VID-M12-05-CT — Continuity and Differentiability — Full Chapter Test | Class 12 Mathematics

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CodeVID-M12-05-CT

Continuity and Differentiability — Full Chapter Test

Chapter: Continuity and Differentiability

Topic: Complete chapter — all topics

Maximum Marks: 35

Time: 80 minutes

Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

This is a full-length, CBSE board-hard test covering the whole chapter. All questions are compulsory.
Section A carries 1 mark each (including an assertion-reason item), Section B 2 marks, Section C 3 marks and Section D 5 marks (long answer / case based).
Show all working for Sections B, C and D. Final answers are given in the Answer Key.

Section A — Multiple Choice Questions 5 × 1 = 5 marks

The function $f(x)=\dfrac{1-\cos4x}{x^2}$ for $x\ne0$, and $f(0)=k$, is continuous at $x=0$ for $k$ equal to:

A.$4$
B.$8$
C.$16$
D.$2$

If $y=\sin^{-1}\left(2x\sqrt{1-x^2}\right)$ where $-\dfrac1{\sqrt2}

A.$\dfrac{2}{\sqrt{1-x^2}}$

B.$\dfrac{1}{\sqrt{1-x^2}}$

C.$\dfrac{-2}{\sqrt{1-x^2}}$

D.$\dfrac{2x}{\sqrt{1-x^2}}$

Assertion (A): $f(x)=|x-2|+|x-3|$ is not differentiable at $x=2$ and $x=3$.
Reason (R): The modulus function $|x-a|$ is not differentiable at $x=a$.

A.Both A and R are true and R is the correct explanation of A
B.Both A and R are true but R is not the correct explanation of A
C.A is true but R is false
D.A is false but R is true

If $x=t^2$ and $y=t^3$, then $\dfrac{d^2y}{dx^2}$ equals:

A.$\dfrac{3}{4t}$
B.$\dfrac32$
C.$\dfrac{3t}{2}$
D.$\dfrac{3}{4t^2}$

If $y=x^x$, then the value of $\dfrac{dy}{dx}$ at $x=e$ is:

A.$e^e$
B.$2e^e$
C.$e^e(1+e)$
D.$e^{e+1}$

Section B — Short Answer I 4 × 2 = 8 marks

If $x^y=e^{x-y}$, show that $\dfrac{dy}{dx}=\dfrac{\log x}{(1+\log x)^2}$.

Differentiate $\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$ with respect to $x$, for $|x|<1$.

Find $\dfrac{dy}{dx}$ if $y=(\sin x)^x$, $0

If $x=a(\theta-\sin\theta)$ and $y=a(1+\cos\theta)$, find $\dfrac{dy}{dx}$.

Section C — Short Answer II 4 × 3 = 12 marks

10.

Differentiate $y=(\log x)^x+x^{\log x}$ with respect to $x$.

11.

If $y=e^{a\cos^{-1}x}$, $-1\le x\le1$, show that $(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}-a^2y=0$.

12.

Find the values of $a$ and $b$ such that $f(x)=\begin{cases}5,&x\le2\\ax+b,&2

13.

Verify the Mean Value Theorem for $f(x)=x^3-5x^2-3x$ on $[1,3]$ and find the value of $c$.

Section D — Long Answer / Case Based 2 × 5 = 10 marks

14.

If $x=a(\cos t+t\sin t)$ and $y=a(\sin t-t\cos t)$, find $\dfrac{d^2y}{dx^2}$ at $t=\dfrac{\pi}{4}$.

15.

Let $f(x)=\begin{cases}x^2\sin\dfrac1x,&x\ne0\\0,&x=0.\end{cases}$
(a) Show that $f$ is continuous at $x=0$. [1]
(b) Show that $f$ is differentiable at $x=0$ and find $f'(0)$. [2]
(c) Show that $f'$ is not continuous at $x=0$. [2]

Answer Key

Section A — Multiple Choice Questions

(B) $8$
(A) $\dfrac{2}{\sqrt{1-x^2}}$
(A) Both A and R are true and R is the correct explanation of A
(A) $\dfrac{3}{4t}$
(B) $2e^e$

Section B — Short Answer I

Taking logs, $y\log x=x-y\Rightarrow y=\dfrac{x}{1+\log x}$. Then $\dfrac{dy}{dx}=\dfrac{(1+\log x)\cdot1-x\cdot\frac1x}{(1+\log x)^2}=\dfrac{\log x}{(1+\log x)^2}$.
Put $x=\tan\theta$, $|\theta|<\frac{\pi}{4}$. Then $\dfrac{2x}{1+x^2}=\sin2\theta$ and $\sin^{-1}(\sin2\theta)=2\theta=2\tan^{-1}x$. Hence the derivative is $\dfrac{2}{1+x^2}$.
$\log y=x\log\sin x\Rightarrow\dfrac1y\dfrac{dy}{dx}=\log\sin x+x\cot x$. So $\dfrac{dy}{dx}=(\sin x)^x\left[\log\sin x+x\cot x\right]$.
$\dfrac{dx}{d\theta}=a(1-\cos\theta)$, $\dfrac{dy}{d\theta}=-a\sin\theta$. So $\dfrac{dy}{dx}=\dfrac{-\sin\theta}{1-\cos\theta}=-\cot\dfrac{\theta}{2}$.

Section C — Short Answer II

Let $u=(\log x)^x$: $\log u=x\log(\log x)$, so $\dfrac{u'}{u}=\log(\log x)+\dfrac{1}{\log x}$. Let $v=x^{\log x}$: $\log v=(\log x)^2$, so $\dfrac{v'}{v}=\dfrac{2\log x}{x}$. Hence $\dfrac{dy}{dx}=(\log x)^x\left[\log(\log x)+\dfrac1{\log x}\right]+x^{\log x}\cdot\dfrac{2\log x}{x}$.
$y_1=e^{a\cos^{-1}x}\cdot\dfrac{-a}{\sqrt{1-x^2}}$, so $\sqrt{1-x^2}\,y_1=-ay$. Differentiate: $\sqrt{1-x^2}\,y_2-\dfrac{x}{\sqrt{1-x^2}}\,y_1=-ay_1$. Multiply by $\sqrt{1-x^2}$: $(1-x^2)y_2-xy_1=-a\sqrt{1-x^2}\,y_1=-a(-ay)=a^2y$. Hence $(1-x^2)y_2-xy_1-a^2y=0$.
Continuity at $x=2$: $\lim_{x\to2^+}(ax+b)=2a+b=5$. Continuity at $x=10$: $\lim_{x\to10^-}(ax+b)=10a+b=21$. Subtracting, $8a=16$, so $a=2$ and $b=1$.
$f$ is a polynomial, hence continuous on $[1,3]$ and differentiable on $(1,3)$. $f(1)=-7$, $f(3)=-27$, so $\dfrac{f(3)-f(1)}{3-1}=-10$. $f'(c)=3c^2-10c-3=-10\Rightarrow3c^2-10c+7=0\Rightarrow c=1$ or $c=\dfrac73$. Only $c=\dfrac73$ lies in $(1,3)$, so MVT is verified with $c=\dfrac73$.

Section D — Long Answer / Case Based

$\dfrac{dx}{dt}=a\,t\cos t$ and $\dfrac{dy}{dt}=a\,t\sin t$, so $\dfrac{dy}{dx}=\tan t$. Then $\dfrac{d^2y}{dx^2}=\dfrac{d}{dt}(\tan t)\cdot\dfrac{dt}{dx}=\sec^2t\cdot\dfrac{1}{a\,t\cos t}=\dfrac{\sec^3t}{a\,t}$. At $t=\dfrac{\pi}{4}$: $\sec^3\dfrac{\pi}{4}=2\sqrt2$, so the value is $\dfrac{2\sqrt2}{a\cdot\frac{\pi}{4}}=\dfrac{8\sqrt2}{\pi a}$.
(a) $\left|x^2\sin\frac1x\right|\le x^2\to0$, so $\lim_{x\to0}f(x)=0=f(0)$. (b) $f'(0)=\lim_{h\to0}\dfrac{h^2\sin\frac1h-0}{h}=\lim_{h\to0}h\sin\dfrac1h=0$ (since $\left|h\sin\frac1h\right|\le|h|$). So $f'(0)=0$. (c) For $x\ne0$, $f'(x)=2x\sin\dfrac1x-\cos\dfrac1x$. As $x\to0$, the first term tends to $0$ but $\cos\dfrac1x$ oscillates between $-1$ and $1$, so $\lim_{x\to0}f'(x)$ does not exist and hence is not equal to $f'(0)=0$. Thus $f'$ is not continuous at $0$.

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