MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 7

If $$x=\sqrt { { a }^{ \sin ^{ -1 }{ t } } } ;\quad y=\sqrt { { a }^{ \cos ^{ -1 }{ t } } } $$, then $$\cfrac { dy }{ dx } =\cfrac { -y }{ x } $$

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True
0%
False

$$f(x)=\dfrac {1-\cos(1-\cos x)}{x^4}$$ is continuous at $$ x= 0 $$, then $$f(0)=$$

0%
$$\dfrac{1}{2}$$
0%
$$\dfrac{1}{4}$$
0%
$$\dfrac{1}{6}$$
0%
$$\dfrac{1}{8}$$

Let $$f(x)$$ be a function defined on$$(-a,a)$$ with $$a > 0$$. Amuse that $$f(x)$$ is continuous at $$x=0$$ and $$\underset{x\to 0}{\lim}\dfrac {f(x)-f(kx)}{x}=\alpha$$, where $$k \in (0,1)$$ then compute $$f'(0^{+})$$ and $$f'(0^{-})$$, and comment upon the differentiablity of $$f$$ at $$x=0$$? Denote $$\alpha$$.

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$$\underset{x\to o}{\lim}\dfrac{f(x) - f(0)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$
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$$\underset{x\to o}{\lim}\dfrac{f(kx) - f(0)}{kx} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$
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$$\underset{x\to 0}{\lim}\dfrac{f(x) - f(kx)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$
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$$\underset{x\to o}{\lim}\dfrac{f(kx) - f(0)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$

If $$x=a\left(\cos t+\log \tan\dfrac{t}{2}\right), y=a\sin t$$, then evaluate $$\dfrac{d^2y}{dx^2}$$ at $$t=\dfrac{\pi}{3}$$.

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$$\cfrac{4\sqrt3}{a}$$
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$$\cfrac{\sqrt3}{a}$$
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$$\cfrac{8\sqrt3}{a}$$
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$$\cfrac{8\sqrt3}{5a}$$

Let $$f(x)$$ be a continuous funct and $$g(x)$$ be a discontinuous $$funct^{n}$$, then $$f(x) + g(x)$$ is continuous $$funct^{n}$$. ? it is .

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True
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False

If $$x=\cos ec \theta-\sin \theta,y=\cos ec^{n}\theta-\sin^{n}\theta$$ then $$(x^{2}+4)\left(\dfrac {dy} {dx}\right)^{2}-n^{2}y^{2}=$$

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$$n^{2}$$
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$$2n^{2}$$
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$$3n^{2}$$
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$$4n^{2}$$

If $$x=a\sec^{2}\theta$$, $$y=a\tan^{3}\theta$$ then $$\dfrac {d^{3}y}{dx^{3}}$$

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$$\dfrac {-3}{8a^{2}}\cot^{3}\theta$$
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`$$\dfrac {3}{8a^{2}}\cot^{3}\theta$$
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$$3\sec^{2}\theta \tan \theta$$
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$$\dfrac {3}{4a^{2}}\cot^{3}\theta$$

Explanation

$$\alpha =a\sec^2 \theta , \ y=a\tan 3 \theta ,\ \dfrac {d^3y}{dx^3}=?$$

$$\dfrac {d^3y}{dx^3}=\dfrac {}d^3\theta{d\theta ^3} \dfrac {d^3\theta}{dx^3}$$

$$\dfrac {dy}{d\theta} =\dfrac {d}{d\theta} (a\tan ^3\theta) =a.3\tan^2 \theta \sec^2 \theta$$

$$\dfrac {d^2 y}{d\theta ^2}=a^3 (\sec^2 \theta. \tan \theta .\sec^2 \theta +\tan^2 \theta .2\sec \theta (\sec \theta \tan \theta))$$

$$=3a (2\tan \theta \sec^4\theta +2\tan^3 \theta \sec^2\theta )$$

$$\dfrac {dy}{d\theta^2} =60 [(\tan \theta .4\sec^3\theta \sec \theta +\sec^4 \theta .\sec^2 \theta) +(\tan^3\theta .2\sec \theta \sec \theta \tan \theta +\sec^2 \theta .3\tan^2 \theta \sec^2 \theta)]$$

$$=60 [5\sec^4\theta +\sec^6 \theta +2\tan^4 \theta \sec^2\theta +3\tan^2 \theta \sec^2 \theta]$$

$$=60[7\sec^4\theta \tan^2 \theta +2\tan^4 \theta \sec^2 \theta +\sec^6 \theta]$$

$$\dfrac {dx}{d\theta} \Rightarrow \ \dfrac {d}{d\theta}(a\sec^2 \theta)=a.2\sec \theta \sec \theta \tan \theta$$

$$=2a(\sec^2\theta \tan \theta)$$

$$\dfrac {d^2x}{d\theta} =2a(\tan \theta .2\sec \theta \tan \theta +\sec^2 \theta -\sec^2 \theta )$$

$$=2a((2\tan^2 \theta \sec^2 \theta)+\sec ^4 \theta)$$

$$\dfrac {d^3 x}{d\theta ^3}=2a[(2\tan^2\theta 32\sec \theta \sec \theta \tan \theta+2\sec^2 \theta +2\sec^2\theta 2\tan \theta \sec^2 \theta )+ 4\sec^3\theta \sec \theta \tan \theta]$$

$$=2a [4\tan^3\theta \sec^2\theta +4\sec \theta \sec^4 \theta+4\tan \theta \sec^3\theta]$$

$$=2a.4 [\tan^3\theta \sec^2\theta +2\tan \theta \sec^4 \theta]$$

$$\dfrac {d^3y}{dx^3}\ \Rightarrow \ \dfrac {6a (7\sec^4 \theta \tan^2 \theta +2\tan^4 \theta \sec^2\theta +\sec^6 \theta)}{8a (\tan \theta \sec^2 \theta [\tan^2\theta +2\sec^2 \theta])}$$

$$\Rightarrow \ \dfrac {3}{4} \dfrac {\sec^2\theta [7 \sec^2 \theta \tan ^2 \theta + 2\tan ^4 \theta +\sec^4 \theta]}{\tan \theta \sec^4 \theta (\tan ^2\theta +2 \sec^2 \theta)}$$

$$\Rightarrow \ (Since\ 1+\tan^2\theta =\sin^2\theta)$$

$$\Rightarrow \ \dfrac {3}{4} \left [\dfrac {7(1+\tan^2\theta) (\tan^2 \theta)+2\tan^2 \theta +(1+\tan^2\theta)}{\tan \theta (\tan \theta+2(1+\tan \theta))}\right]$$

on further simplyfying we get

$$\Rightarrow \ 3\sec^2 \theta \tan \theta$$

The value of k which makes $$f(x)=\left\{\begin{matrix} \sin\dfrac{1}{x}, x\neq 0\\ k, x=0\end{matrix}\right.$$ continuous at $$x=0$$ is?

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$$8$$
0%
$$1$$
0%
$$-1$$
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None

Let $$f(x) = \underset{0}{\overset{x}{\int}} |2t - 3| dt$$, then $$f$$ is

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continuous at $$x = \dfrac{3}{2}$$
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continuous at $$x = 3$$
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differentiable at $$x = \dfrac{3}{2}$$
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differentiable at $$x = 0$$

Let $$f(x) = \left\{\begin{matrix}x &x < 1 \\ 2 - x & 1 \leq x \leq 2\\ -2 + 3x - x^{2} & x > 2\end{matrix}\right.$$ then $$f(x)$$ is

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Differentiable at $$x = 0$$
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Differentiable at $$x = 2$$
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Differentiable at $$x = 1$$ and $$x = 2$$
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Not differentiable at $$x = 0$$

Range of function $$f(x)-g(x)$$ in [-$$\pi, \pi$$] is-

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$$[0, e^{\pi}+1]$$
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$$[0, e^{\pi}-1]$$
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$$e^{\pi}-1, e^{\pi}+1]$$
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$$[e^{-\pi}, e^{\pi}]$$

Which of the following is NOT CORRECT-

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$$b=d$$
0%
$$a=d$$
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$$a\neq b$$
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$$c\neq d$$

If $$f\left( x \right) = \left( {x - a} \right)g\left( x \right)$$ and $$g\left( x \right)$$ is continuous $$x = a$$ then $${f^{'}}\left( 1 \right) = $$

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$$g\left( 1 \right)$$
0%
$${g^{'}}\left( 1 \right)$$
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$$ - {g^{'}}\left( 1 \right)$$
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$${g^{'}}\left( { - 1} \right)$$

Function $$f(x)-g(x)$$ is-

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continuous & differentiable at $$x=0$$
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continuous but not differentiable at $$x=0$$
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discontinuous at $$x=0$$
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non differentiable because it is discontinuous at $$x=0$$

Differentiate $$\log(1+x^{2})$$ with respect $$\tan^{-1}x$$.

0%
$$2x$$
0%
$$x^2$$
0%
$$x$$
0%
$$x^3$$

An angle $$\theta$$ through which a pulley turns with time $$'t'$$ is completed by $$\theta = t^{2} + 3t - 5\ sq. cms/ min$$. Then the angular velocity for $$t = 5\ sec$$.

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$$5^{c}/ sec$$
0%
$$13^{c}/ sec$$
0%
$$23^{c}/ sec$$
0%
$$35^{c}/ sec$$

If the following function is continuous at $$x=0$$, find the value of $$k$$:

$$f\left( x \right) = \left\{ {\begin{array}{ccccccccccccccc}{\dfrac{{\sin \frac{{3x}}{2}}}{x}}&{,x \ne 0}\\k&{,x = 0}\end{array}} \right.$$

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$$\dfrac{1}{3}$$
0%
$$\dfrac{2}{3}$$
0%
$$\dfrac{4}{3}$$
0%
$$\dfrac{3}{2}$$

Let $$g\left(x\right)=\dfrac{f\left(x\right)}{x+1}$$ where $$f\left(x\right) $$ is differentiable on $$\left[0,5\right]$$ such that $$f\left(0\right)=4,f\left(5\right)=-1$$. There exists $$c\in \left(0,5\right)$$ such that $$g^{ ' }\left( c \right) $$ is ?

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$$-\dfrac{1}{6}$$
0%
$$\dfrac{1}{6}$$
0%
$$-\dfrac{5}{6}$$
0%
$$-1$$

Let $$f(x)=3x^{10}-7x^8+5x^6-21x^3+3x^2-7$$.
The value of $$\lim _{ h\rightarrow 0 }{ \dfrac { f\left( 1-h \right) -f\left( 1 \right) }{ { h }^{ 3 }+3h } } $$ is

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$$\dfrac{50}{3}$$
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$$\dfrac{22}{3}$$
0%
$$13$$
0%
None of these

If $$f$$ is differentiable on $$(a,b)$$ and if $$f(a)=f(b)=0$$ then for any $$\alpha $$ there is an $$x\in (a,b)$$ such that $$\alpha f(x)+f'(x)-1=0$$

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True
0%
False

A function $$f:R\rightarrow R$$ is such that $$f(1)=3$$ and $$f'(1)=6$$. Then $$\rightarrow { \displaystyle \lim _{ x\rightarrow 0 }{ \left[ \dfrac { f\left( 1+x \right) }{ f\left( 1 \right) } \right] } }^{ 1/x }=$$ ?

0%
$$1$$
0%
$$e^{2}$$
0%
$$e^{1/2}$$
0%
$$e^{3}$$

If $$y=y(x)$$ and it follows the relation $$e^{xy^{2}}+y\cos(x^{2})=5$$ then $$y'(0)$$ is equal to

0%
$$4$$
0%
$$-16$$
0%
$$-4$$
0%
$$16$$

If $$f$$ is a real valued differentiable function satisfying $$\left| f\left( x \right) -f\left( y \right) \right| \le { \left( x-y \right) }^{ 2 },x,y\epsilon R\ and f\left( 0 \right) ,\ then f\left( 1 \right)$$ equals ?

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

If $$f (x)$$ is differentiable everywhere, then:

0%
$$| f |$$ is differentiable everywhere
0%
$$| { f } | ^ { 2 }$$ is differentiable everywhere
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$$ f | f |$$ is not differentiable at some point
0%
$$ f + | f |$$ is differentiable everywhere

Let $$f$$ differentiable at $$x=0$$ and $$f'\left( 0 \right) = 1$$. Then $$\mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( h \right) - f\left( { - 2h} \right)}}{h} = $$=

0%
3
0%
2
0%
1
0%
6

Let $$f(x)$$ be differentiable function such that $$f\left(\dfrac{x+y}{1-xy}\right)=f(x)+f(y) \forall x$$ and $$y$$. If $$\underset { x\rightarrow 0 }{ lt } \dfrac { f\left( x \right) }{ x } =\dfrac { 1 }{ 3 }$$ then $$f(1)$$ equals

0%
$$\dfrac{1}{4}$$
0%
$$\dfrac{1}{6}$$
0%
$$\dfrac{1}{12}$$
0%
$$\dfrac{1}{8}$$

If $$f(x)=x^{n}\sin \dfrac {1}{x},f(0)=0$$

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$$f$$ is continuous differentiable at $$x=0$$ for any $$n< 2, n\in N$$
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$$f$$ is continuous differentiable for all $$x$$ for any $$n\ge 2, n\in N$$
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$$f$$ is discountinuous at $$x=0$$ if $$n=0$$
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$$f'(x)$$ is discountinuous at $$x=0$$ if $$n=0$$

If $$x=\dfrac{e^t+e^{-t}}{2}, y=\dfrac{e^t-e^{-t}}{2}$$, then $$\dfrac{dx}{dy}=$$

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

If $$y=a^{{a}^{{x}}}$$, then $$\dfrac {dy}{dx}=$$

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$$y.a^{x}{(\log a)}^{2}$$
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$$y.a^{x}.\log a$$
0%
$${(y.a^{x})}^{2}$$
0%
$$(y.a^{x})$$

Solve:

$$\dfrac { d } { d x } \tan ^ { - 1 } ( \sec x + \tan x )$$

0%
$$1$$
0%
$$1 / 2$$
0%
$$\cos x$$
0%
sec $$x$$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 7 - 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 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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