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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 13 - MCQExams.com
CBSE
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 ?
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curve of degree $$2$$
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curve passing through origin
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a straight line with slope $$2$$
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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
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$$\dfrac{1}{{n}^{2}-1}$$
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$$\dfrac{1}{{n}^{2}+1}$$
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$$\dfrac{n}{{n}^{2}-1}$$
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$$\dfrac{n}{{n}^{2}+1}$$
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
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Continuous but bounded
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Differentiable
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Continuous but not bounded
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Continuous and bounded
If $$f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$$, then $$f(x)$$ is differentiable on
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$$\left(-\infty,\infty\right)$$
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$$\left[2,\infty\right)-\left\{4\right\}$$
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$$\left[2,\infty\right)$$
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$$\left(0,\infty\right)$$
If $$y=\sec^{-1}\left(\dfrac {\sqrt {x}+1}{\sqrt {x}-1}\right)+\sin^{-1}\left(\dfrac {\sqrt {x}-1}{\sqrt {x}+1}\right)$$, then $$\dfrac {dy}{dx}$$ is equal to
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$$0$$
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$$\dfrac {1}{\sqrt {x}+1}$$
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$$1$$
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$$None\ of\ these$$
If $$f(x)$$ 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
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$$0$$
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$$1$$
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$$2$$
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$$3$$
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
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$$-1$$
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$$0$$
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$$1$$
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$$2$$
If $$y=\tan^{-1}\left(\dfrac{\ell n\dfrac{e}{x^{2}}}{\ell nx^{2}}\right)+\tan^{-1}\dfrac{3+2\ell nx }{1-6\ell nx}$$ then $$\dfrac{d^{2}y}{dx^{2}}=$$
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$$2$$
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$$1$$
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$$0$$
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$$-1$$
If $$f(x)=p\left|\sin x\right|+qe^{\left|x\right|}+r\left|x\right|^{3}$$ and $$f(x)$$ is differentiable at $$x=0$$, then
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$$q+r=0;p$$ is any real number
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$$p+q=0;r$$ is any real number
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$$q=0,r=0;p$$ is any real number
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$$r=0,0\p=o\0;q$$ is any real number
The derivative of $$f (\tan^{-1}x)$$, where $$f(x)=\tan x$$ is
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$$1$$
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$$\dfrac{1}{1+x^{2}}$$
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$$2$$
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$$\dfrac{-1}{1+x^{2}}$$
If $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$, then $$\dfrac{d^{2}y}{dx^{2}}=$$
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$$-\dfrac{b^{4}}{a^{2}y^{3}}$$
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$$-\dfrac{b^{2}}{ay^{2}}$$
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$$-\dfrac{-b^{3}}{a^{2}y^{3}}$$
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$$-\dfrac{b^{3}}{a^{2}y^{2}}$$
If $$f(x) = \sqrt{x+2\sqrt{2x-4}} + \sqrt{x-2\sqrt{2x-4}}$$, then f(x) is differentiable on
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$$(-\infty, \infty)$$
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$$[2, \infty) - $$ { $$4$$}
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$$[2, \infty)$$
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$$[0, \infty)$$
$$f(x)=min {1,cos\,x,1-sin\,x}, -\pi \leq x \leq \pi$$ then
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f(x) is not differentiable at '0'
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f(x) is differentiable at $$\pi/2$$
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f(x) has local maximum at '0'
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None of these
If y = tan $$^{-1}$$ $$\dfrac{\sqrt{1 + x^2} -1}{x}$$, then$$\dfrac{dy}{dx}$$=
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$$\dfrac{1}{1 + x^2}$$
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$$\dfrac{- 1}{1 + x^2}$$
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$$\dfrac{1}{2(1 + x^2)}$$
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$$\dfrac{2}{1 + x^2}$$
Let $$f$$ be a differentiable function for all $$x$$. If $$f(1) = -2$$ and $$f'(x) \ge 2$$ for $$x \in [1, 6]$$ then
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$$f(6) < 5$$
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$$f(6) = 5$$
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$$f(6) \ge 8$$
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$$f(6) < 8$$
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
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$$1$$
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$$2$$
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$$3$$
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$$4$$
If $$y=cos^{-1}(\frac {x-x^{-1}}{x+x^{-1}}),$$ then $$\frac {dy}{dx}=$$
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$$\frac {-2}{1+x^{2}}$$
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$$\frac {2}{1+x^{2}}$$
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$$\frac {1}{1+x^{2}}$$
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$$\frac {-1}{1+x^{2}}$$
If $$Cos^{-1}\left ( \frac{x^2-y^2}{x^2+y^2} \right )=a$$, then $$\frac{dy}{dx}=$$
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$$\frac{x}{y}$$
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$$\frac{-x}{y}$$
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$$\frac{y}{x}$$
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$$\frac{-y}{x}$$
The derivate of $$\tan^{-1}[\dfrac {\sin x}{1+\cos x}]$$ with respect to $$\tan^{-1}[\dfrac {\cos x}{1+\sin x}]$$ is
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$$2$$
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$$-1$$
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$$0$$
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$$-2$$
If $$y = \cos^{-1}\left(\dfrac{x - x^{-1}}{x + x^{-1}}\right)$$, then $$\dfrac{dy}{dx}$$ is equal to
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$$\dfrac{-2}{1 +x^2}$$
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$$\dfrac{2}{1 + x^2}$$
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$$\dfrac{1}{1 + x^2}$$
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$$\dfrac{-1}{1 + x^2}$$
if $$f(x)=a\left| sinx \right| +{ be }^{ x }+c\left| { x }^{ 3 } \right| $$, where $$ a,b,c \epsilon$$ R, is differentiable at $$x=0$$ then
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$$a=0,b and c$$ are real numbers
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$$c=0$$,$$a=0$$, be is any real numbers
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$$b=0,c=0$$ a is any real numbers
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$$a=0,b=0,c$$ is any real numbers
If $$f(x)$$ 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
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$$1$$
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$$2$$
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$$3$$
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$$4$$
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
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2 f(x)
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$$\dfrac { f\left( x \right) }{ x } $$
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2x f(x)
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$$\dfrac { 2f\left( x \right) }{ x } $$
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
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f(x) =x $$\ell nx+3x-\frac { { x }^{ 2 } }{ 2 } $$
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$$f(x)\quad =\quad x\ell nx$$
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$$x={ e }^{ -3 }$$ is the abscissa of the point of inflection of f(x)
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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
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$$3$$
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$$1$$
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$$6$$
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$$5$$
$$S$$ 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
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$$2$$
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$$3$$
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$$4$$
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$$\phi$$
If $$y = \sin ^ { - 1 } \dfrac { 2 x } { 1 + x ^ { 2 } } ,$$ then which of the following is not correct ?
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$$\dfrac { d y } { d x } = \dfrac { 2 } { 1 + x ^ { 2 } }$$ for $$| x | < 1$$
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$$\dfrac { d y } { d x } = \dfrac { - 2 } { 1 + x ^ { 2 } }$$ for $$| x | > 1$$
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$$\dfrac { d y } { d x } = 2$$ for $$x = - 1$$
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$$\dfrac { d y } { d x }$$ does not exist at $$| x | = 1$$
$$\frac { d }{ dx } (\sin ^{ -1 }{ \{ \frac { \sqrt { 1+x } +\sqrt { 1-x } }{ 2 } \} } )=$$
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$$\frac { -1 }{ 2\sqrt { 1-{ x }^{ 2 } } } $$
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$$\frac { 1 }{ 2\sqrt { 1-{ x }^{ 2 } } } $$
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$$\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } $$
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$$\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } $$
If $$y = {\cot ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right]\;then\;\dfrac{{dy}}{{dx}}\;is\;equal\;to\;$$
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$$\dfrac{1}{2}$$
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$$\dfrac{2}{3}$$
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$$3$$
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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
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1
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2
0%
3
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none of these
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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