NCERT Exemplar Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

The NCERT Exemplar for Class 12 questions is important for students, as it helps them to get acquainted with the different question variations and hence, develop problem-solving abilities. Students can make use of theNCERT Exemplar Solutions available subject-wise to get assistance in solving the exercise problems in every chapter. All the solutions are developed by following the latest CBSE guidelines, so that students can score more marks.

Continuity and Differentiability is the 5th chapter of NCERT Exemplar for Class 12. This is an important chapter as it lays a foundation for Differential Calculus. The topics covered in this chapter are continuity, differentiability, algebra of continuous functions, derivatives of composite functions, implicit functions and inverse trigonometric functions, exponential and logarithmic functions, logarithmic differentiation, derivatives of functions in parametric forms, second-order derivative and mean value theorem. The solutions to the exercise problems for this chapter are available in PDF format asNCERT Exemplar Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability and can be accessed by the students from the link given below.

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Exercise 5.3 Page No: 107

1. Examine the continuity of the function f (x) = x3 + 2x2 – 1 at x = 1

Solution:

We know that, y = f(x) will be continuous at x = a if,

Thus, f(x) is continuous at x = 1.

Find which of the functions in Exercises 2 to 10 is continuous or discontinuous at the indicated points:

2.
at x = 2

Solution:

Checking the continuity of the given function, we have

Thus, f(x) is discontinuous at x = 2.

3.
at x = 0

Solution:

Checking the right hand and left hand limits of the given function, we have

Therefore, the given function f(x) is discontinuous at x = 0.

4.
at x = 2

Solution:

Thus, f(x) is continuous at x = 2.

5.
at x = 4

Solution:

Checking the right hand and left hand limits for the given function, we have

Thus, f(x) is discontinuous at x = 4.

6.
at x = 0

Solution:

Checking the right hand and left hand limits for the given function, we have

Thus, the given function f(x) is continuous at x = 0.

7.
at x = a

Solution:

Checking the right hand and left hand limits for the given function, we have

Thus, the given function f(x) is continuous at x = 0.

8.
at x = 0

Solution:

Checking the right hand and left hand limits for the given function, we have

Thus, f(x) is discontinuous at x = 0.

9.
at x = 1

Solution:

Checking the right hand and left hand limits for the given function, we have

Thus, the given function f(x) is continuous at x = 1.

10. f (x) = |x| + |x – 1| at x = 1

Solution:

Checking the right hand and left hand limits for the given function, we have

Thus, f(x) is continuous at x = 1.

Find the value of k in each of the Exercises 11 to 14 so that the function f is continuous at the indicated point:

11.

Solution:

Finding the left hand and right hand limits for the given function, we have

So,

7 = 2k

k = 7/2 = 3.5

Therefore, the value of k is 3.5

12.

Solution:

The given function f(x) can be rewritten as,

So, k = ½

Therefore, the value of k is ½

13.

Solution:

Finding the left hand and right hand limits for the given function, we have

Therefore, the value of k is -1

14.

Solution:

Finding the left hand and right hand limits for the given function, we have

Therefore, the value of k is ± 1

15. Prove that the function f defined by

remains discontinuous at x = 0, regardless the choice of k.

Solution:

Finding the left hand and right hand limit for the given function, we have

Now, as the left hand limit and the right hand limit are not equal and the value of both the limits are a constant.

Hence, regardless the choice of k, the given function remains discontinuous at x = 0.

16. Find the values of a and b such that the function f defined by

is a continuous function at x = 4.

Solution:

Finding the left hand and right hand limit for the given function, we have

So, -1 + a = a + b = 1 + b

-1 + a = a + b and 1 + b = a + b

We get, b = -1 and 1 + -1 = a + -1 ⇒ a = 1

Therefore, the value of a = 1 and b = -1

17. Given the function f (x) = 1/(x + 2) . Find the points of discontinuity of the composite function

y = f (f (x)).

Solution:

Given,

Now, the function will not be defined and continuous where

2x + 5 = 0 ⇒ x = -5/2

Therefore, x = -5/2 is the point of discontinuity.

18. Find all points of discontinuity of the function

Solution:

Now,

if f(t) is discontinuous, then 2 – x = 0 ⇒ x = 2

And, 2x – 1 = 0 ⇒ x = ½

Therefore, the required points of discontinuity for the given function are 2 and ½.

19. Show that the function f (x) = |sin x + cos x| is continuous at x = p. Examine the differentiability of f, where f is defined by

Solution:

Given,

f(x) = |sin x + cos x| at x = π

Now, put g(x) = sin x + cos x and h(x) = |x|

Hence, h[g(x)] = h(sin x + cos x) = |sin x + cos x|

Now,

g(x) = sin x + cos x is a continuous function since sin x and cos x are two continuous functions at x = π.

We know that, every modulus function is a common function is a continuous function everywhere.

Therefore, f(x) = |sin x + cos x| is continuous function at x = π.

20.
at x = 2.

Solution:

We know that, a function f is differentiable at a point ‘a’ in its domain if

Lf’(c) = Rf’(c)

Therefore, f(x) is not differentiable at x = 2.

21.

Solution:

Given,

Therefore, f(x) is differentiable at x = 0

22.

Solution:

We know that, f(x) is differentiable at x = 2 if

Thus, f(x) is not differentiable at x = 2.

23. Show that f (x) = |x – 5| is continuous but not differentiable at x = 5.

Solution:

Therefore, f(x) is not differentiable at x = 5.

24. A function f : R ® R satisfies the equation f ( x + y) = f (x) f (y) for all x, y ÎR, f (x) ¹ 0. Suppose that the function is differentiable at x = 0 and f ¢ (0) = 2. Prove that f ¢(x) = 2 f (x).

Solution:

Given,

f : R ® R satisfies the equation f ( x + y) = f (x) f (y) for all x, y ÎR, f (x) ¹ 0

Let us take any point x = 0 at which the function f(x) is differentiable.

Therefore, f’(x) = 2f(x).

Differentiate each of the following w.r.t. x (Exercises 25 to 43) :

25.

Solution:

26.

Solution:

27.

Solution:

28.

Solution:

29.

Solution:

30. sinn (ax2 + bx + c)

Solution:

31.

Solution:

32. sin x2 + sin2 x + sin2 (x2)

Solution:

33.

Solution:

34. (sin x)cos x

Solution:

35. sinm x . cosn x

Solution:

36. (x + 1)2 + (x + 2)3 + (x + 3)4

Solution:

37.

Solution:

38.

Solution:

39.

Solution:

40.

Solution:

41.

Solution:

42.

Solution:

43.

Solution:

Find dy/dx of each of the functions expressed in parametric form in Exercises from 44 to 48.

44.

Solution:

Given,

x = t + 1/t, y = t – 1/t

Differentiating both the parametric functions w.r.t θ

45.

Solution:

Given,

46. x = 3cosq – 2cos3q, y = 3sinq – 2sin3q.

Solution:

Given, x = 3cosq – 2cos3q, y = 3sinq – 2sin3q.

Differentiating both the parametric functions w.r.t. q

47.

Solution:

Given,

sin x = 2t/(1 + t2), tan y = 2t/ (1 – t2)

48.

Solution:

On differentiating both the given parametric functions w.r.t. t, we have

49. If x = ecos2t and y = esin2t, prove that dy/ dx = y log x/ x log y.

Solution:

Given,

x = ecos2t and y = esin2t

So, cos 2t = log x and sin 2t = log y

Now, differentiating both the parameter functions w.r.t t, we have

50. If x = asin2t (1 + cos2t) and y = b cos2t (1–cos2t), show that

Solution:

Given,

x = asin2t (1 + cos2t) and y = b cos2t (1–cos2t)

Differentiating both the parametric equations w.r.t t, we have

51. If x = 3sint – sin 3t, y = 3cost – cos 3t, find

Solution:

Given,

x = 3sint – sin 3t, y = 3cost – cos 3t

Now, differentiating both the parametric functions w.r.t t, we have

52. Differentiate x/sinx w.r.t sin x.

Solution:

53. Different w.r.t tan-1 x when x ≠ 0.

Solution: