If $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$, then $$\dfrac{d^{2}y}{dx^{2}}=$$

$$-\dfrac{-b^{3}}{a^{2}y^{3}}$$

MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 13

If f is twice differentianle such that f"

$\left( x \right) = - f\left( x \right),f'\left( x \right) = g\left( x \right),h'\left( x \right) = {\left( {f\left( x \right)} \right)^2} + {\left( {g\left( x \right)} \right)^2}$ and

$h\left( 0 \right) = 2,h\left( 1 \right) = 4$ ,then equation of

$h\left( x \right)$ represents ?

0%
curve of degree $2$
0%
curve passing through origin
0%
a straight line with slope $2$
0%
a straight line with slope $-2$

The value of

$\displaystyle \int _{ 0 }^{ \pi /2 }{ \frac { \sec ^{ 2 }{ x } dx }{ \left( \sec { x } +\tan { x } \right) ^{ n } } , } n>1$ is equal to

0%
$\dfrac{1}{{n}^{2}-1}$
0%
$\dfrac{1}{{n}^{2}+1}$
0%
$\dfrac{n}{{n}^{2}-1}$
0%
$\dfrac{n}{{n}^{2}+1}$

Let

$f$ be the function on

$[0,\,1]$ given by

$f\left( x \right) = x\sin \frac{\pi }{x}\,for\,x \ne 0$ and

$f(0)=0$ . Then

0%
Continuous but bounded
0%
Differentiable
0%
Continuous but not bounded
0%
Continuous and bounded

$f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ , then

$f(x)$ is differentiable on

0%
$\left(-\infty,\infty\right)$
0%
$\left[2,\infty\right)-\left\{4\right\}$
0%
$\left[2,\infty\right)$
0%
$\left(0,\infty\right)$

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

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

0%
$0$
0%
$\dfrac {1}{\sqrt {x}+1}$
0%
$1$
0%
$None\ of\ these$

$f(x)$ is a non constant polynomial function

$f:R\rightarrow R$ such that

$7\dfrac{d}{dx}(xf(x))=3f(x)+4f(x+1),\ f(-1)+f(0)=2$ , then number of such function is

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

The function

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

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

$y=\tan^{-1}\left(\dfrac{\ell n\dfrac{e}{x^{2}}}{\ell nx^{2}}\right)+\tan^{-1}\dfrac{3+2\ell nx }{1-6\ell nx}$ then

$\dfrac{d^{2}y}{dx^{2}}=$

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

$f(x)=p\left|\sin x\right|+qe^{\left|x\right|}+r\left|x\right|^{3}$ and

$f(x)$ is differentiable at

$x=0$ , then

0%
$q+r=0;p$ is any real number
0%
$p+q=0;r$ is any real number
0%
$q=0,r=0;p$ is any real number
0%
$r=0,0\p=o\0;q$ is any real number

The derivative of

$f (\tan^{-1}x)$ , where

$f(x)=\tan x$ is

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

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

$\dfrac{d^{2}y}{dx^{2}}=$

0%
$-\dfrac{b^{4}}{a^{2}y^{3}}$
0%
$-\dfrac{b^{2}}{ay^{2}}$
0%
$-\dfrac{-b^{3}}{a^{2}y^{3}}$
0%
$-\dfrac{b^{3}}{a^{2}y^{2}}$

$f(x) = \sqrt{x+2\sqrt{2x-4}} + \sqrt{x-2\sqrt{2x-4}}$ , then f(x) is differentiable on

0%
$(-\infty, \infty)$
0%
$[2, \infty) -$ { $4$ }
0%
$[2, \infty)$
0%
$[0, \infty)$

$f(x)=min {1,cos\,x,1-sin\,x}, -\pi \leq x \leq \pi$ then

0%
f(x) is not differentiable at '0'
0%
f(x) is differentiable at $\pi/2$
0%
f(x) has local maximum at '0'
0%
None of these

If y = tan

$^{-1}$

$\dfrac{\sqrt{1 + x^2} -1}{x}$ , then

$\dfrac{dy}{dx}$ =

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

Let

$f$ be a differentiable function for all

$x$ . If

$f(1) = -2$ and

$f'(x) \ge 2$ for

$x \in [1, 6]$ then

0%
$f(6) < 5$
0%
$f(6) = 5$
0%
$f(6) \ge 8$
0%
$f(6) < 8$

The number of points at which

$g(x)= \dfrac { 1 }{ 1+\dfrac { 2 }{ f(x) } }$ is not differentiable where

$f(x)= \dfrac { 1 }{ 1+\dfrac { 1 }{ x } }$ is

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

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

$\frac {dy}{dx}=$

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

$Cos^{-1}\left ( \frac{x^2-y^2}{x^2+y^2} \right )=a$ , then

$\frac{dy}{dx}=$

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

The derivate of

$\tan^{-1}[\dfrac {\sin x}{1+\cos x}]$ with respect to

$\tan^{-1}[\dfrac {\cos x}{1+\sin x}]$ is

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

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

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

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

$f(x)=a\left| sinx \right| +{ be }^{ x }+c\left| { x }^{ 3 } \right|$ , where

$a,b,c \epsilon$ R, is differentiable at

$x=0$ then

0%
$a=0,b and c$ are real numbers
0%
$c=0$ , $a=0$ , be is any real numbers
0%
$b=0,c=0$ a is any real numbers
0%
$a=0,b=0,c$ is any real numbers

$f(x)$ is differentiable function and

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

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

$(1, 3)$ is

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

Let f be a differentiable function satisfying the condition

$f\left( \dfrac { x }{ y } \right) =\dfrac { f\left( x \right) }{ f\left( y \right) } ,$ for all x, y

$\left( \neq 0 \right) \epsilon R$ and f(y)

$\neq$ If f'(1)=2, then f'(x) is equal to

0%
2 f(x)
0%
$\dfrac { f\left( x \right) }{ x }$
0%
2x f(x)
0%
$\dfrac { 2f\left( x \right) }{ x }$

Let f be a differentiable function satisfying the relation f(xy) = xf(y) - 2xy+yf(x) (where x,y > 0) and f(1) = 3, then

0%
f(x) =x $\ell nx+3x-\frac { { x }^{ 2 } }{ 2 }$
0%
$f(x)\quad =\quad x\ell nx$
0%
$x={ e }^{ -3 }$ is the abscissa of the point of inflection of f(x)
0%
the equation f(x) = k has two solution if x $\in \quad (-{ e }^{ -3 },0)$

Let

$f : R \rightarrow R$ be a twice differentiable function satisfying

$f^{ { \prime \prime } }(x)=5f^{ { \prime } }(x)+6f\geq 0,\forall x\geq 0,f(0)=1$ and

$f ^ { \prime } ( 0 ) = 0 .$ if

$f ( x )$ satisfies

$f(x)\geq a.h(bx)-b\cdot \overline { h } (ax),\forall x\geq 0,$ then

$a + b$ is equal to

0%
$3$
0%
$1$
0%
$6$
0%
$5$

$S$ be the set in which function defined by

$f(x)=\sin |x|- |x|+2(x-\pi)\cos |x|$ is differtentiable then no of element in

$S$ is

0%
$2$
0%
$3$
0%
$4$
0%
$\phi$

$y = \sin ^ { - 1 } \dfrac { 2 x } { 1 + x ^ { 2 } } ,$ then which of the following is not correct ?

0%
$\dfrac { d y } { d x } = \dfrac { 2 } { 1 + x ^ { 2 } }$ for $| x | < 1$
0%
$\dfrac { d y } { d x } = \dfrac { - 2 } { 1 + x ^ { 2 } }$ for $| x | > 1$
0%
$\dfrac { d y } { d x } = 2$ for $x = - 1$
0%
$\dfrac { d y } { d x }$ does not exist at $| x | = 1$

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

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

$y = {\cot ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right]\;then\;\dfrac{{dy}}{{dx}}\;is\;equal\;to\;$

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

The number of point at which the function F(x)= max

$\left\{ a-x,a+x,b \right\} -\infty <x<\infty ,0<a<b$ cannot be differentiable is

0%
1
0%
2
0%
3
0%
none of these

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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