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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 14 - MCQExams.com
CBSE
Class 12 Commerce Maths
Continuity And Differentiability
Quiz 14
If $$f\left(x\right)$$ 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:
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$$\dfrac{25}{9}$$
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$$\dfrac{13}{6}$$
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$$\dfrac{23}{18}$$
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$$\dfrac{31}{18}$$
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
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$$h'\left( x \right) $$is differentiable at x=0
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$$h'\left( x \right) $$ is continuous at x=0 but is not differentiable at x=0
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$$h'\left( x \right) $$ is differentiable at x=0 but $$h'\left( x \right) $$ differentiable is not continuous at x=0
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$$h'\left( x \right) $$ is not differentiable at x=0
If $$f(x)=0$$ for $$x<0$$ and $$f(x)$$ is differentiable at $$x=0$$, then for $$x\ge 0, f(x)$$ may be
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$$x^2$$
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$$x$$
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$$-x$$
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$$-x^{3/2}$$
The value of $$f(4)$$ is
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$$160$$
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$$240$$
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$$200$$
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$$None\ of\ these $$
If $$y = {\sin ^{ - 1}}x + {\sin ^{ - 1}}\sqrt {1 - {x^2}} $$, then $$\frac{{dx}}{{dx}}$$
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$$0$$
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$$1$$
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$$-2x$$
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$$\frac{{y}}{{x}}$$
$$y = {x^{20}},{\text{find}}\dfrac{{{d^2}y}}{{d{x^2}}}$$
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$$381{x^{18}}$$
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$$380{x^{18}}$$
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$$389{x^{18}}$$
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$$370{x^{18}}$$
If $$x=\theta-\frac{1}{\theta},y=\theta+\frac{1}{\theta}$$ then $$\dfrac{dy}{dx}$$=
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$$\dfrac{xy}{y^2+2}$$
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y/x
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-x/y
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-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 $${y}^{2}={\left(\theta+\dfrac{1}{\theta}\right)}^{2}$$
$$\Rightarrow {y}^{2}={\theta}^{2}+\dfrac{1}{{\theta}^{2}}-2\theta\dfrac{1}{\theta}$$
$$\Rightarrow {y}^{2}={\theta}^{2}+\dfrac{1}{{\theta}^{2}}-2$$
$$\Rightarrow {y}^{2}+2={\theta}^{2}+\dfrac{1}{{\theta}^{2}}$$
$$\therefore \dfrac{dy}{dx}=\dfrac{1-\dfrac{1}{{\theta}^{2}}}{1+\dfrac{1}{{\theta}^{2}}}$$
$$=\dfrac{xy}{{y}^{2}+2}$$
If $$y=\cos ^{ -1 }{ (\frac { x-{ x }^{ -1 } }{ x+{ x }^{ -1 } } ) } \quad then\quad \frac { dy }{ dx } =$$
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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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$$\frac { -1 }{ 1+{ x }^{ 2 } } $$
$$dx\left\{ sin{ }^{ -1 }(\frac { 5x+12\sqrt { 1-x{ }^{ 2 } } }{ 13 } ) \right\} =$$
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$$\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } $$
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$$\frac { -1 }{ \sqrt { 1+{ x }^{ 2 } } } $$
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$$\frac { 2 }{ \sqrt { 1-{ x }^{ 2 } } } $$
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$$0$$
Let fbe twice differentiable function such that $${ g }^{ 1 }\left( x \right) =-f\left( x \right) and\quad \quad \quad \quad \quad { f }^{ 1 }\left( x \right) =g\left( x \right) ,\\ h(x)={ \left( f\left( x \right) \right) }^{ 2 }+{ \left( g\left( x \right) \right) }^{ 2 }.\quad Ifh(5)=11,\quad thenh(10)\quad is$$
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22
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11
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0
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8
If $$y=\tan ^{ -1 }{ [x+\sqrt { 1+{ x }^{ 2 } } ] } $$ then $$\frac { dy }{ dx } =$$
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$$\frac { 1 }{ 1+{ x }^{ 2 } } $$
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$$\frac { 1 }{ 2(1+{ x }^{ 2 }) } $$
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$$\frac { 2 }{ 1+{ x }^{ 2 } } $$
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None of these
$$\frac { d[\sec ^{ -1 }{ (\sin { x } +{ x }^{ 2 }) } ] }{ dx } =$$
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$$\frac { \sin { 2x } +{ x }^{ 2 } }{ \sin { x } +{ x }^{ 2 }()\sqrt { { (\sin { x } +{ x }^{ 2 }) }^{ 2 }-1 } } $$
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$$\frac { \cos { x } +2x }{ \sin { x } +{ x }^{ 2 }()\sqrt { { (\sin { x } +{ x }^{ 2 }) }^{ 2 }-1 } } $$
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$$\frac { 2x }{ \sin { x } \sqrt { \sin { x } -1 } } $$
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None of these
$$f(x) = \log_{1 - 2x}(1 + 2x)$$ for $$ x \ne 0$$
$$= k$$ for $$x = 0$$
is continuous at $$x = 0$$, find $$k.$$
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$$1$$
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$$-1$$
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$$0$$
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$$\frac{1}{2}$$
$$The\, \, function\, \, g(x)=\left| { \begin{array} { *{ 20 }{ c } }{ x+b } & { x<0 } \\ { \cos \, x } & { x\ge 0 } \end{array} } \right. \, \, can\, \, be\, \, made\, differentiable\, at\, x=0$$
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$$if\,b\,\,is\,\,equal\,to\,zero\,$$
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$$if\,b\,\,is\,\,not\,equal\,\,to\,zero\,$$
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$$if\,b\,\,is\,\,not\,equal\,\,to\,zero\,$$
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$$for\,no\,value\,of\,b$$
If $$y={ tan }^{ -1 }\left( \dfrac { 1-{ cos }^{ 2 }x }{ 1+{ cos }^{ 2 } } \right) ,$$, then $$\dfrac { dy }{ xy } =$$
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$$\dfrac { sinx }{ 1+{ cos }^{ 4 }x } $$
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$$\dfrac { sinx }{ 1+{ cos }^{ 2 }x } $$
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$$\dfrac { sin2x }{ 1+{ cos }^{ 4 }x } $$
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$$\dfrac { sin2x }{ 1+{ cos }^{ 2 }x } $$
Solve $$\frac{d\tan ^{-1}}{dx} (\frac{5x + 1}{3 - x - 6x^2}) = $$
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0
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1
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$$\frac{1}{1 + (3x + 2)^2} + \frac{1}{1 + (2x - 1)^2}$$
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$$\frac{3}{1 + (3x + 2)^2} + \frac{2}{1 + (2x - 1)^2}$$
If $$\cos ^{-1}x + \cos ^{-1}y = \frac{\pi}{2}$$ then $$\frac{dy}{dx}$$
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$$\frac{x}{y}$$
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$$\frac{-2x}{y}$$
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$$\frac{-x}{y}$$
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$$\frac{-y}{x}$$
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
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set of all natural numbers
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set of all irrational number
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$$\left\{0,1,-1\right\}$$
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$$\left\{1,-1\right\}$$
If $$f\left( x \right) =\sqrt { 1-\sqrt { 1-{ x }^{ 2 } } } $$, then $$f(x)$$ is
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continuous on$$[-1,1]$$
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differentiable on $$(-1,0)\cup (0,1)$$
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both (a) and (b)
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None of the above
$$y={ cos }^{ -1 }\cfrac { (9-{ x }^{ 2 }) }{ (9+{ x }^{ 2 }) } $$ then $$Y'(-1)$$ is equal to:
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$$\cfrac {3 }{ 5 } $$
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$$\cfrac { 3 }{ 5 } $$
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$$\cfrac { 2 }{ 7 } $$
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$$\cfrac { 3 }{ 8 } $$
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
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$$2$$
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$$3$$
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$$4$$
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$$1$$
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 solution 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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$$4$$
Solve : $$\dfrac { d } { d x } \left[ \tan ^ { - 1 } \sqrt { 1 + x ^ { 2 } } - \cot ^ { - 1 } \left( - \sqrt { 1 + x ^ { 2 } } \right) \right] =?$$
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$$\pi$$
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$$1$$
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$$0$$
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$$\dfrac { 2 x } { \sqrt { 1 + x ^ { 2 } } }$$
The derivative of $$\cos^{-1} x$$ w.r.t. $$\sqrt{1 - x^2}$$ is
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$$\dfrac{1}{\sqrt{1 - x^2}}$$
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$$\sqrt{1 - x^2}$$
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$$\dfrac{1}{2x}$$
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$$\dfrac{1}{x}$$
let f be the differeniable at x=0 and f'(0)=1 then $$\underset { b\rightarrow 0 }{ lim } \frac { f(h)-f(-2h) }{ h } $$
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3
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2
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1
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-1
If $$f$$ 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 :
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a curve of degree $$2$$
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a curve passing through the origin
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a straight line with slope $$2$$
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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)
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$$0$$
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$$-1$$
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$$-2$$
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$$None\ of\ these$$
If $$y = {\cot ^{ - 1}}\left( {\frac{{\sin x}}{{1 + \cos x}}} \right)$$ then $$\frac{{dy}}{{dx}}$$ is
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$$\frac{{ - 1}}{2}$$
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$$\frac{{ 1}}{2}$$
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$$0$$
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$$1$$
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)
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$$0$$
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$$4$$
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$$9$$
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$$None\ of\ these$$
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)
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There is atleast one point in each of the intervals (-1, 0) and (0, 1) where $$\left| f'(x) \right| \le 2$$
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there is atleast one point in a each of the interval (-1, 0) and (0, 1) where f(x)$$\le S$$
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there is no point of local maximum of F(x) in (-1, 1)
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For some $$c\in \left( -1,1 \right) $$, $$F'\left( c \right) \ge 6,F"\left( c \right) \le 0$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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