If $$y = |\cos x| + |\sin x|$$, then $$\dfrac {dy}{dx}$$ at $$x = \dfrac {2\pi}{3}$$ is

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

If $$f(x)=\left| \log { \left| x \right| } \right| $$, then

$$f(x)$$ is continuous for all $$x$$ in its domain but not differentiable at $$x=\pm 1$$

$$f(x)$$ is continuous and differentiable for all $$x$$ in its domain

$$f(x)$$ is neither continuous nor differentiable at $$x=\pm 1$$

MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 6

$\displaystyle y=\frac{1}{1+x^{\beta -\alpha}+x^{\gamma -\alpha}}+\frac{1}{1+x^{\alpha-\beta}+x^{\gamma -\beta }}+\frac{1}{1+x^{\alpha -\gamma }+x^{\beta-\gamma }}$
then

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

0%
$0$
0%
$1$
0%
$\displaystyle (a+\beta +\gamma )X^{\alpha +\beta +\gamma -1}$
0%
None of these

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

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

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

0%
$\displaystyle \frac{1-\sqrt{3}}{2}$
0%
$0$
0%
$\displaystyle \frac{\sqrt{3}-1}{2}$
0%
None of these

Explanation

In the neighborhood of

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

$\Rightarrow y=-\cos\:x+\sin\:x$

$\Rightarrow \left ( \cfrac{dy}{dx} \right )=\sin\:x+\cos\:x$
Thus at

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

$\cfrac{dy}{dx}=\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}=\dfrac{1}{2}\left ( \sqrt{3}-1 \right )$

Evaluate

$\displaystyle \lim_{n\rightarrow \infty }\left [ \left ( 1+\frac{1}{n^{2}} \right )\left ( 1+\frac{2^{2}}{n^{2}} \right )\left ( 1+\frac{3^{2}}{n^{2}} \right )......\left ( 1+\frac{n^{2}}{n^{2}} \right ) \right ]^{1/n}$

0%
$2e ^\left(\dfrac{\pi - 4}{2}\right)$
0%
$2e ^\left(\dfrac{\pi - 2}{2}\right)$
0%
$2e^ \left(\dfrac{\pi - 4}{4}\right)$
0%
$2e^ \left(\dfrac{\pi - 4}{3}\right)$

Explanation

Let

$\displaystyle S=\frac { lim }{ n\rightarrow \infty } { \left[ \left( 1+\frac { 1 }{ { n }^{ 2 } } \right) \left( 1+\frac { { 2 }^{ 2 } }{ { n }^{ 2 } } \right) ...\left( 1+\frac { { n }^{ 2 } }{ { n }^{ 2 } } \right) \right] }^{ \frac { 1 }{ n } }$

$\displaystyle \Rightarrow logS=\frac { lim }{ n\rightarrow \infty } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ log } \left( 1+\frac { { r }^{ 2 } }{ { n }^{ 2 } } \right)$

$\displaystyle =\int _{ 0 }^{ 1 }{ log } \left( 1+{ x }^{ 2 } \right) dx.=\left( xlog\left( 1+{ x }^{ 2 } \right) \right) _{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ \frac { { 2x }^{ 2 } }{ 1+{ x }^{ 2 } } dx }$

$\displaystyle \Rightarrow log\left( \frac { S }{ 2 } \right) =2\left[ \int _{ 0 }^{ 1 }{ \frac { { dx } }{ 1+{ x }^{ 2 } } } -\int _{ 0 }^{ 1 }{ dx } \right] =2\left[ \frac { \pi }{ 4 } -1 \right] =\frac { \pi -4 }{ 2 }$

$\displaystyle \Rightarrow S=2{ e }^{ \left( \frac { \pi -4 }{ 2 } \right) }$

The limit of

$x\sin { \left( { e }^{ \frac { 1 }{ x } } \right) }$ as

$x\rightarrow 0$

0%
is equal to $0$
0%
is equal to $1$
0%
is equal to $\cfrac { e }{ 2 }$
0%
does not exist

Explanation

$\displaystyle \lim _{ x\rightarrow 0 }{ x\sin { { e }^{ (1/x) } } }$
LHL

$=f(0-0)=\displaystyle \lim _{ x\rightarrow 0 }{ (-h)\sin { { e }^{ (-1/h) } } }$

$=-0\times \sin { \left( { e }^{ -\infty } \right) }$

$=-0\times \sin { (0)=0 }$

$=0\times$ (a finite number between

$-1$ to

$+1$ )

$\quad \left( \because -1\le \sin { x } \le 1 \right)$

$\because$ LHL

$=$ RHL

$\because$

$\displaystyle \lim _{ x\rightarrow 0 }{ x\sin { { e }^{ (1/x) } } }$ exist and equal to

$0$

$r=\left[2\phi +\cos^2\left(2\phi +\dfrac{\pi}4\right)\right]^{\tfrac12},$ then what is the value of the derivative of

$\dfrac{dr}{d\phi}$ at

$\phi=\dfrac{\pi}4?$

0%
$2\left(\displaystyle\frac{1}{\pi+1}\right)^{\tfrac12}$
0%
$2\left(\displaystyle\frac{2}{\pi+1}\right)^{2}$
0%
$\left(\displaystyle\frac{2}{\pi+1}\right)^{\tfrac12}$
0%
$2\left(\displaystyle\frac{2}{\pi+1}\right)^{\tfrac12}$

Explanation

$r=\sqrt { 2\phi +{ cos }^{ 2 }\left( 2\phi +\dfrac { \pi }{ 4 } \right) }$

$\phi =\dfrac { \pi }{ 4 }$

$\Rightarrow r=\sqrt { 2\times \dfrac { \pi }{ 4 } +{ cos }^{ 2 }\left( 2\times \dfrac { \pi }{ 4 } +\dfrac { \pi }{ 4 } \right) }$

$=\sqrt { \dfrac { \pi }{ 2 } +{ cos }^{ 2 }\left( \dfrac { 3\pi }{ 4 } \right) }$

$r\left( \dfrac { \pi }{ 4 } \right) =\sqrt { \dfrac { \pi }{ 2 } +\dfrac { 1 }{ 2 } } =\dfrac { \sqrt { \pi +1 } }{ \sqrt { 2 } }$

${ r }^{ 2 }=2\phi +{ cos }^{ 2 }\left( 2\phi +\dfrac { \pi }{ 4 } \right)$

Differentiate w.r.t

$\phi$

$\Rightarrow 2r\dfrac { dr }{ d\phi } =2+2\cos { \left( 2\phi +\dfrac { \pi }{ 4 } \right) } \left( -\sin { \left( 2\phi +\dfrac { \pi }{ 4 } \right) } \right) \left( +2 \right)$

$2r\left( \dfrac { \pi }{ 4 } \right) \dfrac { dr }{ d\phi } =2+2\cos { \dfrac { 3\pi }{ 4 } \left( -\sin { \left( \dfrac { 3\pi }{ 4 } \right) } \right) } \left( +2 \right)$

$=2+2\times \dfrac { -1 }{ \sqrt { 2 } } \times \dfrac { -1 }{ \sqrt { 2 } } \times \left( +2 \right)$

$2\times \left( \sqrt { \dfrac { \pi +1 }{ 2 } } \right) \times \dfrac { dr }{ d\phi } =4$

$\dfrac { dr }{ d\phi } =\dfrac { 2\sqrt { 2 } }{ \sqrt { \pi +1 } }$

The value of the constant

$\alpha$ and

$\beta$ such that

$\displaystyle \lim_{x\rightarrow \infty}\left(\displaystyle\frac{x^2+1}{x+1}-\alpha x-\beta\right)=0$ are respectively.

0%
$(1, 1)$
0%
$(-1, 1)$
0%
$(1,-1)$
0%
$(0, 1)$

Explanation

Simplifying the above expression gives us

$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{x^{2}+1-\alpha x(x+1)-\beta(x+1)}{x+1})$

$=\displaystyle \lim_{x\rightarrow \infty}(\dfrac{x^{2}(1-\alpha)-x(\alpha +\beta)+1-\beta)}{x+1})$

$=0$ implies that

$Numerator<Denominator$ .
Hence coefficient of

$x^{2}$ will be 0. Therefore

$\alpha=1$ .
Hence

$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{-x(1 +\beta)+1-\beta)}{x+1})=0$

Applying L'Hopital's Rule we get

$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{-(1 +\beta)}{1})=0$
Or

$\beta+1=0$
Or

$\beta=-1$ .
Hence

$(\alpha,\beta)=1,-1$

If the function

$f(x)$ satisfies

$\displaystyle \lim_{x\rightarrow 1}\frac{f(x)-2}{x^2-1}=\pi$ , then

$\displaystyle \lim_{x\rightarrow 1}f(x)=$

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

Explanation

It is given that,

$\displaystyle \lim_{x\rightarrow 1}\frac{f(x)-2}{x^2-1}=\pi$ exists and equal to

$1$ .

Clearly the denominator is zero at

$x=1$ , so for above limit to exist

numerator should also be zero, as x approaches

$1$ .

Therefore,

$\displaystyle \lim_{x\to 1}f(x)=2$

Note: If

$\displaystyle \lim_{x\to 1}f(x) \neq 1$ , given limit will not exist

$f(x) = \log \left (e^{x} \left (\dfrac {x - 2}{x + 2}\right )^{\dfrac {3}{4}} \right ) \Rightarrow f'(0) =$

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

Explanation

$f(x) = \log \left (e^{x} \left (\dfrac {x - 2}{x + 2}\right )^{\dfrac {3}{4}}\right )$

$f(x) = x + \dfrac {3}{4} [\log (x - 2) - \log (x + 2)]$

$f'(x) = 1 + \dfrac {3}{4} \left [\dfrac {1}{x - 2} - \dfrac {1}{x + 2} \right ]$

$f'(x) = \left [1 + \dfrac {3}{x^{2} - 4}\right ]$

$f'(0) = 1 -\dfrac {3}{4} = \dfrac {1}{4}$

$f(x) = \sec (3x)$ , then

$f'\left (\dfrac {3\pi}{4}\right ) =$

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

Explanation

Given,

$f(x) = \sec(3x)$
We get

$f'\left( x \right) =3\sec(3x)\tan(3x)$
Therefore, we get

$f'\left( \dfrac { 3\pi }{ 4 } \right) =3\sec\left (\dfrac { 9\pi }{ 4 }\right )\tan\left (\dfrac { 9\pi }{ 4 } \right)$

$=3\sec\left (\dfrac { \pi }{ 4 } \right)\tan\left (\dfrac { \pi }{ 4 }\right )$

$=3\times \sqrt { 2 } \times 1$

$=3\sqrt 2$

$y = f(x^{2} + 2)$ and

$f'(3) = 5$ , then

$\dfrac {dy}{dx}$ at

$x = 1$ is _____

0%
$5$
0%
$25$
0%
$15$
0%
$10$

Explanation

$y=f(x^2+2)$

Using chain rule of differentiation

$\dfrac{dy}{dx}=\dfrac{d}{d(x^2+2)}[f(x^2+2)]\cdot \dfrac{d}{dx}(x^2+2)$

$\quad =f'(x^2+2)\cdot 2x=2xf'(x^2+2)$

Hence

$\dfrac{dy}{dx}\bigg|_{x=1}=2\cdot 1\cdot f'(1^2+2)=2\cdot f'(3)=2\cdot 5=10$

If f(x) is a function such that

$f^{\prime \prime}(x)+f(x)=0$ and

$g(x)=[f(x)]^2+[f'(x)]^2$ and g(3)=8, then

$g(8)=$ _____

0%
$0$
0%
$3$
0%
$5$
0%
$8$

Explanation

We have,

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

Differentiate the function g(x)

$\Rightarrow g'(x)=2f(x)f'(x)+2f'(x)f''(x)$ , use chain rule

$=2f'(x)[f(x)+f''(x)]=2f'(x)(0)=0$ , use the given condition

Hence

$g(x)$ is a constant function

$\Rightarrow g(x)=c$ , constant

But

$g(3)=8$ , so

$g(x)=8$ , for all real x

Hence

$g(8)=8$

$\displaystyle \lim_{x\rightarrow \infty}\dfrac{x^3+1}{x^2+1}-(ax+b)=2$ , then

0%
$a=2$ and $b=-1$
0%
$a = 1$ and $b = 1$
0%
$a = 1$ and $b = -1$
0%
$a = 1$ and $b = -2$

Explanation

Given

$\displaystyle \lim _{ x\rightarrow \infty }{ \frac { { x }^{ 3 }+1 }{ { x }^{ 2 }+1 } -(ax+b)=2 }$

$\Rightarrow \displaystyle \lim _{ x\rightarrow \infty }{ (\frac { { (1-a)x }^{ 3 }-b{ x }^{ 2 }-ax-b+1 }{ { x }^{ 2 }+1 } )=2 }$

For the limit to exist, the coefficient of

$x^{3}$ must be zero because if it is not zero then the limit is infinite

$\Rightarrow a=1$

So the given one will reduce to

$\displaystyle \lim _{ x\rightarrow \infty }{ (\frac { -b{ x }^{ 2 }-ax-b+1 }{ { x }^{ 2 }+1 } )=2 }$

If we apply the limit , we get

$-b=2$

$\Rightarrow b=-2$

Therefore the correct option is

$D$

$y = \tan^{-1} \left (\dfrac {1}{1 + x + x^{2}}\right ) + \tan^{-1} \left (\dfrac {1}{x^{2} + 3x + 2}\right ) + \tan^{-1} \left (\dfrac {1}{x^{2} + 5x + 6}\right ) + .... +$ upto

$n$ terms then

$\dfrac {dy}{dx}$ at

$x = 0$ and

$n = 1$ is equal to

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

Explanation

$y= \tan^{-1} \left (\dfrac {1}{1 + x(x + 1)}\right ) + \tan^{-1}\left (\dfrac {1}{1 + (x + 1)(x + 2)}\right ) +\\ \quad\tan^{-1} \left (\dfrac {1}{1 + (x + 2)(x + 3)}\right ) + ..... + \tan^{-1} \left (\dfrac {1}{1 + (x + n - 1) (x + n)}\right )$

when

$n = 1$

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

$\Rightarrow \dfrac {dy}{dx} = \dfrac {1}{1 + (x + 1)^{2}} - \dfrac {1}{1 + x^{2}} \\ \therefore \dfrac {dy}{dx}]_{x = 0} = \dfrac {1}{1 + 1} - \dfrac {1}{1} = \dfrac {1}{2} - 1 = -\dfrac {1}{2}$

$\displaystyle \lim_{x\rightarrow \pi/4} \dfrac {\tan x - 1}{\cos 2x}$ is equal to

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

Explanation

$\displaystyle \lim_{x\rightarrow \pi/4} \dfrac {\tan x - 1}{\cos 2x} = \displaystyle \lim_{h\rightarrow 0}\dfrac {\tan \left (\dfrac {\pi}{4} + h\right ) - 1}{\cos 2\left (\dfrac {\pi}{4} + h\right )}$

$\left [\because x = \dfrac {\pi}{4} + h\right ]$

$= \displaystyle \lim_{h\rightarrow 0} \dfrac {\left (\dfrac {1 + \tan h}{1 - \tan h}\right ) - 1}{\cos \left (\dfrac {\pi}{2} + 2h\right )}$

$= \displaystyle \lim_{h\rightarrow 0} \dfrac {1 + \tan h - 1 + \tan h}{-\sin 2h (1 - \tan h)}$

$= \displaystyle \lim_{h\rightarrow 0} \dfrac {-2\tan h}{2\sin h \cos h (1 - \tan h)}$

$= \displaystyle \lim_{h\rightarrow 0} \dfrac {-1}{\cos^{2} h (1 - \tan h)} = -1$

Hence, option D is correct.

$\displaystyle \lim_{x\rightarrow 3} = \dfrac {\sqrt {x} -\sqrt {3}}{\sqrt {x^{2} - 9}}$ is equal to

0%
$1$
0%
$3$
0%
$\sqrt {3}$
0%
$-\sqrt {3}$
0%
$0$

Explanation

The value of

$\displaystyle \lim_{x\rightarrow 3} = \dfrac {\sqrt {x} -\sqrt {3}}{\sqrt {x^{2} - 9}}$ is

$= \displaystyle \lim_{x\rightarrow 3} \dfrac {\sqrt {x} - \sqrt {3}}{\sqrt {x - 3}\sqrt {x} + 3} \times \dfrac {\sqrt {x} + \sqrt {3}}{\sqrt {x} + \sqrt {3}}$

$= \displaystyle \lim_{x\rightarrow 3} \dfrac {x - 3}{\sqrt {x - 3} \sqrt {x + 3}}\times \dfrac {1}{\sqrt {x} + \sqrt {3}}$

$= \displaystyle \lim_{x\rightarrow 3} \dfrac {\sqrt {x - 3}}{\sqrt {x + 3}} \times \dfrac {1}{\sqrt {x} + \sqrt {3}} = 0$

What is

$\displaystyle \lim_{x \rightarrow 0 } x^2 \sin \left(\frac{1}{x}\right)$ equal to ?

0%
0
0%
1
0%
1/2
0%
Limit does not exist.

Explanation

$\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ 2 }\sin { \left( \cfrac { 1 }{ x } \right) } }$

$={ 0 }^{ 2 }\times \sin { \left( \cfrac { 1 }{ 0 } \right) } \quad \left[ -1\le \sin { \left( \cfrac { 1 }{ x } \right) } \le 1 \right]$

$=0$

The value of

$\displaystyle \lim _{ x\rightarrow \pi /6 }{ \cfrac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } -1 } }$ is

0%
$3$
0%
$-3$
0%
$6$
0%
$0$

Explanation

$L=\lim _{ x\rightarrow { \pi }/{ 6 } }{ \dfrac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } -1 } } \\ L=\dfrac { 2\sin ^{ 2 }{ \left( \dfrac { \pi }{ 6 } \right) } +\sin { \left( \dfrac { \pi }{ 6 } \right) } -1 }{ 2\sin ^{ 2 }{ \left( \dfrac { \pi }{ 6 } \right) } -3\sin { \left( \dfrac { \pi }{ 6 } \right) } -1 } \\ L=\dfrac { 2.\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 2 } -1 }{ 2.\dfrac { 1 }{ 4 } -3.\dfrac { 1 }{ 2 } -1 } =\dfrac { 0 }{ -2 }= 0$

So option $D$ is correct

$f\left( x \right) =\begin{vmatrix} \sin { x } & \cos { x } & \tan { x } \\ { x }^{ 3 } & { x }^{ 2 } & x \\ 2x & 1 & 1 \end{vmatrix}$ , then

$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { f\left( x \right) }{ { x }^{ 2 } } }$ is

0%
$-1$
0%
$3$
0%
$1$
0%
Zero

Explanation

$f(x)=\begin{vmatrix} sinx & cosx & tanx \\ { x }^{ 3 } & { x }^{ 2 } & x \\ 2x & 1 & 1 \end{vmatrix}\\ \Longrightarrow f(x)=sinx({ x }^{ 2 }-x)-cosx({ x }^{ 3 }-2{ x }^{ 2 })+tanx({ x }^{ 3 }-2{ x }^{ 3 })\\ \Longrightarrow f(x)={ x }^{ 2 }sinx-xsinx-{ x }^{ 3 }cosx+2{ x }^{ 2 }cosx-{ x }^{ 3 }tanx\\ \Longrightarrow \dfrac { f(x) }{ { x }^{ 2 } } =sinx-xcosx+2cosx-xtanx-\dfrac { sinx }{ x } \\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =\underset { x\longrightarrow 0 }{ lim } sinx-\underset { x\longrightarrow 0 }{ lim } xcosx+\underset { x\longrightarrow 0 }{ lim } 2cosx-\underset { x\longrightarrow 0 }{ lim } xtanx-\underset { x\longrightarrow 0 }{ lim } \dfrac { sinx }{ x } \\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =0-0+2-0-1\\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =1\\ \\$

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

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

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

Explanation

Given,

$y = \dfrac {1}{1 + x + x^{2}} \Rightarrow 1 + x + x^{2} = \dfrac {1}{y}$
On differentiating w.r.t.

$x$ , we get

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

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

$y = |\cos x| + |\sin x|$ , then

$\dfrac {dy}{dx}$ at

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

0%
$\dfrac {1 - \sqrt {3}}{2}$
0%
$0$
0%
$\dfrac {1}{2}(\sqrt {3} - 1)$
0%
None of these

Explanation

According to question,

$|\cos x| = -\cos x$

$|\sin x| = \sin x$

$\therefore \dfrac {dy}{dx} = \sin x + \cos x$

$\therefore \left (\dfrac {dy}{dx}\right )_{\dfrac {2\pi}{3}} = \dfrac {\sqrt {3}}{2} - \dfrac {1}{2} = \dfrac {\sqrt {3} - 1}{2}$ .

Which one of the following statements is correct?

0%
$\displaystyle \lim_{x \rightarrow 0} (fog) (x)$ exists.
0%
$\displaystyle \lim_{x \rightarrow 0} (gof) (x)$ exists.
0%
$\displaystyle \lim_{x \rightarrow 0-} (fog) (x) = \displaystyle \lim_{x \rightarrow 0-} (gof) (x)$
0%
$\displaystyle \lim_{x \rightarrow 0+} (fog) (x) =\displaystyle \lim_{x \rightarrow 0-} (gof) (x)$

Explanation

$(fog) (x) = [\sin x]$

$\displaystyle \lim_{x\to 0^+} [\sin x]=0$ and

$\displaystyle \lim_{x\to 0^-} [\sin x]=-1$ . Hence,

$\displaystyle \lim_{x\to 0} fog(x)$ does not exist.

Now,

$gof(x)=\sin [x]$

$\displaystyle \lim_{x\to 0^+} \sin [x]=0$ and

$\displaystyle \lim_{x\to 0^-} \sin [x]=\sin (-1)$ . Hence,

$\displaystyle \lim_{x\to 0} gof(x)$ does not exist.

But,

$\displaystyle \lim_{x\to 0^+} fog(x)=\displaystyle \lim_{x\to 0^-} gof(x)$

Let

$C(\theta)=\displaystyle\sum _{n=0}^{\infty}\dfrac{\cos(n\theta)}{n!}$
Which of the following statements is FALSE?

0%
$C(0).C(\pi)=1$
0%
$C(0).C(\pi) > 2$
0%
$C(\theta) > 0$ for all $\theta \in R$
0%
$C(\theta) \neq 0$ for all $\theta \in R$

Explanation

$C(0)=e$

$C(\pi)=\dfrac{1}{e}$
Hence, option A and B are correct

$C'(\theta)= -\displaystyle \sum_{n=0}^{\infty }\dfrac{\sin\theta }{(n-1)!}$

$C'(\theta)=0$
Therefore, option D is false because

$C'(\theta) =0$ for

$n=0$

$\displaystyle\lim_{x\rightarrow\frac{\pi}{6}}\frac{\sin\left(x-\displaystyle\frac{\pi}{6}\right)}{\sqrt{3-2cos x}}$ is equal to :

0%
$0$
0%
$\displaystyle\frac{1}{(\sqrt{3}-2)}$
0%
$1$
0%
$\infty$

Explanation

$\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \dfrac { \sin { (x-\cfrac { \pi }{ 6 } } ) }{ \sqrt { 3-2\cos { x } } } }$

$\cfrac { \pi }{ 6 }$ we get the numerator as

$\sin { (x-\cfrac { \pi }{ 6 } ) } = 0$

and the denominator as

$\sqrt { 3-2\cos { x } } =\sqrt { 3-\sqrt { 3 } }$

So,it is not an indeterminant form as we have

$\dfrac{0}{\sqrt{3 - \sqrt{3}} }$

Hence the answer will be zero.

$\underset{x\to 0}{\lim}\dfrac{x^a\sin^b x}{\sin(x^c)}, a, b, c, \in R \sim \{0\}$ exists and has non-zero value, then

0%
$a,b,c$ are in A.P
0%
$a,b,c$ are in G.P
0%
$a,b,c$ are in H.P
0%
none of these

Explanation

Given, limit is

$\lim_{x\rightarrow 0}\dfrac {x^a\sin^bx}{\sin (x^c)}$

$\underset{x\to 0}{\lim} x^a.\dfrac{\left(\dfrac{\sin x}{x}\right)^b}{\left(\dfrac{\sin(x^c)}{x^c}\right)}.\dfrac{x^b}{x^c}$

$=\underset{x\to 0}{\lim} x^{a+b-c} \dfrac{\left(\dfrac{\sin x}{x}\right)^b}{\left(\dfrac{\sin(x^c)}{x^c}\right)}$

The above limit non-zero valu only when

$a+b-c = 0$

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

0%
$f(x)$ is continuous and differentiable for all $x$ in its domain
0%
$f(x)$ is continuous for all $x$ in its domain but not differentiable at $x=\pm 1$
0%
$f(x)$ is neither continuous nor differentiable at $x=\pm 1$
0%
None of the above

Explanation

It is evident from the graph of

$f(x)=\left| \log { \left| x \right| } \right| \quad$ that

$f(x)$ is everywhere continuous but not differentiable at

$x=\pm 1$

$y=a\cos { \left( \sin { 2x } \right) } +b\sin { \left( \sin { 2x } \right) }$ , then

$y''+\left( 2\tan { 2x } \right) y'=$

0%
$0$
0%
$4\left( \cos ^{ 2 }{ 2x } \right) y$
0%
$-4\left( \cos ^{ 2 }{ 2x } \right) y$
0%
$-\left( \cos ^{ 2 }{ 2x } \right) y$

Explanation

$y=acos(sin2x)+bsin(sin2x)$ ....[1]

$y^{'}=a(-sin(sin2x))(cos2x)(2)+b(cos(sin2x))(cos2x)(2)$

$y^{'}=2cos2x(b(cos(sin2x))-a(sin(sin2x)))$ ...[2]

$y^{''}=2cos2x(b(-sin(sin2x))(cos2x)(2)-a(cos(sin2x))(cos2x)(2))+2(-sin2x)(2)(b(cos(sin2x))-a(sin(sin2x)))$

$y^{''}=-4cos^22x(b(sin(sin2x))+a(cos(sin2x)))+4(-sin2x)(b(cos(sin2x))-a(sin(sin2x)))$

$y^{''}=-(4cos^22x)y+4(-sin2x)(b(cos(sin2x))-a(sin(sin2x)))$ ...(Using [1])

$y^{''}=-(4cos^22x)y+4(-sin2x)(\dfrac{cos2x}{cos2x})(b(cos(sin2x))-a(sin(sin2x)))$

$y^{''}=-(4cos^22x)y+2(-tan2x)(2cos2x))(b(cos(sin2x))-a(sin(sin2x)))$

$y^{''}=-(4cos^22x)y+2(-tan2x)y^{'}$ ...(Using [2])

$y^{''}+(2tan2x)y^{'}=-4(cos^22x)y$

Hence the answer is option (C)

$\lim _{ x\rightarrow 3 }{ \left( { x }^{ 3 }-4 \right) /\left( x+1 \right) } =$

0%
$(4/23)$
0%
$(2/23)$
0%
$(1/8)$
0%
$(23/4)$

$f(x) = \begin{vmatrix} \cos x& x & 1\\ 2\sin x & x^{2} & 2x\ \\ \tan x & x & 1\end{vmatrix}$ , then

$\displaystyle \lim_{x\rightarrow 0} \dfrac {f'(x)}{x}$ .

0%
Exists and is equal to $-2$
0%
Does not exist
0%
Exist and is equal to $0$
0%
Exists and is equal to $2$

Explanation

$f(x) = \begin{vmatrix} \cos x& x & 1\\ 2\sin x & x^{2} & 2x\ \\ \tan x & x & 1\end{vmatrix}$

$=\cos x(x^{2}-2x^{2})-x(2\sin x-2x\tan x)+1(2x\sin x-x^{2}\tan x)$

$=-x^{2}\cos x-2x\sin x+2x^{2}\tan x+2x\sin x-x^{2}\tan x$

$=x^{2}\tan x-x^{2}\cos x$

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

$f'(x)=2x(\tan x-\cos x)+x^{2}(\sec^{2}x+\sin x)$

$\therefore \displaystyle \lim_{x\rightarrow 0} \dfrac{f'(x)}{x}=\lim_{x\rightarrow 0}\dfrac{2x(\tan x-\cos x)+x^{2}(\sec^2{x}+\sin x)}{x}$

$=\displaystyle \lim_{x\rightarrow 0} 2(\tan x-\cos x)+x(\sec^{2}x+\sin x)$

$=2(0-1)+0=-2$

Hence,

$\therefore \displaystyle \lim_{x\rightarrow 0} \dfrac{f'(x)}{x}=-2$

$\displaystyle\frac{d}{dx}\tan^{-1}\left(\displaystyle\frac{1-x}{1+x}\right)=$ ____________.

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

Explanation

Let

$x=tan\theta$

$\dfrac{d}{dx}tan^{-1}(\dfrac{1-x}{1+x})$

$=\dfrac{d}{dx}tan^{-1}(\dfrac{1-tan\theta}{1+tan\theta})$

$=\dfrac{d}{dx}tan^{-1}(tan(\dfrac{\pi}{4}-\theta))$

$=\dfrac{d}{dx}(\dfrac{\pi}{4}-\theta))$

$=\dfrac{d}{dx}(\dfrac{\pi}{4}-tan^{-1}x))$

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

Hence, the answer is option (B).

Differentiate the following w.r.t.

$x$ .

$\sin x\ log x$ .

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$\dfrac{\sin x}{x}-\cos x\,\, log \,x$
0%
$\dfrac{\sin x}{x}+\cos x\,\, log \,x$
0%
$\dfrac{\cos x}{x}+\cos x\,\, log \,x$
0%
$\dfrac{\tan x}{x}+\cos x\,\, log \,x$

Explanation

$\cfrac { d }{ dx } \sin { x } \log { x }$

$=\left\{ \cfrac { d }{ dx } \sin { (x) } \right\} \log { x } +\sin { x } .\cfrac { d }{ dx } \left( \log { \left( x \right) } \right)$

$=\log { x } .\cos { x } +\sin { x } .\cfrac { 1 }{ x }$

$=\cfrac { \sin { x } }{ x } +\log { x } .\cos { x }$

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 6 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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