If $$y=\cos ^{ -1 }{ (\frac { x-{ x }^{ -1 } }{ x+{ x }^{ -1 } } ) } \quad then\quad \frac { dy }{ dx } =$$

$$\frac { -2 }{ 1+{ x }^{ 2 } } $$

$$\frac { 2 }{ 1+{ x }^{ 2 } } $$

$$\frac { 1 }{ 1+{ x }^{ 2 } } $$

$$\frac { -1 }{ 1+{ x }^{ 2 } } $$

If $$y=\tan ^{ -1 }{ [x+\sqrt { 1+{ x }^{ 2 } } ] } $$ then $$\frac { dy }{ dx } =$$

$$\frac { 1 }{ 2(1+{ x }^{ 2 }) } $$

$$\frac { 1 }{ 1+{ x }^{ 2 } } $$

$$\frac { 2 }{ 1+{ x }^{ 2 } } $$

If $$f\left( x \right) =\sqrt { 1-\sqrt { 1-{ x }^{ 2 } } } $$, then $$f(x)$$ is

differentiable on $$(-1,0)\cup (0,1)$$

MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 14

$f\left(x\right)$ is a differentiable function in the interval

$\left(0,\infty\right)$ such that

$f\left(1\right)=1$ and

$\lim_{t\rightarrow x}\dfrac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1$ , for each

$x>0$ , then

$f\left(\dfrac{3}{2}\right)$ is equal to:

0%
$\dfrac{25}{9}$
0%
$\dfrac{13}{6}$
0%
$\dfrac{23}{18}$
0%
$\dfrac{31}{18}$

Let

$f\left( x \right)=x\left| x \right| ,g\left( x \right)=sinx$ and

$h\left( x \right) =\left( gof \right) \left( x \right) .$ Then

0%
$h'\left( x \right)$ is differentiable at x=0
0%
$h'\left( x \right)$ is continuous at x=0 but is not differentiable at x=0
0%
$h'\left( x \right)$ is differentiable at x=0 but $h'\left( x \right)$ differentiable is not continuous at x=0
0%
$h'\left( x \right)$ is not differentiable at x=0

$f(x)=0$ for

$x<0$ and

$f(x)$ is differentiable at

$x=0$ , then for

$x\ge 0, f(x)$ may be

0%
$x^2$
0%
$x$
0%
$-x$
0%
$-x^{3/2}$

The value of

$f(4)$ is

0%
$160$
0%
$240$
0%
$200$
0%
$None\ of\ these$

$y = {\sin ^{ - 1}}x + {\sin ^{ - 1}}\sqrt {1 - {x^2}}$ , then

$\frac{{dx}}{{dx}}$

0%
$0$
0%
$1$
0%
$-2x$
0%
$\frac{{y}}{{x}}$

$y = {x^{20}},{\text{find}}\dfrac{{{d^2}y}}{{d{x^2}}}$

0%
$381{x^{18}}$
0%
$380{x^{18}}$
0%
$389{x^{18}}$
0%
$370{x^{18}}$

$x=\theta-\frac{1}{\theta},y=\theta+\frac{1}{\theta}$ then

$\dfrac{dy}{dx}$ =

0%
$\dfrac{xy}{y^2+2}$
0%
y/x
0%
-x/y
0%
-y/x

Explanation

$x=\theta-\dfrac{1}{\theta}$

$\dfrac{dx}{d\theta}=1-\left(\dfrac{-1}{{\theta}^{2}}\right)$

$\therefore \dfrac{dx}{d\theta}=1+\dfrac{1}{{\theta}^{2}}$

$y=\theta+\dfrac{1}{\theta}$

$\dfrac{dy}{d\theta}=1+\left(\dfrac{-1}{{\theta}^{2}}\right)$

$\therefore \dfrac{dy}{d\theta}=1-\dfrac{1}{{\theta}^{2}}$

$\therefore \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}=\dfrac{1-\dfrac{1}{{\theta}^{2}}}{1+\dfrac{1}{{\theta}^{2}}}$

We have

$x=\theta-\dfrac{1}{\theta}$ and

$y=\theta+\dfrac{1}{\theta}$

$\Rightarrow xy={\theta}^{2}-\dfrac{1}{{\theta}^{2}}$

and

${y}^{2}={\left(\theta+\dfrac{1}{\theta}\right)}^{2}$

$\Rightarrow {y}^{2}={\theta}^{2}+\dfrac{1}{{\theta}^{2}}-2\theta\dfrac{1}{\theta}$

$\Rightarrow {y}^{2}={\theta}^{2}+\dfrac{1}{{\theta}^{2}}-2$

$\Rightarrow {y}^{2}+2={\theta}^{2}+\dfrac{1}{{\theta}^{2}}$

$\therefore \dfrac{dy}{dx}=\dfrac{1-\dfrac{1}{{\theta}^{2}}}{1+\dfrac{1}{{\theta}^{2}}}$

$=\dfrac{xy}{{y}^{2}+2}$

$y=\cos ^{ -1 }{ (\frac { x-{ x }^{ -1 } }{ x+{ x }^{ -1 } } ) } \quad then\quad \frac { dy }{ dx } =$

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

$dx\left\{ sin{ }^{ -1 }(\frac { 5x+12\sqrt { 1-x{ }^{ 2 } } }{ 13 } ) \right\} =$

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

Let fbe twice differentiable function such that

${ g }^{ 1 }\left( x \right) =-f\left( x \right) and\quad \quad \quad \quad \quad { f }^{ 1 }\left( x \right) =g\left( x \right) ,\\ h(x)={ \left( f\left( x \right) \right) }^{ 2 }+{ \left( g\left( x \right) \right) }^{ 2 }.\quad Ifh(5)=11,\quad thenh(10)\quad is$

0%
22
0%
11
0%
0
0%
8

$y=\tan ^{ -1 }{ [x+\sqrt { 1+{ x }^{ 2 } } ] }$ then

$\frac { dy }{ dx } =$

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

$\frac { d[\sec ^{ -1 }{ (\sin { x } +{ x }^{ 2 }) } ] }{ dx } =$

0%
$\frac { \sin { 2x } +{ x }^{ 2 } }{ \sin { x } +{ x }^{ 2 }()\sqrt { { (\sin { x } +{ x }^{ 2 }) }^{ 2 }-1 } }$
0%
$\frac { \cos { x } +2x }{ \sin { x } +{ x }^{ 2 }()\sqrt { { (\sin { x } +{ x }^{ 2 }) }^{ 2 }-1 } }$
0%
$\frac { 2x }{ \sin { x } \sqrt { \sin { x } -1 } }$
0%
None of these

$f(x) = \log_{1 - 2x}(1 + 2x)$ for

$x \ne 0$

$= k$ for

$x = 0$
is continuous at

$x = 0$ , find

$k.$

0%
$1$
0%
$-1$
0%
$0$
0%
$\frac{1}{2}$

$The\, \, function\, \, g(x)=\left| { \begin{array} { *{ 20 }{ c } }{ x+b } & { x<0 } \\ { \cos \, x } & { x\ge 0 } \end{array} } \right. \, \, can\, \, be\, \, made\, differentiable\, at\, x=0$

0%
$if\,b\,\,is\,\,equal\,to\,zero\,$
0%
$if\,b\,\,is\,\,not\,equal\,\,to\,zero\,$
0%
$if\,b\,\,is\,\,not\,equal\,\,to\,zero\,$
0%
$for\,no\,value\,of\,b$

$y={ tan }^{ -1 }\left( \dfrac { 1-{ cos }^{ 2 }x }{ 1+{ cos }^{ 2 } } \right) ,$ , then

$\dfrac { dy }{ xy } =$

0%
$\dfrac { sinx }{ 1+{ cos }^{ 4 }x }$
0%
$\dfrac { sinx }{ 1+{ cos }^{ 2 }x }$
0%
$\dfrac { sin2x }{ 1+{ cos }^{ 4 }x }$
0%
$\dfrac { sin2x }{ 1+{ cos }^{ 2 }x }$

Solve

$\frac{d\tan ^{-1}}{dx} (\frac{5x + 1}{3 - x - 6x^2}) =$

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

$\cos ^{-1}x + \cos ^{-1}y = \frac{\pi}{2}$ then

$\frac{dy}{dx}$

0%
$\frac{x}{y}$
0%
$\frac{-2x}{y}$
0%
$\frac{-x}{y}$
0%
$\frac{-y}{x}$

Let

$f$ be a function defined by

$2f\left(\sin x\right) + f\left(\cos x\right) = x ~\forall x\in R$ , then set of points where

$f$ is not differentiable is

0%
set of all natural numbers
0%
set of all irrational number
0%
$\left\{0,1,-1\right\}$
0%
$\left\{1,-1\right\}$

$f\left( x \right) =\sqrt { 1-\sqrt { 1-{ x }^{ 2 } } }$ , then

$f(x)$ is

0%
continuous on $[-1,1]$
0%
differentiable on $(-1,0)\cup (0,1)$
0%
both (a) and (b)
0%
None of the above

$y={ cos }^{ -1 }\cfrac { (9-{ x }^{ 2 }) }{ (9+{ x }^{ 2 }) }$ then

$Y'(-1)$ is equal to:

0%
$\cfrac {3 }{ 5 }$
0%
$\cfrac { 3 }{ 5 }$
0%
$\cfrac { 2 }{ 7 }$
0%
$\cfrac { 3 }{ 8 }$

The number of values of

$x$ at which

$f(x) = \left| x - \frac { 3 } { 2 } \right| + | x - 2 | + ( x - 1 ) | x - 1 |$ is not differentiable

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

$f(x)$ is differentiable function and

$f(1) = \sin 1, f(2) = \sin 4, f(3) = \sin 9$ , then the minimum number of distinct solution of equation

$f'(x) = 2x \cos x^2$ in

$(1, 3)$ is

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

Solve :

$\dfrac { d } { d x } \left[ \tan ^ { - 1 } \sqrt { 1 + x ^ { 2 } } - \cot ^ { - 1 } \left( - \sqrt { 1 + x ^ { 2 } } \right) \right] =?$

0%
$\pi$
0%
$1$
0%
$0$
0%
$\dfrac { 2 x } { \sqrt { 1 + x ^ { 2 } } }$

The derivative of

$\cos^{-1} x$ w.r.t.

$\sqrt{1 - x^2}$ is

0%
$\dfrac{1}{\sqrt{1 - x^2}}$
0%
$\sqrt{1 - x^2}$
0%
$\dfrac{1}{2x}$
0%
$\dfrac{1}{x}$

let f be the differeniable at x=0 and f'(0)=1 then

$\underset { b\rightarrow 0 }{ lim } \frac { f(h)-f(-2h) }{ h }$

0%
3
0%
2
0%
1
0%
-1

$f$ is twice differentiable such that

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

$f ^ { \prime } ( x ) = g ( x ).$ If

$h ( x )$ is a twice differentiable function that

$h ^ { \prime } ( x ) = ( f ( x ) ) ^ { 2 } + ( g ( x ) ) ^ { 2 } .$ If

$h ( 0 ) = 2 , h ( 1 ) = 4 ,$ then the equation

$y = h ( x )$ represents :

0%
a curve of degree $2$
0%
a curve passing through the origin
0%
a straight line with slope $2$
0%
a straight line with $y$ intercept equal to $2$

$f(x)=\begin{cases} x\left( \dfrac { a{ e }^{ \frac { 1 }{ \left| x \right| } }+3{ e }^{ \frac { -1 }{ x } } }{ \left( a+2 \right) { e }^{ \frac { 1 }{ \left| x \right| } }-{ e }^{ \frac { -1 }{ x } } } \right),\ \ x\neq 0 \\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0 \end{cases}$ is differentiable at

$x=0$ then

$[a]=......$ (

$[\ ]$ denotes greatest integer function)

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

$y = {\cot ^{ - 1}}\left( {\frac{{\sin x}}{{1 + \cos x}}} \right)$ then

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

0%
$\frac{{ - 1}}{2}$
0%
$\frac{{ 1}}{2}$
0%
$0$
0%
$1$

Number of points in

$[0,\pi]$ , where

$f(x)=\left[\dfrac{x\tan{x}}{\sin{x}+\cos{x}}\right]$ is non-differentiable is/are (where

$[.]$ denotes the greatest integer function)

0%
$0$
0%
$4$
0%
$9$
0%
$None\ of\ these$

Let F(x) =

$\left( f\left( x \right) \right) ^{ 2 }+\left( f\left( x \right) \right) ^{ 2 },F\left( 0 \right) -6$ where f(x) is a differential function such that

$\left| f\left( x \right) \right| \le 1\forall x\notin \left[ -1,1 \right]$ then choose the correct statement (s)

0%
There is atleast one point in each of the intervals (-1, 0) and (0, 1) where $\left| f'(x) \right| \le 2$
0%
there is atleast one point in a each of the interval (-1, 0) and (0, 1) where f(x) $\le S$
0%
there is no point of local maximum of F(x) in (-1, 1)
0%
For some $c\in \left( -1,1 \right)$ , $F'\left( c \right) \ge 6,F"\left( c \right) \le 0$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 14 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 14 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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