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CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 10 - MCQExams.com
CBSE
Class 11 Engineering Maths
Limits And Derivatives
Quiz 10
Differentiate the following function with respect to x.
$$(2x^2-3)\sin x$$.
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$$4x\sin x+(2x^2+3)\cos x$$.
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$$4x\sin x+(2x^2-3)\sin x$$.
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$$4x\sin x+(2x^2-3)\cos x$$.
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$$4x\cos x+(2x^2-3)\cos x$$.
Explanation
Given expression
$$(2x^2-3)\sin x$$
differentiating w.r.t. $$x$$, we get
$$=\dfrac d{dx} ((2x^2-3)\sin x)$$
$$=\dfrac d{dx}(2x^2-3)\sin x+2x^2-3\dfrac d{dx}(\sin x)$$
$$= 4x\sin x+(2x^2-3)\cos x$$
If $$f$$ is differentiable at $$x = 1$$ and $$\underset{h \rightarrow 0}{\lim} \dfrac{1}{h} f (1 + h) = 5, f'(1) = $$
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$$0$$
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$$1$$
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$$3$$
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$$4$$
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$$5$$
Explanation
$$f'(1)=\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)-f(1)}{h}$$; function is differentiable.
and $$\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)}{h}=5$$ [Given function is continuous]
$$\Rightarrow f(1)=0$$
Hence, $$f'(1)=\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)}{h}=5$$
If $$y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+}}}.....\infty$$ then $$\dfrac{dy}{dx}=?$$
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$$\dfrac{\sin x}{(2y-1)}$$
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$$\dfrac{\cos x}{(y-1)}$$
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$$\dfrac{\cos x}{(2y-1)}$$
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$$none\ of\ these$$
If $$y=(\tan x)^{\cot x}$$ then $$\dfrac{dy}{dx}=?$$
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$$\cot x.(\tan x)^{\cot x-1}.\sec^2x$$
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$$-(\tan x)^{\cot x}.\csc^2x$$
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$$(\tan x)^{\cot x}.\csc^2 x(1-\log \tan x)$$
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$$none\ of\ these$$
If $$y=\sqrt{x\sin x}$$ then $$\dfrac{dy}{dx}=?$$
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$$\dfrac{(x\cos x+\sin x)}{2\sqrt{x\sin x}}$$
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$$\dfrac{1}{2}(x\cos x+\sin x).\sqrt{x\sin x}$$
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$$\dfrac{1}{2\sqrt{x\sin x}}$$
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$$none\ of\ these$$
If $$x=a(\cos\theta+\theta\sin\theta)$$ and $$y=a(\sin\theta-\theta\cos\theta)$$ then $$\dfrac{dy}{dx}=?$$
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$$\cot\theta$$
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$$\tan\theta$$
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$$a\cot\theta$$
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$$a\tan\theta$$
If $$x=a\sec\theta, y=b\tan\theta$$ then $$\dfrac{dy}{dx}=$$
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$$\dfrac{b}{a}\sec\theta$$
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$$\dfrac{b}{a} \ cosec\theta$$
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$$\dfrac{b}{a}\cot\theta$$
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$$none\ of\ these$$
If $$y=\sqrt{\dfrac{1+\sin x}{1-\sin x}}$$ then $$\dfrac{dy}{dx}=?$$
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$$\dfrac{1}{2}\sec^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$$
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$$\dfrac{1}{2}\csc^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$$
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$$\dfrac{1}{2}\csc \left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\cot \left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$$
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$$none\ of\ these$$
Explanation
$$y=\left\{\dfrac{1+\cos \left(\dfrac{\pi}{2}-x\right)}{1-\cos \left(\dfrac{\pi}{2}-x\right)}\right\}^{\dfrac{1}{2}}=\left\{\dfrac{2\cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}{2\sin^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}\right\}^{\dfrac{1}{2}}=\cot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$$.
$$\Rightarrow \dfrac{dy}{dx}=-\csc^2\left({\dfrac{\pi}{4}-\dfrac{x}{2}}\right)\times\dfrac{dy}{dx}\left({\dfrac{\pi}{4}-\dfrac{x}{2}}\right)$$
$$\therefore \dfrac{dy}{dx}=\dfrac12\csc^2\left({\dfrac{\pi}{4}-\dfrac{x}{2}}\right)$$
If $$\lim _{ x\rightarrow 0 }{ \cfrac { x\left( 1+a\cos { x } \right) -b\sin { x } }{ { x }^{ 3 } } } =1$$ then
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$$a=-5/2$$, $$b=-1/2$$
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$$a=-3/2$$, $$b=-1/2$$
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$$a=-3/2$$, $$b=-5/2$$
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$$a=-5/2$$, $$b=-3/2$$
The value of $$ \displaystyle \lim _{x \rightarrow \pi} \dfrac{1+\cos ^{3} x}{\sin ^{2} x} $$ is
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1/3
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2/3
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-1/4
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3/2
Explanation
We have $$ \displaystyle \lim _{x \rightarrow \pi} \dfrac{1+\cos ^{3} x}{\sin ^{2} x} $$
$$=\displaystyle \lim _{x \rightarrow \pi} \dfrac{(1+\cos x)\left(1-\cos x+\cos ^{2} x\right)}{(1-\cos x)(1+\cos x)}$$
$$=\displaystyle \lim _{x \rightarrow \pi} \dfrac{1-\cos x+\cos ^{2} x}{1-\cos x}=\dfrac{1+1+1}{1+1}=\dfrac{3}{2}$$
If $$\displaystyle {f}'(x) = sin\,x + sin\,4x .\, cos \,x $$ then $$\displaystyle {f}'(x) \left (2x^{2} + \dfrac{\pi}{2} \right ) $$ at $$ x = \sqrt{\dfrac{\pi}{2}} $$ is equal to
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$$ -1 $$
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$$ 0 $$
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$$\displaystyle -2\sqrt{2\pi} $$
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None of these
If $$\displaystyle sin\, y = x\, sin ( a + y) $$ and
$$\displaystyle \dfrac{dy}{dx} = \dfrac{A}{ 1 + x^{2} - 2x \, cos a } $$ then the value of $$ A $$ is
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$$ 2 $$
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$$ cos\, a $$
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$$ sin\,a $$
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None of these
$$ \displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x^{n}}{(\sin x)^{m}},(m<n) $$ is equal to
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1
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0
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n/m
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None of these
Explanation
$$ \displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x^{n}}{(\sin x)^{m}}=\lim _{x \rightarrow 0}\left(\dfrac{\sin x^{n}}{x^{n}}\right)\left(\dfrac{x^{n}}{x^{m}}\right)\left(\dfrac{x}{\sin x}\right)^{m} $$
$$ =\displaystyle \lim _{x \rightarrow 0} x^{n-m}=0 \quad[\because m<n] $$
$$ The \ value \ of \displaystyle \lim _{x \rightarrow 1}(2-x)^{\tan \dfrac{\pi x}{2}} $$ is
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$$e^{-2 \pi} $$
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$$ e^{1 / \pi} $$
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$$ e^{2 /\pi} $$
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$$ e^{-1 / \pi} $$
Explanation
$$ \displaystyle \lim _{x \rightarrow 1}(2-x)^{\tan \dfrac{\pi x}{2}} $$
$$ =\displaystyle \lim _{x \rightarrow 1}\{1+(1-x)\}^{\tan \dfrac{\pi x}{2}} $$
$$ =e^{\displaystyle \lim _{x \rightarrow 1}(1-x) \tan \frac{\pi x}{2}} $$
$$ =e^{\displaystyle \lim _{x \rightarrow 1}(1-x) \cot \left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)} $$
$$ =e^{\displaystyle \lim _{x \rightarrow 1} \dfrac{(1-x)}{\tan \left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)}} $$
$$=e^{\dfrac{2}{\pi}\displaystyle \lim_{x \to 1}{\dfrac{\dfrac{\pi}{2}(1-x)}{tan \left(\dfrac{\pi}{2}(1-x)\right)}}}$$
$$=e^{\dfrac{2}{\pi}}$$
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{x\left(e^{x}-1\right)}{1-\cos x} $$ is equal to
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0
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$$\infty$$
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-2
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2
Explanation
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{x\left(e^{x}-1\right)}{1-\cos x}=\lim _{x \rightarrow 0} \dfrac{2 x\left(e^{x}-1\right)}{4 \sin ^{2} \dfrac{x}{2}} $$
$$ =2 \displaystyle \lim _{x \rightarrow 0}\left[\dfrac{(x / 2)^{2}}{\sin ^{2} \dfrac{x}{2}}\right]\left(\dfrac{e^{x}-1}{x}\right)=2 $$
$$ \displaystyle \lim _{x \to \pi / 2}\left[x \tan x-\left(\dfrac{\pi}{2}\right) \sec x\right] $$ is equal to
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1
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-1
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0
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$$None \ of \ these$$
Explanation
$$ \displaystyle \lim _{x \rightarrow \pi / 2}\left[x \tan x-\left(\dfrac{\pi}{2}\right) \sec x\right] $$
$$ =\displaystyle \lim _{x \rightarrow \pi / 2} \dfrac{2 x \sin x-\pi}{2 \cos x} \quad\quad\quad \left(\dfrac{0}{0}form\right)$$
$$ =\displaystyle \lim _{x \rightarrow \pi / 2} \dfrac{[2 \sin x+2 x \cos x]}{-2 \sin x} $$
(Applying L'Hopital's rule) $$ =-1 $$
$$ \displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{\sqrt{8 x^{2}+7 x+1}} $$ is equal to
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$$ -\dfrac{1}{2 \sqrt{2}} $$
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$$ \dfrac{1}{2 \sqrt{2}} $$
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$$ \dfrac{1}{\sqrt{2}} $$
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Does not exist
Explanation
a. $$\displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{\sqrt{8 x^{2}+7 x+1}}= \displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{-x \sqrt{8+\dfrac{7}{x}+\dfrac{1}{x^{2}}}} $$
$$ =-\displaystyle \lim _{x \rightarrow-\infty} \dfrac{\tan \dfrac{1}{x}}{\dfrac{1}{x} \sqrt{8+\dfrac{7}{x}+\dfrac{1}{x^{2}}}}=-\dfrac{1}{2 \sqrt{2}} $$
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)} $$ is equal to
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1
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0
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2
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None of these
Explanation
$$ \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)} $$
$$ =\dfrac{x^{4}\left(1+\tan ^{2} x+\tan ^{4} x\right)}{\tan ^{4} x\left(\tan ^{4} x-\tan ^{2} x+1\right)}=\dfrac{x^{4}}{\tan ^{4} x}, x \neq 0 $$
$$ \Rightarrow \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)}=\displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}}{\tan ^{4} x}=1 $$
If $$ f(x)=\dfrac{\cos x}{(1-\sin x)^{1 / 3}}, $$ then
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$$\lim _{ x\rightarrow \dfrac { \pi^- }{2 } }{ f(x)=-\infty } $$
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$$\lim _{ x\rightarrow \dfrac { \pi^+ }{2 } }{ f(x)=\infty } $$
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$$\lim _{ x\rightarrow \dfrac { \pi }{2 } }{ f(x)=\infty } $$
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none of these
Explanation
d. $$\lim _{x \rightarrow \dfrac{\pi}{2}} \dfrac{\cos x}{(1-\sin x)^{1 / 3}}=\lim _{t \rightarrow 0} \dfrac{-\sin t}{(1-\cos t)^{1 / 3}}$$
$$=-\lim _{t \rightarrow 0} \dfrac{2 \sin \dfrac{t}{2} \cos \dfrac{t}{2}}{\left(2 \sin ^{2} \dfrac{t}{2}\right)^{1 / 3}}$$
$$ =-\lim _{t \rightarrow 0} 2^{2 / 3} \cos \dfrac{t}{2}\left(\sin \dfrac{t}{2}\right)^{1 / 3}=0 $$
$$ \displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}} $$ is equal to
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2
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-2
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1/2
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-1/2
Explanation
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{4 \sin ^{4} x}$$
$$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x}{4 \sin ^{4} x}\left[\dfrac{2 \tan x}{1-\tan ^{2} x}-2 \tan x\right]$$
$$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan ^{3} x}{2 \sin ^{4} x\left(1-\tan ^{2} x\right)}$$
$$=\dfrac{1}{2} \lim _{x \rightarrow 0} \dfrac{x}{\sin x} \dfrac{1}{\cos ^{3} x} \dfrac{1}{1-\tan ^{2} x}$$
$$=\dfrac{1}{2} \times 1 \times \dfrac{1}{1^{3}} \times \dfrac{1}{1-0}=\dfrac{1}{2}$$
$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1+\sin \pi\left(\dfrac{3 x}{1+x^{2}}\right)}{1+\cos \pi x} $$ is equal to
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0
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1
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2
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4
Explanation
$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1+\sin \pi\left(\dfrac{3 x}{1+x^{2}}\right)}{1+\cos \pi x} $$
$$ =\displaystyle \lim _{x \rightarrow 1} \dfrac{1-\cos \left(\dfrac{3 \pi}{2}-\dfrac{3 \pi x}{1+x^{2}}\right)}{1-\cos (\pi-\pi x)} $$
$$ =\lim _{x \rightarrow 1} \dfrac{2 \sin ^{2}\left(\dfrac{3 \pi}{4}-\dfrac{3 \pi x}{2\left(1+x^{2}\right)}\right)}{2 \sin ^{2}\left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)} $$
$$ =\displaystyle \lim _{x \rightarrow 1}\left(\dfrac{\dfrac{3 \pi}{4}-\dfrac{3 \pi x}{2\left(1+x^{2}\right)}}{\dfrac{\pi}{2}-\dfrac{\pi x}{2}}\right)^{2} $$
$$ =\displaystyle \lim _{x \rightarrow 1} 9\left(\dfrac{\dfrac{1}{2}-\dfrac{x}{1+x^{2}}}{1-x}\right)^{2}=\lim _{x \rightarrow 1} 9\left(\dfrac{x-1}{2\left(1+x^{2}\right)}\right)^{2}=0 $$
$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1-x^{2}}{\sin 2 \pi x} \text { is equal to }$$
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$$ \dfrac{1}{2 \pi} $$
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$$ \dfrac{-1}{\pi} $$
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$$ \dfrac{-2}{\pi} $$
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None of these
Explanation
$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1-x^{2}}{\sin 2 \pi x}$$
$$=-\displaystyle \lim _{x \rightarrow 1} \dfrac{2 \pi(1-x)(1+x)}{2 \pi \sin (2 \pi-2 \pi x)}$$
$$=-\displaystyle \lim _{x \rightarrow 1} \dfrac{(2 \pi-2 \pi x)}{\sin (2 \pi-2 \pi x)} \dfrac{1+x}{2 \pi}=\dfrac{-1}{\pi}$$
Let $$f(x)= \sin x+ax+b$$, then which of the following is/are true.
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$$f(x)=0$$ has on;ly one root which is possitive if $$a>1, b<0$$
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$$f(x)=0$$ has on;ly one root which is negative if $$a>1, b<0$$
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$$f(x)=0$$ has on;ly one root which is negaitive if $$a < -1, b<0$$
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None of these
Explanation
$$f(x)=\sin x +ax+b$$
$$f(x)=0$$
$$\sin x +ax+b$$
$$f^{'}(x)=\cos x +G \rightarrow $$ when $$a > l$$
$$3^{'}(x)> o$$
$$f(x) \uparrow $$ increasing
$$f^{''}(x)=-\sin x$$
$$f(x)=\sin x +ax+b$$
$$f(0)=0+0+b$$
but $$b < 1$$
when $$a> \bot$$
$$\cos x +a > 0$$
$$f^{'}(x) > 0$$
hence $$f(x)$$ increasing function
$$f(x)=\sin x +ax+b$$
$$y=f(0)=\sin 0+a \times a +b$$
$$y=b$$
when, $$ b < 0$$
$$y=b$$
when $$b > 0$$ and $$a> 1$$
But
$$y=\sin x +ax+b$$
$$y=ax+\dfrac{(b+\sin x)}{\cos t}$$
$$y=mx+C$$
A right line.
A right line cut $$X-$$axis
The value of $$ \displaystyle \lim _{x \rightarrow a} \sqrt{a^{2}-x^{2}} \cot \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}} $$ is
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$$\dfrac{2 a}{\pi} $$
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$$-\dfrac{2 a}{\pi} $$
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$$ \dfrac{4 a}{\pi} $$
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$$-\dfrac{4 a}{\pi} $$
Explanation
$$ \displaystyle \lim _{x \rightarrow a} \sqrt{a^{2}-x^{2}} \cdot \cot \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}$$
$$=\displaystyle \lim _{x \rightarrow a} \dfrac{\sqrt{a^{2}-x^{2}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}$$
$$=\dfrac{2}{\pi} \displaystyle \lim _{x \rightarrow a} \dfrac{\dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}(a+x)=\dfrac{4 a}{\pi}$$
Which of the following is not true about $$y=f(x)$$?
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It is an increasing function
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It is a monotonic function
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It has infinite points of inflections
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None of these
Explanation
$$f(x)=x+\cos x -a\Rightarrow f'(x)=1-\sin x \ge 0 \forall x\in R$$.
Thus $$f(x)$$ is increasing in $$(-\infty , \infty)$$, as for $$f'(x)=0, x$$ is not forming an interval.
Also $$f"(x)=-\cos x=0$$
$$\Rightarrow x=(2n+1)\dfrac {\pi}{2}, n\in Z$$
Hence infinite points of inflection
Now $$f(0)=1-a$$.
For positive root $$1-a < 0 \Rightarrow a > 1$$. For negative root
$$1-a > 0\Rightarrow a < 1$$.
$$ \displaystyle \lim _{n \rightarrow \infty} \sum_{x=1}^{20} \cos ^{2 n}(x-10) $$ is equal to
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0
0%
1
0%
19
0%
20
Explanation
$$ \because \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n} x=\begin{cases}1, x=r \pi, r \in I \\ 0, x \neq r \pi, r \in I\end{cases}$$
Here, for $$ x=10, \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n}(x-10)=1 $$
and in all other cases it is zero.
$$ \therefore \displaystyle \lim _{n \rightarrow \infty} \sum_{x=1}^{\infty} \cos ^{2 n}(x-10)=1 $$
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin \left(x^{2}\right)}{\ln \left(\cos \left(2 x^{2}-x\right)\right)} $$ is equal to
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2
0%
-2
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1
0%
-1
Explanation
$$\displaystyle \lim_{x \to 0}\dfrac{sin(x^2)}{In(cos(2x^2-x))}$$
$$\displaystyle \
lim_{x \to 0}\dfrac{sin(x^2)}{log \left(1-2 sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}$$
$$=\displaystyle \lim_{x \to 0} \dfrac{sin(x^2)x^2}{\dfrac{x^2log \left(1-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}{-2sin^2 \left(\dfrac{2x^2-x}{2}\right)}\left[-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right]}$$
$$=\displaystyle \lim_{x \to 0} \dfrac{x^2}{\dfrac{2sin^2 \left(\dfrac{2x^2-x}{2}\right)}{\left(\dfrac{2x^2-x}{2}\right)^2}\left(\dfrac{2x^2-x}{2}\right)^2}$$
$$\displaystyle \lim_{x \to 0}-\dfrac{2x^2}{(2x^2-x)^2}=\displaystyle \lim_{x \to 0}-\dfrac{2}{(2x-1)^2}=-2$$
Differential coefficient of $$\sec (\tan^{-1} x)$$ w.r.t. x is
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$$\dfrac{x}{\sqrt{1+x^2}}$$
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$$\dfrac{x}{1+x^2}$$
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$$x\sqrt{1+x^2}$$
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$$\dfrac{1}{\sqrt{1+x^2}}$$
Explanation
a
The value of $$\displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin x}{x}$$ is
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$$0$$
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$$\infty$$
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$$1$$
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$$-1$$
Explanation
$$\displaystyle \lim_{x \rightarrow \infty} \dfrac {\sin x}{x}$$
Let $$x = 1/y$$ or $$y = \dfrac {1}{x}$$
Som $$x\rightarrow \infty \Rightarrow y \rightarrow 0$$
$$\therefore \displaystyle \lim_{x \rightarrow \infty} \left (\dfrac {\sin x}{x}\right ) = \displaystyle \lim_{y \rightarrow 0} y \sin \left (\dfrac {1}{y}\right )$$
$$= \displaystyle \lim_{y \rightarrow 0} y \cdot \displaystyle \lim_{y \rightarrow 0} \sin \dfrac {1}{y}$$
$$= 0\times \text {(any variable quantity)}$$
$$= 0$$
Hence, option (A) is correct.
The value of $$\displaystyle \lim_{x\rightarrow 0} \dfrac {1 - \cos x}{x^{2}}$$ is
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$$0$$
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$$1/2$$
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$$-1/2$$
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$$-1$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0} \dfrac {1 - \cos x}{x^{2}}$$
$$= \displaystyle \lim_{x\rightarrow 0} \dfrac {1 - 1 + 2\sin^{2} x/2}{x^{2}}$$
$$= \displaystyle \lim_{x\rightarrow 0} \dfrac {2\sin^{2} x/2}{x^{2}}$$
$$= \displaystyle \lim_{x\rightarrow 0} 2\left (\dfrac {\sin x/2}{x/2}\right )^{2} \times \dfrac {1}{4}$$
$$= \dfrac {1}{2}\times \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin x/2}{x/2}\right )^{2}$$
$$= \dfrac {1}{2} \times 1 = \dfrac {1}{2}$$
Hence option (B) is correct.
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