Question

Class 12 Maths MCQ Questions of Continuity and Differentiability with Answers?

Answer 1

In the Class 12 Mathematics exam, you can expect a few questions on this topic. The topic “Continuity and Differentiability” makes a significant contribution to CBSE Class 12 Mathematics syllabus. You can easily solve the MCQ Questions related to Continuity and Differentiability in board exams if you develop a stronghold on the topics covered in the Class 12 Maths MCQ Questions of Continuity and Differentiability with Answers. 

You at Sarthaks eConnect have attempted to provide Important MCQ Questions for Class 12 Maths. These Continuity and Differentiability Class 12 Important Questions are prepared by subject expert teachers to help you to prepare for the chapter in a better way and enhance your Class 12 Mathematics scores. 

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Practice MCQ Question for Class 12 Maths chapter-wise

1. A function /is said to be continuous for x ∈ R, if

(a) it is continuous at x = 0
(b) differentiable at x = 0
(c) continuous at two points
(d) differentiable for x ∈ R

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

(a) π/2
(b) π/4
(c) π/3
(d) π/6

3. The value of c in Rolle’s Theorem for the function f(x) = e^x sin x, x ∈ [0, π] is

(a) π/6
(b) π/4
(c) π/2
(d) 3π/4

4. The derivative of f(tan x) w.r.t. g(sec x) at x = π/4, where f'(1) = 2 and g'(√2) = 4, is

(a) 1/√2
(b) √2
(c) 1
(d) 0

5. If x² + y² = 1, then

(a) yy'' – (2y')² + 1 = 0
(b) yy'' + (y')² + 1 = 0
(c) yy'' – (y')² – 1 = 0
(d) yy'' + (2y')² + 1 = 0

6. A value of c for which the Mean value theorem holds for the function f(x) = log_ex on the interval [1, 3] is

(a) 2log₃e
(b) 1/2log_e3
(c) log₃e
(d) log_e3

7. The value of c in mean value theorem for the function f(x) = (x – 3)(x – 6)(x – 9) in [3, 5] is

(a) 6 ± √(13/3)
(b) 6 + √(13/3)
(c) 6 – √(13/3)
(d) None of these

8. The value of c in Mean value theorem for the function f(x)=x² in [2, 4] is

(a) 4
(b) 2
(c) 7/2
(d) 3

9. The set of points where the function f given by f (x) =| 2x – 1| sin x is differentiable is

(a) R
(b) R = {1/2}
(c) (0, ∞)
(d) None of these

10. The function f(x) = e^|x| is

(a) continuous everywhere but not differentiable at x = 0
(b) continuous and differentiable everywhere
(c) not continuous at x = 0
(d) None of these

11. If f(x) = x² sin 1/x, where x ≠ 0, then the value of the function f(x) at x = 0, so that the function is continuous at x = 0 is

(a) 0
(b) -1
(c) 1
(d) None of these

12. If y = ax² + b, then dy/dx at x = 2 is equal to ax

(a) 4a
(b) 3a
(c) 2a
(d) None of these

13. If y = (1 + x) (1 + x²) (1 + x⁴) …….. (1 + x²ⁿ), then the value of dy/dx at x = 0 is

(a) 0
(b) -1
(c) 1
(d) None of these

14. If f(x) = 5x/(1−x)^2/3 + cos² (2x + 1), then f'(0) =

(a) 5 + 2 sin 2
(b) 5 + 2 cos 2
(c) 5 – 2 sin 2
(d) 5 – 2 cos 2

15. If sec ((x-y)/(x+y)) = a, then dy/dx is equal to

(a) y/x
(b) -y/x
(c) x/y
(d) -x/y

16. If y = tan^-1 [ sin x + cos x]/[cos x - sin x], then dy/dx is

(a) 0
(b) π/4
(c) 1
(d) 1/2

17. If y = tan^-1(√x−x/1+x^3/2), then y'(1) is equal to

(a) 0
(b) (√x−x/1+x^3/2)
(c) -1
(d) –1/4

18. If y = ae^x + be^-x where a, b are parameters then y'' =

(a) y
(b) y'
(c) -y'
(d) 0

19. If sin y + e^{-xcos y} = e, then dy/dx at (1, π) is equal to

(a) sin y
(b) -x cos y
(c) e
(d) sin y – x cos y

20. If y = 2^x3^2x-1 then dy/dx is equal to dx

(a) (log 2) (log 3)
(b) (log lg)
(c) (log 18²) y²
(d) y (log 18)

Answer 2

1. Answer: (d) differentiable for x ∈ R

2. Answer: (b) π/4

Explanation: Here, f′(c)=0

⇒2cos2c=0

⇒2c=π/2

⇒c=π/4

3. Answer: (d) 3π/4

Explanation: Since Rolle’s theorem is satisfied

∴f′(c)=0⇒e^csinc+eose^c=0

⇒e^c{sinc+eosec}=0

∴sinc+cosc=0(∵e^c≠0)

⇒tanc=−1⇒c=tan⁻¹(−1)=π−π/4=3π/4

4.  Answer: (a) 1/√2

Explanation: y = f(tan x) and z = g(sec x)

\(\frac{dy}{dx}=f'(tan\,x).sec^2x\) \(and\,\frac{dz}{dx}=g'(sec\,x).sec\,x\,tan\,x\)

\(\therefore\frac{dy}{dz}=\frac{f'(tan\,x).sec^2x}{g'(sec\,x).sec\,x\,tan\,x}\)

\(\left.\frac{dy}{dz}\right|_{x=\frac{\pi}4}=\frac{f'(1).(\sqrt2)^2}{g'(\sqrt2).(\sqrt2)}=\frac1{\sqrt2}\)

5. Answer: (b) yy'' + (y')² + 1 = 0

Explanation:

The given expression is x² + y² = 1

Differentiating w.r.t. x, we get 

2x + 2y(dy/dx) = 0

x + y(dy/dx) = 0

Again differentiating, w.r.t. x, we get 

1 + (dy/dx)² + y(d²y/dx²) = 0

1 + y'² + y.y'' = 0

y.y'' + (y')² + 1 = 0

6.  Answer: (a) 2log₃e

Explanation: \(f(x)=log_ex\)

\(f'(x)=\frac1x\)

By Lagrange's mean value theorem, for c in (1, 3)

\(f'(c)=\frac{f(3)-f(1)}{3-1}\)

\(\Rightarrow\frac1c=\frac{log_e3}2\)

\(\Rightarrow c=\frac2{log_e3}\)

\(\Rightarrow c=2log_3e\)

7. Answer: (c) 6 – √(13/3)

Explanation: Here, \(f'(c)=\frac{f(5)-f(3)}{5-3}\)

\(\Rightarrow 3c^2-36c+99=\frac{8-0}2=4\)

\(\Rightarrow c=6\pm\sqrt{\frac{13}3}\)

Hence, \(c=6-\sqrt{\frac{13}3}\) (since other value is not permissible)

8. Answer: (d) 3

Explanation: 

Mean value theorem states that if f(x) is defined and continuous on interval [a,b] and diffrentiable on (a,b)then there is atleast one number in c in interval (a,b) i.e. a f′(c)= \(\frac{f(b)-f(a)}{b-a}\)

Graph of f(x) \(=x^2\)

\(\therefore\) Function is continuous

f'(x) = 2x

\(f'(c)=\frac{f(b)-f(a)}{b-a}\)

\(f'(c)=\frac{(4)^2-(2)^2}{4-2}\)

\(f'(c)=\frac{16-4}2=6\)

2x = 6

x = 3

\(\therefore\) c = 3 and 2 < 3 < 4

9.  Answer: (b) R = {1/2}

Explanation: f(x) = |2x - 1| sin x

\(f(x)=\begin{cases}-(2x-1)sin\,x,&x<1/2\\(2x-1)sin\,x,&x\geq1/2\end{cases}\)

\(Lf'(\frac12)=\underset{h\to0}{lim}\frac{f(1/2+h)-f(1/2)}h\)

\(=\underset{h\to0}{lim}\frac{-[2(1/2+h)-1]\,sin(1/2+h)-0}h\)

\(=\underset{h\to0}{lim}\frac{-2h\,sin(1/2+h)}h=-2\,sin(1/2)\)

\(Rf'(\frac12)=\underset{h\to0^+}{lim}\frac{f(1/2+h)-f(1/2)}h\)

\(=\underset{h\to0^+}{lim}\frac{[2(1/2+h)-1]\,sin(1/2+h)-0}h\)

\(=\underset{h\to0^+}{lim}\frac{2h\,sin(1/2+h)}h=2\,sin(1/2)\)

\(Lf'(\frac12)\neq Rf'(\frac12)\)

So, f(x) is not differentiable at x = 1/2

Hence f(x) is differentiable \(\forall\,x\in R-\{\frac12\}\)

10. Answer: (a) continuous everywhere but not differentiable at x = 0

11.  Answer: (a) 0

Explanation:

\(f(0)=\underset{x\to0}{lim}\,x\,sin\frac1x\)

We know \(\forall\,x\in R,sin\frac1x\in[-1,1]\)

Hence, f(0) \(=\underset{x\to0}{lim}\,x\,sin\frac1x=0\)

12.  Answer: (a) 4a

Explanation: 

We have, \(y=ax^2+b\)

\(\Rightarrow\frac{dy}{dx}=2ax\)

\(\left.\frac{dy}{dx}\right|_{x=2}=2a\times2=4a\)

13. Answer: (c) 1

Explanation: 

We have, \(y=(1+x)(1+x^2)(1+x^4)...(1+x^4)\)

take natural logarithm both sides

\(In\,y=In(1+x)+In(1+x^2)\) \(+In(1+x^4)+...+In(1+x^{2n})\)

Now differentiate both sides w.r.t. x

\(\Rightarrow \frac1y\frac{dy}{dx}=\frac1{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+...+\frac{2nx^{2n-1}}{1+x^{2n}}\)

\(\Rightarrow\frac{dy}{dx}=y(\frac1x+\frac{2x}{1+x^2}+...+\frac{2nx^{2n-1}}{1+x^{2n}})\)

Now substitute x = 0 to get the required value

\(\Rightarrow\left.\frac{dy}{dx}\right|_{x=0}=1(1+0+0+...+0)=1\)

Note that at x = 0, valu of y is also 1

14. Answer: (c) 5 – 2 sin 2

Explanation: \(f(x)=5x(1-x)^{-\frac23}+cos^2(2x+1)\)

\(f'(x)=5\{x\times{\frac{-2}3}(1-x)^{-5/3}\) \((-1)+(1-x)^{-2/3}\times1\}\)

\(+2\,cos(2x+1)\{-sin(2x+1)\times2\}\)

\(\Rightarrow f'(x)=5(1-x)^{-\frac23}\) \(+\frac{10x}3(1-x)^{-\frac53}-2\,sin(4x+2)\)

\(\therefore\) f'(0) = 5 - 2 sin 2.

15. Answer: (a) y/x

Explanation: Given sec ((x-y)/(x+y)) = a

(x-y)/(x+y) = sec^-1 a

Differentiate w.r.t.x

[(x+y)(1-dy/dx) – (x-y)(1+dy/dx)]/(x+y)² = 0
(x+y)(1-dy/dx) – (x-y)(1+dy/dx) = 0

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

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

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

dy/dx = y/x

Answer 3

16. Answer: (c) 1

Explanation: Given y = tan^-1 [ (sin x + cos x)/(cos x – sin x)]

Divide numerator and denominator by cos x

y = tan^-1 [ (tan x + 1)/(1 – tan x)]

= tan^-1 [ (1 + tan x)/(1 – tan x)]

= tan^-1 [ (tan π/4 + tan x)/(1 – tan π/4 tan x)]

= tan^-1 tan (π/4 + x)

= (π/4 + x)

dy/dx = 1

17.  Answer: (d) –14

Explanation: \(y=tan^{-1}(\frac{\sqrt x-x}{1+x^{3/2}})\)

\(=tan^{-1}(\frac{\sqrt x-x}{1+\sqrt x.x})\)

\(=tan^{-1}(\sqrt x)-tan^{-1}(x)\)

Differentiating w.r.t. x, we get

\(y'=\frac1{1+x}.\frac1{2\sqrt x}-\frac1{1+x^2}\)

\(\Rightarrow y'(1)=\frac12.\frac12-\frac12\)

\(=-\frac14\)

18. Answer: (a) y

Explanation: Given y = ae^x + be^-x

Differentiate w.r.t.x

y' = ae^x – be^-x

Again differentiate w.r.t.x

y'' = ae^x + be^-x

= y

19. Answer: (c) e

Explanation: Given,

\(sin\,y+e^{-x\,cos\,y}=e\)

\(cos\,y\frac{dy}{dx}+e^{-x\,cos\,y}\left[x\,sin\,y\frac{dy}{dx}-cos\,y\right]=0\)

\(\Rightarrow cos\,y\frac{dy}{dx}+e^{-x\,cos\,y}\,x\,sin\,y\frac{dy}{dx}\) \(-cos\,ye^{-x\,cos\,y}=0\)

\(\frac{dy}{dx}[cos\,y+x\,sin\,ye^{-x\,cos\,y}]=cos\,ye^{-x\,cos\,y}\)

\(\frac{dy}{dx}=\frac{cos\,ye^{-x\,cos\,y}}{cos\,y+x\,sin\,ye^{-x\,cos\,y}}\)

\(\left[\frac{dy}{dx}\right]_{(1,\pi)}=\frac{cos\,\pi e^{-cos\,\pi}}{cos\,\pi+1\,sin\,\pi e^{-cos\,\pi}}\)

\(\left[\frac{dy}{dx}\right]_{(1,\pi)}=\frac{-1\times e}{-1+0\times e}=e\)

20. Answer: (d) y (log 18)

Explanation: Given, \(y=2^x.3^{2x-1}\)

Differentiating w.r.t. x, we get

\(\frac{dy}{dx}=2^x.\frac d{dx}(3^{2x-1})\) \(+(3^{2x-1})+(3^{2x-1})\frac d{dx}(2^x)...(i)\)

Let \(3^{2x-1}=u\)

\(\Rightarrow log\,u=(2x-1)\,log3\)

\(\Rightarrow\frac{du}{dx}=3^{2x-1}\times2.\,log3\)

\(\therefore\) From (i), we have

\(\frac{dy}{dx}=2^x.3^{2x-1}(2)\,log\,3+2^x.3^{2x-1}\,log\,2\)

\(\Rightarrow\frac{dy}{dx}=2^x.3^{2x-1}[2\,log\,3+log\,2]\)

\(\Rightarrow\frac{dy}{dx}=y\,log\,18\)

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