If $$y=\log_{3}x+3 \log_{e} x+2 \tan x$$, then $$\displaystyle \frac{dy}{dx}=$$

$$\displaystyle \frac {1}{x \log_e 3}+\displaystyle \frac {3}{x}+ \sec^2 x$$

$$\displaystyle \frac {1}{\log_e 3}+\displaystyle \frac {3}{x}+2 \sec^2 x$$

If $$y=x^2+sin^{-1}x+log_ex$$, find $$\dfrac {dy}{dx}$$

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

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

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

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

MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 4

$[\mathrm{x}]$ denotes the greatest integer contained in

$\mathrm{x},$ then for 4

$<\mathrm{x}<5,\ \displaystyle \frac{d}{dx}\{[x]\}=$

0%
$[x - 4, 5]$
0%
$[x]$
0%
$0$
0%
$1$

Explanation

$[\ ]$ denote greatest integer function ,i.e

$[n+f]=n$ ,

where

$n=integer\$ and

$f= fraction$ ,

$4<x<5$ , then

$[x] =4 =$ constant
Hence

$\dfrac{d}{dx}\{[x]\} =\dfrac{d}{dx}(4)=0$

For the function

$f(x) = \displaystyle \frac{x^{100}}{100} + \frac{x^{99}}{99} + ........... + \frac{x^2}{2} + x+1$ ,

$f'(1) =$

0%
$x^{100}$
0%
$100$
0%
$101$
0%
None of these

Explanation

$Given:\cfrac { d }{ dx } (x^{ n })=nx^{ n-1 }\\ \therefore for\; f(x)=\cfrac { x^{ 100 } }{ 100 } +\cfrac { x^{ 99 } }{ 99 } +...............+\cfrac { x^{ 2 } }{ 2 } +1\\ f'(x)=\cfrac { 100x^{ 99 } }{ 100 } +99\cfrac { x^{ 98 } }{ 99 } +.............+\cfrac { 2x }{ 2 } +1\\ Here\; f'(1)=1+1+.........to\; 100term=\; 100\\ Hence\; f(x)=\cfrac { x^{ 100 } }{ 100 } +\cfrac { x^{ 99 } }{ 99 } +...........+\cfrac { x^{ 2 } }{ 2 } +x+1=100$

$f(x)=\left\{\begin{matrix}\displaystyle \frac{1-\cos x}{x} &x\neq 0 \\ 0 & x=0\end{matrix}\right.$ , then

$f^{'}(0)=$

0%
$\dfrac{1}{2}$
0%
$\dfrac {1}{4}$
0%
$\dfrac{3}{4}$
0%
Does not exist

Explanation

Using fundamental theorem,

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

$=\displaystyle \lim_{h\to 0}\frac{\dfrac{1-\cos h}{h}-0}{h}$

$=\displaystyle \lim_{h\to 0}\frac{1-\cos h}{h^2}$

$=\displaystyle \lim_{h \rightarrow 0} \dfrac{1}{2}.\frac{\sin^2 \frac{h}{2}}{\frac{h^2}{4}}= \dfrac{1}{2}$

Derivative of which function is

$f'(x) = x \sin x$ ?

0%
$x \sin x + \cos x$
0%
$x \cos x+ \sin x$
0%
$x\sin \left ( \dfrac{\pi}{2}-x \right ) + \cos \left ( \dfrac{\pi}{2}-x \right )$
0%
$x \cos \left ( \dfrac{\pi}{2}-x \right ) + \sin \left ( \dfrac{\pi}{2}-x \right )$

Explanation

We know

$\frac { d }{ dx } \left( f.g \right) =\frac { df }{ dx } .g+\frac { dg }{ dx } .f\\ \Rightarrow \frac { d }{ dx } \left( x.sin(x) \right) =\frac { d(x) }{ dx } .sin(x)+\frac { d(sin(x)) }{ dx } .x\\ \Rightarrow 1\times sin(x)+cos(x).x=x.cos(x)+sin(x)\\$

We know that

$cos(x)=sin\left( \frac { \pi }{ 2 } -x \right) sin(x)=cos\left( \frac { \pi }{ 2 } -x \right) \\\Rightarrow x.sin(\left( \frac { \pi }{ 2 } -x \right) +cos\left( \frac { \pi }{ 2 } -x \right)$

Both B and C are correct

$y=x^{-\tfrac12}+\log_5x+\displaystyle \frac {\sin x}{\cos x}+2^x$ , then find

$\dfrac {dy}{dx}$

0%
$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$
0%
$\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$
0%
$-\displaystyle \frac {3}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$
0%
$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\cos^2x+2^x\log 2$

Explanation

Here, we have function

$y=x^{-1/2}+\log _5x+\tan x+2^x$

On differentiating w.r.t x, we get

$\dfrac {dy}{dx}=\dfrac {d}{dx}(x)^{-1/2}+\dfrac {d}{dx}(\log _5x)+\dfrac {d}{dx} \tan x+\dfrac {d}{dx}(2^x)$

$=-\dfrac {1}{2}(x)^{-1/2-1}+\dfrac {1}{x \log _e5}+\sec ^2x+2^x \log 2$

$=-\dfrac {1}{2}x^{-3/2}+\dfrac {1}{x\log _e5}+\sec ^2x+2^x\log 2$

$\displaystyle y=5^{3-x^2}+(3-x^2)^5$ , then

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

0%
$-2x\left \{5^{3-x^2}\cdot \log_e5+5(3-x^2)^4\right \}$
0%
$-x\left \{5^{3-x^2}\cdot \log_e5+5(3-x^2)^4\right \}$
0%
$-2x\left \{5^{3-x^2}\cdot \log_e5+(3-x^2)^4\right \}$
0%
$-2x\left \{5^{3-x^2}+5(3-x^2)^4\right \}$

Explanation

$\displaystyle \frac { d }{ dx } \left( 5^{ 3-x^{ 2 } }+(3-x^{ 2 })^{ 5 } \right)$

$={ 5 }^{ 3-{ x }^{ 2 } }\log _{ e }{ 5\displaystyle \frac { d }{ dx } \left( 3-{ x }^{ 2 } \right) +5{ \left( 3-{ x }^{ 2 } \right) }^{ 4 }\displaystyle \frac { d }{ dx } { \left( 3-{ x }^{ 2 } \right) } }$

$={ 5 }^{ 3-{ x }^{ 2 } }\log _{ e }{ 5\left( -2x \right) } +5{ \left( 3-{ x }^{ 2 } \right) }^{ 4 }\left( -2x \right)$

$=-2x\left( { 5 }^{ 3-{ x }^{ 2 } }\log _{ e }{ 5+5\left( 3-{ x }^{ 2 } \right) } \right)$

$y=\log_{3}x+3 \log_{e} x+2 \tan x$ , then

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

0%
$\displaystyle \frac {1}{x \log_e 3}+\displaystyle \frac {3}{x}+2 \sec^2 x$
0%
$\displaystyle \frac {1}{x \log_e 3}+\displaystyle \frac {3}{x}+ \sec^2 x$
0%
$\displaystyle \frac {1}{\log_e 3}+\displaystyle \frac {3}{x}+2 \sec^2 x$
0%
$\displaystyle \frac {1}{x \log_e 3}-\displaystyle \frac {3}{x}+2 \sec^2 x$

Explanation

$\displaystyle \frac { d }{ dx } \left( \log _{ 3 }{ x } +3\log _{ e }{ x } +2\tan { x } \right)$

$=\displaystyle \frac { d }{ dx } \left( \displaystyle \frac { \log_e { x } }{ \log { 3 } } +3\log_e { x } +2\tan { x } \right)$

$=\displaystyle \frac { 1 }{ x\log _{ e }{ 3 } } +\displaystyle \frac { 3 }{ x } +2\sec ^{ 2 }{ x }$

$y=x^2+sin^{-1}x+log_ex$ , find

$\dfrac {dy}{dx}$

0%
$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$
0%
$\displaystyle \frac {dy}{dx}=x+\displaystyle \frac{1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$
0%
$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}-\displaystyle \frac {1}{x}$
0%
$\frac {dy}{dx}=2x-\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$

Explanation

$y=x^2+sin^{-1}+log_ex$
On differentiating, we get

$\dfrac {dy}{dx}=\dfrac {d}{dx}(x^2)+\dfrac {d}{dx}sin^{-1}x)+\dfrac {d}{dx}(log_ex)$
or

$\dfrac {dy}{dx}=2(x)^{2-1}+\dfrac {1}{\sqrt {1-x^2}}+\dfrac {d}{dx}(log_ex)$

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

$\displaystyle y=e^{x \log a}+e^{a \log x}+e^{a \log a}$ , then

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

0%
$a^x \log a+x^{a-1}$
0%
$a^x \log a+ax$
0%
$a^x \log a+ax^{a-1}$
0%
$a^x \log a+ax^{a}$

Explanation

Let

$y=\displaystyle e^{x log a}+e^{a log x}+e^{a log a}$

$={ e }^{ \log _{ e }{ { a }^{ x } } }+{ e }^{ \log _{ e }{ { x }^{ a } } }+{ e }^{ \log _{ e }{ { a }^{ a } } }$

$={ a }^{ x }+{ x }^{ a }+{ a }^{ a }$ ......... Since

${ e }^{ \log _{ e }{ x } }=x$

$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left( { a }^{ x }+{ x }^{ a }+{ a }^{ a } \right)$

$={ a }^{ x }\log { a } +a.{ x }^{ a-1 }$

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

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

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

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

Explanation

$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left( |cosx|+|sinx| \right)$

$=-\sin { x } +\cos { x }$ at

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

$=\left| -\sin { \displaystyle \frac { 2\pi }{ 3 } } \right| +\left| \cos { \displaystyle \frac { 2\pi }{ 3 } } \right|$

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

$=\displaystyle \frac { 1 }{ 2 } \left( \sqrt { 3 } -1 \right)$

Find the derivative of

$\sec^{-1}\left (\displaystyle \frac {x+1}{x-1}\right )+\sin^{-1}\left (\displaystyle \frac {x-1}{x+1}\right )$

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

Explanation

Let

$y=\sec^{-1}\left (\displaystyle \frac {x+1}{x-1}\right )+\sin^{-1}\left (\displaystyle \frac {x-1}{x+1}\right )$

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

$\therefore \displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left(\displaystyle \frac { \pi }{ 2 } \right) =0$

$\because \sin ^{ -1 }{ \theta +\cos ^{ -1 }{ \theta } =\displaystyle \frac { \pi }{ 2 } }$

$y=\log_{10}x+\log_x 10+\log_xx+\log_{10} 10$ , then

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

0%
$\displaystyle \frac {1}{x \log_e 10}-\displaystyle \frac {\log_e 10}{x(\log_ex)^2}$
0%
$\displaystyle \frac {1}{\log_e 10}-\displaystyle \frac {\log_e 10}{x(\log_ex)^2}$
0%
$\displaystyle \frac {1}{x \log_e 10}-\displaystyle \frac {\log_e 10}{x^2(\log_ex)^2}$
0%
None of these

Explanation

$\displaystyle \dfrac { d }{ dx } \left( \log _{ 10 }{ x+\log _{ x }{ 10+\log _{ x }{ x+\log _{ 10 }{ 10 } } } } \right)$

$=\displaystyle \dfrac { 1 }{ x\log _{ e }{ 10 } } +\displaystyle \dfrac { \left( -\log _{ e }{ 10\left( \displaystyle \dfrac { 1 }{ x } \right) } \right) }{ { \left( \log_e { x } \right) }^{ 2 } }$

$=\displaystyle \dfrac { 1 }{ x\log _{ e }{ 10 } } -\displaystyle \dfrac { \log _{ e }{ 10 } }{ x{ \left( \log_e { x } \right) }^{ 2 } }$

$\displaystyle \lim_{x\to0}\left( x^{-3}\sin{3x} + ax^{-2} + b \right)$ exists and is equal to 0, then

0%
$a = -3$ and $b = \dfrac{9}{2}$
0%
$a = 3$ and $b = \dfrac{9}{2}$
0%
$a = -3$ and $b = -\dfrac{9}{2}$
0%
$a = 3$ and $b = -\dfrac{9}{2}$

Explanation

$\displaystyle\lim_{x\to\infty}\displaystyle\frac{\sin3x}{x^3}+\displaystyle\frac{a}{x^2}+b = \lim_{x\to0}\displaystyle\frac{\sin3x+ax+bx^3}{x^3}$

$=\displaystyle\lim_{x\to0}\displaystyle\frac{3\displaystyle\frac{\sin3x}{3x}+a+bx^2}{x^2}$

$\mbox{For existence, }$

$(3+a)=0 \mbox{ or } a=-3$

$\therefore L=\displaystyle\lim_{x\to0}\displaystyle\frac{\sin3x-3x+bx^3}{x^3}$

$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{\sin t - t}{t^3}+b = 0 ........ (3x=t )$
The left hand side reduces to

$\dfrac{0}{0}$ form by substituting the limit

Then, using L'Hospital's rule

$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{\cos t - 1}{3t^2}+b = 0$

Again, applying L'Hospital's rule

$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{-\sin t}{6t}+b = 0$

$\Rightarrow -\displaystyle\frac{27}{6}+b=0 \mbox{ or } b=\frac{9}{2}$

$y=logx^3+3 sin^{-1}x+kx^2$ , then find

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

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

Explanation

Here,

$y=\log x^3+3 \sin^{-1} x+kx^2$

On differentiating we get

$\displaystyle \dfrac {dy}{dx}=\dfrac {d}{dx}[\log x^3]+\dfrac {d}{dx}[3 \sin^{-1}x]+\dfrac {d}{dx}[kx^2]$

$=3\displaystyle \dfrac {d}{dx}[\log x]+3\dfrac {d}{dx}(\sin^{-1}x)+k\dfrac {d}{dx}(x^2)$

$=3\cdot \displaystyle \dfrac {1}{x}+3\cdot \dfrac {1}{\sqrt {1-x^2}}+k(2x)$

The value of

$\displaystyle\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$ for

$(a>1)$ is equal to?

0%
$1$
0%
$0$
0%
$\displaystyle\frac{\pi}{2}$
0%
Does not exist

Explanation

$\mathrm{L} \displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$

$\displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{\log{x}}{\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log a}{\log x}}}}=\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{1}{a\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log^2 a. x}{1}}}}=\frac{\cot^{-1}0}{\sec^{-1} \infty}=\frac{\pi/2}{\pi/2}=1$
Note: L-Hospital's rule has been used in step two in both numerator and denominator alone.

The value of

$\displaystyle \lim_{x \rightarrow \pi/6} \frac{2 \sin^2 x + \sin x-1}{2 \sin^2 x - 3 \sin x + 1}$

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

Explanation

We have

$\displaystyle \lim_{x \rightarrow \pi/6} \frac{2 sin^2 x + sin x -1}{2 sin^2 x-3 sin x + 1}$

$\displaystyle = \lim_{x \rightarrow \pi/6} \frac{(2 sin x - 1)(sin x + 1)}{(2 sin x - 1)(sin x - 1)}$

$\displaystyle = \lim_{x \rightarrow \pi/6} \frac{sin x + 1}{sin x - 1} = - 3$

$y = sec^{-1}\left(\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}\right) + \sin^{-1}\left(\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}\right)$ , then

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

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

Explanation

$y = sec^{-1}\left(\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}\right) + \sin^{-1}\left(\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}\right)$

$y=\cos ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) +\sin ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) } } =\displaystyle \frac { \pi }{ 2 }$ .............

$\because \sin ^{ -1 }{ \theta +\cos ^{ -1 }{ \theta } =\displaystyle \frac { \pi }{ 2 } }$

$\therefore \displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left(\displaystyle \frac { \pi }{ 2 } \right) =0$

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

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

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

Explanation

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

Differentiating both the sides

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

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

$\Rightarrow \log { 2\left( { 2 }^{ x }+{ 2 }^{ y }\displaystyle \dfrac { dy }{ dx } \right) =\log { 2 } \left( { 2 }^{ x+y }+{ 2 }^{ x+y }\displaystyle \dfrac { dy }{ dx } \right) }$

$\Rightarrow \left( { 2 }^{ y }-{ 2 }^{ x+y } \right)\displaystyle \dfrac { dy }{ dx } ={ 2 }^{ x+y }-{ 2 }^{ x }$

$\Rightarrow \displaystyle \dfrac { dy }{ dx } =\displaystyle \dfrac { { 2 }^{ x }{ 2 }^{ y }-{ 2 }^{ x } }{ { 2 }^{ y }-{ 2 }^{ x }{ 2 }^{ y } } =\displaystyle \dfrac { { 2 }^{ x }\left( { 2 }^{ y }-1 \right) }{ { 2 }^{ y }\left( 1-{ 2 }^{ x } \right) }$

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

$f'(x)=\sin x+\sin 4x\cdot \cos x$ , then

$f'\left (2x^2+\displaystyle \frac {\pi}{2}\right )$ is

0%
$4x\left \{\cos(2x^2)-sin 8x^2\cdot \sin 2x^2\right \}$
0%
$4x\left \{\cos(2x^2)+\sin 8x^2\cdot \sin 2x^2\right \}$
0%
$\left \{\cos (2x^2)-\sin 8x\cdot \sin 2x^2\right \}$
0%
none of the above

Explanation

$f'(x)=\sin x+\sin 4x\cdot \cos x$

$f'\left (2x^2+\displaystyle \frac {\pi}{2}\right )$ =

$\displaystyle \frac { d }{ dx } \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) \left[ \sin { \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) +\sin { 4 } \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) .\cos { \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) } } \right]$

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

The solution set of

${f}'(x)>{g}'(x)$ where

$f(x)=\displaystyle \frac{1}{2}(5^{2x+1})$ &

$g(x)= 5^x+4x(\ln 5)$ is

0%
$x>1$
0%
$0< x< 1$
0%
$x \leq 0$
0%
$x>0$

Explanation

Given,

$f(x)= \dfrac {1}{2}(5^{2x+1})$

$f'(x)= 5^{2x+1}\times \log 5$

$g'(x)=5^{x}l\log 5 +4\ln 5=\ln 5(5^{x}+4)$

$f'(x)>g'(x)$

$=>5\times 5^{2x}>(5^{x}+4)$
Let

$t=5^{x}$

$=> 5t^{2}-t-4>0$

$=>(t-1)(t+\frac{4}{5})>0$

$=>t>1$ or

$t>-\frac{4}{5}$

$=>5^{x}>5^{0} =>x>0$
so,

$x>0$

$f:R\rightarrow R$ and

$\displaystyle f(x)=\frac {x(x^4+1)(x+1)+x^4+2}{x^2+x+1}$ , then

$f(x)$ is

0%
one-one ito
0%
many-one onto
0%
one-one onto
0%
many-one into

Explanation

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

$\displaystyle \quad \quad =\frac {x(x^4+1)(x+1)+(x^4+1)+1}{x^2+x+1}$

$\displaystyle \quad \quad =\frac {(x^4+1)(x^2+x+1)+1}{x^2+x+1}$

$\displaystyle \quad \quad =(x^4+1)+\frac {1}{x^2+x+1}$

$\displaystyle f'(x) =4x^3-\frac {2x+1}{(x^2+x+1)^2}=$ not always positive or negative
Thus,

$f$ is many one.

Also range and co-domain of

$f$ are not same,
Hence is many-one into function

Suppose the function

$f(x)-f(2x)$ has the derivative

$5$ at

$x=1$ and derivative

$7$ at

$x=2$ .The derivative of the function

$f(x)-f(4x)$ at

$x=1$ , has the value equal to

0%
$19$
0%
$9$
0%
$17$
0%
$14$

Which one of the following statements is true?

0%
If $\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$ and $\displaystyle\lim_{x\rightarrow c}{f(x)}$ exist, then $\displaystyle\lim_{x\rightarrow c}{g(x)}$ exists.
0%
If $\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$ exists, then $\displaystyle\lim_{x\rightarrow c}{f(x)}$ and $\displaystyle\lim_{x\rightarrow c}{g(x)}$ exist.
0%
If $\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$ and $\displaystyle\lim_{x\rightarrow c}{f(x)}$ exist, then $\displaystyle\lim_{x\rightarrow c}{g(x)}$ also exists.
0%
If $\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$ exists, then $\displaystyle\lim_{x\rightarrow c}{f(x)}$ and $\displaystyle\lim_{x\rightarrow c}{g(x)}$ also exist.

Explanation

For option A and B, take

$f(x)=x$ and

$g(x)=\dfrac { 1 }{ x }$ .
Limit of

$g(x)$ doesn't exist at

$x=0$ , but for

$\displaystyle \lim _{ x\to 0} f(x).g(x)$ and

$f(x)=x$ limit exists.

For option C,
Lets assume

$h(x) = f(x)+g(x)$ ,
Take limit on both sides as

$x\to c$

$\lim_{x\to c}h(x) = \lim_{x\to c}(f(x)+g(x))$
And it is given that

$\lim_{x\to c}f(x)$ exists.
Using sum law of limits:
If

$\lim_{x\to c}F(x)$ and

$\lim_{x\to c}G(x)$ exists, then

$\lim_{x\to c}(F(x)\pm G(x))$ also exists.

$\therefore \lim_{x\to c}(h(x) - f(x)) = \lim_{x\to c}(f(x) + g(x) - f(x)) = \lim_{x\to c}g(x)$ exists.

For option D, take

$f(x)=-\dfrac{cosx}{x^2}$ and

$g(x)=\dfrac { 1 }{ x^2 }$ .
Limit of

$g(x)$ and

$f(x)=-\dfrac{cosx}{x^2}$ doesn't exist at

$x=0$ , but for

$\displaystyle \lim _{ x\to 0} f(x)+g(x)$ limit exists.

Hence option C.

$\displaystyle y=\frac { x }{ a+\displaystyle\frac { x }{ b+\displaystyle\frac { x }{ a+\displaystyle\frac { x }{ b+.....\infty } } } }$ , then

$\cfrac{dy}{dx} =$

0%
$\displaystyle\frac{a}{ab+2ay}$
0%
$\displaystyle\frac{b}{ab+2by}$
0%
$\displaystyle\frac{a}{ab+2by}$
0%
$\displaystyle\frac{b}{ab+2ay}$

Explanation

$\displaystyle y = \frac{x}{a+\frac{x}{b+y}}$

$\Rightarrow \displaystyle y =\frac{x(b+y)}{ab+ay+x}$

$\Rightarrow aby+ay^2+xy=bx+xy\Rightarrow aby+ay^2=bx$
Differentiating both sides w.r.t

$x$

$\Rightarrow\displaystyle (ab+2ay)\frac{dy}{dx}=b\therefore \frac{dy}{dx}=\frac{b}{ab+2ay}$

Given :

$f(x)=4x^3-6x^2\cos2a+3x \sin 2a.\sin 6a+\sqrt{\ln (2a-a^2)}$ then

0%
$f(x)$ is not defined at $x=\displaystyle \frac{1}{2}$
0%
${f}'(\displaystyle \frac{1}{2})<0$
0%
$f'(x)$ is not defined at $x=\displaystyle \frac{1}{2}$
0%
${f}'(\displaystyle \frac{1}{2})>0$

Explanation

$f'(x)=12x^{2}-12xcos2a+3sin2a.sin6a+0$

$f'(1/2)=\frac{12}{4}-\frac{12}{2}cos2a+3sin2a.sin6a$

$=3-6cos2a+1.5(cos4a-cos8a)$
So, it will alwyas be greater than

$0$

Let

$\displaystyle f\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ n } }{ n } \right) =\frac { f\left( { x }_{ 1 } \right) +f\left( { x }_{ 2 } \right) +...+f\left( { x }_{ n } \right) }{ n }$ where all

${ x }_{ i }\in R$ are independent to each other and

$n\in N$ . if

$f(x)$ is differentiable and

$f'\left( 0 \right) =a,f\left( 0 \right) =b$ and

$f'\left( x \right)$ is equal to

0%
$a$
0%
$0$
0%
$b$
0%
None of these

Explanation

Differentiating the given equation w.r.t.

${ x }_{ 1 },$ we get

$\displaystyle \dfrac { 1 }{ n } f'\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ n } }{ n } \right) =\dfrac { f'\left( { x }_{ 1 } \right) }{ n }$

$[$ Since all

${ x }_{ i }'s$ are independent to each other,

$\displaystyle \therefore \dfrac { d{ x }_{ i } }{ { dx }_{ j } } =0$ if

$i\neq j$ and

$\displaystyle \dfrac { { dx }_{ i } }{ { dx }_{ j } } =1$ if

$(i=j)]$

On putting

${ x }_{ 1 }={ x }_{ 2 }=...={ x }_{ n-1 }=0$ and

${ x }_{ n }=x,$ we get

$\displaystyle f'\left( \dfrac { x }{ n } \right) =f'\left( 0 \right) =a.$

On integrating, we get

$\displaystyle nf'\left( \dfrac { x }{ n } \right) =ax+c$

Since

$f(0)=b,$ we have

$c=nb$

$\displaystyle \therefore nf'\left( \dfrac { x }{ n } \right) =ax+nb\Rightarrow nf'\left( x \right) =nax+nb\Rightarrow f\left( x \right) =ax+b.$

$\displaystyle \therefore f'\left( x \right) =a,\forall x\in R$

$5f(x)+3f\left ( \displaystyle \frac{1}{x} \right )=x+2$ and

$y=xf(x)$ then

$\left (\displaystyle \frac{dy}{dx} \right )_{x=1}$ is equal to ?

0%
$14$
0%
$\displaystyle \frac{7}{8}$
0%
$1$
0%
none of these

Explanation

$5f(x)+3f\left ( \displaystyle \frac{1}{x} \right )=x+2\Rightarrow (1)$
Replace

$x$ by

$\dfrac{1}{x}$

$\Rightarrow 5f\left ( \displaystyle \frac{1}{x} \right )+3f(x)=\dfrac{1}{x}+2\Rightarrow (2)$

$\Rightarrow 5\times(1)-3\times (2)\Rightarrow 16f(x) =5x-\dfrac{3}{x}+4=\dfrac{5x^2+4x-3}{x}$

$\therefore y = xf(x) =\dfrac{5x^2+4x-3}{16}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{10x+4}{16}$

$\therefore \left(\dfrac{dy}{dx}\right)_{x=1}=\dfrac{10+4}{16}=\dfrac{7}{8}$

$\displaystyle f\left( x \right) =\sqrt { 1+\sqrt { x } } , x > 0,$ then

$\displaystyle f\left ( x \right )\cdot f'\left ( x \right )$ is equal to

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

Explanation

Let

$y=f\left( x \right) =\sqrt { 1+\sqrt { x } }$

Then,

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

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

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

$\Rightarrow f(x)f'(x)=\displaystyle \frac{1}{4\sqrt{x}}$

$y=\sqrt{\sin x+\sqrt{\sin x +\sqrt{\sin x+-\infty }}}$ then

$\displaystyle \frac{dy}{dx}$ equals:

$(\sin x> 0)$

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

Explanation

$y=\sqrt{\sin x\sqrt{\sin x \sqrt{\sin x+-\infty }}}$
Taking square on both the sides, we get

${ y }^{ 2 }=\sin { x } +\sqrt { \sin x\sqrt { \sin x\sqrt { \sin x+-\infty } } }$

${ y }^{ 2 }=\sin { x } +y$ ............ Since

$y=\sqrt{\sin x\sqrt{\sin x \sqrt{\sin x+-\infty }}}$

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

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

$f\left( x \right)=\begin{cases} \sin { x } \qquad ;\qquad x\neq n\pi ,n=0,\pm 1,\pm 2,\pm 3..... \\ 2\qquad \qquad ;\qquad otherwise \end{cases}$ and

$g\left( x \right) =\begin{cases} { x }^{ 2 }+1\qquad ;\qquad x\neq 0 \\ 4\qquad \qquad ;\qquad x=0 \end{cases}.$

Then

$\lim _{ x\rightarrow 0 }{ g\left( f\left( x \right)\right)}$ is

0%
$1$
0%
$4$
0%
$5$
0%
non-existent

Explanation

$\lim _{ x\rightarrow 0 }{ g\left( f\left( x \right) \right) } =g\left( f\left( 0 \right) \right)$

$=g(2)={ 2 }^{ 2 }+1=5$

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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