Q: What is the derivative of x |x| with respect to x, where x<0 ?

Given function is f(x) = x |x| . For x Substituting this into the function gives f(x) = x -x = -1. Differentiating with respect to x gives d dx (-1) = 0. Answer: 0

Q: Passage: Let f(x)= cases ax(x-1), & x 3 cases Given that f(x) is continuous for all x but not differentiable at x=1. Further f'(x) is continuous at x=3. Question: What is the value of q ?

For f(x) to be continuous at x=3, we must have _ x 3^- f(x) = _ x 3^+ f(x). 3 - 1 = p(3)^2 + q(3) + 2 2 = 9p + 3q + 2 9p + 3q = 0 3p + q = 0 For f'(x) to be continuous at x=3, the left-hand derivative must equal the right-hand derivative at x=3. f'(x) = 1 for 1 f'(x) = 2px + q for x > 3 Equating the derivatives at x=3: 1 = 2p(3) + q 6p + q = 1 Solving the two equations: 3p + q = 0 6p + q = 1 Subtracting the first equation from the second gives 3p = 1 p = 1 3 . Substituting p into the first equat

Question 1

What is the derivative of x |x| with respect to x, where x<0 ?

Accepted Answer

Given function is f(x) = x |x| . For x Substituting this into the function gives f(x) = x -x = -1. Differentiating with respect to x gives d dx (-1) = 0. Answer: 0

Question 2

Passage: Let f(x)= cases ax(x-1), & x 3 cases Given that f(x) is continuous for all x but not differentiable at x=1. Further f'(x) is continuous at x=3. Question: What is the value of q ?

Accepted Answer

For f(x) to be continuous at x=3, we must have _ x 3^- f(x) = _ x 3^+ f(x). 3 - 1 = p(3)^2 + q(3) + 2 2 = 9p + 3q + 2 9p + 3q = 0 3p + q = 0 For f'(x) to be continuous at x=3, the left-hand derivative must equal the right-hand derivative at x=3. f'(x) = 1 for 1 f'(x) = 2px + q for x > 3 Equating the derivatives at x=3: 1 = 2p(3) + q 6p + q = 1 Solving the two equations: 3p + q = 0 6p + q = 1 Subtracting the first equation from the second gives 3p = 1 p = 1 3 . Substituting p into the first equat

Question 3

Passage: Let f(x)= (x^2) and g(x)=x|x| for |x| Question: If q(x)=f g(x), then which of the following statements is/are correct ? I. q(x) is continuous at x=0. II. q(x) is differentiable at x=0. Select

Accepted Answer

Given f(x) = (x^2) and g(x) = x|x|. The composite function q(x) is given by: q(x) = f(g(x)) = f(x|x|) = ((x|x|)^2) Since (x|x|)^2 = x^2|x|^2 = x^4, we have: q(x) = (x^4) Checking continuity at x=0: _ x 0 q(x) = _ x 0 (x^4) = 0 Also, q(0) = (0) = 0. Since _ x 0 q(x) = q(0), q(x) is continuous at x=0. Statement I is correct. Checking differentiability at x=0: q'(0) = _ x 0 q(x) - q(0) x - 0 = _ x 0 (x^4) x q'(0) = _ x 0 ( (x^4) x^4 x^3 ) = 1 0 = 0 Since the limit exists and is finite, q(x) is diff

Question 4

Passage: Let f(x)= (x^2) and g(x)=x|x| for |x| Question: If p(x)=f(x)g(x), then which of the following statements is/are correct ? I. p(x) is continuous at x=0. II. p(x) is differentiable at x=0. Sele

Accepted Answer

Given f(x) = (x^2) and g(x) = x|x|. The function p(x) is given by p(x) = f(x)g(x) = x|x| (x^2). We can redefine p(x) as: p(x) = x^2 (x^2) for x 0 p(x) = -x^2 (x^2) for x Checking continuity at x = 0: _ x 0^+ p(x) = _ x 0^+ x^2 (x^2) = 0 _ x 0^- p(x) = _ x 0^- -x^2 (x^2) = 0 p(0) = 0 Since _ x 0^+ p(x) = _ x 0^- p(x) = p(0), p(x) is continuous at x = 0. Statement I is correct. Checking differentiability at x = 0: Right Hand Derivative (RHD) at x = 0: _ h 0^+ p(h) - p(0) h = _ h 0^+ h^2 (h^2) - 0

Question 5

If f(x) is differentiable at x=a, then consider the following statements : I. f(x) is continuous at x=a II. _ x a f(x)=f(a) Which of the statements given above is/are correct ?

Accepted Answer

If a function f(x) is differentiable at a point x=a, then it is necessarily continuous at that point. This makes Statement I correct. By the definition of continuity at a point x=a, the limit of the function as x approaches a must be equal to the value of the function at a. Thus, _ x a f(x)=f(a). This makes Statement II correct. Therefore, both statements I and II are correct. Answer: Both I and II

Question 6

Passage: Let f(x)= cases ax(x-1), & x 3 cases Given that f(x) is continuous for all x but not differentiable at x=1. Further f'(x) is continuous at x=3. Question: What is the value of p ?

Accepted Answer

For f(x) to be continuous at x=3, the left-hand limit and right-hand limit at x=3 must be equal. _ x 3^- f(x) = 3 - 1 = 2 _ x 3^+ f(x) = p(3)^2 + q(3) + 2 = 9p + 3q + 2 Equating them, we get: 9p + 3q + 2 = 2 9p + 3q = 0 q = -3p The derivative of the function for x (1, 3) is f'(x) = 1, and for x > 3 is f'(x) = 2px + q. Since f'(x) is continuous at x=3, the left-hand derivative and right-hand derivative at x=3 must be equal. _ x 3^- f'(x) = 1 _ x 3^+ f'(x) = 2p(3) + q = 6p + q Equating them, we ge

Question 7

Consider the following for the two (02) items that follow: Let f(x) = cases x^3, & x^2 What is _ x 0 f'(x) equal to?

Accepted Answer

Given f(x) = cases x^3, & x^2 For x 0, the relevant domain is x^2 In this interval, the function is f(x) = x^3. Differentiating with respect to x, we get: f'(x) = 3x^2 Taking the limit as x 0: _ x 0 f'(x) = _ x 0 3x^2 = 0 Answer: 0

Question 8

For the following two (02) items: Let the function f(x) = |x - 3| + |x - 4| be defined on the interval [0, 5]. What is dy dx at x = 3.5 equal to?

Accepted Answer

Given f(x) = |x - 3| + |x - 4| on the interval [0, 5]. At x = 3.5, we have 3 In the interval (3, 4), x - 3 > 0 and x - 4 Therefore, f(x) = (x - 3) - (x - 4) = 1. Since f(x) is a constant in the neighborhood of x = 3.5, its derivative is zero. dy dx = 0 at x = 3.5.

Question 9

For the following two (02) items: Consider the function f(x) = cases 4(5^x), & x If the function is continuous, then what is the value of k?

Accepted Answer

For the function to be continuous at x = 0, the left-hand limit must be equal to the right-hand limit and the value of the function at x = 0. Left-hand limit at x = 0 is given by: _ x 0^ - f(x) = _ x 0^ - 4(5^x) = 4(5^0) = 4 Right-hand limit at x = 0 is given by: _ x 0^ + f(x) = _ x 0^ + (8k + x) = 8k + 0 = 8k Equating the left-hand limit and the right-hand limit: 8k = 4 k = 4 8 = 0.5 Answer: 0.5

Question 10

For the following two (02) items: Let the function f(x) = |x - 3| + |x - 4| be defined on the interval [0, 5]. Consider the following statements: I. The function is differentiable at x = 3. II. The fu

Accepted Answer

The given function is f(x) = |x - 3| + |x - 4| defined on [0, 5]. We can redefine the function by removing the absolute value signs in different intervals: f(x) = cases -(x - 3) - (x - 4), & 0 x Simplifying this, we get: f(x) = cases 7 - 2x, & 0 x Differentiating f(x) with respect to x in the open intervals: f'(x) = cases -2, & 0 At x = 3: Left Hand Derivative (LHD) = -2 Right Hand Derivative (RHD) = 0 Since LHD RHD, f(x) is not differentiable at x = 3. At x = 4: Left Hand Derivative (LHD) = 0 R

Question 11

Consider the following for the two (02) items that follow: Let f(x) = cases x^3, & x^2 Consider the following statements: I. The function is continuous at x = -1. II. The function is differentiable at

Accepted Answer

The given function can be rewritten as: f(x) = cases x^2, & x -1 \ x^3, & -1 Checking continuity at x = -1: Left Hand Limit (LHL) = _ x -1^- x^2 = (-1)^2 = 1 Right Hand Limit (RHL) = _ x -1^+ x^3 = (-1)^3 = -1 Since LHL RHL, f(x) is not continuous at x = -1. Statement I is incorrect. Checking differentiability at x = 1: First, check continuity at x = 1: LHL = _ x 1^- x^3 = 1 RHL = _ x 1^+ x^2 = 1 f(1) = 1^2 = 1 The function is continuous at x = 1. Now, check the derivatives at x = 1: Left Hand

Continuity and Differentiability — Previous Year Questions & Solutions