1. An object AB has been placed at the centre of curvature ‘C’ on the concave mirror.
  2. A ray of light AD which is parallel to principle axis passes through the focus ‘F after reflection as DA’.
    TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 11
  3. A ray of light passing through the focus of the concave mirror becomes parallel to the principle axis after reflection.
  4. Here AE ray passing through focus and reflected as EA’.
  5. The reflected rays DA’ and EA’ meet at A’ point. So the real image formed at point A’ of the object.
    We get couple image AB perpendicular to the principal axis.
  6. Thus A’B’ is the real inverted image of the object AB.

IV. Multiple choices questions

Question 1.
If an object is placed at ‘C’ on the principal axis in front of a concave mirror, the position of the image is ………………. . [ ]
(a) at infinity
(b) between F and C
(c) at C
(d) beyond C
Answer:
(c) at C

Question 2.
We get a diminished image with a concave mirror when the object is placed ………………………. . [ ]
(a) at F
(b) between the pole and F
(c) at C
(d) beyond C
Answer:
(d) beyond C

Question 3.
We get a virtual image in a concave mirror when the object is placed …………………….. . [ ]
(a) at F
(b) between the pole and F
(c) at C
(d) beyond C
Answer:
(b) between the pole and F

Question 4.
Which of the following represents Magnification? [ ]
(i) \(\frac{v}{u}\)
(ii) \(\frac{-v}{u}\)
(iii) \(\frac{h_i}{h_0}\)
(iv) \(\frac{h_0}{h_i}\)
(a) i, ii
(b) ii, iii
(c) iii, iv
(d) iv, i
Answer:
(d) iv, i

Question 5.
ray which seems to be traveling through the focus of a convex mirror, path of the reflected ray of an incident ……………… . [ ]
(a) parallel to the axis
(b) along the same path in opposite direction
(c) through F
(d) through C
Answer:
(a) parallel to the axis

TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces

Question 6.
Size of image formed by a convex mirror is always …………………… .[ ]
(a) enlarged
(b) diminished
(c) equal to the size of object
(d) depends on position of object
Answer:
(b) diminished

Question 7.
An object is placed at a certain distance on the principal axis of a concave mirror. The image is formed at a distance of 30 cm from the mirror. Find the object distance if radius of curvature R = 15cm. [ ]
(a) 15 cm
(b) 10 cm
(c) 30 cm
(d) 7.5 cm
Answer:
(c) 30 cm

Question 8.
All the distances related to spherical mirror will be measured from …………………… .[ ]
(a) object to image
(b) focus of the mirror
(c) pole of the mirror
(d) image to object
Answer:
(c) pole of the mirror

Question 9.
The minimum distance from real object to a real image in a concave mirror is ………………….. . [ ]
(a) 2f
(b) f
(c) 0
(d) f/2
Answer:
(c) 0

Suggested Experiments

Question 1.
Conduct an experiment to find the focal length of concave mirror.
(or)
How can you find out the focal length of concave mirror experimentally when there is no sunlight?
Answer:
Aim: To find the focal length of a concave mirror,
Materials required : (i) A concave mirror (ii) V-shape stand (iii) A candle (iv) A meter scale.
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 12
Procedure :

  1. Place the concave mirror on the V-shape stand.
  2. Keep a burning candle in front of the concave mirror.
  3. Place a thick white paper behind the candle. This acts as a screen.
  4. Adjust distances between candle and mirror, screen and mirror by moving them either forward or backward till a clear well-defined image appears on the screen.
  5. Measure the distance between the mirror and candle (object distance u) and the distance between mirror and screen (image distance v).
  6. Using the mirror formula, \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v} \) or f = \(\frac{u v}{u+v} \)
    This gives the focal length of the concave mirror.

Question 2.
Find the nature and position of images when an object is placed at different places on the principal axis of a concave mirror.
Answer:
Aim: Observing the types of images and measuring the object distance and image distance from the concave mirror.
Material required: A candle, paper, concave mirror (known focal length), V- stand, measuring tape or meter scale.

Procedure:

  1. Place the concave mirror on V-stand, a candle and meter scale as shown in figure.
    TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 13
  2. Keep the candle at different distances from the mirror (10 cm to 80 cm) along the axis and by moving the paper screen find the position where you get the sharp image on paper.
  3. Note down your observations in the following table.
  4. Since we know the focal point and centre of curvature, we can classify our above observations as shown in the following table.

TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 14

Suggested Project Works

Question 1.
Collect information about the history of spherical mirrors in human civilization, write a report on it.
Answer:

  1. The first mirrors used by people were most likely pools of water or still water. The earliest manufactured mirrors were pieces of polished stones.
  2. Parabolic mirrors were described and studied in classical antiquity by the mathematician Archimedes in his work on burning mirrors.
  3. Ptolemy conducted a number of experiments with curved polished iron mirrors. He also discussed plane, convex, concave, and spherical mirrors in his optics.
  4. In China, people began making mirrors with the use of silver mercury amalgams as early as 500 AD.
  5. In 16th century, Venice, a big city popular for its glass-making expertise, became a centre of mirror production using this new technique.
  6. The invention of the silvered-glass mirror is credited to German Chemist Justus Von Liebig in 1835.

TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces

Question 2.
Think about the objects which act as concave or convex mirrors in your surroundings. Make a table of these objects and display in your classroom.
Answer:
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 15

Question 3.
Collect photographs from your daily life where you use convex and concave mirrors and display in your classroom.
Answer:
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 16

TS 10th Class Physical Science Reflection of Light at Curved Surfaces Intext Questions

Page 1

Question 1.
Is the image formed by a bulged surface same as the image formed by a plane mirror?
Answer:
No, the image formed by a bulged surface is virtual, created, and diminished image.

Question 2.
Is the mirror used in automobiles a plane mirror? Why it is showing small images?
Answer:
No, the mirror used in automobile is convex mirror. At it bulging outwards it forms small images.

Question 3.
Why does our image appear thin or bulged out in some mirrors?
Answer:
The image in a mirror appears thin or bulged out because the thickness of the mirror may vary or the reflecting surface may not be flat.

TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces

Question 4.
Can be see inverted image in any mirror?
Answer:
Can we see inverted image in any mirror? Yes, we can see inverted image in concave mirror for a distant objects.

Question 5.
Can we focus the sunlight at a point using a mirror instead of a magnifying glass?
Answer:
Yes. By using a black paper with a tiny hole at its centre.

Question 6.
Are the angle of reflection and angle of incidence also equal for reflection by curved surfaces?
Answer:
No.

Page 4

Question 7.
Does this help you to verify the conclusions you arrived at with your drawing?
Answer:
Yes.

Question 8.
What happens if you hold the paper at a distance shorter than the focal length from the mirror and move it away?
Answer:
We find there is no point at which the reflected rays converge at a point. But as we move the paper away from the focal point, we find images formed at different distances from the mirror.

Question 9.
Does the image of the sun become smaller or bigger?
Answer:
We notice that the image of the sun keeps on becoming smaller. Beyond the focal point, it will become bigger.

Page 5

Question 10.
Do we get an image with a concave mirror at the focus every time?
Answer:
We get the images not only at the focus every time, we get different ¡mages by keeping the object at different points depending on focal length of mirror.

Page 6

Question 11.
It is inverted or erect, enlarged or diminished?
Answer:
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 19

Question 12.
What do you infer from the table?
Answer:
From the table 2. I infer that images can be formed at other positions different from focal point.

Page 7

Question 13.
Why only at point A?
Answer:
If we hold the screen at any point before or beyond point A (for example at point B), we see that the rays will meet the screen at different points due to these rays. If we draw more rays emanating from the same tip we will see that
at point A they will meet but at point B they do not. So the image of the tip of the flame will be sharp at point A.

Page 9

Question 14.
Where is the base of the candle expected to be in the image when the candle is placed on the axis of the mirror?
Answer:
The base of the candle is going to be on the principle axis ¡n the image when the object is placed on the axis of the mirror.

Question 15.
During the experiment, did you get any positions where you could not get an image on the screen?
Answer:
Yes. When the object is placed at a distance less than the focal length of the mirror we do not get an ¡mage on the screen.

Page 11

Question 16.
Have you observed the rearview mirrors of a car?
Answer:
Yes. I have observed rear view mirror of a car.

TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces

Question 17.
What type of surface do they have?
Answer:
A concave mirror will be like the rubber sole bent inwards and the reflecting surface will be curved inwards.

Question 18.
Can we draw ray diagrams for convex surface?
Answer:
We can draw ray diagrams with convex surface by making use of “easy” rays that we have identified, with small modifications.

Think and discuss 

Question 1.
See the figure – 5 In text. A set of parallel rays are falling on a convex mirror. What conclusions can you draw from this?
Answer:
On seeing figure 5, the following conclusions can be drawn;

  1. The parallel beam of rays meet at infinity.
  2. So the Image Is not visible. :
  3. This parallel beam of rays forms a virtual image which is smaller than the size of object.

TS 10th Class Physical Science Important Questions Chapter 1 Reflection of Light at Curved Surfaces 17

Question 2.
Will you get a point Image If you place a paper at the focal point?
Answer:
No. These parallel beam of rays do not meet at a visible point and we do not get a point image.

Question 3.
Do you get an image when object is placed at F? Draw a ray diagram. Do the experiment.
Answer:
We did not get the image when the object is placed at Focus ‘F’ of concave mirror.
Experiment:
Aim: Observing the Image formed by the object which is placed at ‘F’ of concave mirror.
Materal required: A candle, paper, a concave mirror (known focal length), V-stand, measuring
tape or meter scale.
TS 10th Class Physical Science Important Questions Chapter 1 Reflection of Light at Curved Surfaces 18
Procedure:
1. Place the concave mirror on V-stand, a candle and meter scale as shown in figure.
TS 10th Class Physical Science Important Questions Chapter 1 Reflection of Light at Curved Surfaces 19
2. Keep the candle at a distance equal to the focal length of the mirror. (which Is known)
3. Now move the paper screen away from the mirror along the axis to observe the image.
4. You will notice that the image cannot be seen because it Is formed at infinity.

TS 10th Class Physical Science Reflection of Light at Curved Surfaces Activities

Activity 1

Question 1.
Explain an activity to find the normal to a curved surface.
Answer:

  1. Take a small piece of thin foam or rubber.
  2. Put some pins in a straight line on the foam.
  3. All these pins are perpendicular to the foam.
  4. If the foam is considered as a mirror, each pin would represent the normal at that point.
  5. Any ray incident at the point where the pin makes contact with the surface will reflect at the same angle the incident ray made with the pin-normal.
  6. Now bend the foam piece inwards.
  7. The pins still represent the normal at various poInts.
  8. You will observe that all the pins tend to coverage at a point.
  9. This will be appear like a concave mirror.
  10. Now bend the foam piece outwards.
  11. The pins seem to move away from each other that means they diverge.
  12. The pins still represent the normal at various points.
  13. This will be appear like a convex mirror.
    TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 20

Activity 2

Question 2.
How do you Identify the focal point and foal length of a concave mirror?
Answer:
Hold a concave mirror perpendicular to me direction of sunlight

TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 21

  • Take a small paper and slowly move it in front of the mirror.
  • Find the point where you get smallest and brigh – test spot, which is the image of the sun.
  • The rays coming from the sun parallel to the concave mirror are converging at a point. This point is called focus or focal point (F) of the concave mirror.
  • Measure the distance of this spot from the pole (P) of the mirror.
  • ThIs distance Is the focal length (f) of the mirror

Lab Activity

Question 1.
Describe an experiment to observe types of mages formed by e concave mirror and measure the object end image distances.
(OR)
Write the experimental method in measuring the distances of object and image using concave mirror. And write the table For observations.
Answer:
Aim: Observing the types of images and reasoning the object distance and image distance from the concave mirror.
Material required: A candle, paper, concave minor (known focal length), V- stand, moesunng tape or meter scale.
Procedure
1. Place the concave mirror on V-stand, a candle and meter scale as shown in figure.
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 22
2. Keep the candle at different distances from the mIrror (10 cm to 80 cm) along the axis and by moving the paper screen fd the position where you get the sharp mage on paper.
3. Note down your observations in the following table.
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 23
4. Since we know the focal point and centre of curvature, we can re-classify our above observations as shown in the following table.
TS 10th Class Physical Science Solutions Chapter 1 Reflection of Light at Curved Surfaces 24

TS 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Ex 1.1

Students can practice TS 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Ex 1.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Exercise 1.1

Question 1.
Which is the greatest and the smallest among the following numbers ?
(i) 15432, 15892, 15370, 15524
Answer:
All the four numbers are of 5 digits.
The digits in ten thousands place in all the given numbers are same.
The digits in thousands place are also same.
So we move to hundreds place to compare them.
The digits in hundreds place are 4, 8, 3 and 5 respectively.
∴ 3 < 4 < 5 < 8
∴ The greatest number is 15892
The smallest number is 15370

TS 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Ex 1.1

(ii) 25073, 25289, 25800, 25623
Answer:
All the four numbers are of 5 digits. The digits in ten thousands place in all the given numbers are same. The digits in thousands place are also same. So we move to hundreds place to compare them.
The digits in hundreds place are 0, 2, 8, 6 respectively.
∴ 0 < 2 < 6 < 8
The greatest number is 25800
The smallest number is 25073

(iii) 44687, 44645, 44670, 44602
Answer:
All the four numbers are of 5 digits.
The digits in ten thousands place, thousands place and hundreds place are same in all the numbers.
So we move to tens place to compare them.
The digits in tens place are 8, 4, 7, 0 respectively.
∴ 0 < 4 < 7 < 8
∴ The greatest number is 44687.
The smallest number is 44602.

(iv) 75671, 75635, 75641, 75610
Answer:
All the four numbers are of 5 digits. The digits in ten thousands place, thousands place and hundreds place are same in all the numbers. So we move to tens place to compare them. The digits in tens place are 7, 3, 4, 1 respectively.
∴ 1 < 3 < 4 < 7
∴ The greatest number is 75671 The smallest number is 75610

v) 34895, 34891, 34899, 34893
Sol. All the four numbers are of 5 digits. The digits in ten thousands place, thousands place, hundreds place and tens place are same in all the numbers. So we move to units place to compare them.
The digits in units place are 5, 1, 9, 3 respectively.
∴ 1 < 3 < 5 < 9
∴ The greatest number is 34899
The smallest number is 34891

Question 2.
Write the numbers in ascending (in-creasing) order.
(i) 375, 1475, 15951, 4713
Answer:
We can say that 15951 is the greatest number and 375 is the smallest number by counting the digits in the numbers. Regarding 1475 and 4713 it is clear that 1475 < 4713
∴ 375 < 1475 < 4713 < 15951
The ascending order is 375, 1475, 4713, 15951

(ii) 9347, 19035, 22570, 12300
Answer:
The number with four digits (i.e.,) 9347 is the smallest number because the other numbers are of five digits.
In 19035, 22570, 12300 numbers, the greatest number is 22570.
Again 12300 < 19035
∴ 9347 < 12300 < 19035 < 22570
The ascending order is 9347, 12300, 19035, 22570

Question 3.
Write the numbers in descending (decreasing) order.
(i) 1876, 89715, 45321, 89254
Answer:
1876 is the smallest number because it has four digits while the other numbers are of five digits.
The other numbers are 89715, 45321, 89254.
45321 < 89254 < 89715 (∵ 2 < 7) ∴ The descending order is 89715, 89254, 45321, 1876. (∵ 89715 > 89254 > 45321 > 1876)

TS 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Ex 1.1

(ii) 3000, 8700, 3900, 18500
Answer:
18500 is the greatest number because it has five digits while the other numbers are of four digits.
The other numbers are 3000,8700,3900
3000 < 3900 < 8700 ∴ The descending order is 18500, 8700, 3900, 3000. (∵ 18500 > 8700 > 3900 > 3000)

Question 4.
Compare the numbers by placing appropriate symbol (< or >) in the space given.
(i) 3854 ………………… 15200
Answer:
3854 < 15200 (ii) 4895 …………………. 4864 Answer: 4895 > 4864

(iii) 99454 ………………….. 99445
Answer:
99454 > 99445

(iv) 14500 ………………….. 14499
Answer:
14500 > 14499

Question 5.
Write the numbers in words :
(i) 72642
Answer:
72,642 = Seventy two thousand six hundred forty two

(ii) 55345
Answer:
55,345 = Fifty five thousand three hundred forty five

(iii) 66600
Answer:
66,600 = Sixty six thousand six hundred

(iv) 30301
Answer:
30,301 = Thirty thousand three hundred one

Question 6.
Write the numbers in figures:
(i) Forty thousand two hundred seventy.
Answer:
40,270

(ii) Fourteen thousand sixty four.
Answer:
14,064

(iii) Nine thousand seven hundred.
Answer:
9,700

(iv) Sixty thousand.
Answer:
60,000

Question 7.
Form four digit numbers with the digits 4, 0, 3, 7 and find which is the greatest and the smallest among them ?
Answer:
The digits given are 4, 0, 3, 7.
The four digit numbers formed with the digits are
4037, 4073, 4370, 4307, 4730, 4703
3047, 3074, 3470, 3407, 3704, 3740
7034, 7043, 7304, 7340, 7403, 7430
The greatest number is 7430
The smallest number is 3047

TS 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Ex 1.1

Question 8.
Write the following numbers.
(i) the smallest four digit number
Answer:
1000

(ii) the greatest four digit number
Answer:
9999

(iii) the smallest five digit number
Answer:
10000

(iv) the greatest five digit number
Answer:
99999

TS Inter 1st Year Maths 1A Functions Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 1 Functions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Functions Important Questions

Very Short Answer Questions

Question 1.
If \(A=\left\{0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}\right\}\) surjection defined by f(x) = cos x, then find B
Solution:
f: A → B is a surjection
⇒ Codomain of B = Range f(A)
Given f(x) = cos x
TS Inter 1st Year Maths 1A Functions Important Questions 2

TS Inter 1st Year Maths 1A Functions Important Questions

Question 2.
find the domain of the real-valued function f(x)=\(\frac{1}{\log (2-x)}\)
Solution:
f(x)=\(\frac{1}{\log (2-x)}\) is defined for 2 – x > 0 and 2 – x ≠ 1
⇒ x-2<0 and 2-x
⇒ x<2 and x ≠ 1 ⇒ x ∈(-∞, 2)- {1}
∴ Domain of f = {x / x ∈ (-∞, 2)-{1}}

Question 3.
If f : A → B, g: B → C are two bijective functions, then prove that gof = A → C is also a bijective function.
Solution:
i) Given f, g are bijections, f, g are both one one and onto.
To prove that gof : A → C is one one:
TS Inter 1st Year Maths 1A Functions Important Questions 3
(∵ f : A → B is one one, and
g : B → C are one one)
∴ gof : A → C is one one

ii) To Prove that gof = A → C is onto:
Let C∈C; since g : B → C is on to ∀ c ∈ C ∃
b E B such that g(b) = c …………………….. (1)
Also f: A → B is on to for b∈B ∃ a∈A such that f (a) = b …………………….. (2)
∴ bc = g(b) = g[f(a)]
= (gof) (a)
Hence for c ∈ C, ∃ a ∈ A such that
gof : A → C is onto
Hence from the above two results
gof : A → C is a Bijection.

Question 4.
If f : A → B is a function and lA,IB are identity functions on A, B respectively then prove that folA lBof = f
Solution:
i) To prove that folA = f
Since f : A → B and ‘A : A → A, we have folA:
A → B defined on the same domain A
TS Inter 1st Year Maths 1A Functions Important Questions 4

ii) To prove that lBof = f
Since f : A → B and lB : B → B we have
lBof : A → B defined In the same domain A
TS Inter 1st Year Maths 1A Functions Important Questions 5

Question 5.
If f : A → B is a bijective function, then prove that (i) fof-1 = IB (ii) f-1of = IA
Solution:
To prove that fof-1 = IB
Given f: A → B is a bijection then we have
f-1 : B → A is also a bijection
TS Inter 1st Year Maths 1A Functions Important Questions 6

TS Inter 1st Year Maths 1A Functions Important Questions

Question 6.
lf f : A → B, g : B→C are two bijective functions then prove that (gof)-1 = f-1 og-1
Solution:
Given that f : A→ B and g: B → C are bijections
we have gof = A → C is a bijection.
∴ (gof)-1 : C → A is also a bijection.
Also since f : A→B and g : B → C are bijections then f-1: B → A and g-1: c → B are bijections and hence f-1og-1; c → A is also a bijection.
(gof)-1 and f-1og-1 are two functions defined or the same domain C.
let c E C and g : B → C is a bijection ∃ unique b E B. Such that g(b)=c ⇒ b=g-1 (c)
Also b ∈ B and f: A → B is a bijection, ∃ a unique a ∈ A such that f (a) b ⇒ a = f-1(b)
TS Inter 1st Year Maths 1A Functions Important Questions 7

Question 7.
If f : A → B and g : B→ A are two functions such that gof = IA and fog = IB then g = f-1
Solution:
i) To prove that f is one one.
TS Inter 1st Year Maths 1A Functions Important Questions 8
So these exists a reimage g(b) A for ‘b’ Under ‘f’
∴ f is onto.
Hence ‘f’ is one one, onto and hence a bijection.
∴ f : B → A exists and is also one one onto

TS Inter 1st Year Maths 1A Functions Important Questions

iii) To prove g = f-1
Now g : B → A and f-1: B → A
We have g and f-1 are del med in the same domain B.
Let a ∈ A and b be the f image of ‘a’ where b∈B.
TS Inter 1st Year Maths 1A Functions Important Questions 9

Question 8.
If f : A→ B, g : B→C and h : C→ Dare functions then ho (gof) = (hog) of
Solution :
Given f : A → B and g : B → C we have
gof : A→ C
Now gof: A → C and h: C → D we have
ho(gof) : A→ D, Also hog : B→D and f : A → B
We have (hog) of : A → D
Hence (hog) of and ho(gof) are defined in the same domain A.
Let a ∈ A then (ho(gof)] (a) = h [(gof) (a)]
= h [g [f(a)]
= (hog) [f(a)] = [(hog) of] (a)
∴ ho (gof) = (hog) of.

Question 9.
On what domain the functions f(x) = x2– 2x and g(x) = – x+6 are equal?
Solution:
f(x) = g(x)
x2– 2x = – x+6
= x2-x-6-0
= (x-3) (x+2) = 0 = x = -2,3
∴ f(x) and g(x) are equal on the domain { -2,3}

Question 10.
Find the inverse of the function f(x) = 5x
Solution:
Let y = 5x = f(x) then x = f-1(y)
Also x = log5y
∴ f1(y) = log5(y)
∴ f1(y) log5y ⇒ f-1(x) = log5x

Question 11.
If f : R – {o} → R is deflued by f(x) = x+\(\frac{1}{x}\)!,then prove that [f(x)]2 = f(x2) + f(1)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 11

Question 12.
If the function of defined by
TS Inter 1st Year Maths 1A Functions Important Questions 12
then find the values If exist of f(4); f(2.5), f(-2), f(-4), f(0), f(-7)
Solution :
i) Since f(x)=3x-2 for x>3
f(4) = 3(4) – 2 = 10
Domain of f is
(- ∞, – 3)∪[-2,2] u(3, ∞)

TS Inter 1st Year Maths 1A Functions Important Questions

ii) f(2.5) does not exist since 2.5 does not belong to the domain of f.

iii) f(x) = x2– 2 for x [-2,2]
We have f(-2) = (2) -2 = 2

iv) f(x) = 2x + I for x < – 3
f( – 4)=2(- 4)+ 1= – 7

v)f(x)=x2 – 2 for x∈[-2,2] and f(0) = – 2.

vi) f(x) = 2x + 1, for x < – 3
f(-7) = 2(-7)+ 1 = – 14+ 1 = – 13

Question 13.
Determine whether the function f: R → R defined by
TS Inter 1st Year Maths 1A Functions Important Questions 17
is an injection or a surjection or a bijection?
Solution :
By definition of the function f(3) = 3
and f(1 )= 5(1) – 2 =3
∴ 1 and 3 have same f image
Hence f is not an injection.
Let y ∈ R then y>2 or y≤2
if y>2 take x = y ∈ R so that 1(x) = x = y
TS Inter 1st Year Maths 1A Functions Important Questions 13
∴ f is a surjection.
∴ Since f is not an injection it is not a bijection.

Question 14.
Find the domain of definition of the function y(x), given by the equation 2x +2y = 2.
Solution :
TS Inter 1st Year Maths 1A Functions Important Questions 14

Question 15.
If f : R→ R defined as f(x+y)=f(x)+f(y)∀x, y ∈ R and f(1) = 7, then find \(\sum_{r=1}^n f(r)\)
Solution :
Consider
f(2)=f(1+1) = f(1)+f(1)=2f(1)
f(3) = 1(2 + 1) f(2) + f(1) = 2f(1) + f(1) = 3f(1)
Sìmilarly f(r) = r f(1)
TS Inter 1st Year Maths 1A Functions Important Questions 15

Question 16.
If \(f(x)=\frac{\cos ^2 x+\sin ^4 x}{\sin ^2 x+\cos ^4 x}, \forall x \in R\) then show that f(2012) = 1.
Solution :
TS Inter 1st Year Maths 1A Functions Important Questions 16

TS Inter 1st Year Maths 1A Functions Important Questions

Question 17.
If f : R→ R, g: R → R defined by f(x) = 4x -1 and g(x)= x2+2 then find
(i) (gof) (x)
(ii) (gof) \(\left(\frac{a+1}{4}\right)\)
(iii) (fof) (x)
(iv) go (fof) (0)
Solution :
Given f(x) = 4x – 1 and g(x) = x2+2
Where f: R → R and g: R → R then

¡) (gof)(x) = g[f(x)] = g[4x – 1]
=(4x – 1 )2+2= 16 x 2- 8x+3

TS Inter 1st Year Maths 1A Functions Important Questions 18
iii) (fof) (x) = f [f(x)] = f[4x -1]
= 4 (4x – 1) -1 = 16x – 5

iv) [go (fof)] (0) = go [f(f(0)]
= go [f (-1)] = g[f(-1)]
= g[-5] = 25 + 2 = 27

Question 18.
If f : [0, 3] – [0, 3] is defined by
TS Inter 1st Year Maths 1A Functions Important Questions 19
then show that f [0, 3] ⊆ [0, 3] and find fof
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 20

Question 19.
If f, g : R→ R are defined by
TS Inter 1st Year Maths 1A Functions Important Questions 21
then find (fog)π + (gof) (e)
Solution:
We have g(π) = 0, and f(e) = 1
∴ (fog) π = f [g(π)] = f(0) = 0
(gof)(e)=g[f(e)]2g(1) = – 1
∴ (fog)(π)+(gof)(e) = 0 – 1 = – 1

TS Inter 1st Year Maths 1A Functions Important Questions

Question 20.
Let A = {1, 2,3} ,B = {a,b,c}, C = {p,q,r}.
If f : A+B, g : B → C are defined by
f= ((1, a), (2, c), (3, b))
g = ((a, q), (b, r), (c, p)) then show that f-1og-1 = (gof)-1
Solution:
Given f : A → Band g: B → C we have
f-1 {(a, 1), (c, 2), (b, 3)}
and g-1 = {(q, a), (r, b), (p, c)}
f-1og-1 {(q, 1), (r, 3), (p, 2)}
gof = {(1, q), (2, p), (3, r)}
(gof)-1 = ((q, 1), (p, 2), (r, 3))
∴ (gof)-1= f-1og-1

Question 21.
If f : Q → Q defined by f(x) = 5x + 4 ∀ x∈Q show that f is a bijection and find f-1.
Solution:
Let x1, x2 ∈ Q then f(x1) = f(x2)
⇒ 5x1 + 4 = 5x2 + 4
⇒ 5x1 = 5x2 ⇒ x1 = x2
∴ f is an injection.
TS Inter 1st Year Maths 1A Functions Important Questions 22

Question 22.
Find the domains of the following real-valued functions.

(i) \(f(x)=\frac{1}{6 x-x^2-5}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 23

TS Inter 1st Year Maths 1A Functions Important Questions

(ii) \(f(x)=\frac{1}{\sqrt{x^2-a^2}},(a>0)\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 24

(iii) \(f(x)=\sqrt{(x+2)(x-3)}\)
Solution:
\(f(x)=\sqrt{(x+2)(x-3)} \in R\)
TS Inter 1st Year Maths 1A Functions Important Questions 25

(iv) \(\mathbf{f}(\mathbf{x})=\sqrt{(\mathbf{x}-\alpha)(\beta-\mathbf{x})}, \quad(0<\alpha<\beta)\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 26

(v) \(f(x)=\sqrt{2-x}+\sqrt{1+x}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 27

(vi) \(f(x)=\sqrt{x^2-1}+\frac{1}{\sqrt{x^2-3 x+2}}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 28

TS Inter 1st Year Maths 1A Functions Important Questions

(vii) \(f(\mathbf{x})=\frac{1}{\sqrt{|\mathbf{x}|-\mathbf{x}}}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 29

(viii) \(\mathbf{f}(\mathbf{x})=\sqrt{|\mathbf{x}|-\mathbf{x}}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 30

Question 23.
If f = {(4, 5), (5, 6), (6, – 4) and g = ((4, -4), (6, 5), (8, 5)} then find
i) f+g
ii) f – g
iii) 2f + 4g
iv) f+4
v) fg
vi) \(\frac{f}{g}\)
vii) \(|\mathbf{f}|\)
viii) \(\sqrt{f}\)
ix) f2
x) f3
Solution:
Given f {(4, 5), (5, 6), (6, – 4)] and g = ((4,-4), (6, 5), (8, 5)) then domain of f = {4, 5, 6) and Range of f = {4, 6, 8)
Domain of f ± g = A B = (4, 6)
= (domain of f) ∩ (domain of g)

i) f+g={(4,5,-4)(6,-4+5))
= {(4, 1), (6, 1)}

ii) f-g= {(4,5+4),(6,-4-5)}
= {(4,9), (6,-9)}

iii) Domain of 2f = {4, 5, 6}
Domain of 4g (4, 6, 8)
Domain of 2f + 4g = (4, 6)
∴2f = {(4, 10), (5, 12), (6, -8)}
4g = {(4, – 16), (6, 20), (8, 20)}
∴2f + 4g = {(4,-6),(6,12)}

TS Inter 1st Year Maths 1A Functions Important Questions

iv) Domain of f + 4 = {4,5, 6}
f + 4 = {(4, 9), (5, 10), (6, 0)}

v) Domain of fg = (domain of f) n (domain of g)
A∩B = {4, 6}
= {(4, (5) (-4), (6, (- 4), (5)}
= {(4, – 20), (6, – 20)}

TS Inter 1st Year Maths 1A Functions Important Questions 31

ix) Domain of f2 = (Domain of f(x)] (4, 5, 6)
∴ f2 = ((4, 25), (5, 36), (6, 16))

x) Domain of f3 = (4, 5, 6)
∴ f3 = ((4, 125), (5, 216), (6, -64))

TS Inter 1st Year Maths 1A Functions Important Questions

Question 24.
Find the domains and ranges of the following real valued functions.
(i) \(f(x)=\frac{2+x}{2-x}\)
(ii) \(f(x)=\frac{x}{1+x^2}\)
(iii)\(f(x)=\sqrt{9-x^2}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 35
TS Inter 1st Year Maths 1A Functions Important Questions 33

(ii) \(f(x)=\frac{x}{1+x^2}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 34

TS Inter 1st Year Maths 1A Functions Important Questions

(iii) \(f(x)=\sqrt{9-x^2}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 35
TS Inter 1st Year Maths 1A Functions Important Questions 36
But f(x) posses only non negative values Range of f = [0, 3]

Question 25.
If f(x) = x2 and g(x) = I x find the following functions.
i) f+g
ii) f- g
iii) fg
iv) 2f
v) f2
vi) f+3
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 37

Question 26.
Determine whether the following functions are even or odd

(i) f(x) = ax a -x + sin x
Solution :
A function f is said to be even if [(-x) = f(x) and odd If f(-x) = – f(x)
f(x) = ax a -x – sinx
= (ax a -x + sin x) = – f(x)
∴ f is an odd function.

TS Inter 1st Year Maths 1A Functions Important Questions

(ii) \(f(x)=x\left(\frac{e^x-1}{e^x+1}\right)\)
Solution :
Given \(f(x)=x\left(\frac{e^x-1}{e^x+1}\right)\)
TS Inter 1st Year Maths 1A Functions Important Questions 38

(iii) f(x) = log \(\left(x+\sqrt{x^2+1}\right)\)
Solution :
TS Inter 1st Year Maths 1A Functions Important Questions 39

Question 27.
Find the domains of the following real-valued functions.

(i) \(f(\mathbf{x})=\frac{1}{\sqrt{[\mathbf{x}]^2-[\mathbf{x}]-2}}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 40
TS Inter 1st Year Maths 1A Functions Important Questions 41

(ii) f(x) = log (x – [x])
Solution:
f(x) ∈ R
⇔ x – [x] >0 ⇔ x>[x]
⇔ x is not an integer.
∴ Domain of f is R – Z

TS Inter 1st Year Maths 1A Functions Important Questions

(iii) \(f(x)=\sqrt{\log _{10}\left(\frac{3-x}{x}\right)}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 42

(iv) \(f(x)=\sqrt{x+2}+\frac{1}{\log _{10}(1-x)}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 43

(v) \(f(x)=\frac{\sqrt{3+x}+\sqrt{3-x}}{x}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 44

TS Inter 2nd Year Economics Notes Chapter 1 Economic Growth and Economic Development

Here students can locate TS Inter 2nd Year Economics Notes Chapter 1 Economic Growth and Economic Development to prepare for their exam.

TS Inter 2nd Year Economics Notes Chapter 1 Economic Growth and Economic Development

→ The increase in the real output of goods and services is called economic growth. It is progressive changes in the socio-economic structure of a country.

→ Economic development is a process where an economy’s real national income increases over a long period of time.

→ Objectives of Economic development – High rate of growth, Economic self-reliance, Social justice, Modernization, Economic stability, Inclusive growth.

→ Indicators of Economic Development, National income, per Percapita income, Real National income, PQLI, HDI, GDI, GEM, SPI, and MPI.

TS Inter 2nd Year Economics Notes Chapter 1 Economic Growth and Economic Development

→ Factors Hindering Economic development – Lack of national resources, lower rate of growth of human capital, poor infrastructure, vicious circle of poverty.

→ Factors promoting economic development – Economic factors, non – Economic factors.

→ Characteristics of developed economies – the significance of the industrial sector, High rate of capital formation, use of High production Techniques, low growth of population, per capita gross National income.

→ Characteristics of developing economies – Low level of income, the predominance of agriculture, capital deficiency, technological backwardness, inadequate infrastructure, high population, high rate of literacy, High infant mortality rate, traditions, joint family system.

TS Inter 2nd Year Economics Notes Chapter 1 ఆర్థిక వృద్ధి, ఆర్థికాభివృద్ధి

→ దీర్ఘకాలంలో వస్తు సేవల వాస్తవిక ఉత్పత్తి పెరుగుదలను తెలియజేయడాన్ని ఆర్థికవృద్ధి అంటారు.

→ ఆర్థికాభివృద్ధి అనేది ఆర్థిక వృద్ధి కంటే చాలా విస్తృతమైన భావన. ఇది ఒక దేశంలోని ఆర్థిక వృద్ధితో పాటు, సాంఘిక, ఆర్థిక, వ్యవస్థాపూర్వక మార్పులను సూచించును.

→ ఆర్థికాభివృద్ధి లక్ష్యాలు:

  1. అధిక వృద్ధి రేటు,
  2. ఆర్థిక స్వావలంబన,
  3. సామాజిక న్యాయం,
  4. ఆధునికీకరణ,
  5. ఆర్థిక స్థిరత్వం,
  6. సమ్మిళిత వృద్ధి.

→ ఆర్థికాభివృద్ధి సూచికలు:

  1. జాతీయాదాయం
  2. తలసరి ఆదాయం
  3. నిజ జాతీయాదాయం
  4. భౌతిక జీవన ప్రమాణ సూచిక
  5. మానవ అభివృద్ధి సూచిక
  6. లింగ సంబంధ అభివృద్ధి సూచిక
  7. బహుపార్శ్వ పేదరిక సూచిక.

TS Inter 2nd Year Economics Notes Chapter 1 Economic Growth and Economic Development

→ ఆర్థికాభివృద్ధి నిరోధకాలు:

  1. సహజ వనరుల కొరత,
  2. అల్పమానవ మూలధన వృద్ధి రేటు,
  3. అవస్థాపనా సదుపాయాల కొరత,
  4. పేదరిక విషవలయాలు.

→ ఆర్థికాభివృద్ధిని ప్రోత్సహించే కారకాలు: ఆర్థిక కారకాలు, ఆర్థికేతర కారకాలు.
TS Inter 2nd Year Economics Notes Chapter 1 Economic Growth and Economic Development 1

→ అభివృద్ధి చెందిన ఆర్థిక వ్యవస్థ లక్షణాలు: పారిశ్రామిక రంగ ప్రాధాన్యత, అధిక స్థాయిలో మూలధన కల్పన, ఆధునిక ఉత్పత్తి, సాంకేతికత, నైపుణ్యం, తక్కువ జనాభా వృద్ధి, తలసరి స్థూల జాతీయాదాయం.

TS Inter 2nd Year Economics Notes Chapter 1 Economic Growth and Economic Development

→ అభివృద్ధి చెందుతున్న దేశాల లక్షణాలు: అల్పఆదాయం, వ్యవసాయ రంగ ప్రాధాన్యత, మూలధన లోటు, సాంకేతికంగా వెనుకబడి ఉండటం, తక్కువ అవస్థాపనా సదుపాయాలు, అధిక స్థాయిలో జనాభా వృద్ధి, నిరక్షరాస్యత, శిశు మరణాల రేటు అధికం, సంప్రదాయ హద్దులు, దృక్పథాలు, ఉమ్మడి కుటుంబ వ్యవస్థ.

TS 10th Class Physics Guide Pdf | TS 10th Class Physical Science Study Material Pdf Telangana

TS Telangana SCERT Class 10 Physics Solutions | TS 10th Class Physics Study Material Telangana

TS 10th Class Physical Science Study Material Pdf Telangana | TS 10th Physics Study Material Telangana

TS 10th Class Study Material

Maths 1B Important Questions Chapter Wise with Solutions Pdf 2022 TS | TS Inter 1st Year Maths 1B Important Questions

TS Inter 1st Year Maths 1B Important Questions with Solutions Pdf 2022 | Maths 1B Important Questions 2022 TS

TS Inter First Year Maths 1B Important Questions | Maths 1B Important Questions Pdf 2022 TS

  1. Maths 1B Locus Important Questions
  2. Maths 1B Transformation of Axes Important Questions
  3. Maths 1B Straight Lines Important Questions
  4. Maths 1B Pair of Straight Lines Important Questions
  5. Maths 1B Three-Dimensional Coordinates Important Questions
  6. Maths 1B Direction Cosines and Direction Ratios Important Questions
  7. Maths 1B The Plane Important Questions
  8. Maths 1B Limits and Continuity Important Questions
  9. Maths 1B Differentiation Important Questions
  10. Maths 1B Applications of Derivatives Important Questions

TS Inter 1st Year Maths 1B Blue Print Weightage

TS 9th Class Telugu Important Questions 2022-2023

TS 9th Class Telugu Important Questions 2022-2023

TS 9th Class Study Material

Maths 1A Important Questions Chapter Wise with Solutions Pdf 2022 TS | TS Inter 1st Year Maths 1A Important Questions

TS Inter 1st Year Maths 1A Important Questions with Solutions Pdf 2022 | Maths 1A Important Questions 2022 TS

TS Inter First Year Maths 1A Important Questions | Maths 1A Important Questions Pdf 2022 TS

  1. Maths 1A Functions Important Questions
  2. Maths 1A Mathematical Induction Important Questions
  3. Maths 1A Matrices Important Questions
  4. Maths 1A Addition of Vectors Important Questions
  5. Maths 1A Products of Vectors Important Questions
  6. Maths 1A Trigonometric Ratios up to Transformations Important Questions
  7. Maths 1A Trigonometric Equations Important Questions
  8. Maths 1A Inverse Trigonometric Functions Important Questions
  9. Maths 1A Hyperbolic Functions Important Questions
  10. Maths 1A Properties of Triangles Important Questions

TS Inter 1st Year Maths 1B Blue Print Weightage

TS 10th Class Physical Science Chapter Wise Important Questions 2022-2023

TS 10th Class Physics Chapter Wise Important Questions 2023 | TS SSC Physical Science Important Questions

TS 10th Class Physical Science Important Questions Pdf | TS SSC Physics Important Questions

TS 10th Class Study Material

TS 10th Class Telugu Guide Pdf Download Telangana 2022-2023

TS 10th Class Telugu Guide Telangana | 10th Class Telugu Study Material Telangana Pdf

10th Class Telugu Guide Pdf Download Telangana | 10th Class Telugu Study Material Telangana

TS 10th Class Telugu Grammar Telangana

TS 10th Class Study Material

TS 10th Class Maths Bits with Answers Pdf in English & Telugu Medium | TS Maths Bits for 10th Class

TS 10th Class Maths Important Bits | TS 10th Class Maths Bit Bank Pdf

  1. Real Numbers Bits for 10th Class
  2. Sets Bits for 10th Class
  3. Polynomials Bits for 10th Class
  4. Pair of Linear Equations in Two Variables Bits for 10th Class
  5. Quadratic Equations Bits for 10th Class
  6. Progressions Bits for 10th Class
  7. Coordinate Geometry Bits for 10th Class
  8. Similar Triangles Bits for 10th Class
  9. Tangents and Secants to a Circle Bits for 10th Class
  10. Mensuration Bits for 10th Class
  11. Trigonometry Bits for 10th Class
  12. Applications of Trigonometry Bits for 10th Class
  13. Probability Bits for 10th Class
  14. Statistics Bits for 10th Class

TS 10th Class Study Material