If $$y = x^{x^{x^{x^{.^{.^{\infty}}}}}}$$, then $$y' =$$

$$\dfrac {y^{2}}{1 - y\log x}$$

$$\dfrac {-y^{2}}{1 - y\log x}$$

If $$\int_{0}^{x}(t^{2}+2t+2)$$ dt, $$2\leq x\leq 4$$

The maximum value of f(x) is $$\dfrac{136}{3}$$

The minimum value of f(x) is 10

The maximum value of f(x) is 26

If $$x^{2} + y^{2} = a^{2}$$ and $$k=1/a$$, then $$k$$ is equal to ?

$$\dfrac { y\prime \prime }{ \sqrt { 1+y\prime } }$$

$$\dfrac { \left| y\prime \prime \right| }{ \sqrt { { \left( 1+{ y\prime }^{ 2 } \right) }^{ 3 } } }$$

$$\dfrac {2 y\prime \prime }{ \sqrt { 1+y\prime } }$$

$$\dfrac { y\prime \prime }{ 2\sqrt { { \left( 1+{ y\prime }^{ 2 } \right) }^{ 3 } } } $$

If $$x ^ { y } + y ^ { x } = a ^ { b }$$ then find that $$\dfrac { d y } { d x }$$

$$ -\dfrac{yx^{x-1}+y^x\log y}{x^y\log x+xy^{x-1}}$$

$$ \dfrac{yx^{x-1}+y^x\log y}{x^y\log x+xy^{x-1}}$$

$$ -\dfrac{yx^{x-1}+y^x\log y}{x^y\log x+xy^{x}}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 16

Given a function '

$g$ ' whcih has a derivative

$g'(x)$ for every real '

$x$ ' and which satisfy

$g'(0)=2$ and

$g(x+y)={e}^{y}. g(x)+{e}^{x}.g(y)$ for all

$x, y$ . Find

$g(x)$ .

0%
$2x{e}^{x}$
0%
$xe^x$
0%
$x+e^x$
0%
$x-e^x$

Explanation

We have

$g'(0) = 2$ and

$g(x + y) = e^y.g(x) + e^x.g(y)$

Substituting

$y = -x$ , we get

$g(0) = e^{-x}.g(x) + e^x.g(-x)$

$x = 0, g(0) = 2g(0) \Rightarrow g(0) = 0$ and thus,

$g(x) + e^{2x}.g(-x) = 0$ ... (i)

Now, differentiating the given equation w.r.t.

$x$ , we have

$g'(x+y) = e^y\cdot g'(x) + e^x\cdot g(y)$ ... (ii)

Also, differentiating the equation (i) w.r.t.

$x$ , we have

$g'(x) = e^{2x}(g'(-x) -2g(-x))$ ... (iii)

Substituting (iii) in (ii), we get

$g'(x+y) = e^ye^{2x}(g'(-x)-2g(-x)) + e^xg(y)$ ...(iv)

Assuming

$y=-x$ in (iv), we get

$g'(0) = e^xg'(-x) - e^xg(-x)$

$\Rightarrow 2e^{-x} = g'(-x) - g(-x)$

Assuming

$-x = X$ , we get

$2e^X = g'(X) - g(X)$

$\Rightarrow g'(X) = g(X) + 2e^X$

Using property of linear differential equation, we get

$g(X) =\cfrac{\displaystyle \int 2e^X\cdot e^{\int -1\ dX}}{\displaystyle e^{\int-1\ dX}}$

$\Rightarrow g(X) = 2Xe^X$

Hence,

$g(x) = 2xe^x$

Constant of integration is

$0$ as

$g(0) = 0$

Let

$f$ be a differential function such that

$f(x)=f(4-x)$ and

$g(x)=f(2+x)$ for all

$x\in R,$ then

0%
graph of $f(x)$ is symmetric about the line $x=2$
0%
$f'(2)=0$
0%
graph of $g(x)$ is symmetric about x-axis
0%
$g'(0)=0$

Explanation

Given

$f(x)=f(4-x)$ ....(1)

And

$g(x)=f(2+x)$ ,,,(2) Putting 2+x in place of

$x$ in (1),

We get

$f(2+x)=f(2-x)$ ........3

$\Rightarrow$ graph of

$f(x)$ is symmetric about the line

$x=2$

From (2) and (3),

$g(x)=f(2+x)$ ........4

Putting

$(-x)$ in place of

$x$ in (2),

we get

$g(-x)=f(2-x)$ .......5

From (3), (4) and (5)

$g(-x)=g(x),$

Therefore

$g(x)$ is an even funcction

Hence gragh of

$g(x)$ is symmetric about y-axis

Differentiating both sides of (3) w.r.t

$x$ , we get

$f'(2+x)=-f'(2-x)$

Putting

$x=0,$ we get

$f'(2)=0$

Again

$g(x)=g(-x)$

Differentiating w.r.t

$x$ we get

$g'(x)=-g'(-x)$

Putting

$x=0,$ we get

$2g'(0)=0\Rightarrow g'(0)=0$

Length of the subtangent at

$(x_l, y_l)$ on

$x^n y^m = a^{m+n}, m, n > 0,$ is

0%
$\dfrac{n}{m}x_l$
0%
$\dfrac{m}{n}|x_l|$
0%
$\dfrac{n}{m}|y_l|$
0%
$\dfrac{n}{m}|x_l|$

Explanation

The subtangent is given by

$\cfrac{y}{\tfrac{dy}{dx}}$

Here,

$x^ny^m = a^{m + n}$

$\therefore y^m = a^{m + n}x^{-n}$

$\therefore y =\left( a^{\tfrac{m + n}{m}}\right)\left(x^{\tfrac{-n}{m}}\right)$

$\Rightarrow \cfrac{dy}{dx} = \left(a^{\tfrac{m + n}{m}}\right) \times \left(\tfrac{-n}{m} \right)\times \left(x^{\tfrac{-n - m}{m}}\right)$

$\Rightarrow$ subtangent at

$(x_l,y_l) = a^{\tfrac{m + n}{m}}x_l^{\tfrac{-n}{m}} \times \cfrac{1}{\left(a^{\tfrac{m + n}{m}} \right)\times \left( \tfrac{-n}{m}\right) \times ( x_l^{\tfrac{-n - m}{m}})}$

$= \cfrac{-m}{n} \times x_l$

$= \cfrac{m}{n}|x_l|$ , since length cannot be negative

The slope(s) of common tangent(s) to the curves $\displaystyle y={ e }^{ -x }$ and $\displaystyle y={ e }^{ -x }\sin { x }$ can be -

0%
$\displaystyle -{ e }^{ -\tfrac{\pi}2 }$
0%
$\displaystyle -{ e }^{ -\pi }$
0%
$\displaystyle \frac { \pi }{ 2 }$
0%
$1$

Explanation

$\displaystyle y={ e }^{ -x }\ \ \&\ y={ e }^{ -x }\sin { x }$

$\displaystyle { y }^{ ' }=-{ e }^{ -x }...(i)\quad \& \quad { y }^{ ' }=-{ e }^{ -x }\left( \sin { x } -\cos { x } \right) ...(ii)$

equating (i) and (ii)

$\Rightarrow \displaystyle { e }^{ -x }\left( 1-\sin { x } +\cos { x } \right) =0$

$\displaystyle { e }^{ -x }\neq 0\quad \Rightarrow \quad 1-\sin { x } +\cos { x } =0$

$\displaystyle \Rightarrow \quad 2\cos ^{ 2 }{ \frac { x }{ 2 } } \quad =\quad 2\sin { \frac { x }{ 2 } } \cos { \frac { x }{ 2 } }$

$\displaystyle \Rightarrow \quad 2\cos { \frac { x }{ 2 } } \left( \sin { \frac { x }{ 2 } -\cos { \frac { x }{ 2 } } } \right) =0 \Rightarrow x=\frac { \pi }{ 2 } ,\pi$

Slope can be $\displaystyle -{ e }^{ -\tfrac{\pi}2 }$ and $\displaystyle -{ e }^{ -\pi }$

The derivative of

$\displaystyle (\tan x)^{x}$ is equal to-

0%
$\displaystyle x(\tan x)^{x-1}$
0%
$\displaystyle (\tan x)^{x}\left [ \sec x+\tan x \right ]$
0%
$\displaystyle (\tan x)^{x}\left [ x\sec x \csc x +\log \tan x\right ]$
0%
$\displaystyle(\tan x)^{x}\left [ \sec ^{2}x+x\tan x \right ]$

Explanation

$\displaystyle y=(\tan x)^{x}$

$\Rightarrow \log y = x \log \tan x$

$\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx}=\log { \tan { x } } +\frac { x\times 1 }{ \tan x } \times \sec ^{ 2 } x$

$\Rightarrow \displaystyle \frac{dy}{dx}=y\left [ \log \tan x+x\sec x \, cosec x \right ]$

$\Rightarrow \displaystyle \frac{dy}{dx}=(\tan x)^{x}\left [ \log \tan x+x\sec x\, cosec x \right ]$

$x=\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } }$ and

$y=\dfrac { 2at }{ 1+{ t }^{ 2 } }$ , then

$\dfrac { dy }{ dx }$ is equal to:

0%
$\dfrac { a\left( 1-{ t }^{ 2 } \right) }{ 2t }$
0%
$\dfrac { a\left( { t }^{ 2 }-1 \right) }{ 2t }$
0%
$\dfrac { a\left( { t }^{ 2 }+1 \right) }{ 2t }$
0%
$\dfrac { a\left( { t }^{ 2 }-1 \right) }{ t }$

Explanation

Given,

$x=\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } }$ and

$y=\dfrac { 2at }{ 1+{ t }^{ 2 } }$

On differentiating with respect to

$t$ , respectively, we get

$\dfrac { dx }{ dt } =\dfrac { \left( 1+{ t }^{ 2 } \right) \left( 0-2t \right) -\left( 1-{ t }^{ 2 } \right) \left( 0+2t \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } }$

$=\dfrac { -4t }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } }$

and

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

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

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

$\displaystyle e^{\sin (x^{2}+y^{2})}=\tan \frac{y^{2}}{4}+\sin ^{-1}x$ , then

$y' (0)$ can be-

0%
$\displaystyle \frac{1}{3\sqrt{\pi }}$
0%
$\displaystyle -\frac{1}{3\sqrt{\pi }}$
0%
$\displaystyle -\frac{1}{5\sqrt{\pi }}$
0%
$\displaystyle -\frac{1}{3\sqrt{5\pi }}$

Explanation

Given,

$\displaystyle e^{\sin \left ( x^{2}+y^{2} \right )}=\tan \frac{y^{4}}{4}+\sin ^{-1}x$
at x = 0

$\displaystyle \Rightarrow e^{\sin y^{2}}=\tan \frac{y^{2}}{4}\Rightarrow y=\pm \sqrt{\pi }\pm, \sqrt{5\pi },....... etc$
on differentiation

$\displaystyle e^{\sin \left ( x^{2}+y^{2} \right )}\left [ \cos \left ( x^{2}+y^{2} \right )\left ( 2x+2y\frac{dy}{dx} \right ) \right ]$ =

$\displaystyle \frac{2y}{4}\sec ^{2}\frac{y^{2}}{4}\frac{dy}{dx}+\frac{1}{\sqrt{1-x^{2}}}$
put x = 0

$\displaystyle \Rightarrow e^{\sin y^{2}}\left [ \cos y^{2}\left ( 2y\frac{dy}{dx} \right ) \right ]=\frac{y}{2}\sec ^{2}\frac{y^{2}}{4}\frac{dy}{dx}+1$
at x = 0, y =

$\displaystyle \sqrt{\pi }\Rightarrow \frac{dy}{dx}=\frac{-1}{3\sqrt{\pi }}$
at x = 0, y = -

$\displaystyle \sqrt{\pi }\Rightarrow \frac{dy}{dx}=\frac{1}{3\sqrt{\pi }}$
at x = 0, y =

$\displaystyle \sqrt{5\pi }\Rightarrow \frac{dy}{dx}=\frac{-1}{3\sqrt{5\pi }}$
at x = 0, y = -

$\displaystyle \sqrt{5\pi }\Rightarrow \frac{dy}{dx}=\frac{1}{3\sqrt{5\pi }}$

The solution of differential equation

$\displaystyle ydx+\left( x-{ y }^{ 2 } \right) dy=0$

0%
$\displaystyle { e }^{ \frac { y }{ x } }=\sin { x } +c$
0%
$\displaystyle y=cx\log { x }$
0%
$\displaystyle x=\frac { { y }^{ 2 } }{ 3 } +\frac { c }{ y }$
0%
$\displaystyle \cos { \left( \frac { y-2 }{ x } \right) } =a$

Explanation

$ydx + (x - y^2) dy = 0$

$ydx + xdy - y^2 dy = 0$

$d(xy) - y^2 dy = 0$

Integrating, we have

$xy - \dfrac{y^3}{3} = c$

This can also be written as

$x = \dfrac{y^3}{3} + \dfrac{c}{y}$

The derivative of

$y=x^{2^x}$ w.r.t x is :

0%
$x^{2^x}2^x\left(\displaystyle \frac{1}{x} + \ln x \ln 2 \right)$
0%
$x^{2^x}\left(\displaystyle\frac{1}{x} \ln x \ln 2\right)$
0%
$x^{2^x}2^x\left(\displaystyle\frac{1}{x} \ln x\right)$
0%
$x^{2^x}2^x\left(\displaystyle\frac{1}{x}+\frac{\ln x}{\ln 2}\right)$

Explanation

We have,

$y={ x }^{ { 2 }^{ x } }$

take logarithm on the both sides we get,

$\ln { y } ={ 2 }^{ x }\ln { x }$

diff. w.r.t.

$x$

$\cfrac { 1 }{ y } \cfrac { dy }{ dx } =\left[ { 2 }^{ x }\log { 2. } \ln { x } +\cfrac { { 2 }^{ x } }{ x } \right]$

$\cfrac { dy }{ dx } ={ 2 }^{ x }.{ x }^{ { 2 }^{ x } }\left[ \ln { 2 } .\ln { x } +\cfrac { 1 }{ x } \right]$

$={ 2 }^{ x }.{ x }^{ { 2 }^{ x } }\left[ \cfrac { 1 }{ x } + \ln x \ln 2 \right]$

$x=\sqrt{a^{\sin^{-1}t}}$ ,

$y=\sqrt{a^{\cos^{-1}t}}$ , then

$\dfrac{dy}{dx}=$ ________.

0%
$\dfrac{-y}{x}$
0%
$\dfrac{x}{y}$
0%
$\dfrac{y}{x}$
0%
$\dfrac{-x}{y}$

Explanation

$x = \sqrt{a^2{\sin^{-1} t}} ; y = \sqrt{a^{\cos^{-1} t}}$

$\Rightarrow\ x^2 = a^{\sin^{-1} t} ; y^2 = a^{\cos^{-1} t}$

$\Rightarrow\ \log(x^2) = (\sin^{-1} t) \log a$ ; .....

$(1)$

$\log (y^2) = (\cos^{-1} t) \log a$ ....

$(2)$

Adding eq.

$(1)$ and eq.

$(2)$ ;

$\Rightarrow\ \log (x^2) + \log(y^2) = (\log a) (\sin^{-1} t + \cos^{-1} t)$

$\Rightarrow\ \log (x^2 . y^2) = (\log a) \times \dfrac{\pi}{2}$

$\left [\because \sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2} \right ]$

$\Rightarrow\ x^2.y^2 = e^{\frac{\pi}{2} \log a}$

Differentiating both sides w.r.t

$x$

$\Rightarrow\ 2x. y^2 + 2y. x^2 \dfrac{dy}{dx} = 0$

$\Rightarrow\ y + x \dfrac{dy}{dx} = 0$

$\therefore\ \dfrac{dy}{dx} = \dfrac{-y}{x}$

$\Rightarrow\ x^2 = a^{\sin^{-1} t} ; y^2 = a^{\cos^{-1} t}$

$\Rightarrow\ \log(x^2) = (\sin^{-1} t) \log a$ ; .....

$(1)$

$\log (y^2) = (\cos^{-1} t) \log a$ ....

$(2)$

Adding eq.

$(1)$ and eq.

$(2)$ ;

$\Rightarrow\ \log (x^2) + \log(y^2) = (\log a) (\sin^{-1} t + \cos^{-1} t)$

$\Rightarrow\ \log (x^2 . y^2) = (\log a) \times \dfrac{\pi}{2}$

$\left [\because \sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2} \right ]$

$\Rightarrow\ x^2.y^2 = e^{\frac{\pi}{2} \log a}$

Differentiating both sides w.r.t

$x$

$\Rightarrow\ 2x. y^2 + 2y. x^2 \dfrac{dy}{dx} = 0$

$\Rightarrow\ y + x \dfrac{dy}{dx} = 0$

$\therefore\ \dfrac{dy}{dx} = \dfrac{-y}{x}$

$x = \sec \theta - \cos \theta$ and

$y = \sec^{n}\theta - \cos^{n}\theta$ , then

$\left (\dfrac {dy}{dx}\right )^{2}$ is equal to

0%
$\dfrac {n^{2}(y^{2} + 4)}{x^{2} + 4}$
0%
$\dfrac {n^{2}(y^{2} - 4)}{x^{2}}$
0%
$n\left (\dfrac {y^{2} - 4}{x^{2} - 4}\right )$
0%
$\left (\dfrac {ny}{x}\right )^{2} - 4$

Explanation

$\dfrac {dy}{d\theta} = n \sec^{n - 1}\theta \cdot \sec \theta \cdot \tan \theta - n\cdot \cos \theta^{n - 1} (-\sin \theta)$

$= n\tan \theta (\sec^{n} \theta + \cos^{n} \theta)$

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

$= \tan \theta(\sec \theta + \cos \theta)$

$\therefore \dfrac {dy}{dx} = \dfrac {n\tan \theta (\sec^{n}\theta + \cos^{n}\theta)}{\tan \theta(\sec \theta + \cos \theta)}$

$\left (\dfrac {dy}{dx}\right )^{2} = \dfrac {n^{2}(\sec^{n}\theta + \cos^{n}\theta)^{2}}{(\sec \theta + \cos \theta)^{2}}$

$= \dfrac {n^{2}\left \{(\sec^{n}\theta - \cos^{n}\theta)^{2} + 4\right \}}{(\sec \theta - \cos \theta)^{2} + 4} = \dfrac {n^{2}(y^{2} + 4)}{x^{2} + 4}$

$y = x^{x^{x^{x^{.^{.^{\infty}}}}}}$ , then

$y' =$

0%
$\dfrac {-y^{2}}{x(1 - y\log x)}$
0%
$\dfrac {y^{2}}{1 - y\log x}$
0%
$\dfrac {y^{2}}{x(1 - y\log x)}$
0%
$\dfrac {-y^{2}}{1 - y\log x}$

Explanation

Given

$y=x^{x^{x^{x^{.^\infty}}}}$

We can write

$y=x^y$

Taking log on both sides we get,

$log \:y=log (x^y)$

$\Rightarrow log\:y=y\: log \: x$

Differentiate both sides with respect to

$x$ , we get

$\Rightarrow \dfrac{d}{dx}(log \:y)=\dfrac{d}{dx}(y\: log\:x)$

$\Rightarrow \dfrac{1}{y} \dfrac{dy}{dx}=y \times \dfrac{d}{dx}(log\: x)+log \:x \times \dfrac{dy}{dx}$

$\Rightarrow \dfrac{1}{y} \dfrac{dy}{dx}=\dfrac{y}{x}+log \: x \times \dfrac{dy}{dx}$

$\Rightarrow \dfrac{dy}{dx} \left(\dfrac{1}{y}-log \: x \right)=\dfrac{y}{x}$

$\Rightarrow \dfrac{dy}{dx} \left(\dfrac{1-y\: log \: x}{y} \right)=\dfrac{y}{x}$

$\Rightarrow \dfrac{dy}{dx} =\dfrac{y^2}{x(1-y\: log\: x)}$

$\int_{0}^{x}(t^{2}+2t+2)$ dt,

$2\leq x\leq 4$

0%
The maximum value of f(x) is $\dfrac{136}{3}$
0%
The minimum value of f(x) is 10
0%
The maximum value of f(x) is 26
0%
None of these

$x\sqrt {1 + y} + y\sqrt {1 + x} = 0$ , then

$\dfrac {dy}{dx}$ is equal to

0%
$\cfrac{1}{(1+x)^{2}}$
0%
$\cfrac{-1}{(1+x)^{2}}$
0%
$\cfrac{1}{(1-x)^{2}}$
0%
None of these

Explanation

Given

$x\sqrt{1+y}+y\sqrt{1+x}=0$

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

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

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

$\cfrac{1+y}{y^{2}}=\cfrac{1+x}{x^{2}}$

$x^{2}+x^{2}y=y^{2}+y^{2}x$

$x^{2}-y^{2}=y^{2}x-x^{2}y$

$(x+y)(x-y)=xy(y-x)$

$\therefore x+y+xy=0\Rightarrow y=\cfrac{-x}{1+x}$

$(\because x\neq y)$

Differentiate on both sides w.r.t x

$1+\cfrac{\mathrm{d} y}{\mathrm{d} x}+x\cfrac{\mathrm{d} y}{\mathrm{d} x}+y=0$

$\therefore \cfrac{\mathrm{d} y}{\mathrm{d} x}=\cfrac{-(1+y)}{1+x}$

$=\cfrac{-\left ( 1-\cfrac{x}{1+x} \right )}{1+x}$

$=\cfrac{-1}{(1+x)^{2}}$

$\therefore \cfrac{\mathrm{d} y}{\mathrm{d} x}=\cfrac{-1}{(1+x)^{2}}$

$y^{y^{y^{....{^\infty}}}} = \log_e(x+\log_e(x+....))$ , then

$\dfrac{dy}{dx}$ at

$(x= e^2-2, y= \sqrt2)$ is

0%
$\dfrac{\log\left(\dfrac{e}{2}\right)}{2\sqrt2(e^2-1)}$
0%
$\dfrac{\log2}{2\sqrt2(e^2-1)}$
0%
$\dfrac{\sqrt2 \log\dfrac{e}{2}}{(e^2-1)}$
0%
None of these

Let

$y=x^{x^x}$ , then differentiate

$y$ w.r.t

$x$ .

0%
$x^{x^x}\left(\dfrac{1}{x}+\log x+(\log x)^2\right)$
0%
$x^{x^x}(x^x)\left(\dfrac{1}{x}+\log x-(\log x)^2\right)$
0%
$x^{x^x}(x^x)\left(\dfrac{1}{x}+\log x+(\log x)^2\right)$
0%
$x^{x^x}(x^x)\left(\dfrac{1}{x}-\log x-(\log x)^2\right)$

Explanation

$y=x^{x^x}$

diffrentiate it w.r.t.

$x$

$\dfrac{dy}{dx}=(e^{lnx^{x^x}})^{'}$

$\dfrac{dy}{dx}=(e^{x^x lnx})^{'}$

$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})^{'}$

$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}lnx)^{'}$

$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}(xlnx)^{'}ln x$ +

$\dfrac{e^{xlnx}}{x})$

$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}(xlnx)^{'}ln x$ +

$\dfrac{e^{xlnx}}{x})$

$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}(lnx+\dfrac{x}{x})ln x$ +

$\frac{e^{xlnx}}{x}$

$\dfrac{dy}{dx}=x^{x^{x}}$

$(x^x(lnx+1)lnx+\dfrac{x^x}{x}$

$\dfrac{dy}{dx}=x^{x^{x}}$

$(x^x)(\dfrac{1}{x}+lnx+(lnx)^2)$

so option C is correct.

Let

$y=f(x)$ be a differentiable curve satisfying

$\int _{ 2 }^{ x }{ f\left( t \right) dt+2=\dfrac { { x }^{ 2 } }{ 2 } + } \int _{ x }^{ 2 }{ { t }^{ 2 }f\left( t \right) dt }$ ,then

$\displaystyle \int _{ -\pi /4 }^{ \pi /4 }{ \dfrac { f\left( x \right) +{ x }^{ 9 }-{ x }^{ 3 }+x+1 }{ \cos ^{ 2 }{ x } } dx }$ equals-

0%
$0$
0%
$1$
0%
$2$
0%
$4$

Explanation

$\int_{2}^{x} f(t) dt + 2 = \dfrac{x^2}{2} + \int_{x}^{2} t^2 f(t) dt$

By differentiate on both sides, we get

$\implies f(x) + 0 = x - x^2 f(x)$

$\therefore f(x) = \dfrac{x}{x^2 + 1}$

$I = \displaystyle \int_{\dfrac{-\pi}{4}}^{\dfrac{\pi}{4}} \dfrac{(\dfrac{x}{x^2 + 1}) + x^9 - x^3 + x + 1}{\cos^2 x} dx$

$\dfrac{x}{x^2 + 1}, x^9, x^3, x$ are odd functions

$I = 0 + 0 + 0 + 0 + \displaystyle \int_{\dfrac{-\pi}{4}}^{\dfrac{\pi}{4}} \dfrac{1}{\cos^2 x} dx$

$I = \displaystyle \int_{\dfrac{-\pi}{4}}^{\dfrac{\pi}{4}} \sec^2 x dx$

$I = 2 \displaystyle \int_{0}^{\dfrac{\pi}{4}} \sec^2 x dx$

$I = 2 \left[\tan x\right]_{0}^{\dfrac{\pi}{4}}$

$I = 2 [1 - 0]$

$I = 2$ (Ans)

Option C is correct.

$\int \dfrac {\cos 4x+1}{\cos x-\tan x}dx=A \cos 4x+B;$ where

$A$ &

$B$ are constants, then

0%
$A=-1/4$ & $B$ may have any value
0%
$A=-1/8$ & $B$ may have any value
0%
$A=-1/2$ & $B=-1/4$
0%
$A=B=1/2$

Consider the following statements:

$S_1: \lim_\limits{x \to 0} \dfrac{[x]}{x}$ is an indeterminate form (where [.] denotes greatest integer function).

$S_2: \lim_\limits{x\to\infty}\dfrac{sin(3^x)}{3^x}=0$

$S_3: \lim_\limits{x \to \infty}\sqrt{\dfrac{x- sinx}{x+cos^2x}}$ does not exist.

$S_4: \lim_\limits{n\to \infty}\dfrac{(n+2)!+(n+1)!}{(n+3)! }(n \in N=0$
State, in order, whether

$S_1, S_2, S_3, S_4$ are true or false

0%
FTFT
0%
FTTT
0%
FTFF
0%
TTFT

$x^{2} + y^{2} = a^{2}$ and

$k=1/a$ , then

$k$ is equal to ?

0%
$\dfrac { y\prime \prime }{ \sqrt { 1+y\prime } }$
0%
$\dfrac { \left| y\prime \prime \right| }{ \sqrt { { \left( 1+{ y\prime }^{ 2 } \right) }^{ 3 } } }$
0%
$\dfrac {2 y\prime \prime }{ \sqrt { 1+y\prime } }$
0%
$\dfrac { y\prime \prime }{ 2\sqrt { { \left( 1+{ y\prime }^{ 2 } \right) }^{ 3 } } }$

Explanation

${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$

$2x+2y{ y }_{ 1 }=0$

${ y }_{ 1 }=\frac { -x }{ y }$ \\

$y{ y }_{ 1 }+x=0$

$y{ y }_{ 2 }+{ { y }_{ 1 } }^{ 2 }+1=0$

$y=-\frac { 1+{ { y }_{ 1 } }^{ 2 } }{ { y }_{ 2 } } \Longrightarrow (1)$

$k=\frac { 1 }{ a } =\left| \frac { 1 }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } \right| =\left| \frac { 1 }{ y\sqrt { 1+\frac { { x }^{ 2 } }{ { y }^{ 2 } } } } \right|$

$=\frac { 1 }{ y } \sqrt { 1+{ { y }_{ 1 } }^{ 2 } }$

$=\frac { { y }^{ \prime \prime } }{ \left( 1+{ { y }_{ 1 } }^{ 2 } \right) \sqrt { 1+{ { y }_{ 1 } }^{ 2 } } }$

$=\frac { \left| { y }^{ \prime \prime } \right| }{ { \left( 1+{ { y }_{ 1 } }^{ 2 } \right) }^{ \frac { 3 }{ 2 } } }$

$x ^ { y } + y ^ { x } = a ^ { b }$ then find that

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

0%
$-\dfrac{yx^{x-1}+y^x\log y}{x^y\log x+xy^{x-1}}$
0%
$\dfrac{yx^{x-1}+y^x\log y}{x^y\log x+xy^{x-1}}$
0%
$-\dfrac{yx^{x-1}+y^x\log y}{x^y\log x+xy^{x}}$
0%
None of these

Explanation

Let

$u=x^{y}$ and

$v=y^{x}$

$u=x^{y}$

$\log x=y\log x$

Differentiating we get

$\dfrac{1}{u}.\dfrac{du}{dy}=\dfrac{y}{x}+\log\dfrac{dy}{dx}$

$\dfrac{du}{dx}=u\left(\dfrac{y}{x}+\log x.\dfrac{dy}{dx}\right)$

$v=y^{x}$

$\log v=x\log y$

Differentiating we get

$\dfrac{1}{v}.\dfrac{dv}{dx}=\log y+\dfrac{x}{y}.\dfrac{dy}{dx}$

$\dfrac{dv}{dx}=v\left(\log y+\dfrac{x}{y}.\dfrac{dy}{dx}\right)$

$u+v=a^{2}$

Differentiating, we egt

$\dfrac{du}{dx}+\dfrac{dv}{dx}=0$

$\dfrac{x^{y}.y}{x}+x^{y}\log x.\dfrac{dy}{dx}+y^{x}\log y+y^{x}\dfrac{x}{y}\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}(x^{y}\log x+xy^{x-1})=-(yx^{y-1}+y^{x}\log y)$

$\dfrac{dy}{dx}=-\dfrac{yx^{y-1}+y^{x}\log y}{x^{y}\log x+xy^{x-1}}$

$\frac{d}{{dx}}\left( {\frac{{1 + {x^2} + {x^4}}}{{1 + x + {x^2}}}} \right) = ax + b$ , then

$\left( {a,b} \right) =$

0%
$\left( { - 1,2} \right)$
0%
$\left( { - 2,1} \right)$
0%
$\left( {2, - 1} \right)$
0%
$\left( {1,2} \right)$

Explanation

$\dfrac{d}{dx}\left\{\dfrac{1+{x}^{2}+{x}^{4}}{1+x+{x}^{2}}\right\}=ax+b$

$\dfrac{d}{dx}\left\{\dfrac{1+2{x}^{2}+{x}^{4}-{x}^{2}}{1+x+{x}^{2}}\right\}=ax+b$

$\Rightarrow\,\dfrac{d}{dx}\left\{\dfrac{{\left(1+{x}^{2}\right)}^{2}-{x}^{2}}{1+x+{x}^{2}}\right\}=ax+b$

$\Rightarrow\,\dfrac{d}{dx}\left\{\dfrac{\left(1+{x}^{2}+x\right)\left(1+{x}^{2}-x\right)}{1+x+{x}^{2}}\right\}=ax+b$

$\Rightarrow\,\dfrac{d}{dx}\left(1+{x}^{2}-x\right)=ax+b$

$\Rightarrow\,2x-1=ax+b$

$\Rightarrow\,a=2,\,b=-1$

$\therefore\,\left(a,b\right)=\left(2,-1\right)$

$f(x+y)=f(x)f(y)\forall\ x,y\ \in\ R$ and

$f'(0)=5,f(2)=6$ , then

$f'(2)$

0%
$20$
0%
$10$
0%
$30$
0%
$40$

$x \sin y=\sin (y+a)$ and

$\dfrac{dy}{dx}=\dfrac{A}{1+x^{2}-2 x \cos a}$ then the value of

$A$ is

0%
$2$
0%
$\cos a$
0%
$-\sin a$
0%
$-2$

Explanation

$\displaystyle x sin y = sin (y+a) ; \frac{dy}{dx} = \frac{A}{1+x^{2}-2x\,cos\,a}$

L.H.S

$\displaystyle x sin\,y = sin(y+a)$

$\displaystyle \Rightarrow x\, sin\, y = sin\, a\, cos\, y+cos\,a \,sin\,y$

$\displaystyle sin\,y (a-cos\,a) = cos\,y sin\,a$

$\displaystyle \frac{sin\,y}{cos\,y} = \frac{sin\,a}{x-cos\,a}$

$\displaystyle tan\,y = \frac{sin\,a}{x-cos\,a};y = tan^{-1}(\frac{sin\,a}{x-cos\,a})$

$\displaystyle \therefore \frac{dy}{dx} = \frac{1}{1+(\frac{sin\,a}{x-cos\,a})^{2}}\times [\frac{(x-cosa)(0)(sina)(1-0)}{(x-cosa)^{2}}]$

$\displaystyle \Rightarrow \frac{(x-cos\,a)^{2}}{(x^{2}+cos^{2}a+sin^{2}a-2x\,cos\,a)}$

$\displaystyle \Rightarrow \frac{-sin}{1+x^{2}-2xcos\,a}$

RHS

$\displaystyle \Rightarrow \frac{A}{1+x^{2}-2x\,cos\,a}$

$\displaystyle\therefore A = -sin\,a,$ option C is correct

1126104_1140161_ans_a8085044ee054e7a9b93a1d896a09505.jpg

Let

$g(x)$ be a continuous function for all

$x$ , and

$f(x)=f(\alpha)+(x-\alpha).g(x)\ \forall \ x\ =\epsilon \ R$ . Then;

0%
$f(x)$ is necessarily differentiable at $x=\alpha$
0%
$f(x)$ is not necessarily differentiable at $x=\alpha$
0%
$f(x)$ is not necessarily continuous at $x=\alpha$
0%
$None\ of\ these$

For the function,

$f(x) = (x - \frac{1}{x})^2$ , the first derivative with respect to x is

0%
$2 (x - \frac{1}{x^3})$
0%
$2 (x - \frac{1}{x})$
0%
$2 (x + \frac{1}{x^2})$
0%
$2 (x - \frac{1}{x^2})$

$y = y ( x )$ is an implicit function of

$x$ given by

$y \cos x + x \cos y = \pi,$ the

$y ^ { \prime \prime } ( 0 )$ is equal to

0%
$\pi$
0%
$- \pi$
0%
$0$
0%
$2 \pi$

$x^{m}y^{n}=(x+y)^{m+n}$ , then

$xy^{3}=y$ , if

$\dfrac {x}{y} \neq \dfrac {m}{n}$

0%
True
0%
False

The value of

$\int _{ 0 }^{ \left[ x \right] }{ \left( x-\left[ x \right] \right) dx\quad is\left( \left[ . \right] denotes\quad greatest\quad integer\quad function \right) }$

0%
$\left[ x \right]$
0%
$2\left[ x \right]$
0%
$\dfrac { \left[ x \right] }{ 2 }$
0%
$3\left[ x \right]$

Let

$f ' (x)= -f(x)$ , where f(x) is a continous double differentiable function and

$g(x)=f '(x)$ . If

$F(x)=\left[ f\left( \dfrac { x }{ 2 } \right) \right] ^{ 2 }+\left[ g\left( \dfrac { x }{ 2 } \right) \right] ^{ 2 }$ and F(5)=5,then F(10) =

0%
0
0%
5
0%
10
0%
25

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 16 - 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 16 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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