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Class 12 Maths Chapter 5 Continuity and Differentiability MCQs

Class 12 Maths Chapter 5 Continuity and Differentiability MCQs are provided online here to help students improve their test scores. Answers and complete explanations are provided for these multiple-choice questions. The continuity and differentiability class 12 MCQs are framed as per the CBSE syllabus (2022–2023) and NCERT guidelines. Click here to access all chapter-by-chapter MCQs for class 12 Maths.

MCQs on Class 12 Maths Chapter 5 Continuity and Differentiability

Check out the MCQs for Class 12 Maths Chapter 5 Continuity and Differentiability. Each MCQ has four options, only one of which is correct. Students must select the correct option and compare their answers with the provided solutions. We have provided a detailed explanation to the questions of continuity and differentiability class 12 MCQs, which will help students to understand the concept very well. Also, find important questions for class 12 Maths here.

Practice the below-given continuity and differentiability class 12 MCQs to score good marks in the examination.

Download PDF – Chapter 5 Continuity and Differentiability

\(\begin{array}{l}\text{1) If the function f(x) is differentiable at x=a, then} \lim_{x \to a }\frac{x^{2}f(a)-a^{2}f(x)}{x-a} \text{ is}\end{array} \)
  1. a2 f(a)
  2. a f(a)-a2 f’(a)
  3. 2a f(a)-a2 f’(a)
  4. 2a f(a)+a2 f’(a)

Answer: Option (c) 2a f(a)-a2 f’(a)

Explanation:

Given: The function f(x) is differentiable at x=a, so f’(a) exists.

By using L’ Hospital rule,

\(\begin{array}{l}\lim_{x \to a }\frac{x^{2}f(a)-a^{2}f(x)}{x-a} = lim_{x \to a }\frac{2x f(a)-a^{2}f'(x)}{1}\end{array} \)

= 2a f(a) – a2f’(a)

Hence, option (c) is the correct answer.

\(\begin{array}{l}\text{2) The point(s) at which the function f is given by  }f(x) = \left\{\begin{matrix} \frac{x}{|x|} & x<0\\ -1 & x\geq 0 \end{matrix}\right. \text{ is continuous, is/are}\end{array} \)
  1. x ∈ R
  2. x = 0
  3. x ∈ R-{0}
  4. x = -1 and 1

Answer: Option (a) x ∈ R

Explanation:

Given:

\(\begin{array}{l}f(x) = \left\{\begin{matrix} \frac{x}{|x|} & x<0\\ -1 & x\geq 0 \end{matrix}\right.\end{array} \)

Since, |x| = -x for x<0,

\(\begin{array}{l}f(x) = \left\{\begin{matrix} \frac{x}{(-x)} & x<0\\ -1 & x\geq 0 \end{matrix}\right.\end{array} \)
\(\begin{array}{l}f(x) = \left\{\begin{matrix} -1 & x<0\\ -1 & x\geq 0 \end{matrix}\right.\end{array} \)

Hence, f(x) = -1 is a constant function and the constant function is continuus for all real numbers.

Therefore, x ∈ R is the correct answer.

3) If x =t2, y=t3, then d2y/dx2 =

  1. 3/2
  2. 3/4t
  3. 3/2t
  4. 3t/2

Answer: Option (b) 3/4t

Explanation:

x=t2 ⇒dx/dt = 2t

y=t3⇒dy/dt = 3t2

⇒dy/dx = 3t2/2t = 3t/2

Hence, d2y/dx2 = 3/4t.

4) The value of c in Rolle’s theorem for the function, f(x) = sin 2x in [0, π/2] is

  1. π/4
  2. π/6
  3. π/2
  4. π/3

Answer: Option (a) π/4

Explanation:

For the function, f(x) = sin 2x in [0, π/2], f’(c) = 0

Hence, 2 cos 2c =0

2c =π/2

c = π/4

Hence, option (a) π/4 is the correct answer.

5) If y = ax2+b, then dy/dx at x = 2 is equal to

  1. 2a
  2. 3a
  3. 4a
  4. None of these

Answer: Option (c) 4a

Explanation:

Given that, y=ax2+b

Then dy/dx = 2ax

At x=2, dy/dx = 2(a)(2) = 4a

Hence, the correct answer is option (c) 4a

6) The function f(x) = [ln(1+ax)-ln(1-bx)]/x, not defined at x=0. The value should be assigned to f at x=0, so that it is continuous at x =0, is

  1. a+b
  2. a-b
  3. b-a
  4. ln a + ln b

Answer: Option (a) a+b

For f(x) to be continuous at x =0,

f(0) = limx→0f(x)

Therefore, f(0) = limx→0[ln(1+ax)-ln(1-bx)]/x

By using limx→0[ln(1+x)]/x = 1, we get

f(0) = a+b.

7) If x sin(a+y) = sin y, then dy/dx is equal to

  1. [sin2(a+y)]/sin a
  2. sin a /[sin2(a+y)]
  3. [sin(a+y)]/sin a
  4. sin a /[sin(a+y)]

Answer: Option (a) [sin2(a+y)]/sin a

Given: x sin(a+y) = sin y

⇒x = sin y/sin (a+y)

Now, differentiate both sides with respect to x, we get

\(\begin{array}{l}1 = \frac{sin(a+y)cos y\frac{dy}{dx}-cos(a+y)sin y\frac{dy}{dx}}{sin^{2}(a+y)}\end{array} \)

Simplifying the above equation, we get

⇒sin2(a+y) = sin (a+y-y)dy/dx

⇒ dy/dx = [sin2(a+y)]/sin a

8) If xy. yx = 16, then dy/dx at (2, 2) is

  1. 0
  2. 1
  3. -1
  4. None of these

Answer: Option (c) -1

Explanation:

Given: xy. yx = 16

Now, take log on both sides, we get

log xy + log yx = log 16

y log x + x log y = log 16

Differentiate with respect to x, we get

(y/x) + log x (dy/dx) + (x/y)(dy/dx) + log y = 0

Hence, dy/dx = – (y/x)[(y+xlogy)/((x+ylogx)]

Therefore, dy/dx at (2,2) = -1.

9) If y = √[sinx + y], then dy/dx =

  1. cos x /(2y-1)
  2. cos x/(1-2y)
  3. sin x/(2y-1)
  4. sin x/(1-2y)

Answer: Option (a) cos x /(2y-1)

Explanation:

Given:

y = √[sinx + y]

Now, take square on both sides,

y2 = sin x+y

Differntiating with respect to x, we get

2y(dy/dx) = cosx + dy/dx

Hence, 2y(dy/dx) – (dy/dx) = cos x

(dy/dx) (2y-1) = cos x

Hence, dy/dx = cos x /(2y-1).

10) For x∈R, f(x) = |log 2-sinx| and g(x)= f(f(x)), then

  1. g is not differentiable at x =0
  2. g’(0) = cos(log 2)
  3. g’(0) = – cos (log 2)
  4. g is diffentiable at x =0 and g’(0)= -sin (log 2)

Answer: Option (b) g’(0) = cos(log 2)

Explanation:

Given: f(x) = |log 2 – sinx|

g(x) = f(f(x))

g’(x) = f’(f(x))f’(x)

Hence, g’(0) = f’(f(0))f’(0)

When x tends to 0, log 2 > sinx

Thus, f(x) = log 2 -sinx

f’(x) = -cos x

f’(0) = – cos 0 = -1

f’(log 2) = – cos (log 2)

⇒g’(0) = (-cos(log 2)) (-1) = cos (log 2)

Hence, option (b) g’(0) = cos (log 2) is the correct answer.

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