MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 15

Let

$h(x)$ be differentiable for all

$x$ and let

$f(x)=(kx+e^x) h(x)$ where

$k$ is some constant If

$h(0)=5$ ,

$h'(0)=-2$ and

$f'(0)=18$ then the value of

$k$ is equal to

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

Explanation

$f(x)=(kx+e^{x})h(x)$

$f'(x)=(kx+e^{x})h'(x)+h(x)(k+e^{x})$

So,

$f'(0)=h'(0)+h(0)(k+1)=-2+5(k+1)=18$

$4=k+1 => k=3$

$\sin(xy)+\cos(xy)=0$ then

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

0%
$\displaystyle \frac{y}{x}$
0%
$-\displaystyle \frac{y}{x}$
0%
$-\displaystyle \frac{x}{y}$
0%
$\displaystyle \frac{x}{y}$

Explanation

$\cos xy(xy'+y)-\sin xy(xy'+y)=0$

$(xy'+y)(\cos xy-\sin xy)=0$

$So, y'=\displaystyle \dfrac{-y}{x}$ (Since,

$\sin xy \neq \cos xy$ )

Given

$f(x)=-\displaystyle \frac{x^3}{3}+x^2\sin 1.5a-x\sin a.\sin 2a-5 \arcsin (a^2-8a+17)$ then :

0%
$f(x)$ is not defined at $x=\sin 8$
0%
${f}'(\sin 8)>0$
0%
$f'(x)$ is not defined at $x=\sin 8$
0%
${f}'(\sin 8)<0$

Explanation

Given:

$f(x) = \dfrac{-x^3}{3} + x^2 \sin 1.5\, a - x \sin a . \sin 2a - 5\, arc \sin (a^2 - 8a + 17)$

$f(x) = \dfrac{-x^3}{3} + x^2\sin 1.5\, a - x \sin a \times \sin 2a-5\sin^{-1}(a^2-8a+17)$

$\left(\sin c + \sin d = 2\sin \dfrac{c+d}{2} \cos \dfrac{c-d}{2}\right)$

For the function to be defined

$-1\le a^2 - 8a + 17 \le 1$

$-1 \le (a-4)^2 + 1 \le 1$

$-1-1\le (a-4)^2 \le 1-1$

$-2 \le (a-4)^2 \le 0$

$a -4 = 0 \Rightarrow a = 4$

Putting value of '

$a$ ' in the +ve

$f(x)$

$f(x) = \dfrac{-x^3}{3} + x^2 \sin 6 - x\sin 4 \times \sin 8 - 5\dfrac{\pi}{2}$

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

domain

$x \in R \because$ polynomial function

$f'(x) = -\dfrac{3x^2}{3} + 2x\sin 6 - \sin 4 \times \sin 8$

$= -x^2 + 2\sin 6x - \sin 4 \times \sin 8$

Putting

$x = \sin 8$

$f'(\sin 8) = -\sin^28 + 2\sin 6 \sin 8 - \sin 4 \sin 8$

$= \sin 8 [2\sin 6 - \sin 4 - \sin 8]$

$=\sin 8 [ 2\sin 6 - (\sin 4 + \sin 8)]$

$=\sin 8[2\sin 6 - 2\sin 6\cos 2]$

$= 2\sin 6 \sin 8 [ 1 - \cos 2]$

Value of

$\cos 2 = 0.999$

$\therefore 1 - \cos 2 = +ve$

$6\, rad \approx 344^o$

$344^o$ is in

$IV^{th}$ quadrant

$\therefore \sin 6$ is (-ve)

$8\, rad \approx 548.4^o$

$548.4^o$ is in

$II^{th}$ quadrant

$\therefore \sin 8$ is (+ve)

$\therefore f'(\sin 8) = 2\,\underbrace{\sin 6}_{-ve} \,\underbrace{\sin 8}_{+ve} \,\underbrace{(1-\cos 2)}_{+ve}$

$\therefore f'(\sin 8)$ is +ve

$\therefore f'(\sin 8) < 0$

$\to$ option D

The equation

${ \left( x-n \right) }^{ m }+{ \left( x-{ n }^{ 2 } \right) }^{ m }+{ \left( x-{ n }^{ 3 } \right) }^{ m }+...+{ \left( x-{ n }^{ m } \right) }^{ m }=0$

$(m$ is odd positive integer

$),$ has

0%
all real roots
0%
one real and $\left( n-1 \right)$ imaginary roots
0%
one real and $\left( m-1 \right)$ imaginary roots
0%
No real roots

Explanation

Let

$f\left( x \right) ={ \left( x-n \right) }^{ m }+{ \left( x-{ n }^{ 2 } \right) }^{ m }+{ \left( x-{ n }^{ 3 } \right) }^{ m }+...+{ \left( x-{ n }^{ m } \right) }^{ m }$

$\Rightarrow f'\left( x \right) =m{ \left( x-n \right) }^{ m-1 }+m{ \left( x-{ n }^{ 2 } \right) }^{ m-1 }+m{ \left( x-{ n }^{ 3 } \right) }^{ m-1 }+...+m{ \left( x-{ n }^{ m } \right) }^{ m-1 }$

Since

$m$ is odd,

$\left( m-1 \right)$ is even.

Therefore,

$f'\left( x \right) =0$ has no real root.

$\Rightarrow f\left( x \right) =0$ has one real and

$\left( m-1 \right)$ imaginary roots.

Suppose

$A=\displaystyle \frac{dy}{dx}$ when

$x^2+y^2=4$ at

$(\sqrt{2},\sqrt{2})$ ,

$B=\displaystyle \frac{dy}{dx}$ when

$\sin y+ \sin x=\sin x-\sin y$ at

$(\pi,\pi)$ and

$C=\displaystyle \frac{dy}{dx}$ when

$2e^{xy}+e^x e^y-e^x-e^y=e^{xy+1}$ at

$(1,1)$ , then

$(A+B+C)$ has the value equal to

0%
$-1$
0%
$e$
0%
$-3$
0%
$0$

Explanation

$A: \displaystyle \frac {d}{dx} (x^2+y^2=4)$ at

$(\sqrt 2, \sqrt 2))$

$2x+2y\displaystyle \frac { dy }{ dx } =0$

$\displaystyle \frac { dy }{ dx } =-\displaystyle \frac { x }{ y } =-\displaystyle \frac { \sqrt { 2 } }{ \sqrt { 2 } } =-1$

$B: \displaystyle \frac {d}{dx}(\sin y+\sin x=\sin x\cdot \sin y)$ at

$(\pi, \pi)$

$\cos { y\displaystyle \frac { dy }{ dx } } +\cos { x } =\sin { x } \cos { y\displaystyle \frac { dy }{ dx } +\sin { y } \cos { x } }$

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

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

$B=\displaystyle \frac { -1\left( 0-1 \right) }{ -1\left( 1-0 \right) } =-1$

$C: \displaystyle \frac {d}{dx}(2e^{xy}+e^xe^y-e^x-{ e }^{ y }=e^{xy+1})$ at

$(1, 1)$

$\left[ 2{ e }^{ xy }\left( x\displaystyle \frac { dy }{ dx } +y \right) \right] +{ e }^{ x }{ e }^{ y }\displaystyle \frac { dy }{ dx } +{ e }^{ y }{ e }^{ x }-{ e }^{ x } -{ e }^{ y }\frac { dy }{ dx } ={ e }^{ xy+1 }\left( x\displaystyle \frac { dy }{ dx } +y \right)$

$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { y{ e }^{ xy+1 }-2y{ e }^{ xy }-{ e }^{ x+y }+{ e }^{ x } }{ 2x{ e }^{ xy }+{ e }^{ x+y }-x{ e }^{ xy+1 }-{ e }^{ y } }$ ..........

since

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

$C=\displaystyle \frac { { e }^{ 2 }-2{ e }^{ 1 }-{ e }^{ 2 }+{ e }^{ 1 } }{ 2{ e }^{ 1 }+{ e }^{ 2 }-{ e }^{ 2 }-{ e }^{ 1 } }=-{1}$

$A+B+C=-1-1-1=-3$

$\displaystyle y=x^{\ln x^{\ln \left ( \ln x \right )}}$ , then

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

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

Explanation

Given

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

Taking log on both sides,

$\displaystyle lny=ln(lnx)(lnx)^{2}$

$\dfrac{y'}{y}=ln(lnx)(2lnx.\dfrac{1}{x})+(lnx)^{2}(\dfrac{1}{xlnx})$

$\Rightarrow \displaystyle \dfrac{y'}{y}=ln(lnx)(2lnx.\dfrac{1}{x})+(lnx)(\dfrac{1}{x})$

$\Rightarrow \displaystyle y'=y(lnx)(\dfrac{1}{x})(2ln(lnx)+1)$

$\Rightarrow \displaystyle y'=\dfrac{y}{x}(lnx^{lnlnx})(2ln(lnx)+1)$

We know,

$\Rightarrow \displaystyle lny=lnx^{lnlnx}.lnx$

So,

$y'=\dfrac{ylny}{xlnx}(2ln(lnx)+1)$

If for a continuous function

$f,f(0)=f(1)=0,f'(1)=2$ and

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

$g'(0)$ is equal to

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

Explanation

Given

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

Differentiating both sides w.r.t

$x$ , we get

$g'( x ) =f'( { e }^{ x } ) .{ e }^{ x } e^{f(x)}+f\left( { e }^{ x } \right) .{ e }^{ f\left( x \right) }.f'\left( x \right)$ ...(1)

Now, since we already know that

$f(0)=0,f'(1)=2$

Substitute

$x=0$ in (1), we get

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

$=f'(1)1.e^{0}+f(1)e^{0}f'(0)$

$=2.1.1+0=2$

Let

$f(x)$ be a polynomial function of degree

$2$ and

$f(x)>0$ for all

$x\in R$ . If

$g(x)=f(x)+f^{'}(x)+f^{"}(x)$ , then for any

$x$

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

Explanation

Given,

$g(x)=f(x)+f^{'}(x)+f^{"}(x)$ ....(1)
Let

$f(x)=ax^{2}+bx+c a\ne 0$

$f'(x)=2ax+b$

$\Rightarrow f''(x)=2a$

Substitute these values in (1)

$g(x)=ax^{2}+(2a+b)x+(c+b+2a)$ ....(2)
Since given,

$f(x)>0$

$\Rightarrow a>0$ and

$D<0$

$\Rightarrow a>0$ and

$b^{2}-4ac<0$

Now, we will find

$D$ for

$g(x)$ ,

$D=(2a+b)^{2}-4a(c+b+2a)$

$\Rightarrow D=(b^{2}-4ac)-4a^{2}$

$\Rightarrow D<0 (\because b^{2}-4ac <0$ and

$4a^{2}>0)$
So, for

$g(x), a>0$ and

$D<0$
Hence,

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

$\displaystyle f(x-y), f(x).f(y)$ and

$f(x+y)$ are in AP for all

$x, y$ and

$f(0)\neq 0$ , then

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

Explanation

Given,

$f(x-y), f(x),f(y)$ and

$f(x+y)$ are in AP

$\Rightarrow f(x-y)+f(x+y)=2f(x)f(y)$ ....(1)
Substitute

$x=0, y=0$ in (1)

$\Rightarrow 2f(0)=2f(0)f(0)$

$\Rightarrow f(0)(f(0)-1)=0$

$\Rightarrow f(0)=1 (\because f(0)\ne 0$ given)

Substitute

$x=0, y=x$ in (1), we get

$f(-x)+f(x)=2f(0)f(x)$

$\Rightarrow f(-x)=f(x)$ ....(2)

$\Rightarrow f(-2)=f(2), f(-3)=f(3)$
Differentiating (2) w.r.t x

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

$f'(-x)+f'(x)=0$

$f'(-2)+f'(2)=0$ and

$f'(-3)+f'(3)=0$

Consider the functions defined implicitly by the equation

$\displaystyle y^{3}-3y+x=0$ on various intervals in the real line. If

$\displaystyle x\epsilon \left ( -\infty , -2 \right )\cup \left ( 2, \infty \right ),$ the equation implicitly defines a unique real valued differentiable function

$\displaystyle y= f\left ( x \right ).$ If

$\displaystyle x\epsilon \left ( -2 , -2 \right )$ the equation implicitly defines a unique real valued differentiable function

$\displaystyle y= g\left ( x \right )$ satisfying

$\displaystyle g= g\left ( 0\right )= 0.$

$\displaystyle f\left ( -10\sqrt{2} \right )= 2\sqrt{2}$ then

$\displaystyle f''\left ( -10\sqrt{2} \right )=$

0%
$\displaystyle \frac{4\sqrt{2}}{7^{3}\cdot 3^{2}}$
0%
$\displaystyle -\frac{4\sqrt{2}}{7^{3}\cdot 3^{2}}$
0%
$\displaystyle \frac{4\sqrt{2}}{7^{3}\cdot 3^{3}}$
0%
$\displaystyle \frac{4\sqrt{2}}{7^{}\cdot 3^{}}$

Explanation

$y^{3}-3y+x=0$
Or

$3y^{2}.y'-3y'+1=0$
Or

$3y'(y^{2}-1)=-1$
Or

$y'=\dfrac{1}{3(1-y^{2})}$
Now

$y"=\dfrac{1}{3(1-y^{2})^{2}}[-2y'y]$

$=3(y')^{2}(-2y'y)$

$=-6yy'^{3}$
Now

$y'=\dfrac{1}{3(1-y^{2})}$ at

$y=2\sqrt{2}$

$=\dfrac{1}{3(-7)}$
Hence

$y"=-12\sqrt{2}\dfrac{1}{3^{3}.7^{3}}$

$=\dfrac{-4\sqrt{2}}{3^{2}.7^{3}}$

If the prime sign (') represents differentiation w.r.t.

$x$ and

$f^{'}=\sin x+\sin 4x.\cos x$ , then

$f^{'}\left ( 2x^{2}+\cfrac{\pi }{2} \right )$ at

$x=\sqrt{\dfrac{\pi }{2}}$ is equal to

0%
$0$
0%
$-1$
0%
$-2\sqrt{2\pi }$
0%
none of these

Explanation

Given

$f^{'}(x)=\sin x+\sin 4x.\cos x$

Now,

$f^{'}\left ( 2x^{2}+\dfrac{\pi }{2} \right )$

$=f^{ ' }\left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) \dfrac { d }{ dx } \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right)$

$=\left[ \sin \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) +\sin 4\left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) .\cos \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) \right] 4x$

$=\left[ \cos 2x^{ 2 }-\sin 8x^{ 2 }.\sin 2x^{ 2 } \right] 4x$

$=\left[ \cos \pi -\sin 4\pi .\sin \pi \right] 4\sqrt { \dfrac { \pi }{ 2 } }$

$=-2\sqrt{2\pi}$

$f(x)=x.\left | x \right |$ , then its derivative is:

0%
$2x$
0%
$-2x$
0%
$2|x|$
0%
${2}{x} sgn x$

Explanation

$x|x|=x.\sqrt{x^{2}}$
Its derivative will be,

$\displaystyle =\frac{x^{2}}{\sqrt{x^{2}}}+\sqrt{x^{2}}$

$\displaystyle =\frac{2x^{2}}{\sqrt{x^{2}}}$

$=2|x|$

$\sqrt{y+x}+\sqrt{y-x}=c$ (where

$c\neq 0$ ), then

$\displaystyle \frac{dy}{dx}$ has the value equal to

0%
$\displaystyle \frac{2x}{c^{2}}$
0%
$\displaystyle \frac{x}{y+\sqrt{y^{2}-x^{2}}}$
0%
$\displaystyle \frac{y-\sqrt{y^{2}-x^{2}}}{x}$
0%
$\displaystyle \frac{c^{2}}{2y}$

Explanation

$\displaystyle \sqrt{y+x}+\sqrt{y-x}=c$
Differentiating,

$\displaystyle \frac{y'+1}{2\sqrt{y+x}}+\frac{1-y'}{2\sqrt{y-x}} =0$

$\Rightarrow y'\sqrt{y-x}+\sqrt{y-x}=\sqrt{y+x}-y'\sqrt{y+x}$

$\displaystyle y'=\frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y-x}+\sqrt{y+x}}$ multiplying and dividing by

$\sqrt{y-x}-\sqrt{y+x},$

$\displaystyle =\frac{y+x+y-x-2\sqrt{y+x}\sqrt{y-x}}{y-x-y-x}$

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

$\displaystyle =\frac{y-\sqrt{y^{2}-x^{2}}}{x}$
multiplying and dividing by

$\sqrt{y-x}+\sqrt{y+x},$

$\displaystyle =\frac{y+x-y+x}{y-x+y+x+2\sqrt{y^{2}-x^{2}}}$

$\displaystyle =\frac{x}{y+\sqrt{y^{2}-x^{2}}}$
We know,

$\displaystyle y+x+y-x+2\sqrt{y^{2}-x^{2}}=c^{2}$

$\displaystyle y+\sqrt{y^{2}-x^{2}}=\frac{c^{2}}{2}$
so, the differentiation would be,

$\displaystyle \frac{2x}{c^{2}}$

$y=\left | \cos x \right |+\left | \sin x \right |$ then

$\frac{dy}{dx}$ at

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

0%
$\frac{1-\sqrt{3}}{2}$
0%
$0$
0%
$\frac{1}{2}\left ( \sqrt{3}-1 \right )$
0%
none of these

Explanation

$\displaystyle \dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{-cosxsinx}{|cosx|}+\dfrac{cosxsinx}{|sinx|}$

$\displaystyle \dfrac{\mathrm{d} y}{\mathrm{d} x}|_{x=\dfrac{2\pi }{3}}$

$\displaystyle =\dfrac{-cos\dfrac{2\pi }{3}sin\dfrac{2\pi }{3}}{|cos\dfrac{2\pi }{3}|}+\dfrac{cos\dfrac{2\pi }{3}sin\dfrac{2\pi }{3}}{|sin\dfrac{2\pi }{3}|}$

$\displaystyle =\dfrac{sin\dfrac{2\pi }{3}}{1}+\dfrac{cos\dfrac{2\pi }{3}}{1}$

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

$\displaystyle =\dfrac{\sqrt{3}-1}{2}$

$\displaystyle y=\left ( 1+\frac{1}{x} \right )^{x}+x^{1+\frac{1}{x}}.$
Differentiate

0%
$\displaystyle \left ( 1+\frac{1}{x} \right )^{x}\left [ \log \left ( 1+\frac{1}{x} \right )-\frac{1}{x+1} \right ]+x^{1+1/x}\left [ \frac{x+1-\log x}{x^{2}} \right ].$
0%
$\displaystyle \left ( -1+\frac{1}{x} \right )^{x}\left [ \log \left ( 1+\frac{1}{x} \right )-\frac{1}{x+1} \right ]+x^{1+1/x}\left [ \frac{x+1+\log x}{x^{2}} \right ].$
0%
$\displaystyle \left ( 1+\frac{1}{x} \right )^{x}\left [ \log \left ( 1+\frac{1}{x} \right )-\frac{1}{x+1} \right ]+x^{1-1/x}\left [ \frac{x+1-\log x}{x^{2}} \right ].$
0%
$\displaystyle \left ( -1+\frac{1}{x} \right )^{x}\left [ \log \left ( 1+\frac{1}{x} \right )-\frac{1}{x+1} \right ]+x^{1-1/x}\left [ \frac{x+1+\log x}{x^{2}} \right ].$

Explanation

Let $y = y_1+y_2$
$\Rightarrow \cfrac{dy}{dx}=\cfrac{dy_1}{dx}+\cfrac{dy_2}{dx}$
Now $y_1 = \left(1+\frac{1}{x}\right)^x \Rightarrow \log y_1 =x \log\left(1+\frac{1}{x}\right)$

Differentiating both side w.r.t $x$

$\Rightarrow \cfrac{1}{y_1}\cfrac{dy_1}{dx} = \log(1+\frac{1}{x})+x.\cfrac{1}{1+\frac{1}{x}}.\left(\frac{-1}{x}^2\right)$

$\Rightarrow \cfrac{dy_1}{dx} = y_1\left[\log(1+\frac{1}{x})-\cfrac{1}{1+x}\right]$
And $y_2 = x^{1+\frac{1}{x}} \Rightarrow \log y_2 = (1+\frac{1}{x}) \log x$

Differentiating both sides,

$\Rightarrow \cfrac{1}{y_2}\cfrac{dy_2}{dx} = (1+\frac{1}{x})\cfrac{1}{x}+(\log x)(\frac{-1}{x}^2)$

$\Rightarrow \cfrac{dy_2}{dx} = y_2\left[\cfrac{(1+x)- \log x}{x^2}\right]$
$\therefore \cfrac{dy}{dx} = \left(1+\cfrac{1}{x}\right)\left[\log(1+\frac{1}{x})-\cfrac{1}{1+x}\right]+x^{1+\frac{1}{x}}\left[\cfrac{(1+x)- \log x}{x^2}\right]$

$\displaystyle f\left ( a \right )=a^{2},\phi \left ( a \right )=b^{2}$ and

$\displaystyle {f}'\left ( a \right )=3{\phi }'\left ( a \right )$ then

$\displaystyle \lim_{x\rightarrow 0}\frac{\sqrt{f\left ( x \right )}-a}{\sqrt{\phi \left ( x \right )}-b}$ is

0%
$\displaystyle b^{2}/a^{2}$
0%
$\displaystyle b/a$
0%
$\displaystyle 2b/a$
0%
None of these

Explanation

$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { f\left( x \right) } -a }{ \sqrt { \phi \left( x \right) } -b } } =\lim _{ x\rightarrow a }{ \frac { f\left( x \right) -{ a }^{ 2 } }{ \phi \left( x \right) -{ b }^{ 2 } } \times \frac { \sqrt { \phi \left( x \right) } +b }{ \sqrt { f\left( x \right) } +a } }$

$\displaystyle =\lim _{ x\rightarrow a }{ \frac { f\left( x \right) -f\left( a \right) }{ \phi \left( x \right) -\phi \left( a \right) } \times \frac { \sqrt { \phi \left( x \right) } +b }{ \sqrt { f\left( x \right) } +a } }$

$\displaystyle =\frac { f'\left( a \right) }{ \phi' \left( a \right) } \times \frac { \sqrt { \phi \left( a \right) } +b }{ \sqrt { f\left( a \right) } +a } =\frac { 3b }{ a }$

Using the ahove approximation, the value

$\displaystyle \sqrt{104}$ is

0%
$10.18$
0%
$10.49$
0%
$10.2$
0%
$10.28$

Explanation

To find the approximation of the value of

$\sqrt{104}$
The nearest number to

$104$ whose square root can be taken is

$100$
so let us consider that

$x=100$ and

$\delta x= dx= 4$
Now consider,

$y=\sqrt{x}$
Differentiating Equation(i) with respect to x we have

$\cfrac { dy }{ dx } =\cfrac { d }{ dx } \left( \sqrt { x } \right)$

$\cfrac { dy }{ dx } =\cfrac { 1 }{ 2\sqrt { x } } \rightarrow \left( i \right)$
Taking the differential of eq(i) we get,

$dy=\cfrac { 1 }{ 2\sqrt { x } } dx$
using values

$x=100$ and

$dx=4$ we have

$dy=\cfrac { 1 }{ 2\sqrt { 100 } } \left( 4 \right) =\cfrac { 4 }{ 20 }$

$dy=0.2$

$\therefore \sqrt { 104 } =y+dy=\sqrt { x } +dy$

$=\sqrt { 100 } +0.2=10+0.2$

$\sqrt { 102 } =10.2$

$\displaystyle y=\sum _{ r=1 }^{ x }{ \tan ^{ -1 }{ \frac { 1 }{ 1+r+{ r }^{ 2 } } } }$ then

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

0%
$\displaystyle \frac { 1 }{ 1+{ x }^{ 2 } }$
0%
$\displaystyle \frac { 1 }{ 1+{ \left( 1+x \right) }^{ 2 } }$
0%
$0$
0%
None of these

Explanation

We have,

$\displaystyle y=\sum _{ r=1 }^{ x }{ \tan ^{ -1 }{ \frac { 1 }{ 1+r+{ r }^{ 2 } } } } =\sum _{ r=1 }^{ x }{ \tan ^{ -1 }{ \left( \frac { \left( r+1 \right) -r }{ 1+\left( r+1 \right) r } \right) } }$

$\displaystyle =\sum _{ r=1 }^{ x }{ \left[ \tan ^{ -1 }{ \left( r+1 \right) } -\tan ^{ -1 }{ r } \right] }$

$=\left[ \tan ^{ -1 }{ 2 } -\tan ^{ -1 }{ 1 } +\tan ^{ -1 }{ 3 } -\tan ^{ -1 }{ 2 } +...+\tan ^{ -1 }{ x } -\tan ^{ -1 }{ \left( x-1 \right) } +\tan ^{ -1 }{ \left( x+1 \right) } -\tan ^{ -1 }{ x } \right]$

$=\left[ \tan ^{ -1 }{ \left( x+1 \right) } -\tan ^{ -1 }{ 1 } \right]$

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

Using the above approximation, the value of

$\displaystyle \cos 40^{\circ}$ is

0%
$\displaystyle \frac{1}{\sqrt{2}}-\frac{\pi }{\sqrt{2}36}$
0%
$\displaystyle \frac{1}{\sqrt{2}}+\frac{\pi }{36\sqrt{2}}$
0%
$0.747$
0%
$0.7267$

Explanation

$\displaystyle { 40 }^{ o }=\frac { \pi }{ 4 } -\frac { \pi }{ 36 }$ radians

Let

$f\left( x \right)=\cos { x }$ and

$x$ decreases from

$\displaystyle \frac { \pi }{ 4 }$ to

$\displaystyle \frac { \pi }{ 4 } -\frac { \pi }{ 36 }$

$df=f'\left( x \right) h=-h\sin { x }$

When

$\displaystyle x=\frac { \pi }{ 4 } ,h=-\frac { \pi }{ 36 }$

We have

$\displaystyle df=\frac { \pi }{ 36 } \sin { \frac { \pi }{ 4 } } =\frac { \pi \sqrt { 2 } }{ 72 }$

$\displaystyle \cos { 40 } =\frac { 1 }{ \sqrt { 2 } } +\frac { \pi }{ 36\sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \left( 1+\frac { \pi }{ 36 } \right)$

$\displaystyle y=x^{2}\cos x$ then

$\displaystyle y_{8}\left ( 0 \right )$ is

0%
$72$
0%
$56$
0%
$0$
0%
$-56$

Explanation

Given,

$y={ x }^{ 2 }\cos x\\ u=\cos x\quad and\quad v={ x }^{ 2 }\\ { y }_{ 8 }={ u }_{ 8 }v+{ 8C }_{ 1 }{ u }_{ 7 }{ v }_{ 1 }+{ 8C }_{ 2 }{ u }_{ 6 }{ v }_{ 2 }+........+{ 8C }_{ 7 }{ u }{ v }_{ 7 }+{ 8C }_{ 8 }{ u }{ v }_{ 8 }$
except

${ v }_{ 2 }=2$ all other derivatives of v are zero at x = 0 i.e

${ v }(0)={ v }_{ 1 }(0)=.......{ v }_{ n }(0)=0\quad \Rightarrow { y }_{ 8 }={ 8C }_{ 2 }{ u }_{ 6 }{ v }_{ 2 }\\ { u }_{ n }=\cos (x+\frac { n\pi }{ 2 } )\quad \quad \quad \quad { u }_{ 6 }=\cos(x+3\pi )=-\cos x\\ { u }_{ 6 }(0)=-1\\ { y }_{ 8 }(0)=28(2)(-1)=-56$

Let

$f(x) = \begin{cases} \overset{x}{\underset{0}{\int}} \{1 + |1 - t|\}dt & if \,x > 2\\ 5x - 7 & if \,x \le 2\end{cases}$ then

0%
$\displaystyle f$ is not continuous at $\displaystyle x=2$
0%
$\displaystyle f$ is continuous but not differentiable at $\displaystyle x=2$
0%
$\displaystyle f$ is differentiable everywhere
0%
$\displaystyle {f}'\left ( 2+ \right )$ doesn't exist

Explanation

Let $\displaystyle f(x)=\begin{Bmatrix} \int _{ 0 }^{ x }{ \{ 1+|1-t| \} dt } & x>2 \\ 5x-7 & x\le 2 \end{Bmatrix}$

Let $\displaystyle y=\int _{ 0 }^{ x }{ \{ 1+|1-t| \} dt }$ as $x>2$

$\displaystyle y=\int _{ 0 }^{ 1 }{ (1+1-t)dt } +\int _{ 1 }^{ x }{ tdt }$

$\displaystyle y=2-\frac { 1 }{ 2 } +\frac { { (x }^{ 2 }-1) }{ 2 }$

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

$\displaystyle f(x)=\begin{Bmatrix} 1+\frac { { x }^{ 2 } }{ 2 } ,\quad \quad & x>2 \\ 5x-7,\quad \quad & x\le 2 \end{Bmatrix}$

$\displaystyle f({ 2 }^{ + })=1+\frac { 4 }{ 2 } =3$

$\displaystyle f({ 2 }^{ - })=10-7=3$

Differentiating with respect to $x$ we get

$\displaystyle { f }^{ 1 }(x)=\begin{Bmatrix} x; & x>2 \\ 5; & x\le 2 \end{Bmatrix}$

We can see that it is not differentiable at $x=2$ but continuous at $x=2$

Let

$y$ be an implicit function of

$x$ defined by

$x^{2x}-2x^{x}\cot y-1=0$ . Then

${y}'\left ( 1 \right )$ equals:

0%
$1$
0%
$\log 2$
0%
$-\log 2$
0%
$-1$

Explanation

$x^{2x}-2x^{x}\cot y-1=0$ ........ (i)

$x=1$ , we have

$1-2\cot y-1=0$

$\Rightarrow$

$\cot y=0$

$\therefore$

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

Differentiating (i) w.r.t. x, we have

$\displaystyle 2x^{2x}\left ( 1+\ln x \right )-2\left [ x^{x}\left ( -\csc^{2} y \right )\frac{dy}{dx}+\cot yx^{x}\left ( 1+\ln x \right ) \right ]=0$

$P(1, \frac{\pi}{2})$ , we have

$\displaystyle 2\left ( 1+\ln 1 \right )-2\left [ 1\left ( -1 \right )\left ( \frac{dy}{dx} \right )_{P}+0 \right ]=0$

$\Rightarrow$

$\displaystyle 2+2\left ( \frac{dy}{dx} \right )_{P}=0$

$\therefore$

$\displaystyle \left ( \frac{dy}{dx} \right )_{P}=-1$

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

${f'}\left( { - \dfrac{\pi }{4}} \right)$ is equals

0%
${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( { - \dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } - \dfrac{{2\sqrt 2 }}{\pi }} \right)$
0%
${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } + \dfrac{{2\sqrt 2 }}{\pi }} \right)$
0%
${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{\pi }{4} + \dfrac{{2\sqrt 2 }}{\pi }} \right)$
0%
None of these.

$\phi (x)=f\left( x \right) g\left( x \right) \quad and\quad f^{ ' }\left( x \right) g^{ ' }\left( x \right) =k,\dfrac { 2k }{ f\left( x \right) g\left( x \right) } =$

0%
$\frac { \phi "(x) }{ \phi (x) } -\frac { f"(x) }{ f(x) } -\frac { g"(x) }{ g(x) }$
0%
$\frac { \phi "(x) }{ \phi (x) } +\frac { f"(x) }{ f(x) } +\frac { g"(x) }{ g(x) }$
0%
$\frac { \phi "(x) }{ \phi (x) } +\frac { f"(x) }{ f(x) } -\frac { g"(x) }{ g(x) }$
0%
$\frac { \phi "(x) }{ \phi (x) } -\frac { f"(x) }{ f(x) } +\frac { g"(x) }{ g(x) }$

$\displaystyle y=x^{4}e^{2x}$ then

$\displaystyle y_{10}\left ( 0 \right )$ is equal to

0%
$\displaystyle 2^{10}$
0%
$\displaystyle 315 \times 2^{10}$
0%
$\displaystyle 195 \times 2^{10}$
0%
$\displaystyle 315 \times 2^{8}$

Explanation

$y={ x }^{ 4 }{e}^{2x}\\ u={e}^{2x}\quad and\quad v={ x }^{ 4 }\\ { y }_{ 10 }={ u }_{ 10 }v+{ 10C }_{ 1 }{ u }_{ 9 }{ v }_{ 1 }+{ 10C }_{ 2 }{ u }_{ 8 }{ v }_{ 2 }+........+{ 10C }_{ 9 }{ u }{ v }_{ 9 }+{ 10C }_{ 10 }{ u }{ v }_{ 10 }$
except

${ v }_{ 4 } = 4! = 24$ all other derivatives of v are zero at x = 0 i.e

${ v }(0)={ v }_{ 1 }(0)=.......{ v }_{ n }(0)=0\quad \Rightarrow { y }_{ 10 }={ 10C }_{ 4 }{ u }_{ 6 }{ v }_{ 4 }\\ { u }_{ n }={2}^{n}{e}^{2x}\quad \quad \quad \quad { u }_{ 6 }={2}^{6}{e}^{2x}\\ { u }_{ 6 }(0)={2}^{6}\\ { y }_{ 10 }(0)={ 10C }_{ 4 } \times 24 \times {2}^{6}$

${ y }_{ 10 }(0)=\frac{10 \times 9 \times 8 \times 7}{2 \times 3 \times 4} \times 24 \times {2}^{6}=315 \times {2}^{10}$

A metal sphere with radius of

$10\ cm$ is to be covered with a

$0.02\ cm$ coating of silver approximately silver required is (in

$cm^3$ )

0%
$\displaystyle 2\pi$
0%
$\displaystyle 10\pi$
0%
$\displaystyle 6\pi$
0%
$\displaystyle 8\pi$

Explanation

$\displaystyle V=\frac { 4 }{ 3 } \pi { r }^{ 3 }\Rightarrow dV=4\pi { r }^{ 2 }h$
when

$r=10$ and

$h=0.02$
We have

$dV=4\pi { \left( 10 \right) }^{ 2 }\left( 0.02 \right) =8\pi$

Let f and g be differentiable function such that

${f}'\left ( x \right )=2g\left ( x \right )$ and

${g}'\left ( x \right )=-f\left ( x \right )$ , and let

$T\left ( x \right )=\left ( f\left ( x \right ) \right )^{2}-\left ( g\left ( x \right ) \right )^{2}$ . Then

${T}'\left ( x \right )$ is equal to

0%
$T(x)$
0%
$0$
0%
$2f(x)g(x)$
0%
$6f(x)g(x)$

Explanation

The given equation is:

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

$\Rightarrow T(x) = (f(x) - g(x))(f(x) + g(x))$

Differentiating once w.r.t to x we get,

$\Rightarrow T'(x) = (f'(x) + g'(x))(f(x)-g(x)) + (f'(x)-g'(x))(f(x)+g(x))$

$\Rightarrow T'(x) = (2g(x)-f(x))(f(x)-g(x)) + (2g(x)+f(x))(f(x)+ g(x))$

$\Rightarrow T'(x) = 2g(x)f(x) - 2g(x)^2 -f(x)^2 + f(x)g(x) +2g(x)f(x) + 2g(x)^2 +f(x)^2 + f(x)g(x)$

$\Rightarrow T'(x)= 6f(x)g(x)$ .....Answer

$y\left ( n \right )=e^{x}e^{x^{2}}...e^{x^{n}}, 0< x< 1$ . Then

$\displaystyle \lim_{n\rightarrow \infty }\frac{dy\left ( n \right )}{dx}$ at

$x=\dfrac12$ is

0%
$e$
0%
$4e$
0%
$2e$
0%
$3e$

Explanation

$\displaystyle y\left ( n \right )=e^{x+x^{2}+...+x^{n}}=e^{\dfrac{x\left ( 1-x^{n} \right )}{1-x}}$
So $\displaystyle \dfrac{dy\left ( n \right )}{dx}=e^{\dfrac{x\left ( 1-x^{n} \right )}{1-x}}\times \dfrac{d}{dx}\left ( \dfrac{x\left ( 1-x^{n} \right )}{1-x} \right )$
$\displaystyle \lim_{n\rightarrow \infty }\dfrac{dy\left ( n \right )}{dx}=\lim_{n}e^{\dfrac{x}{1-x}-\dfrac{x^{n+1}}{1-x}}\dfrac{d}{dx}\left [ \lim_{n\rightarrow \infty }\dfrac{x}{1-x}-\dfrac{x^{n+1}}{1-x} \right ]$
$\displaystyle =e^{\dfrac{x}{1-x}}\dfrac{d}{dx}\left ( \dfrac{x}{1-x} \right )$
$\displaystyle =e^{\dfrac{x}{1-x}}\dfrac{1}{\left ( 1-x \right )^{2}}$
$\displaystyle \lim_{n\rightarrow \infty }\dfrac{dy\left ( n \right )}{dx}|_{x=\frac{1}{2}}=4e$

$\displaystyle \int f(x) dx = \frac {3}{55} \sqrt[3]{\tan^5 x} (5 \tan^2 x + 11) + C$ then

$f(x)$ is equal to

0%
$\displaystyle \sqrt[3] {\sin^2 x \cos^{-14} x}$
0%
$\displaystyle \sqrt[3] {\tan^2 x(1 + \tan^2 x)^6}$
0%
$\displaystyle \sqrt[3] {\cos^2 x \sin^{-14} x}$
0%
$\displaystyle \frac {7}{3} \sqrt[3] {\sin^2 x \cos^{-14} x}$

Explanation

$\displaystyle \int { f\left( x \right) dx } =\dfrac { 3 }{ 55 } \sqrt [ 3 ]{ { \left( \tan { x } \right) }^{ 5 } } \left( 5\tan ^{ 2 }{ x+11 } \right)$

$\displaystyle f\left( x \right) =\dfrac { d }{ dx } \left( \dfrac { 3 }{ 55 } \sqrt [ 3 ]{ { \left( \tan { x } \right) }^{ 5 } } \left( 5\tan ^{ 2 }{ x+11 } \right) \right)$

$\displaystyle =\dfrac { 3 }{ 55 } \dfrac { d }{ dx } \left( 5\tan ^{ \tfrac { 11 }{ 3 } }{ x } +11\tan ^{ \tfrac { 5 }{ 3 } }{ x } \right) \sec ^{ 2 }{ x }$

$\displaystyle =\dfrac { 3 }{ 55 } \left( \dfrac { 55 }{ 3 } \tan ^{ \tfrac { 8 }{ 3 } }{ x } +\dfrac { 55 }{ 3 } \tan ^{ \tfrac { 2 }{ 3 } }{ x } \right) \sec ^{ 2 }{ x }$

$\displaystyle =\left( \tan ^{ \tfrac { 8 }{ 3 } }{ x } +\tan ^{ \tfrac { 2 }{ 3 } }{ x } \right) \sec ^{ 2 }{ x } =\tan ^{ \tfrac { 2 }{ 3 } }{ x } \left( { \left( 1+\tan ^{ 2 }{ x } \right) }^{ 2 } \right) \\ =\sqrt [ 3 ]{ \sin ^{ 2 }{ x } \cos ^{ -14 }{ x } }$

Consider the function:

$f(-\infty,\infty)\rightarrow (-\infty,\infty)$ defined by

$\displaystyle f(x)=\frac{x^{2}-ax+1}{x^{2}+ax+1},0<a<2$

Which of the following is true?

0%
$f(x)$ is decreasing on $(-1,1)$ and has a local minimum at $x=1$
0%
$f(x)$ is increasing on $(-1,1)$ and has a local maximum at $x=1$
0%
$f(x)$ is increasing on $(-1,1)$ and has neither a local maximum nor a local minimum at $x=-1$
0%
$f(x)$ is decreasing on $(-1,1)$ and has neither a local maximum nor a local minimum at $x=1$

Explanation

$(x^{2}+ax+1)f(x)=x^{2}-ax+1$
Differentiating we obtain

$\displaystyle (2x+a)f(x)+(x^{2}+ax+1) f'(x)=2x-a$ ...(i)
Putting

$x=1$ and noting that

$\displaystyle f(1)=\frac{2-a}{2+a},$ we obtain

$(2+a)f(1)+(2+a)f'(1)=2-a\Rightarrow f'(1)=0$
Differentiating (i) again

$2f(x)+(2x+a)f'(x)+(2x+a)f'(x) +(x^{2}+ax+1)f''(x)=2$ ...(ii)
Putting

$x=1$ , we have

$\displaystyle 2\cdot \frac{2-a}{2+a}+(2+a) f''(1)=2$

$\displaystyle \Rightarrow f''(1)=\frac{4a}{(2+a)^{2}}$
Putting

$x=-1$ in (i) and noting that

$\displaystyle f(-1)=\frac{2+a}{2-a} \Rightarrow (-2+a)f(-1)+(2-a) f'(-1)=-(2+a)\Rightarrow f'(-1)=0$
Putting

$x=-1$ in (ii), we obtain

$\displaystyle f''(-1)=\frac{-4a}{(2-a)^{2}}$

$(2+a)^{2} f''(1)+(2-a)^{2}f'(-1)=0$
(i)

$\displaystyle\Rightarrow f'(x)=\frac{1}{x^{2}+ax+1}$

$\displaystyle \left[ (2x-a)-(2x+a)\times \frac{x^{2}-ax+1}{x^{2}+ax+1}\right] =\frac{2a(x^{2}-1)}{(x^{2}+ax+1)^{2}}$
So

$f'(x)<0$ if

$x\in (-1,1)$ and

$f'(x)>0$ for

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

i.e. f increases in

$(\infty,-1) \cup (1,\infty)$ and decreases on

$(-1,1)$

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 15 - 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 15 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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