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Class 12-science RD SHARMA Solutions Maths Chapter 16 - Tangents and Normals

Tangents and Normals Exercise MCQ

Solution 30

Let (p, q) be the point on the given curve   at which the normal passes through the origin.

Therefore,

  

 

Now, the equation of normal at (p, q) is passing through origin.

Therefore,

  

As p = 1 satisfies the equation, therefore the abscissa is 1.

Solution 31

The curves are

  … (i)

  … (ii)

Differentiating (i) w.r.t x, we get

  

Differentiating (ii) w.r.t x, we get

  

Now,

So, both the curves are cut each other at right angle.

Solution 32

Given:

Differentiating 'x' and 'y' w.r.t t, we get

  

Dividing (ii) by (i), we get

  

Hence, the tangent to the curve makes an angle of   with x-axis.

Solution 33

Given curve is

  

Therefore, slope of tangent is 2.

The equation of tangent is y - 1 = 2(x - 0)

i.e. y = 2x + 1

This equation of tangent meets x-axis when y = 0

  

Thus, the required point is

Solution 34

Given curve is

  

The curve crosses x-axis when y = 0

Therefore, x = 2

So, the tangent touches the curve at point (2, 0).

  

The equation of tangent at (2, 0) is

  

Solution 35

Given curve is

  

As the tangents are parallel to x-axis, their slope will be 0.

  

When x = 2, y = 23 - 12 × 2 + 18 = 2

When x = -2, y = (-2)3 - 12(-2) + 18 = 34

So, the points are (2, 2) and (-2, 34).

Solution 36

Given curve is

  

At (0, 0), we have

  

Thus, the given curve   has vertical tangent, which is parallel to y-axis, at (0, 0).

Solution 1

Correct option: (c)

  

Solution 2

Correct option: (d)

  

Solution 3

Correct option: (a)

  

Solution 4

Correct option: (b)

  

Solution 5

Correct option: (c)

  

Solution 6

Correct option: (b)

  

Solution 7

Correct option:(b)

  

Solution 8

Correct option: (d)

  

Solution 9

Correct option: (b)

  

Solution 10

Correct option: (c)

  

Solution 11

Correct option: (b)

  

Solution 12

Correct option: (b)

  

Solution 13

Correct option: (d)

  

Solution 14

Correct option: (c)

  

Solution 15

Correct option: (c)

  

Solution 16

Correct option: (b)

  

Solution 17

Correct option: (b)

  

Solution 18

Correct option: (c)

  

Solution 19

Correct option: (c)

  

Solution 20

Correct option: (a)

  

Solution 21

Correct option: (b)

  

Solution 22

Correct option: (c)

  

Solution 23

Correct option: (c)

  

Solution 24

Correct option: (c)

  

Solution 25

Correct option: (a)

  

 

Solution 26

Correct option: (b)

  

Solution 27

Correct option: (a)

  

Solution 28

Correct option: (b)

  

Solution 29

 

NOTE: Options are incorrect.

Tangents and Normals Exercise Ex. 16.1

Solution 1(i)



Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1(vi)

Solution 1(vii)

Solution 1(viii)

Solution 1(ix)

Solution 1(x)

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18


Solution 19

Solution 20

Solution 21

Tangents and Normals Exercise Ex. 16.2

Solution 1

Solution 2

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 3(v)

Solution 3(vi)

Solution 3(vii)

Solution 3(viii)

Solution 3(ix)

Solution 3(x)

Solution 3(xi)


Solution 3(xii)

Solution 3(xiii)

Solution 3(xiv)

Solution 3(xv)

Solution 3(xvi)

The equation of the given curve is y2 = 4x . Differentiating with respect to x, we have: 

begin mathsize 12px style 2 straight y dy over dx equals 4
rightwards double arrow dy over dx equals fraction numerator 4 over denominator 2 straight y end fraction equals 2 over straight y
therefore right enclose dy over dx end enclose subscript open parentheses 1 comma space 2 close parentheses end subscript equals 2 over 2 equals 1
Now comma space the space slope space at space point space left parenthesis 1 comma space 2 right parenthesis space is space 1 over right enclose begin display style dy over dx end style end enclose subscript open parentheses 1 comma space 2 close parentheses end subscript equals fraction numerator negative 1 over denominator 1 end fraction equals negative 1.
therefore Equation space of space the space tangent space at space left parenthesis 1 comma space 2 right parenthesis space is space straight y minus 2 space equals space minus 1 left parenthesis straight x minus 1 right parenthesis.
rightwards double arrow straight y minus 2 equals negative straight x plus 1
rightwards double arrow straight x plus straight y minus 3 equals 0
Equation space of space the space normal space is comma
straight y minus 2 equals negative left parenthesis negative 1 right parenthesis left parenthesis straight x minus 1 right parenthesis
straight y minus 2 equals straight x minus 1
straight x minus straight y plus 1 equals 0
end style

Solution 3(xix)

Solution 4


Solution 5(i)

Solution 5(ii)

From (A)

Equation of tangent is

begin mathsize 12px style open parentheses straight y minus straight a over 5 close parentheses equals 13 over 16 open parentheses straight x minus fraction numerator 2 straight a over denominator 5 end fraction close parentheses
16 straight y minus fraction numerator 16 straight a over denominator 5 end fraction equals 13 straight x minus fraction numerator 26 straight a over denominator 5 end fraction
13 straight x minus 16 straight y minus 2 straight a equals 0
Equation space of space normal space is comma
open parentheses straight y minus straight a over 5 close parentheses equals 16 over 13 open parentheses straight x minus fraction numerator 2 straight a over denominator 5 end fraction close parentheses
13 straight y minus fraction numerator 13 straight a over denominator 5 end fraction equals negative 16 straight x plus fraction numerator 32 straight a over denominator 5 end fraction
16 straight x plus 13 straight y minus 9 straight a equals 0 end style

Solution 5(iii)

Solution 5(iv)

Solution 5(v)

Solution 5(vi)

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 21

   

Solution 3(xvii)

Given equation curve is

Differentiating w.r.t x, we get

  

Slope of tangent at   is

  

Slope of normal will be

  

Equation of tangent at   will be

  

Equation of normal at   is

  

Solution 3(xviii)

Given equation curve is

Differentiating w.r.t x, we get

  

Slope of tangent at   is

  

Slope of normal will be

  

Equation of tangent at   will be

  

Equation of normal at   is

  

Solution 20

Given equation curve is

Differentiating w.r.t x, we get

  

As tangent is parallel to x-axis, its slope will be m = 0

  

As this point lies on the curve, we can find y

  

Or

  

So, the points are (3, 6) and (2, 7).

Equation of tangent at (3, 6) is

y - 6 = 0 (x - 3)

y - 6 = 0

Equation of tangent at (2, 7) is

y - 7 = 0 (x - 2)

y - 7 = 0

Tangents and Normals Exercise Ex. 16.3

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)


Solution 1(v)


Solution 1(vi)

Solution 1(vii)

Solution 1(viii)

Solution 1 (ix)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 4

Solution 5

Solution 6

 

Solution 7

Solution 8(i)

Solution 8(ii)

Solution 9

Solution 10

Tangents and Normals Exercise Ex. 16VSAQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Given curve is y = sin x

  

Slope of tangent at (0, 0) is

  

So, the equation of tangent at (0, 0) is

y - 0 = 1 (x - 0)

y = x

Solution 19

Given curve is

  

Slope of tangent at (0, 0) is

  

Hence, slope of tangent at   is 0.

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