MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 1

$\mathrm{f}(\mathrm{x})$ is a differentiable function and

$\mathrm{g}(\mathrm{x})$ is a double differentiable function such that

$|\mathrm{f}(\mathrm{x})|\leq 1$ and

$\mathrm{f}'(\mathrm{x})=\mathrm{g}(\mathrm{x})$ . If

$\mathrm{f}^{2}(0)+\mathrm{g}^{2}(0)=9$ such that there exists some

$\mathrm{c}\in(-3, 3)$ such that

$\mathrm{g}(\mathrm{c}).\ \mathrm{g}''(\mathrm{c})<0$ , True or false

0%
True
0%
False

Let f be a polynomial function such that

$f(3x)=f'(x)\cdot f''(x)$ , for all

$x\in R$ . Then.

0%
$f''(2)-f'(2)=0$
0%
$f(2)+f'(2)=28$
0%
$f''(2)-f(2)=4$
0%
$f(2)-f'(2)+f''(2)=10$

Explanation

Let

$f(x)=a_ox^n+a_1x^{n-1}+a_2x^{n-2}+....+a_{n-1}x+a_n$

$f'(x)=a_onx^{n-1}+a_1(n-1)x^{n-2}+.....+a_{n-1}$

$f''(x)=a_on(n-1)x^{n-2}+a_1(n-1)(n-2)x^{n-3}+....+a_{n-2}$

$f(3x)=3^na_ox^n+3^{n-1}a_1x^{n-1}+3^{n-2}a_2x^{n-2}+....+3a_{n-1}x+a_n$

$f'(x)f''(x)=[a_o nx^{n-1}+a_1(n-1)x^{x-2}+....+a_{n-1}][a_o n(n-1)x^{n-2}+a_1(n-1)(n-2)x^{n-3}+....+a_{n-2}]$

Comparing highest powers of

$x$ ,

$3^na_ox^n=a^2_o(n-1)x^{n-1+n-2}=a^2_on^2(n-1)x^{2n-3}$

$\therefore 2n-3=n$

$\Rightarrow n=3$

and

$3^na_o=a^2_on^2(n-1)$

$3^3=a_o9(3-1)$

$a_o=\displaystyle\frac{27}{18}=\frac{3}{2}$

$\therefore f(x)=\displaystyle\frac{3}{2}x^3+a_1x^2+a_2x+a_3$

$f(3x)=\displaystyle\frac{81}{2}x^3+9a_1x^2+3a_2x+a_3$

$f'(x)=\displaystyle\frac{9}{2}x^2+2a_1x+a_2$

$f''(x)=9x+2a_1$

$\displaystyle\frac{81}{2}x^3+9a_1x^2+3a_2x+a_3=(\frac{9}{2}x^2+2a_1x+a_2)(9x+2a_1))$

$\displaystyle\frac{81}{2}x^3+9a_1 x^2+3a_2 x+a_3=\frac{81}{2}x^3+[9a_1+18a_1]x^2+[4a^2_1+9a_2]x+2a_1a_2$
Comparing coefficients,

$aa_1=27a_1$

$\Rightarrow a_1=0$

$3a_2=4a^2_1+9a_2=9a_2$

$\Rightarrow a_2=0$

$a_3=2a_1a_2\Rightarrow a_3=0$

$\therefore f(x)=\displaystyle\frac{3}{2}x^3$

$f'(x)=\displaystyle\frac{9}{2}x^2$

$f''(x)=9x$

$f''(2)-f'(2)-18-18=0$ .

$y=\left[\displaystyle x+\sqrt{x^2-1}\right]^{15}+\left[\displaystyle x-\sqrt{x^2-1}\right]^{15}$ , then

$(x^2-1)\displaystyle\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}$ is equal to.

0%
$125y$
0%
$225y^2$
0%
$225y$
0%
$224y^2$

Explanation

$\displaystyle\frac{dy}{dx}=15(x+\sqrt{x^2-1})^{14}\left(1+\displaystyle\frac{2x}{2\sqrt{x^2-1}}\right)+15{(x-\sqrt{x^2-1}})^{14}\left(1-\displaystyle\frac{2x}{2\sqrt{x^2-1}}\right)$

$=15(x+\sqrt{x^2-1})^{14}$

$\left(\displaystyle\frac{\sqrt{x^2-1}+x}{2\sqrt{x^2-1}}\right)+15{(x-\sqrt{(x^2-1})}^{14}\left(\sqrt{x^2-1}-x\right)$

$\displaystyle\frac{dy}{dx}=15\displaystyle\frac{(x+\sqrt{x^2-1})^{15}}{\sqrt{x^2-1}}-\frac{15{(x-\sqrt{x^2-1}})^{15}}{\sqrt{x^2-1}}$

$\sqrt{x^2-1}\displaystyle\frac{dy}{dx}=15(x+\sqrt{x^2-1})^{15}-15(x- \sqrt{x^2-1})^{15}$

$\displaystyle\frac{2x}{2\sqrt{x^2-1}}$

$\displaystyle \frac{dy}{dx}$

$\displaystyle +\frac{d^2y}{dx^2}$

$\sqrt{x^2-1}=$

$\displaystyle\frac{225{(x+\sqrt{x^2-1})}^{15}+225{(x- \sqrt{x^2-1)}^{15}}}{\sqrt{x^2-1}}$

$x\displaystyle\frac{dy}{dx}+(x^2-1)\frac{d^2y}{dx^2}=225y$ .

Let

$f : R \rightarrow R$ and

$g : R \rightarrow R$ be functions satisfying

$f(x + y) = f(x) + f(y) + f(x)f(y)$ and

$f(x) = xg(x)$ for all

$x, y \in R$ . If

$\underset{x \rightarrow 0}{\lim} g(x) = 1$ , then which of the following statements is/are TRUE?

0%
f is differentiable at every $x \in R$
0%
If $g(0) = 1$ , then g is differentiable at every $x \in R$
0%
The derivative $f'\left( 1 \right)$ is equal to $1$
0%
The derivative ${ f }^{ \prime }\left( 0 \right)$ is equal to $1$

For the curve

$x = t^2 - 1, y = t^2 - t$ , tangent is parallel to

$x$ - axis where,

0%
$t = 0$
0%
$t=\dfrac{1}{\sqrt{3}}$
0%
$t=\dfrac{1}{2}$
0%
$t=-\dfrac{1}{\sqrt{3}}$

Explanation

$x = t^2 - 1, y = t^2 - t$

If tangent is parallel to

$x$ axis,

$\cfrac{dy}{dx} = 0$

$\Rightarrow \cfrac{dy}{dt} \times \cfrac{dt}{dx} = 0$

i.e.

$(2t - 1) \times \cfrac{1}{2t} = 0$

i.e.

$t = \cfrac{1}{2}$

$\displaystyle \frac{d}{dx}(\tan ^{-1}x)$

0%
$\displaystyle \frac{1}{1+x^{2}}.$
0%
$\displaystyle \frac{-1}{1+x^{2}}.$
0%
$\displaystyle \frac{-1}{1-x^{2}}.$
0%
$\displaystyle \frac{1}{1-x^{2}}.$

Explanation

Let

$\displaystyle y=\tan ^{-1}x\therefore x=\tan y$

$\displaystyle y+\delta y=\tan ^{-1}\left ( x+\delta x \right )$

$\displaystyle \therefore x+\delta x=\tan\left ( y+\delta y \right )$

$\therefore \displaystyle \frac{dy}{dx}=\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}$

$\displaystyle =\lim_{\delta x\rightarrow 0}\frac{\delta y}{\tan\left ( y+\delta y \right )-\tan y}= \frac{1}{\sec^{2}y}$

$\displaystyle =\frac{1}{1+\tan ^{2}y}= \frac{1}{1+x^{2}}.$

Say true or false.

Every continuous function is always differentiable.

0%
True
0%
False

Explanation

If a function

$f(x)$ is differentiable at a point

$a$ , then it is continuous at the point

$a$ . But the converse is not true.

$f(x)=|x|$ is contineous but not differentiable at

$x=0.$
Therefore, the given statement is false.

$\displaystyle \frac{d(sin^{-1}x)}{dx}$

0%
$\displaystyle \frac{1}{\sqrt{\left ( 1-x^{2} \right )}}$
0%
$\displaystyle \frac{1}{\sqrt{\left ( 1-x^{4} \right )}}$
0%
$\displaystyle \frac{1}{\sqrt{\left ( 1-x^{3} \right )}}$
0%
$\displaystyle \frac{1}{\sqrt{\left ( 1+x^{2} \right )}}$

Explanation

$\displaystyle y= \sin ^{-1}x,$

$\displaystyle y+\delta y= \sin ^{-1}\left ( x+\delta x \right )$

$\displaystyle \therefore \left ( x+\delta x \right) = \sin \left ( y+\delta y \right )$

$\displaystyle \frac{dy}{dx}= \lim_{x\rightarrow 0}\frac{1}{\delta x}(\sin ^{-1}\left ( x+\delta x \right )-\sin ^{-1}x)$

$\displaystyle = \lim_{x\rightarrow 0}\frac{\delta y}{\sin \left ( y+\delta y \right )-\sin y}= \frac{1}{\cos y}$

$\displaystyle = \frac{1}{\sqrt{\left ( 1-\sin ^{2}y \right )}}= \frac{1}{\sqrt{\left ( 1-x^{2} \right )}}$
Note that when

$\displaystyle :\delta x\rightarrow 0, \delta y$ also tends to zero.

$y$ is expressed in terms of a variable

$x$ as

$y = f(x)$ , then

$y$ is called

0%
Explicit function
0%
Implicit function
0%
Linear function
0%
Identity function

Explanation

$y$ is expressed in terms of a variable

$x$ as

$y=f(x)$ , then

$y$ is called explicit function.

The function

$f(x)=\dfrac{1-\sin x+\cos x}{1+\sin x+\cos x}$ is not defined at

$x=\pi$ . The value of

$f(\pi)$ so that

$f(x)$ is continuous at

$x=\pi$ is

0%
$-\dfrac{1}{2}$
0%
$\dfrac{1}{2}$
0%
$-1$
0%
$1$

Explanation

$f$ is continuous at

$x=\pi$ ,

$f(\pi)=\displaystyle \lim_{x\rightarrow\pi}f(x)=\lim_{x\rightarrow\pi}\cfrac{1-\sin x+\cos x}{1+\sin x+\cos x}$

$=\displaystyle \lim_{x\rightarrow \pi}\cfrac{-\cos x-\sin x}{\cos x-\sin x}$ ..... [L-Hospital's Rule]

$=\cfrac{-\cos\pi-\sin\pi}{\cos\pi-\sin\pi}$

$=\cfrac{-(-1)-0}{-1-0}=\cfrac1{-1}=-1$

$\therefore f(\pi)=-1$

Hence, C is the correct option.

If the tangent to the curve

$x=a \, (\theta + \sin \, \theta), y=a (1+ \cos \,\theta)$ at

$\theta=\dfrac{\pi}{3}$ makes an angle

$\alpha (0 \leq\alpha < \pi)$ with x-axis, then

$\alpha$ =

0%
$\dfrac{\pi}{3}$
0%
$\dfrac{2 \pi}{3}$
0%
$\dfrac{\pi}{6}$
0%
$\dfrac{5 \pi}{6}$

Explanation

$x=a\left( \theta +\sin { \theta } \right)$

$y=a\left( 1+\cos { \theta } \right)$

If tangent to curve at $\theta = { \pi }/{ 3 }$ , makes angle $\alpha$ with x-axis,

$\Rightarrow$ slope of tangent = $\tan { \alpha }$

$\Rightarrow$ $\left. \begin{matrix} \tan { \alpha } ={ { \dfrac { { dy }/{ { d\theta } } }{ { dx }/{ { d\theta } } } } }{ { } } \\ \end{matrix} \right| _{ \theta ={ \pi }/{ 3 } }$

$=\dfrac {-a\sin \theta}{ a\left( 1+\cos { \theta } \right) }$

$\left. \begin{matrix}\dfrac{-\sin \theta} { 1+\cos { \theta } } \\ \end{matrix}\right|_{ \theta ={ \pi }/{ 3 } }$

$=-\dfrac{\sqrt{3}/2} { 1+{ 1 }/{ 2 } }$

$\tan { \alpha } \ =-\dfrac{1}{\sqrt { 3 }}$

Ans. $\alpha \ =\ { 5{ \pi } }/{ 6 }$

$f(x)=\begin{cases} ax &a<1&\\ ax^2+bx+2 &a\ge 1&. \end{cases}$
Then the values of

$a$ ,

$b$ for which

$f(x)$ is differentiable, are

0%
$a=\large{\frac{3}{4}}$ , $b=\large{\frac{1}{4}}$
0%
$a=2$ , $b=-2$
0%
$a=\large{\frac{3}{2}}$ , $b=\large{\frac{1}{4}}$
0%
$a=\large{\frac{3}{4}}$ , $b=-2$

Explanation

Given the function

$f(x)$ is differentiable, then it is differentiable at

$x=1$ and it will be continuous there at.

Then using continuity condition at

$x=1$ ,

$\lim\limits_{x\to 1^-}f(x)=\lim\limits_{x\to 1+}f(x)$

or,

$a=a+b+2$

or,

$b=-2$ .

Now using differentiability at

$x=1$ we get,

$\lim\limits_{x\to 1^-}f'(x)=\lim\limits_{x\to 1^+}f'(x)$

or,

$a=2a+b$

or,

$a=-b$

or,

$a=2$ .

$\therefore a=2,b=-2$ .

so, option (B) is true.

$f(x)=\dfrac{1}{2}\left [ \left | \sin x \right |+\sin x \right ],\ 0 < x \leq 2\pi$ , then

$f$ is

0%
Increasing in $\left ( \dfrac{\pi}{2},\dfrac{3\pi}{2} \right )$
0%
Decreasing in $\left ( 0,\dfrac{\pi}{2} \right )$ and increasing in $\left ( \dfrac{\pi}{2},\pi \right )$
0%
Increasing in $\left ( 0,\dfrac{\pi}{2} \right )$ and decreasing in $\left ( \dfrac{\pi}{2},\pi \right )$
0%
Increasing in $\left ( 0,\dfrac{\pi}{4} \right )$ and decreasing in $\left ( \dfrac{\pi}{4},\pi \right )$

$x=a\cos ^{ 4 }{ t } ,y=b\ cosec^{ 4 }{ t }$ , then

$\cfrac { dx }{ dy }$ at

$t=\cfrac { 3\pi }{ 4 }$

0%
$\cfrac{-b}{a}$
0%
$\cfrac{b}{a}$
0%
$\cfrac{a}{b}$
0%
$\cfrac{-a}{16b}$

Let

$y=\sqrt{(\sin x+\sin 2x+\sin 3x)^2+(\cos x+\cos 2x+\cos 3x)^2}$ then which of the following(s) is correct?

0%
$\dfrac{dy}{dx}$ when $x=\dfrac{\pi}{2}$ is $-2$
0%
Value of y when $x=\dfrac{\pi}{5}$ is $\dfrac{3+\sqrt{5}}{2}$
0%
Value of y when $x=\dfrac{\pi}{12}$ is $\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}}{2}$
0%
y simplifies to $(1+2\cos x)$ in $[0, \pi]$

Explanation

$y=\sqrt{(\sin x+\sin 2 x+\sin 3 x)^2+(\cos x+\cos 2 x+\cos 3 x)^2}$

$=\sqrt{(\sin ^2 x+\cos^2 x)+(\sin ^2 2 x+\cos^2 2 x)+(\sin ^2 3 x+\cos^2 3 x)+2(\sin 2 x\sin x+\cos 2 x\cos x)+2(\sin 2 x \sin 3 x+\cos 2 x\cos 3 x)+2(\sin x\sin 3 x+\cos x\cos 3 x}$

$=\sqrt{1+1+1+2\cos(2 x-x)+2\cos (3 x-2 x)+2\cos (3 x- x)}=\sqrt{3+4 \cos x+2\cos 2 x}=\sqrt{4\cos^2 x+4\cos x+1}$

$=|2\cos x+1|$

$x=\dfrac{\pi}{2},\dfrac{d y}{d x}=-2\sin x=-2\sin \pi/2=-2$

$x=\dfrac{\pi}{5},y=2\cos \dfrac{\pi}{5}+1=2\times\dfrac{\sqrt{5}+1}{4}+1=\dfrac{\sqrt{5}+3}{2}$

$x=\dfrac{\pi}{12},y=2\cos \dfrac{\pi}{12}+1=2\times \dfrac{\sqrt{3}+1}{2\sqrt{2}}+1=\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}}{\sqrt{2}}$

$y=\tan^{-1}\left(\dfrac{2^{x+1}}{1+2^{2x}}\right)$ then

$\dfrac{dy}{dx}$ at

$x=0$ is

0%
$\dfrac{1}{10}log2$
0%
$\dfrac{1}{5}log2$
0%
$-\dfrac{1}{10}log2$
0%
$log2$

Explanation

Given,

$y=\tan^{-1}\left(\dfrac{2^{x+1}}{1+2^{2x}}\right)$

or,

$y=\tan^{-1}\left(\dfrac{2.2^{x}}{1+(2^{x})^2}\right)$

or,

$y=2\tan^{-1}(2^x)$

Now,

$\dfrac{dy}{dx}=\dfrac{2}{1+(2^x)^2}\times 2^x\times \log 2$ .

And

$\left.\dfrac{dy}{dx}\right|_{x=0}=\dfrac{2}{1+1}\times \log 2=\log 2$ .

$\displaystyle \int _0^1 \dfrac {e^x}{1+e^{2x}}dx$

0%

$\tan ^{-1}e-\dfrac \pi 4$
0%

$\tan ^{-1}e+\dfrac \pi 4$
0%

$\tan e-\dfrac \pi 4$
0%
None of these

Explanation

Given

$\displaystyle \int _0^1 \dfrac {e^x}{1+e^{2x}}dx$

Let

$e^x=t \implies e^x dx=dt$

$x \to 0\to 1$

$\implies t \to 1 \to e$

$\displaystyle \int_1^e \dfrac 1{1+t^2}dt$

$\left.tan ^{-1}t \right|_1^e$

$\implies \tan ^{-1}e-\dfrac \pi 4$

$y = \tan ^ { - 1 } \left( \cot \left( \dfrac { \pi } { 2 } - x \right) \right) ,$ then

$\dfrac { d y } { d x } =$

0%
$1$
0%
$-1$
0%
$0$
0%
$\dfrac { 1 } { 2 }$

Explanation

$y={ \tan }^{ -1 }\left( \cot\left( \pi /2-x \right) \right)$

$\Rightarrow y={ \tan }^{ -1 }\left( tanx \right) =x$

$\therefore$

$\dfrac { dy }{ dx } =1$ [A]

$\displaystyle\int \dfrac{f(x)dx}{\log(\sin x)}$

$=\log (\log(\sin x))$ then

$f(x) =$

0%
$sinx$
0%
$cosx$
0%
$\log(sinx)$
0%
$cotx$

Explanation

$\int \frac{f(x)}{log (sin x)}dx= log [log(sin x)]$

on differentiating both side cort x we get

$\frac{d}{dx}\int \frac{f(x)}{log(sin x)} dx=\frac{d}{dx}(log (log sin x))$

$\frac{f(x)}{log (sin x)}=\frac{cos x}{log (in x)}$

$[\therefore \frac{d}{dx}log k=\frac{1}{k}]$

$[f(x)= cos x]$

1199975_1382341_ans_7b260c5913cf4deba5e709f74561da6c.JPG

The set of points of continuity of the following

$\sqrt { \dfrac { 1 }{ 2 } -cos^{ 2 }x }$ contains in the interval

0%
$\left[ \dfrac { \pi }{ 4 } ,\dfrac { 3\pi }{ 4 } \right]$
0%
$\left[ \dfrac { 5\pi }{ 4 } ,\dfrac { 7\pi }{ 4 } \right]$
0%
$\left[ \dfrac { 21\pi }{ 4 } ,\dfrac { 23\pi }{ 4 } \right]$
0%
All above the

$y=\log \sin x$ find

$x$ if

$y=0$

0%
$\dfrac {\pi}2$
0%
$\pi$
0%
${\dfrac {\pi}3}$
0%
$\dfrac {-\pi}2$

Explanation

$y=\log \sin x\\y=0\\\log \sin x=\log 1\\\sin x=1\\\sin x=\sin \dfrac{\pi}2\\x=\dfrac{\pi}2$

$f ( x ) = \tan x$ and

$f$ is inverse of

$g ,$ then

$g ^ { \prime } ( x )$ is equal to

0%
$\cot x$
0%
$\dfrac{1}{1+x^2}$
0%
$\dfrac{1}{1-x^2}$
0%
$\tan x$

Explanation

Given

$f(x)=\tan x$ and

$f$ is inverse of

$g$

So ,

$g(x)$ is

$\tan ^{-1} x$

$g^{\prime}(x)=\dfrac{d}{dx}\tan^{-1}x\\g^{\prime}(x)=\dfrac{1}{1+x^2}$

Find the values of a and b so that the function

$f(x)=\left\{\begin{matrix} x^2+3x+a, & if & x\leq 1\\ bx+2, & if & x > 1\end{matrix}\right.$ is differentiable at each

$x\in R$ .

0%
$a=1,b=3$
0%
$a=5,b=3$
0%
$a=3,b=5$
0%
$a=3,b=1$

Explanation

If the function f(x) is required to be differentiable, it must be continuous too for

$x\in R$

But it is clearly evident that f(x) is continuous and differentiable in interval

$(-\infty,1)\cup(1,\infty)$

So, only point need to be examined is x=1 .

For

$f(x)$ to be continuous at

$x=1$ ,

LHL of

$f(x)$ at

$x=1$ must be equal to RHL of

$f(x)$ at

$x=1$

i.e.,

$(1)^2+3(1)+a =b(1)+2 \rightarrow b-a=2$ ...............................................1

For

$f(x)$ to be differentiable at

$x=1$ ,

LHD of f(x) at x=1 must be equal to RHD of

$f(x)$ at

$x=1$

i.e.,

$2(1) + 3 = b \rightarrow b=5$

by putting

$b$ value in we have

$a=3$

Thus we get

$(a,b) = (3,5)$

The set of points where the functions f given by

$f(x)=|x-3|\cos x$ differentiable is

0%
R
0%
R-{ $3$ }
0%
$(0,\infty)$
0%
None of these

Let

$f(x)$ be differentiable function such that

$f\left (\displaystyle \frac{x+y}{1-xy}\right)=f(x)+f(y)\forall x$ and

$y$ . lf

$\displaystyle \lim_{x \rightarrow 0}\frac{f(x)}{x}=\frac{1}{3}$ , then

$f(1)$ equals

0%
$\displaystyle \frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{12}$
0%
$\displaystyle \frac{\pi}{6}$
0%
$\displaystyle \frac{\pi}{3}$

Explanation

$f(\dfrac{x+y}{1-xy})=f(x)+f(y)$

$f(x)=atan^{-1}x$

$\lim_{x\rightarrow 0} \dfrac{a tan^{-1}x}{x}=\dfrac{1}{3}$

$\Rightarrow a=\dfrac{1}{3}$

$f(x)=\dfrac{1}{3}tan^{-1}x$

$\Rightarrow f(1)=\dfrac{1}{3}tan^{-1}(1)$

$=\dfrac{\pi }{12}$

$\displaystyle \mathrm{A}:\dfrac{d}{dx}(\sin x)\ at\ x=\frac{\pi}{2}$

$\displaystyle \mathrm{B}:\dfrac{d}{dx}(\tan^{-1}{x})$ at

${x}=1$

$\displaystyle \mathrm{C}:\dfrac{d}{dx}(\mathrm{e}^{x})$ at

${x}=0$

$\mathrm{D}:\dfrac{d}{dx}(x^{x})\ at\ x=e$

Arrangement of the above values in the increasing order of the magnitude

0%
B, C, A, D
0%
D, A, B, C
0%
D, B, C, A
0%
A, B, C, D

Explanation

$A.\dfrac{d}{dx}(sinx) = cosx$

Put

$x=\dfrac{\pi }{2}$

$=0$

$B.\dfrac{d}{dx}(tan^{-1}x)=\dfrac{1}{1+x^{2}}$

Put

$\ x=1$

$=\dfrac{1}{2}$

$C. \dfrac{d}{dx}(e^{x})=e^{x}$

Put

$x=0$

$=1$

$D.\dfrac{d}{dx}(x^{x})=\dfrac{d}{dx}(e^{xLogx})=x^{x}(Log x+1)$

$=2e^{e}$

${A} :$ If

$x = ct, y = \dfrac{c}{t}$ , then at

$t=1, \dfrac{dy}{dx} =$

$B:$ If

$x=3\cos\theta -\cos^{3}\theta, y=3\sin\theta-\sin^{3}\theta$ , then at

$\theta = \dfrac{\pi}{3}, \dfrac{dy}{dx} =$

$C:$ If

$x = a\left(t+ \dfrac{1}{t}\right), y = a\left(t-\dfrac{1}{t}\right)$ , then at

$t =2, \dfrac{dy}{dx} =$

$D:$ Derivative of

$\log(\sec x)$ with respect to

$\tan x$ at

$x = \dfrac{\pi}{4}$ is

$\\$
Arrangement of the above values in the increasing order is

0%
$C, D, A, B$
0%
$C, A, D, B$
0%
$A, B, D, C$
0%
$B, D, A, C$

Explanation

For

$A$

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times\dfrac{dt}{dx} =-\dfrac{c}{t^2}\times\dfrac{1}{c} = -\dfrac{1}{t^2}$

$t=1, \dfrac{dy}{dx} = -1$

For

$B$

$\dfrac{dy}{dx}=\dfrac{dy}{d\theta}\times\dfrac{d\theta}{dx}$

$= 3\cos\theta-3\sin^2\theta\times \cos\theta \times \dfrac{1}{-3\sin\theta+3\sin\theta\times \cos^2\theta}$

$=-\dfrac{3\cos\theta(1-\sin^2\theta)}{3\sin\theta(1-\cos^2\theta)}= -\dfrac{\cos^3\theta}{\sin^3\theta}$

$\theta = \dfrac{\pi}{3}, \dfrac{dy}{dx} = -\dfrac{1}{3\sqrt3}$

For

$C$

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times\dfrac{dt}{dx} =a\left(1+\dfrac{1}{t^2}\right)\times\dfrac{1}{a\left(1-\dfrac{1}{t^2}\right)}=\dfrac{t^2+1}{t^2-1}$

$t = 2, \dfrac{dy}{dx} = \dfrac{5}{3}$

For

$D$ ,

$\dfrac{dy}{dx}= \dfrac{\sec x\tan x}{\sec x}\times\dfrac{1}{\sec^2x} = \sin x\cos x$

$x = \dfrac{\pi}{4}, \dfrac{dy}{dx} = \dfrac{1}{2}$

So, the arrangement would be

$A, B, D, C$

$t\left( 1+{ x }^{ 2 } \right) =x$ and

${ x }^{ 2 }+{ t }^{ 2 }=y,$ then at

$x=2,$ the value of

$\displaystyle\frac{dy}{dx}$

0%
$\displaystyle\frac{488}{125}$
0%
$\displaystyle\frac{88}{125}$
0%
$\displaystyle\frac{101}{125}$
0%
None of these

Explanation

$\displaystyle { x }^{ 2 }+{ t }^{ 2 }=y\Rightarrow \frac { dy }{ dx } =2x+2t.\frac { dt }{ dx }$ ...(1)

$\displaystyle t=\frac { x }{ 1+{ x }^{ 2 } } \Rightarrow \frac { dt }{ dx } =\frac { 1-{ x }^{ 2 } }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } }$

Substitute these value of

$t$ and

$\displaystyle\frac{dt}{dx}$ in (1), we get

$\displaystyle \frac { dy }{ dx } =2x+\frac { 2x }{ 1+{ x }^{ 2 } } .\frac { 1-{ x }^{ 2 } }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } }$

On ptiing

$x=2$ in

$\displaystyle\frac{dy}{dx}$ , we get

$\displaystyle \frac { dy }{ dx } =\frac { 488 }{ 125 }$

$\mathrm{x}=\mathrm{a}\mathrm{t}^{2},\ \mathrm{y}=2\mathrm{a}\mathrm{t}$ , then

$\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}$ is

0%
$-\dfrac{1}{t^{2}}$
0%
$-\dfrac{1}{2at^{2}}$
0%
$-\dfrac{1}{t^{3}}$
0%
$-\dfrac{1}{2at^{3}}$

Explanation

We have

$x=at^2, y=2at$

$\Rightarrow \dfrac{dx}{dt}=2at=y, \dfrac{dy}{dt}=2a$
Thus

$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{2a}{y}$

$\Rightarrow \dfrac{d^2y}{dx^2}=2a\left(-\dfrac{1}{y^2}\right)\dfrac{dy}{dx}=2a\left(-\dfrac{1}{y^2}\right)\dfrac{2a}{y}=-\dfrac{(2a)^2}{y^3}=-\dfrac{(2a)^2}{(2at)^3}=-\dfrac{1}{2at^3}$

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[l \mathrm{o}\mathrm{g}(\mathrm{a}\mathrm{x})^{\mathrm{x}}]$ , where

$a$ is a constant, is equal to

0%
$1$
0%
$\log {ax}$
0%
$1/a$
0%
$\log {(ax)+1}$

Explanation

$\frac{d}{dx}\left ( x \log ax \right )= \log (ax) +\dfrac{ax}{ax}=1+\log(ax)$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 1 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 1 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.