MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 13

Statement I: lf

$f(\theta)=\cos\theta_{1}.\cos\theta_{2}\ldots.\cos\theta_{n}$ , then the value of

$\displaystyle\tan\theta_{1}+\tan\theta_{2}+\ldots+\tan\theta_{n}=-\frac{f'(\theta)}{f(\theta)}$
Statement II: Differential coefficient of the function

$f(g(x))$ w.r.t function

$g(x)$ is

$f'(g(x))$ .

Which of the above statements is/are correct?

0%
Statement $I$ only
0%
Statement $II$ only
0%
Both $I$ and $II$
0%
Neither $I$ nor $II$

Explanation

$f(\theta) = \cos\theta_1\cdot\cos\theta_2\cdots\cos\theta_n$

Differentiating both sides,

$f'(\theta)=-\sin\theta _{1}\cos\theta_{2}\cdots\cos\theta_{n} -\cos\theta_{1}\sin\theta_{2}\cdots\cos\theta_{n}\cdots\cdots -\cos\theta_{1}\cos\theta_{2}\cdots sin\theta_n$

Dividing

$f'(\theta)$ by

$f(\theta)$ , we get

$\dfrac{f'(\theta)}{f(\theta)}=-\tan\theta _{1} - \tan\theta _{2} - \tan\theta _{3}-\cdots-\tan\theta_n$

$-\dfrac{f'(\theta)}{f(\theta)}=\tan\theta_{1} +\tan\theta _{2}+\tan\theta _{3}-\cdots\cdots-\tan\theta_n$

II:

$\dfrac{d(f(g(x)))}{d(g(x))}=\dfrac{d(f(g(x)))}{d(x)}\times\dfrac{d(x)}{d(g(x))} = \dfrac{f'(g(x))g'(x)}{g'(x)} = {f}'(g(x))$

Both I and II are correct.

Assertion (A) If

$\sin(\mathrm{x}+\mathrm{y})=\log_{e}(\mathrm{x}+\mathrm{y})$ , then

$\displaystyle \frac{dy}{dx}=-1$

Reason (R): The derivative of an odd function is always an even function

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

Explanation

Differentiating the expression we get

$\cos(x+y).(1+y')=(1+y')/(x+y)$

$\displaystyle \Rightarrow \left( 1+y'\right) \left(\cos \left( x+y \right) - \frac{1}{x+y} \right)=0$

$\Rightarrow y'=-1$ (since

$\displaystyle \cos \left (x+y \right) \neq \frac{1}{x+y}$ when

$\sin (x+y) = \log_{e}(x+y)$ as seen from graph)
Hence, the assertion is true.
Now, for an odd function

$f(x)=-f(-x)$
Differentiating the expression we get

$f'(x)=(-1).f'(-x).(-1)$
Thus we get

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

$\Rightarrow$ differentiation of an odd function is an even function.
Hence reason is also true. But it is not an explanation for the assertion.

$y=\cos 2x\cos 3x$ , then

$y_n$ is equal to

$\\$
Where,

$y_n$ denotes the

$n^{th}$ derivative of

$y$ .

0%
$6^{\mathrm{n}}\displaystyle \cos\left(2x+\frac{\mathrm{n}\pi}{2}\right)\cos\left(3x+\frac{\mathrm{n}\pi}{2}\right)$
0%
$\displaystyle \frac{1}{2}\left [5^{\mathrm {n}}\cos\left ( \frac{\mathrm {n}\pi }{2}+5x \right )+\cos\left ( \frac{\mathrm{n}\pi }{2}+x \right ) \right ]$
0%
$\displaystyle \frac{1}{2}\left[5^{\mathrm{n}}\sin\left(5x+\frac{\mathrm{n}\pi}{2}\right)+\sin\left(x+\frac{\pi}{2}\right)\right]$
0%
$0$

Explanation

$y=\displaystyle \frac{1}{2}(2\cos 2x \cos 3x)$

$y=\displaystyle \frac{1}{2}(\cos 5x+\cos x)$

$y_1=\displaystyle \frac{1}{2}(-5\sin 5x-\sin x)$

$y_1=\displaystyle \frac{1}{2}\left ( 5\cos\left ( \frac{\pi }{2}+5x \right )+\cos\left ( \frac{\pi }{2}+x \right ) \right )$

$y_2=\displaystyle \frac{1}{2}\left ( -5^{2}\sin\left ( \frac{\pi }{2}+5x \right )-\sin\left ( \frac{\pi }{2}+x \right ) \right )$

$y_2=\displaystyle \frac{1}{2}\left ( 5^{2}\cos(\pi +5x)+\cos(\pi +x) \right )$

$y_3=\displaystyle \frac{1}{2}\left ( 5^{3}\cos\left ( \frac{3\pi }{2}+5x \right )+\cos\left ( \frac{3\pi }{2}+x \right ) \right )$

It is observed that with each consecutive derivative,

$\dfrac{\pi}{2}$ can be added to the argument of both terms in the bracket for generalization.

$\therefore y_n=\displaystyle \frac{1}{2}\left ( 5^{n}\cos\left ( \frac{n\pi }{2}+5x \right )+\cos\left ( \frac{n\pi }{2}+x \right ) \right )$

$f(x)=|x-1|+|x-3|$ then

$f^{'}(2)=$

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

Explanation

$f(x) = |x-1|+|x-3| =\begin{cases} -(x-1)-(x-3), x\le 1\\(x-1)-(x-3), 1< x\le 3\\x-1+(x-3), x>3\end{cases}$

$\Rightarrow f(x) =\begin{cases} 4-2x, x\le 1\\2, 1< x\le 3\\2x-4, x>3\end{cases}$

$\therefore f'(x) =\begin{cases} -2, x\le 1\\0, 1< x\le 3\\2, x>3\end{cases}$
Hence

$f'(2) =0$

Statement I: If

${x}= {e}^{ {y}+ {e}^{ {y+e}^{ {y}+\ldots\infty}}}$ , then

$\dfrac{ {d} {y}}{ {d} {x}}=\dfrac{ {x}}{1- {x}}$

Statement II: If

${y}=| \cos {x}|+|\sin {x}|$ , then the value of

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

${x}=\dfrac{2\pi}{3}$ is

$\left( \dfrac{\sqrt{3}-1}{2}\right )$

Which of the above statement(s) is/are correct?

0%
Statement $I$ only
0%
Statement $II$ only
0%
Both $I$ and $II$
0%
Neither $I$ nor $II$

Explanation

Statement I:

$\displaystyle x= e^{ \displaystyle y+e^{ \displaystyle y+e^{ \displaystyle y+ \cdots \infty }}}$

$\Rightarrow \displaystyle x=e^{\displaystyle y+x}$
Differentiating with respect to 'x', we get

$\displaystyle 1=e^{ \displaystyle y+x}(1+{y}')$

$\Rightarrow \displaystyle \frac{1}{x}-1={y}'$ (since

$\displaystyle e^{\displaystyle x+y}=x$ )
Hence, statement I is not correct.
Statement II:

$y = \left| \cos x \right| + \left| \sin x \right |$
In the neighborhood of

$\displaystyle \frac {2 \pi}{3}$ ,

$\cos x$ is negative and

$\sin x$ is positive.
Hence,

$\left| \cos x \right| = - \cos x$ and

$\left| \sin x \right| = \sin x$ .

$y=-\cos x + \sin x$ as

$\displaystyle x \rightarrow \frac{2\pi }{3}$

$\Rightarrow {y}'=\sin x+\cos x$

$\Rightarrow \displaystyle { y'| }_{ x=\tfrac { 2\pi }{ 3 }} =\frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ 2 }$

Hence, only statement II is correct.

$\displaystyle \int _{ 0 }^{ { y } } \frac { 1 }{ \sqrt { 1+9{ u }^{ 2 } } } du=u$ , then

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

0%
$\sqrt{1+9{y}^{2}}$
0%
$\displaystyle \frac{1}{\sqrt{1+9{y}^{2}}}$
0%
$9{y}$
0%
$9{y}^{2}$

Explanation

Use Newton-leibnitz rule to differentiate the given integral with respect to

$u$ .

$\displaystyle \frac { 1 }{ \sqrt { 1+9{ y }^{ 2 } } } \frac { dy }{ du } =1\\ \displaystyle \frac { dy }{ du } =\sqrt { 1+9{ y }^{ 2 } }$
Differentiating again with respect to

$u$

$\displaystyle \frac { { d }^{ 2 }y }{ d{ u }^{ 2 } } =\frac { 18y.\frac { dy }{ du } }{ 2\sqrt { 1+9{ y }^{ 2 } } } \\ \displaystyle \frac { { d }^{ 2 }y }{ d{ u }^{ 2 } } =\frac { 18y.\sqrt { 1+9{ y }^{ 2 } } }{ 2\sqrt { 1+9{ y }^{ 2 } } } \\ \displaystyle \frac { { d }^{ 2 }y }{ d{ u }^{ 2 } } =9y$
Hence, option C is correct.

$g(x+y)=g(x)+g(y)+3xy(x+y)\:\forall\:x,\:y\epsilon R$ and

$g^\prime(0)=-4$ .For which of the following values of

$x$ is

$\sqrt{g(x)}$ not defined?

0%
$[-2,0]$
0%
$[-2,\infty]$
0%
$[-1,1]$
0%
none of these

Explanation

$\displaystyle g'(x) = \lim_{h\to 0} \frac{g(x+h)-g(x)}{h}=\lim_{h\to 0} \frac{g(x)+g(h)+3hx(x+h)-g(x)}{h}$

$\displaystyle \Rightarrow g'(x) = \lim_{h\to 0} \frac{g(h)+3hx(x+h)}{h} = 3x^2+g'(0) = 3x^2-4$
Now

$\sqrt{g(x)}$ will not defined when

$g'(x) < 0$

$\Rightarrow 3x^2-4< 0\Rightarrow x \in \left(-\cfrac{2}{\sqrt{3}}, \cfrac{2}{\sqrt{3}} \right)$

Let

${f}({x})$ be a polynomial of degree two which is positive for all

$x \in R$ . lf

${g}({x})={f}({x})+{f}'({x})+{f}''(x)+xf'''({x})+{x}^{2}{f}^{{iv}}({x})$ , then for any real

$x$

0%
$g(x)<0$
0%
$g(x)>0$
0%
$g(x)=0$
0%
$g(x)\geq$ 0

Explanation

$f(x)$ is a polynomial of degree

$=2$

$f(x)> 0\ \ \ \forall\ x\in R$

$f^{'''}(x)=0$ and

$f^{iv}(x)=0$
Let

$f(x)=ax^{2}+bx+c$

$f^{'}(x)=2ax+b$

$f^{''}(x)=2a$

$g(x)=ax^{2}+(2a+b)x+(c+b+2a)$
Now, discriminant of

$g(x) = (b+2a)^2-4a(b+c+2a)$

$= -4a^2+(b^2-4ac) <0$ ........

$[f(x)$ is positive for all real

$x]$
Therefore,

$g(x) > 0$ for all real

$x$ .

The

$n^{th}$ derivative of

$h(x)=e^{3x+5}x^{2}$ at

$x=0$ is

0%
$e^{5}3^{n-2}n(n-1)$
0%
$e^{5}3^{n+2}n(n-1)$
0%
$e^{5}3^{n}n(n-1)$
0%
$e^{5}3^{n-2}n(n+1)$

Explanation

Given:

$h(x) = e^{3x+5}x^2$

Differentiating

$h(x)$ both sides, we get

$h'(x)=3e^{3x+5}x^{2}+2xe^{3x+5}$

$h'(0) = 0$

Again on differentiating

$h'(x)$ , we get

$h''(x)=3h'(x)+6x e^{3x+5}+2e^{3x+5}$

$h''(x)=3h'(x)+3(h'(x)-h(x))+2e^{3x+5}$

$h''(x)=6h'(x)-3 h(x)+2e^{3x+5}$

$h''(0)=2e^{5}$

Differenting

$h''(x)$ , we get

$h'''(x)=6h''(x)-3h'(x)+6e^{3x+5}$

$h'''(0)=18e^{5}$
... and so on.

According to the pattern, the value of

$n^{th}$ derivative of

$h(x)$ at

$x=0$ is

$h^{n}(0)=e^{5}3^{n-2}n(n-1)$

Hence, option A.

Given that $f (x)$
is a differentiable function of $x$ and that $f(x)$ . $f (y)$ = $f (x)$ + $f (y)$ + $f (xy) -2$ and that
$f (2) =5$ .

Then $f (3)$ is equal to?

0%
$6$
0%
$24$
0%
$15$
0%
$19$

Explanation

Given,

$f\left( x \right) .f\left( y \right) =f\left( x \right) +f\left( y \right) +f\left( xy \right) -2\quad -(1)$

$f\left( 2 \right) =5$

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

Equation

$1 \Rightarrow$

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

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

$\therefore f\left( 3 \right) ={ x }^{ 2 }+1={ 3 }^{ 2 }+1=10$

$\displaystyle \mathrm{f}(\mathrm{x})=log\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)$ and

$\displaystyle \mathrm{g}(\mathrm{x})=\frac{3\mathrm{x}+\mathrm{x}^{3}}{1+3\mathrm{x}^{2}}$ then

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}(\mathrm{f}(\mathrm{g}(\mathrm{x})))$ equals

0%
$- f^{'}(x)$
0%
$-f(x)$
0%
$3(f(x))^{2}$
0%
$3f^{'}(x)$

Explanation

Given

$\displaystyle \mathrm{f}(\mathrm{x})=log\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)$ and

$\displaystyle \mathrm{g}(\mathrm{x})=\frac{3\mathrm{x}+\mathrm{x}^{3}}{1+3\mathrm{x}^{2}}$

$f(g(x))=\log \left ( \dfrac{1+3x^{2}+3x+x^{3}}{1-3x^{2}-3x-x^{3}} \right )$

$=\log \left ( \dfrac{(1+x)^{3}}{(1-x)^{3}} \right )$

$=3 \log \left ( \dfrac{1+x}{1-x} \right )$

$f(g(x))=3 f(x) \Rightarrow\dfrac{d f(g(x))}{dx}=\dfrac{d \ (3f(x))}{dx}=3f^{'}(x)$

Let

$\mathrm{f}(\mathrm{x})$ be a function satisfying

$f(x+y)=\mathrm{f}(\mathrm{x})\mathrm{f}(\mathrm{y})$ for all

$x,y$

$\in \mathrm{R}$ and

$\mathrm{f}(\mathrm{x})=1+\mathrm{x}\mathrm{g}(\mathrm{x})$ , where

$\displaystyle\lim_{x\rightarrow 0}\mathrm{g}(\mathrm{x})=1$ , then

$\mathrm{f}'(\mathrm{x})$ is equal to

0%
$x\mathrm {g(x)}$
0%
$\mathrm{g'(x)}$
0%
$\mathrm{f(x)}$
0%
$0$

Explanation

$\displaystyle\mathrm {f'(x)}=\lim_{h \rightarrow 0}\frac{ \mathrm{f (x+h) - f(x)} }{h}$

$\displaystyle =\lim_{h\to 0}\frac{\mathrm {f(x) f(h)- f(x)}}{h}$

$\displaystyle = \mathrm{f(x)}\lim_{h\to 0}\frac{\mathrm {f(h)-1}}{h}$

$\displaystyle =\mathrm{ f(x)}\lim_{h\to 0}\mathrm{ g(h)}$ , $( \because \mathrm{f(x)}= 1+ \mathrm{x g(x)}\; )$

$= \mathrm{f(x)}$ ( $\displaystyle \because \lim_{x \rightarrow 0} \mathrm{g(x)} =1$ )

$\Rightarrow \displaystyle \mathrm{ f'(x)}= \mathrm {f(x)}$

Suppose for a differentiable function

$f,\>f(0)=0,\>f(1)=1,\>f'(0)=4=f'(1)$ .
If

$g(x)= f(e^{x})e^{f(x)}$ , then

$g'(0)$ is equal to

0%
$4$
0%
$8$
0%
$2$
0%
$0$

Explanation

$g(x) = f(e^x)e^{f(x)}$

Differentiating both sides, we get

$g'(x)=f'(e^x)\cdot e^x\cdot e^{f(x)}+f(e^x)\cdot e^{f(x)}\cdot f'(x)$

$g'(0)=f'(1)\cdot e^{f(0)}+f(1)\cdot e^{f(0)}\cdot f'(0)$

$g'(0)=4+4 = 8$

Assertion (A): lf

$f(x)=\cos^{2}x+\cos^{2}\left(x+\dfrac{\pi}3\right)- \cos x \cos \left(x+\dfrac{\pi}3\right)$ then

$f'(x)=0$

Reason(R): Derivative of constant function is zero

0%
Both A & R are true, R is correct explanation for A
0%
Both A & R are true,R is not correct explanation for A
0%
A is true but R is false
0%
A is false but R is true

Explanation

$f(x)=\cos ^{2}x+\cos ^{2}(x+\cfrac{\pi }{3})-2\cos x \cos (x+\cfrac{\pi }{3})+\cos x \cos (x+\cfrac{\pi }{3})$

$=(\cos x - \cos (x+\cfrac{\pi }{3}))^{2} + \cos x \cos (x+\cfrac{\pi }{3})$

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

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

$=\cfrac{\cos^2x}4 + \cfrac{3\sin^2x}4 +\cfrac{\sqrt3}2 \sin x\cos x + \cos x\left(\cfrac{\cos x}2 - \cfrac{\sqrt3}2\sin x\right)$

$f(x)=\cfrac{3}{4}$

$f{}'(x)=0$

$y^2=p(x)$ , a polynomial of degree

$3$ , then

$\displaystyle 2\frac{d}{dx}\left(y^3\dfrac{d^2y}{dx^2}\right)$ , is equal to

0%
$p'''(x)+p'(x)$
0%
$p''(x)+p'''(x)$
0%
$p(x)p'''(x)$
0%
constant

Explanation

$y=\sqrt{p(x)}$

$\dfrac{dy}{dx}=\dfrac{1}{2}\dfrac{p'(x)}{\sqrt{p(x)}}$

$\dfrac{d^{2}y}{dx^{2}}=\dfrac{1}{2}\left ( \dfrac{p''(x)\sqrt{p(x)}-\dfrac{(p'(x))^{2}}{2\sqrt{p(x)}}}{(p(x))} \right )$

$y^{3}\dfrac{d^{2}y}{dx^{2}}=\dfrac{2p^{''}(x)p(x)-(p^{'}(x))^{2}}{4}$

$2\dfrac{d}{dx}\left (y^3 \dfrac{d^{2}y}{dx^{2}} \right )=\dfrac{1}{2}\left(2p'''(x)p(x)+2p''(x)p'(x)-2p'(x)p''(x)\right)$

$2\dfrac{d}{dx}\left(y^3\dfrac{d^2y}{dx^2}\right)=p^{'''}(x)p(x)$

$\mathrm{f}(\mathrm{x})$ is a quadratic expression which is positive for all real vaues of

$\mathrm{x}$ and

$\mathrm{g}(\mathrm{x})=\mathrm{f}(\mathrm{x})+\mathrm{f}'(\mathrm{x})+\mathrm{f}''(\mathrm{x})$ then for any real value of

$\mathrm{x}$

0%
$\mathrm{g}(\mathrm{x})<0$
0%
$\mathrm{g}(\mathrm{x})>0$
0%
$\mathrm{g}(\mathrm{x})=0$
0%
$\mathrm{g}(\mathrm{x})\underline{>}0$

Explanation

Let

$f(x) =ax^2+bx+c$
According to the given condition

$a>0, b^2-4ac <0$ .........(i)

$\therefore g(x) = ax^2+bx+c+2ax+b+2a=ax^2+(b+2a)x+(b+c+2a)$
Now discriminant of

$g(x)$ is

$D=(b+2a)^2-4a(2a+c+b)=b^2+4a^2+4ab-4ab-4ac-8a^2=(b^2-4ac)-4a^2< 0$ using (i)
Hence

$g(x)> 0$ for any

$x \in R$

Let

$f$ be a twice differentiable function such that

$f''\left( x \right) =-f\left( x \right)$ and

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

$h'\left( x \right) ={ \left[ f\left( x \right) \right] }^{ 2 }+{ \left[ g\left( x \right) \right] }^{ 2 },h\left( 1 \right) =6$ and

$h(0)=4$ then

$h(4)$ is equal to?

0%
$16$
0%
$12$
0%
$13$
0%
None of these

Explanation

Given

$h'\left( x \right) ={ \left[ f\left( x \right) \right] }^{ 2 }+{ \left[ g\left( x \right) \right] }^{ 2 }$

Differentiating both side w.r.t

$x$ , we get

$h''\left( x \right) =2f\left( x \right) f'\left( x \right) +2g\left( x \right) g'\left( x \right) \\ =2f\left( x \right) g\left( x \right) +2g\left( x \right) g''\left( x \right) \quad \quad \quad \left[ \because f'\left( x \right) =g\left( x \right) \right] \\ =2f\left( x \right) g\left( x \right) -2g\left( x \right) f\left( x \right) =0\quad \quad \quad \quad \left[ \because f''\left( x \right) =-f\left( x \right) \right]$

Thus

$h'(x)=k,$ a constant for all

$x\in R$ .

Hence

$h(x)=ax+b,$ so that form

$h(0)=4,$ we get

$b=4$

and from

$h(1)=6$ we get

$a=2$

Therefore

$h(4)=12$

$y=x^{\displaystyle x^{\displaystyle x^{\displaystyle \dots^{\displaystyle\infty}}}}$ , find

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

0%
$\displaystyle\frac{y^2}{x(1-y\log{x})}$
0%
$\displaystyle\frac{y}{x(1-\log{x})}$
0%
$\displaystyle\frac{y^2}{x(y-\log{x})}$
0%
None of these

Explanation

The given function may be written as,

$y=x^{\displaystyle y}$
Taking

$\log$ on both sides,

$\log{y}=\log { { x }^{ y } }$

$\log{y}=y\log{x}$
Differentiate w.r. to

$x$

$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =y\frac { d }{ dx } \left( \log { x } \right) +\log { x } \frac { d }{ dx } y$

$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =\frac { y }{ x } +\log { x } .\frac { dy }{ dx }$

$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } -\log { x } .\frac { dy }{ dx } =\frac { y }{ x }$

$\displaystyle \frac { dy }{ dx } \left( \frac { 1 }{ y } -\log { x } \right) =\frac { y }{ x }$

$\displaystyle \frac { dy }{ dx } \left( \frac { 1-y\log { x } }{ y } \right) =\frac { y }{ x }$

$\therefore \displaystyle \frac { dy }{ dx } =\frac { { y }^{ 2 } }{ x\left( 1-y\log { x } \right) }$

${ e }^{ y }+xy=e$ then at

$x=0$ ,

$\displaystyle \frac{d^2y}{dx^2}=e^{-\lambda,}$ then numerical quantity

$-\lambda$ should be equal to

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

If $e^y+xy=$ e then

$e^y+xy=e$

On substituting $x=$ 0, we get $e^y=e$

$y = 1$ when $x = 0$

On differentiating the relation (i) we get

$e^y\dfrac{dy}{dx}+1.y+x.\dfrac{dy}{dx}=0$

On substituting, $x = 0$ , $y = 1$ we get

$e^y(\dfrac{dy}{dx})+1=0\Rightarrow\dfrac{dy}{dx}=\dfrac{-1}{e}$ on differentiating solution (ii) we get

$e^{y}\left( \dfrac{dy}{dx} \right)^{2}+e^{y}\dfrac{d^{2}y}{dx^{2}}+\dfrac{dy}{dx}+\dfrac{dy}{dx}+x\dfrac{d^{2}y}{dx^{2}}=0$

On substituting

$x = 0$ , $y = 1, \dfrac{dy}{dx}=\dfrac{-1}{e} we get$

$=\dfrac{d^2y}{dx^2}=\dfrac{1}{e^2}\rightarrow \lambda=2$

Hence, option 'A' is correct.

Derivative of

$({\log{x}})^{\displaystyle\cos{x}}$ with respect to

$x$ is

0%
$\displaystyle({\log{x}})^{\displaystyle\cos{x}}\left[\frac{\cos{x}}{x\log{x}}-\sin{x}\log{(\log{x})}\right]$
0%
$\displaystyle({\log{x}})^{\displaystyle\cos{x}}\left[\frac{\cos{x}}{x\log{x}}-\cos{x}\log{(\log{x})}\right]$
0%
$\displaystyle({\log{x}})^{\displaystyle\sin{x}}\left[\frac{\sin{x}}{x\log{x}}-\cos{x}\log{(\log{x})}\right]$
0%
None of these

Explanation

$y={ \log { x } }^{ \cos { x } }$
Taking log on both sides, we get

$\log { y } =\log { \left( { \log { x } }^{ \cos { x } } \right) }$

$\log { y } =\cos { x } .\left[ \log { \left( \log { x } \right) } \right]$
Differentiate w.r. to

$x$

$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =\cos { x } \frac { d }{ dx } \left[ \log { \left( \log { x } \right) } \right] +\left[ \log { \left( \log { x } \right) } \right] \frac { d }{ dx } \cos { x }$

$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =\cos { x } .\frac { 1 }{ \log { x } } \frac { d }{ dx } \left( \log { x } \right) -\left[ \log { \left( \log { x } \right) } \right] \sin { x }$

$\displaystyle \frac { dy }{ dx } =y\left[ \cos { x.\frac { 1 }{ x\log { x } } - } \left[ \log { \left( \log { x } \right) } \right] \sin { x } \right]$

$\therefore \displaystyle \frac { dy }{ dx } ={ \log { x } }^{ \cos { x } }\left[ \cos { x.\frac { 1 }{ x\log { x } } - } \left[ \log { \left( \log { x } \right) } \right] \sin { x } \right]$

Let

$\mathrm{f}$ :

$(-1,1 )\rightarrow \mathrm{R}$ be a differentiable function with

$\mathrm{f}(\mathrm{0})=-1$ and

$\mathrm{f}'(\mathrm{0})=1$ . Let

$\mathrm{g}(\mathrm{x})=[\mathrm{f}(2\mathrm{f}(\mathrm{x})+2)]^{2}$ . Then

$\mathrm{g}'(\mathrm{0})=$

0%
$-4$
0%
$0$
0%
$-2$
0%
$4$

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

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

0%
$\displaystyle -\frac{y}{x}$
0%
$\displaystyle\frac{y}{x}$
0%
$\displaystyle\frac{x}{y}$
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None of these

Explanation

We have

$x^m y^n=(x+y)^{m+n}$ .
Taking log on both sides, we get

$m\log{x}+n\log{y}=(m+n)\log{(x+y)}$
Differentiating both sides w.r.t. x, we get

$m\displaystyle\frac{1}{x}+n\frac{1}{y}\frac{dy}{dx}=\frac{m+n}{x+y}\frac{d}{dx}(x+y)$
or

$\displaystyle\left(\frac{m}{x}+\frac{n}{y}\right)\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right)$
or

$\displaystyle\left\{\frac{n}{y}-\frac{m+n}{x+y}\right\}\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}$
or

$\displaystyle\left\{\frac{nx+ny-my-ny}{y(x+y)}\right\}\frac{dy}{dx}=\left\{\frac{mx+nx-mx-my}{(x+y)x}\right\}$
or

$\displaystyle\frac{nx-my}{y(x+y)}\frac{dy}{dx}=\frac{nx-my}{(x+y)x}$
or

$\displaystyle\frac{dy}{dx}=\frac{y}{x}$

If y is a function of x and

$ln (x + y) - 2xy = 0,$ then the value of y'(0) is equal to

0%
1
0%
2
0%
3
0%
4

Explanation

$ln(x+y)-2xy=0$ ...(1)
put

$x=0$ , we get

$ln(0+y)-0=0$

$\Rightarrow y=1$

now, diff. (1)

$\dfrac{1+y'}{x+y}-2y-2xy'=0$

$y' [1 - 2x (x + y)] = 2y (x + y) - 1$

$y'=\dfrac{2y(x+y)-1}{1-2x(x+y)}$
now,

$y'(0)=\dfrac{2-1}{1-0}=1$

Ans: A

$f(x)={|x|}^{|\sin{x}|}$ , then

$\displaystyle f^\prime\left(-\frac{\pi}{4}\right)=$

0%
$\displaystyle {\left(\frac{\pi}{5}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{5}{\pi}}-\frac{2\sqrt{2}}{\pi}\right)$
0%
$\displaystyle {\left(\frac{\pi}{4}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{4}{\pi}}-\frac{2\sqrt{2}}{\pi}\right)$
0%
$\displaystyle {\left(\frac{\pi}{3}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{3}{\pi}}-\frac{3\sqrt{3}}{\pi}\right)$
0%
None of these

Explanation

In the neighbourhood of

$\displaystyle -\frac{\pi}{4}$ , we have

$f(x)={(-x)}^{\displaystyle-\sin{x}}=e^{\displaystyle-\sin{x}\log{(-x)}}$

$\therefore \displaystyle f^\prime(x)=e^{\displaystyle-\sin{x}\log{(-x)}}\left(-\cos{x}.\log{(-x)}-\frac{\sin{x}}{x}\right)$

$\displaystyle =(-x)^{\displaystyle-\sin{x}}\left(-\cos{x}.\log{(-x)}-\frac{\sin{x}}{x}\right)$

$\therefore \displaystyle f^\prime\left(-\frac{\pi}{4}\right)={\left(\frac{\pi}{4}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{-1}{\sqrt{2}}\log{\frac{\pi}{4}}+\frac{4}{\pi}\times\left(\frac{-1}{\sqrt{2}}\right)\right)$

$\displaystyle={\left(\frac{\pi}{4}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{4}{\pi}}-\frac{2\sqrt{2}}{\pi}\right)$

Let

$y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots\infty}}}$ , then

$\displaystyle\frac{dy}{dx}$ is equal to

0%
$\displaystyle\frac{1}{2y-1}$
0%
$\displaystyle\frac{x}{x+2y}$
0%
$\displaystyle\frac{1}{\sqrt{1+4x}}$
0%
$\displaystyle\frac{y}{2x+y}$

Explanation

$y^2=x+y$ or

$\displaystyle\frac{dy}{dx}=\frac{1}{2y-1}$
Also,

$\displaystyle y=\frac{x}{y}+1$ or

$\displaystyle\frac{dy}{dx}=\frac{y}{2x+y}$
Also,

$y^2-y-x=0$
or

$\displaystyle y=\frac{1\pm\sqrt{1+4x}}{2}$ (as

$y>0$ )
or

$\displaystyle y=\frac{1}{4}\frac{4}{\sqrt{1+4x}}=\frac{1}{\sqrt{1+4x}}$

Let

$\mathrm{f}(\mathrm{x})=\mathrm{x}+\tan^{-1}\mathrm{x}, \displaystyle \mathrm{g}(\mathrm{x})=\frac{\mathrm{x}}{1+\mathrm{x}^{2}}(\mathrm{x}>0)$ Then

0%
$\mathrm{f}(\mathrm{x})<\mathrm{g}(\mathrm{x}), \mathrm{x}>0$
0%
$\mathrm{f}(\mathrm{x})>\mathrm{g}(\mathrm{x}), \mathrm{x}>0$
0%
$\mathrm{f}(\mathrm{x})<\mathrm{g}(\mathrm{x})$ in $[1, \infty)$
0%
None of these

Explanation

Let

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

$\therefore$

$\displaystyle \phi'(\mathrm{x})=1+\frac{1}{1+\mathrm{x}^{2}}-\frac{1-\mathrm{x}^{2}}{(1+\mathrm{x}^{2})^{2}}=1+\frac{2\mathrm{x}^{2}}{(1+\mathrm{x}^{2})^{2}}>0 \ for$ all

$\mathrm{x}$

$\therefore$ f(x) is increasing in the domain

$\phi(0)=0$

$\therefore$

$\displaystyle \mathrm{x}+t\mathrm{a}\mathrm{n}^{-1}\mathrm{x}-\frac{\mathrm{x}}{1+\mathrm{x}^{2}}>0, \mathrm{x}>0$

$\Rightarrow \mathrm{f}(\mathrm{x})>g(\mathrm{x}), \mathrm{x}>0$

$x<1$ , then

$\displaystyle\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots\infty=$

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

Explanation

The given series is in the form

$\displaystyle\frac{{f_1}^\prime(x)}{{f_1}(x)}+\frac{{f_2}^\prime(x)}{{f_2}(x)}+\frac{{f_3}^\prime(x)}{{f_3}(x)}+\cdots\infty$
Then consider the product

${f_1}(x).{f_2}(x).{f_3}(x)\cdots{f_n}(x)$ . Now,

$(1-x)(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^{ n-1}})$ ---------(1)

$=(1-x^2)(1+x^2)(1+x^4)\cdots(1+x^{2^{ n-1}})$

$=(1-x^4)(1+x^4)\cdots(1+x^{2^{ n-1}})$

$\:\vdots$

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

$=1-x^{2^{ n}}$
Now, when

$n\rightarrow\infty$ ,

$x^{2^{ n-1}}\rightarrow 0\:(\because x<1)$ .
Therefore, taking

$n\rightarrow\infty$ , in (1), we get

$(1-x)(1+x)(1+x^2)(1+x^4)\cdots=1$
Taking logarithm, we get

$\log{(1-x)}+\log{(1+x)}+\log{(1+x^2)}+\log{(1+x^4)}+\cdots=0$
Differentiating w.r.t.

$x$ , we get

$\displaystyle -\frac{1}{1-x}+\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots=0$

or

$\displaystyle\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots\infty=\frac{1}{1-x}$

$g(x+y)=g(x)+g(y)+3xy(x+y)\:\forall\:x,\:y\epsilon R$ and

$g^\prime(0)=-4$ .The value of

$g^\prime(1)$ is

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

Explanation

$g(x+y)=g(x)+g(y)+3x^2y+3xy^2$ (1)
or

$g^\prime(x+y)=g^\prime(x)+6yx+3y^2$ (differentiating w.r.t.

$x$ keeping

$y$ as constant)
Put

$x=0$ . Then,
or

$g^\prime(y)=g^\prime(0)+3y^2$

$=-4+3y^2$
or

$g^\prime(y)=-4+3y^2$
or

$g(x)=-4x+x^3+c$
Now, put

$x=y=0$ in (1). Then

$g(0)=g(0)+g(0)+0$
or

$g(0)=0$
or

$g(x)=x^3-4x$

$g(x)=0\implies x^3-4x=0\implies x=0,\:2,\:-2$ .
Hence, three roots,

$\sqrt{g(x)}=\sqrt{x^3-4x}$ is defined if

$x^3-4x\ge 0$ or

$x\epsilon [-2,0]\cup[2,\infty]$ .
Also,

$g^\prime(x)=3x^2-4\implies g^\prime(1)=-1$ .

$y=e^{\displaystyle\sqrt{x}}+e^{\displaystyle -\sqrt{x}}$ , then

$\displaystyle\frac{dy}{dx}$ is equal to

0%
$\displaystyle\frac{e^{\displaystyle\sqrt{x}}-e^{\displaystyle -\sqrt{x}}}{2\sqrt{x}}$
0%
$\displaystyle\frac{e^{\displaystyle\sqrt{x}}-e^{\displaystyle -\sqrt{x}}}{2x}$
0%
$\displaystyle\frac{1}{2\sqrt{x}}\sqrt{y^2-4}$
0%
$\displaystyle\frac{1}{2\sqrt{x}}\sqrt{y^2+4}$

Explanation

We have,

$y=e^{\displaystyle\sqrt{x}}+e^{\displaystyle -\sqrt{x}}$
Differentiate w.r. to

$x$

$\displaystyle\frac{dy}{dx}=\frac{e^{\displaystyle\sqrt{x}}}{2\sqrt{x}}-\frac{e^{\displaystyle -\sqrt{x}}}{2\sqrt{x}}=\frac{e^{\displaystyle\sqrt{x}}-e^{\displaystyle -\sqrt{x}}}{2\sqrt{x}}$ ....................option (A)

We can write

${ e }^{\displaystyle \sqrt { x } }-{ e }^{\displaystyle -\sqrt { x } }=\sqrt { { \left( { e }^{\displaystyle \sqrt { x } }+{ e }^{\displaystyle -\sqrt { x } } \right) }^{ 2 }-4 }$

$\therefore \displaystyle \frac { dy }{ dx } \displaystyle =\frac{\sqrt{{\left(e^{\displaystyle\sqrt{x}}+e^{\displaystyle -\sqrt{x}}\right)}^2-4}}{2\sqrt{x}}=\frac{\sqrt{y^2-4}}{2\sqrt{x}}$ .................option (C)

$g(x+y)=g(x)+g(y)+3xy(x+y)\:\forall\:x,\:y\epsilon R$ and

$g^\prime(0)=-4$ .The number of real roots of the equation

$g(x)=0$ is

0%
$2$
0%
$0$
0%
$1$
0%
$3$

Explanation

Substituting

$y = 0$ , we get

$g(x) = g (0) + g(x)$

$\Rightarrow g(0) = 0$
Let

$y =c$ , where

$c$ is a constant. On substituting we get,

$g(x+c) = g(x) +g(c)+ 3xc(x+c)$
Differentiating with respect to

$x$ , we get,

$g'(x+c) = g'(x) + 6cx+3c^2$
Now, if we consider

$c \rightarrow 0$ , and take limits on both sides and divide by

$c$ , we get

$\lim _{ c\rightarrow 0 }{ \displaystyle \frac { g'(x+c)\quad -\quad g'(x) }{ c } \quad =\quad \lim _{ c\rightarrow 0 }{ \quad 6x\quad +\quad 3c } }$

$\Rightarrow g''(x) = 6x$

$\Rightarrow g'(x) = 3x^2 + k$ , where k is a constant.

$g'(0) = -4$

$\Rightarrow k = -4$

$\Rightarrow g'(x) = 3x^2-4$

$\Rightarrow g(x) = x^3 - 4x + m$ , where m is a constant

$g(0) = 0$

$\Rightarrow g(x) =x^3-4x = x(x^2-4)$

$g(x)=0$ , will have three real roots

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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