If $$xlog_e(log_ex)-x^2+y^2=4(y>0)$$, then dy/dx at $$x=e$$ is equal to:

$$\dfrac{1+2e}{2\sqrt{4+e^2}}$$

If $${ (f\left( x \right) ) }^{ g(y) }={ e }^{ f\left( x \right)-g(y) }$$ then $$\frac { dy }{ dx } $$

$$\frac { { f }^{ \prime }(x)\log { f } (x) }{ { g }^{ 1 }(y){ (1+\log { f } (x)) }^{ 2 } } $$

$$\frac { { f }^{ \prime }(x)\log { f } (x) }{ g(y){ (1+\log { f } (x)) }^{ 2 } } $$

$$\frac { { f }(x)\log { f } (x) }{ { g }^{ \prime }(y){ (1-\log { f } (x)) }^{ 2 } } $$

$$2\frac { { f }^{ \prime }(x)\log { f } (x) }{ g(y){ (1+\log { f } (x)) }^{ 2 } } $$

MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 12

The differential of

$f(x)=\sqrt{\dfrac{2-x}{2+x}}$ at

$x=0$ and

$\delta x=0.15$ is

0%
$0.07$
0%
$0.075$
0%
$-0.075$
0%
$0.15$

Let

$f:[0,2]\rightarrow R$ be a twice differentiable function such that

$f"(x)>0$ , for all

$x\in (0,2)$ If

$\phi (x)=f(x)+f(2-x)$ , then

$\phi$ is:

0%
decreasing on $(0,2)$
0%
decreasing on $(0,1)$ and increasing on $(1,2)$
0%
increasing on $(0,2)$
0%
increasing on $(0,1)$ and decreasing on $(1,2)$

Explanation

$\phi(x)=f(x)+f(2-x)$

$\phi '(x)=f'(x)-f'(2-x)$ ...(1)
Since

$f"(x)>0$

$\Rightarrow f'(x) is increasing \forall x\in (0,2)$
case I when

$x>2-x \Rightarrow x>1$

$\Rightarrow \phi'(x) >0\forall x \in (1,2)$

$\therefore \phi (x) is increasing on (1,2)$
case II: when

$x<2-x \Rightarrow x<1$

$\Rightarrow \phi '(x) <0\forall x\in (0,1)$

$\therefore \phi (x)$ is decreasing on

$(0,1)$

$f(x)=|\cos x-\sin x|$ , then

$f'\left(\dfrac{\pi}{6}\right)$ equal to?

0%
$-\dfrac{1}{2}(1+\sqrt{3})$
0%
$\dfrac{1}{2}(1+\sqrt{3})$
0%
$-\dfrac{1}{2}(1-\sqrt{3})$
0%
$\dfrac{1}{2}(1-\sqrt{3})$

For the curve

$\sqrt{x}+\sqrt{y}=1,\ \dfrac{dy}{dx}$ at

$(1/4,1/4)$ is

0%
$1/2$
0%
$1$
0%
$-1$
0%
$2$

Explanation

We have,

$\sqrt{x}+\sqrt{y}=1$

On differentiating this with respect to x and we get,

$\dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=0$

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{\sqrt{x}}{\sqrt{y}}$

At the point $\left( \dfrac{1}{4},\dfrac{1}{4} \right)$

So,

$\dfrac{dy}{dx}=-\dfrac{\sqrt{\dfrac{1}{4}}}{\sqrt{\dfrac{1}{4}}}$

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

Hence, this is the answer.

$y^{2}=ae^{-2x}+(2/5)(cosx-2sinx)$ then

$y\dfrac{dy}{dx}+y^{2}+sinx$ is equal to

0%
-1
0%
1
0%
0
0%
none of these

Explanation

$\begin{array}{l} { y^{ 2 } }=a{ e^{ -2x } }+\frac { 2 }{ 5 } \left( { \cos x-2\sin x } \right) \\ 2y\frac { { dy } }{ { dx } } =-2a{ e^{ -2x } }+\frac { 2 }{ 5 } \left[ { -\sin x-2\cos x } \right] \\ y\frac { { dy } }{ { dx } } =-a{ e^{ -2x } }+\frac { 1 }{ 5 } \left[ { -\sin x-2\cos x } \right] \\ y\frac { { dy } }{ { dx } } +{ y^{ 2 } }+\sin x \\ =-a{ e^{ -2 } }-\frac { 1 }{ 5 } \left[ { \sin x+2\cos x } \right] +a{ e^{ -2x } }+\frac { 2 }{ 5 } \left( { \cos x-2\sin x } \right) +\sin x \\ =0 \\ Hence, \\ option\, C\, is\, correct\, answer. \end{array}$

$x+y=\sin(x+y)$ , then

$\dfrac{dy}{dx}=$

0%
$\dfrac{1}{2}$
0%
$0$
0%
$-1$
0%
$\dfrac{1}{3}$

The solution of

$\dfrac{dy}{dx} = 1+x+y+xy$ is

0%
$\log (1-y) = x+\dfrac{x^3}{2} + C$
0%
$\log (1+y) = x-\dfrac{x^2}{2} + C$
0%
$\log (1+y) = x+\dfrac{x^2}{2} + C$
0%
none of these

Explanation

$\dfrac {dy}{dx}=(1+x)(1+y)$

$\Rightarrow \dfrac {dy}{1+y}=(1+x)dx$

$\Rightarrow log (1+y)=x+\dfrac {x^2}{2}+C$

$y ( x ):$ Solution of a

$D.E.$

$( x \log x ) \dfrac { d y } { d x } + y = 2 x \log x,$

$( x , 1 )$

$y ( e ) = ? \quad x = e$

0%
$e$
0%
$0$
0%
$2$
0%
$2e$

Explanation

$\begin{array}{l} From\, \, the\, given\, ,\, data \\ { e^{ \int { \frac { { dx } }{ { x\log x } } } } }\alpha \left( { \frac { { dy } }{ { dx } } +\frac { y }{ { x\log x } } =2 } \right) \, \, ---\left( 1 \right) \\ Now,\, solving \\ I=\int { \frac { { dx } }{ { x\log x } } } \, \, \, let\, \, \, \log x=t,\, dt=\frac { { dx } }{ x } \\ I=\int { \frac { { dt } }{ t } } =\log t \\ hence,\, \left( 1 \right) \, becomes\, \\ \frac { d }{ { dx } } \left( { { e^{ \log t } }y } \right) =2{ e^{ \log t } } \\ \frac { d }{ { dx } } \left( { y\log x } \right) =2\log x \\ \int { d\left( { y\log x } \right) } =2\int { \log x } dx \\ y\log x=2\left( { x\log x-x } \right) \\ Hence,\, \\ y\left( x \right) =f\left( x \right) =\frac { { 2\left( { x\log x-x } \right) } }{ { \log x } } \\ Now, \\ y\left( e \right) =\frac { { 2e\left( { \log e } \right) -e } }{ { \log e } } =e \\ Hence,\, the\, option\, A\, is\, the\, \, correct\, answer. \end{array}$

$y^{2}+16=2xy$ , then which of the following is not the value of

$y'(5)$ ?

0%
$-\dfrac {2}{3}$
0%
$\dfrac {8}{3}$
0%
$\dfrac {5}{3}$
0%
None of these

$xlog_e(log_ex)-x^2+y^2=4(y>0)$ , then dy/dx at

$x=e$ is equal to:

0%
$\dfrac{e}{\sqrt{4+e^2}}$
0%
$\dfrac{1+2e}{2\sqrt{4+e^2}}$
0%
$\dfrac{2e-1}{2\sqrt{4+e^2}}$
0%
$\dfrac{1+2e}{\sqrt{4+e^2}}$

Explanation

When

$x=e$ ,then

$0-e^2+y^2=4,y=\sqrt{e^2+4}$

Differentiating with respect to x, We get:

$x.\dfrac{1}{\ell n x}.\dfrac{1}{x}+\ell n(\ell n x)-2x+2y.\dfrac{dy}{dx}=0$ at

$x=e$ we get

$1-2e+2y\dfrac{dy}{dx}=0\Rightarrow\dfrac{dy}{dx}=\dfrac{2e-1}{2y}$

$\Rightarrow\dfrac{dy}{dx}=\dfrac{2e-1}{2\sqrt{4+e^2}}$ as

$y(e)=\sqrt{4+e^2}$

For

$x > 1 ,$ if

$( 2 x ) ^ { 2 y } = 4 e ^ { 2 x - 2 y } ,$ then

$\left( 1 + \log _ { \mathrm { e } } 2 \mathrm { x } \right) ^ { 2 } \dfrac { \mathrm { dy } } { \mathrm { dx } }$ is equal to :

0%
$\log _ { e } 2 x$
0%
$\dfrac { x \log _ { e } 2 x + \log _ { e } 2 } { x }$
0%
$x \log _ { e } 2 x$
0%
$\dfrac { x \log _ { e } 2 x - \log _ { e } 2 } { x }$

Explanation

$( 2 x ) ^ { 2 y } = 4 e ^ { 2 x - 2 y }$

$2 y \ln 2 x = \ln 4 + 2 x - 2 y$

$\mathrm { y } = \dfrac { \mathrm { x } + \mathrm { \ln } 2 } { 1 + \mathrm { \ln } 2 \mathrm { x } }$

$y ^ { \prime } = \dfrac { ( 1 + \ln 2 x ) - ( x + \ln 2 ) \dfrac { 1 } { x } } { ( 1 + \ln 2 x ) ^ { 2 } }$

$\mathrm { y } ^ { \prime } ( 1 + \mathrm { \ln } 2 \mathrm { x } ) ^ { 2 } = \left[ \dfrac { \mathrm { x } \mathrm { \ln } 2 \mathrm { x } - \mathrm { \ln } 2 } { \mathrm { x } } \right]$

Let

$y={ t }^{ 10 }+1$ , and

$x={ t }^{ 8 }+1$ , then

$\dfrac { { d }^{ 2 }y }{ { dx }^{ 2 } }$ is

0%
$\dfrac { 5 }{ 2 } t$
0%
$20{ t }^{ 8 }$
0%
$\dfrac { 5 }{ 16{ t }^{ 6 } }$
0%
$\dfrac { 15 }{ 16{ t }^{ 6 } }$

Explanation

$y={t}^{10}+1$

$\dfrac{dy}{dt}=10{t}^{9}$

$x={t}^{8}+1$

$\dfrac{dx}{dt}=8{t}^{7}$

$\therefore \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{10{t}^{9}}{8{t}^{7}}=\dfrac{10}{8}{t}^{2}$

$\therefore \dfrac{{d}^{2}y}{d{x}^{2}}=\dfrac{10}{8}\dfrac{d}{dx}\left({t}^{2}\right)$

$=\dfrac{10}{8}\left(2t\dfrac{dt}{dx}\right)$

$=\dfrac{10}{8}\left(2t\times \dfrac {1}{8{t}^{7}}\right)$

$=\dfrac{10}{8}\times \dfrac {1}{4{t}^{6}}$

$=\dfrac { 5 }{ 16{ t }^{ 6 } }$

Find

$f^{\prime} (3)$ if

$f(x)=x^3+5x^2-3x+5$

0%
$28$
0%
$54$
0%
$32$
0%
None

Explanation

$f(x)=x^3+5x^2-3x+5\\f^{\prime}(x)=3x^2+10x-3\\f^{\prime }(x)=3(3)^2+10(3)-3=54$

$y = \sin ^ { - 1 } x$ then

$\frac { d y } { d x }$ is equal to

0%
$\sec y$
0%
$\cos x$
0%
$\tan x$
0%
$1$

Explanation

$y=\sin ^{-1}x\\x=\sin y \\\dfrac{dx}{dx}=\dfrac{d}{dx}\sin y\\1=\cos y \dfrac{dy}{dx}\\\dfrac{dy}{dx}=\sec y$

$y=e^{\sin x }$ , then find

$\dfrac{dy}{dx}$

0%
$e^{\sin x}{\cos x}$
0%
$e^{\sin x}$
0%
$e^{\cos x}$
0%
$e^{\sin x \cos x}$

Explanation

Given,

$y=e^{\sin x}$

$\dfrac{dy}{dx}=\dfrac{d}{dx} e^{\sin x}$

$\dfrac{dy}{dx}=e^{\sin x}\dfrac d {dx} \sin x$ .....................(applying chain rule of differentiation)

$\dfrac{dy}{dx}=e^{\sin x}{\cos x}$

Hence, the answer is option

$A$

${ (f\left( x \right) ) }^{ g(y) }={ e }^{ f\left( x \right)-g(y) }$ then

$\frac { dy }{ dx }$

0%
$\frac { { f }^{ \prime }(x)\log { f } (x) }{ g(y){ (1+\log { f } (x)) }^{ 2 } }$
0%
$\frac { { f }^{ \prime }(x)\log { f } (x) }{ { g }^{ 1 }(y){ (1+\log { f } (x)) }^{ 2 } }$
0%
$\frac { { f }(x)\log { f } (x) }{ { g }^{ \prime }(y){ (1-\log { f } (x)) }^{ 2 } }$
0%
$2\frac { { f }^{ \prime }(x)\log { f } (x) }{ g(y){ (1+\log { f } (x)) }^{ 2 } }$

Explanation

Let

$y=f\left(x\right)\Rightarrow\,\dfrac{dy}{dx}={f}^{\prime}{\left(x\right)}$

Given:

${\left(f\left(x\right)\right)}^{g\left(y\right)}={e}^{f\left(x\right)-g\left(y\right)}$

Applying

$\log$ to both sides,we get

$\log{{\left(f\left(x\right)\right)}^{g\left(y\right)}}=\log{{e}^{f\left(x\right)-g\left(y\right)}}$

$\Rightarrow\,g\left(y\right)\log{\left(f\left(x\right)\right)}=f\left(x\right)-g\left(y\right)$

$\Rightarrow\,g\left(y\right)\log{\left(f\left(x\right)\right)}+g\left(y\right)=f\left(x\right)$

$\Rightarrow\,g\left(y\right)\left(1+\log{\left(f\left(x\right)\right)}\right)=f\left(x\right)$

$\Rightarrow\,g\left(y\right)=\dfrac{f\left(x\right)}{\left(1+\log{\left(f\left(x\right)\right)}\right)}$

Differentiating both sides w.r.t

$x$ we get

$\Rightarrow\,{g}^{\prime}{\left(y\right)}\dfrac{dy}{dx}=\dfrac{\left(1+\log{\left(f\left(x\right)\right)}\right){f}^{\prime}{\left(x\right)}-\dfrac{f\left(x\right)}{f\left(x\right)}{f}^{\prime}\left(x\right)}{{\left(1+\log{\left(f\left(x\right)\right)}\right)}^{2}}$

$\Rightarrow\,{g}^{\prime}{\left(y\right)}\dfrac{dy}{dx}=\dfrac{\left(1+\log{\left(f\left(x\right)\right)}\right){f}^{\prime}{\left(x\right)}-{f}^{\prime}\left(x\right)}{{\left(1+\log{\left(f\left(x\right)\right)}\right)}^{2}}$

$\Rightarrow\,{g}^{\prime}{\left(y\right)}\dfrac{dy}{dx}=\dfrac{{f}^{\prime}{\left(x\right)}+\log{f\left(x\right)}{f}^{\prime}{\left(x\right)}-{f}^{\prime}{\left(x\right)}}{{\left(1+\log{f\left(x\right)}\right)}^{2}}$

$\Rightarrow\,{g}^{\prime}{\left(y\right)}\dfrac{dy}{dx}=\dfrac{\log{f\left(x\right)}{f}^{\prime}{\left(x\right)}}{{\left(1+\log{f\left(x\right)}\right)}^{2}}$

$\therefore\,\dfrac{dy}{dx}=\dfrac{{f}^{\prime}{\left(x\right)}\log{f\left(x\right)}}{{g}^{\prime}{\left(y\right)}{\left(1+\log{f\left(x\right)}\right)}^{2}}$

The derivative of

$\tan^{-1} \left (\dfrac {\sin x - \cos x}{\sin x + \cos x}\right )$ , with respect to

$\dfrac {x}{2}$ , where

$\left (x \epsilon \left (0, \dfrac {\pi}{2}\right )\right )$ is

0%
$\dfrac {1}{2}$
0%
$\dfrac {2}{3}$
0%
$1$
0%
$2$

Explanation

Let

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

$z=\frac{x}{2}$

Numerator and denominator divides by cosx

$\Rightarrow y= \tan^{-1} \left (\dfrac {\tan x - 1}{\tan x + 1}\right ) \\ \Rightarrow y=tan^{-1} \left (\tan \left (x - \dfrac {\pi}{4}\right )\right) \because tan(A+B)=\frac{tanA+tanB}{1-tanA \times tanB}$

$\because x - \dfrac {\pi}{4}\epsilon \left (-\dfrac {\pi}{4}, \dfrac {\pi}{4}\right )$

$\therefore y = x - \dfrac {\pi}{4}$

differentiate w.r.t.x.

$\frac{d y}{d x}=1$

and

$\frac{d z}{d x}=\frac{1}{2}$
Since

$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$

$\Rightarrow \frac{dy}{d x}=2$

Hence , Option (D) is correct.

Differentiate the following function with respect to x.

$x^n\tan x$ .

0%
$x^{n-1}(n \tan x+x\sec x)$ .
0%
$x^{n-1}(n \tan x+x\sec^2x)$ .
0%
$x^{n-1}(n \tan x+\sec^2x)$ .
0%
$x^{n-1}(n \tan x+x\sec^{-2}x)$ .

Explanation

Given expression

$x^n \tan x$

differentiating w.r.t.

$x$ , we get

$\dfrac d{dx}(x^n \tan x)$

$=\dfrac d{dx}(x^n)\tan x +x^n \dfrac d{dx}\tan x$

$=nx^{n-1}\tan x+ x^n \sec ^2 x$

$= x^{n-1}(n\tan x+x\sec ^2 x)$

Differentiate the following function with respect to x.

$x^3e^x$ .

0%
$x^2e^x(x)$ .
0%
$e^x(3+x)$ .
0%
$x^2(3+x)$ .
0%
$x^2e^x(3+x)$ .

Explanation

Given expression

$x^3e^x$

differentiating w.r.t.

$x$ , we get

$\dfrac d{dx}x^3e^x$

$=x^3 \dfrac d{dx}e^x +\dfrac d{dx}(x^3)e^x$

$=x^3e^x+3x^2 e^x$

$= x^2e^x(x+3)$

Differentiate the following function with respect to x.

$(2x^2+1)(3x+2)$ .

0%
$9x^2+4x+3$ .
0%
$18x^2-8x-3$ .
0%
$9x^2-4x-3$ .
0%
$18x^2+8x+3$ .

Differentiate the following function with respect to x.

$\dfrac{x^3}{3}-2\sqrt{x}+\dfrac{5}{x^2}$ .

0%
$x^2-x^{-1/2}-5x^{-3}$ .
0%
$x^2-x^{-1/2}-10x^{-3}$ .
0%
$x^2-x^{-1/2}+5x^{-3}$ .
0%
$-x^2+x^{-1/2}+10x^{-3}$ .

Differentiate the following function with respect to x.

$x^2e^xlog x$ .

0%
$xe^x(x log x+2 log x)$ .
0%
$xe^x(1+2 log x)$ .
0%
$xe^x(1+x log x)$ .
0%
$xe^x(1+x log x+2 log x)$ .

Explanation

Given expression

$x^2e^x\log x$

differentiating w.r.t.

$x$ , we get

$\dfrac d{dx}(x^2e^x\log x)$

$=x^2e^x\dfrac d{dx}\log x+x^2\log x\dfrac d{dx}e^x+e^x \log x\dfrac d{dx}(x^2)$

$=x^2e^x\left(\dfrac 1x\right)+x^2e^x\log x+2xe^x\log x$

$=xe^x(1+x\log x+2\log x)$

Differentiate the following function with respect to x.

$3^x+x^3+3^3$ .

0%
$3^xlog 3+3x^2$ .
0%
$3^xlog 3+3x$ .
0%
$3^xlog 3+x^2$ .
0%
$xlog 3+3x^2$ .

Let

$x^k+y^k=a^k,(a,k > 0)$ and

$\dfrac{dy}{dx}+\left(\dfrac{y}{x}\right)^{1/3}=0$ , then

$k$ is :

0%
$\dfrac{1}{3}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{3}{2}$

Explanation

$x^k+y^k=a^k$

differentiating w.r.t. x

$kx^{k-1}+k.y^{k-1}\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\dfrac{kx^{k-1}}{k.y^{k-1}}$

$\dfrac{dy}{dx}=-\left(\dfrac{x}{y}\right)^{k-1}$

$\dfrac{dy}{dx}+\left(\dfrac{x}{y}\right)^{k-1}=0$

$k-1=\dfrac{-1}{3}$

$\boxed{k=\dfrac{2}{3}}$

Differentiate the following function with respect to x.

$\sin x\cos x$ .

0%
$2\cos 2x$ .
0%
$\dfrac {\cos 2x}{2}$ .
0%
$\cos 2x$ .
0%
$\cos x$ .

Explanation

Given expression

$\sin x\cos x$

differentiating w.r.t.

$x$ , we get

$\dfrac d{dx}(\sin x\cos x)$

$=\dfrac d{dx}(\sin x)\cos x+\dfrac d{dx}(\cos x)\sin x$

$=\cos ^2 x-\sin ^2 x$

$=\cos 2x$

$y \sqrt{x^{2}+1}=\log (\sqrt{x^{2}+1}-x),$ then

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

0%
0
0%
1
0%
2
0%
None of these

Explanation

$y \sqrt{x^{2}+1}=\log \{\sqrt{x^{2}+1}-x\}$

Differentiating both sides w.r.t.x we get

$\dfrac{d y}{d x} \sqrt{x^{2}+1}+y \dfrac{1}{2 \sqrt{x^{2}+1}} 2 x=\dfrac{1}{\sqrt{x^{2}+1}-x} \times\left\{\dfrac{1}{2} \dfrac{2 x}{\sqrt{x^{2}+1}}-1\right\}$

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

$\Rightarrow\left(x^{2}+1\right) \dfrac{d y}{d x}+x y+1=0$

Let

$u(x)$ and

$v(x)$ be differentiable functions such that

$\dfrac{u(x)}{v(x)}=7 .$ If

$\dfrac{u^{\prime}(x)}{v^{\prime}(x)}=p$ and

$\left(\dfrac{u(x)}{v(x)}\right)^{\prime}=q,$ then

$\dfrac{p+q}{p-q}$ has the value equal to

0%
$1$
0%
$0$
0%
$7$
0%
$-7$

Explanation

$u(x)=7 v(x) \Rightarrow u^{\prime}(x)=7 v^{\prime}(x) \Rightarrow p=7$ (given)

$\operatorname{Again} \dfrac{u(x)}{v(x)}=7 \Rightarrow\left(\dfrac{u(x)}{v(x)}\right)^{\prime}=0 \Rightarrow q=0$

$\text { Now } \dfrac{p+q}{p-q}=\dfrac{7+0}{7-0}=1$

$f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2)$ and

$g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)$
The value of

$f(3)$ is

0%
1
0%
0
0%
-1
0%
-2

Explanation

$f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2)$ and

$g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)$

$\text { Here put } g^{\prime}(1)=a, g^{\prime \prime}(2)=b \quad\quad\quad(1)$

$\text { Then } f(x)=x^{2}+a x+b, f(1)=1+a+b \Rightarrow f^{\prime}(x)=2 x+a$

$\quad f^{\prime \prime}(x)=2$

$\therefore g(x)=(1+a+b) x^{2}+(2 x+a) x+2=x^{2}(3+a+b)+a x\quad+2$

$\Rightarrow g^{\prime}(x)=2 x(3+a+b)+a \text { and } g^{\prime \prime}(x)=2(3+a+b)$

$\text { Hence, } g^{\prime}(1)=2(3+a+b)+a \quad\quad\quad(2)$

$g^{\prime \prime} (2)=2(3+a+b)\quad\quad\quad(3)$

From (1),(2) and (3) we have

$a=2(3+a+b)+a$

$\text { and } b=2(3+a+b)$

$\Rightarrow 3+a+b=0 \text { and } b+2 a+6=0$

Hence,

$b=0$ and

$a=-3 .$ So,

$f(x)=x^{2}-3 x$ and

$g(x)=-3 x+2$

$f(x)=x^{2}-3 x \Rightarrow f(3)=9-9=0$

${f}'({x})={g}({x})$ and

${g}'({x})=-{f}({x})$ for all

$x$ and

${f}(2)=4= {g}(2)$ , then

${f}^{2}(24)+{g}^{2}(24)$ is

0%
$32$
0%
$24$
0%
$64$
0%
$48$

Explanation

$f(x) = -g'(x)$ (given)
By multiply both sides with

$f'(x)$ , we get

$f(x)f'(x) = -g'(x)f'(x)$ ...(1)

$\because f'(x) = g(x)$

$\therefore$ Eqn (1) becomes

$f(x)f'(x) = -g(x)g'(x)$
Now integrating both sides, we get

$f^2(x) + g^2(x) = c$
Where,

$c$ is the constant of integration.
Now, it is given that

$f(2) = g(2) = 4$

$\therefore c = 32$

$\implies f^2(24) + g^2(24) = 32$

Hence, option A is correct.

Assertion(A): Let

${ f }({ x })$ be twice differentiable function such that

$f^{ '' }(x)=-{ f }({ x })$ and

$f^{ ' }(x)={ g }({ x })$ . lf

${ h }({ x })=[{ f }({ x })]^{ 2 }+[{ g }({ x })]^{ 2 }$ and

${ h }(1)=8$ , then

${ h }(2)=8$

Reason (R): Derivative of a constant function is zero.

0%
Both A and R are true R is correct reason of A
0%
Both A and R are true R is not correct reason of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

Let

$f(x)=c$

${f}'(x)=0$ So Reason is True

Now,

$h(x) = [f(x)]^2 + [g(x)]^2$

${h}'(x)=2f(x){f}'(x)+2g(x){g}'(x)$

$=2f(x)g(x)+2g(x){f}''(x)$ ...since

$f'(x) = g(x) \Rightarrow f''(x) = g'(x)$

$=2g(x)[f(x) + f''(x)]$

$=2g(x)(f(x)-f(x))$

$\Rightarrow {h}'(x)=0$
means

$h(x)=C$ (a constant function)
Given

$h(1)=8$

$\Rightarrow h(x)=8$
Hence,

$f(2)= 8$

So, both A and R are correct and R is the correct explanation of A.

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 12 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 12 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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