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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.
(
2
x
2
−
3
)
sin
x
.
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4
x
sin
x
+
(
2
x
2
+
3
)
cos
x
.
0%
4
x
sin
x
+
(
2
x
2
−
3
)
sin
x
.
0%
4
x
sin
x
+
(
2
x
2
−
3
)
cos
x
.
0%
4
x
cos
x
+
(
2
x
2
−
3
)
cos
x
.
Explanation
Given expression
(
2
x
2
−
3
)
sin
x
differentiating w.r.t.
x
, we get
=
d
d
x
(
(
2
x
2
−
3
)
sin
x
)
=
d
d
x
(
2
x
2
−
3
)
sin
x
+
2
x
2
−
3
d
d
x
(
sin
x
)
=
4
x
sin
x
+
(
2
x
2
−
3
)
cos
x
If
f
is differentiable at
x
=
1
and
lim
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0%
0
0%
1
0%
3
0%
4
0%
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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0%
\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
0%
-(\tan x)^{\cot x}.\csc^2x
0%
(\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}}
0%
\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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0%
\dfrac{b}{a}\sec\theta
0%
\dfrac{b}{a} \ cosec\theta
0%
\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)
0%
\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)
0%
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
0%
a=-3/2
,
b=-1/2
0%
a=-3/2
,
b=-5/2
0%
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
0%
2/3
0%
-1/4
0%
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
0%
\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
0%
cos\, a
0%
sin\,a
0%
None of these
\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x^{n}}{(\sin x)^{m}},(m<n)
is equal to
Report Question
0%
1
0%
0
0%
n/m
0%
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}
0%
e^{1 / \pi}
0%
e^{2 /\pi}
0%
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
0%
\infty
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-2
0%
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
0%
-1
0%
0
0%
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}}
0%
\dfrac{1}{2 \sqrt{2}}
0%
\dfrac{1}{\sqrt{2}}
0%
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
0%
0
0%
2
0%
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 }
0%
\lim _{ x\rightarrow \dfrac { \pi^+ }{2 } }{ f(x)=\infty }
0%
\lim _{ x\rightarrow \dfrac { \pi }{2 } }{ f(x)=\infty }
0%
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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0%
2
0%
-2
0%
1/2
0%
-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
0%
1
0%
2
0%
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}
0%
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
0%
f(x)=0
has on;ly one root which is negative if
a>1, b<0
0%
f(x)=0
has on;ly one root which is negaitive if
a < -1, b<0
0%
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}
0%
-\dfrac{2 a}{\pi}
0%
\dfrac{4 a}{\pi}
0%
-\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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0%
It is an increasing function
0%
It is a monotonic function
0%
It has infinite points of inflections
0%
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
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
Report Question
0%
2
0%
-2
0%
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}
0%
\dfrac{1}{\sqrt{1+x^2}}
Explanation
a
The value of
\displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin x}{x}
is
Report Question
0%
0
0%
\infty
0%
1
0%
-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
Report Question
0%
0
0%
1/2
0%
-1/2
0%
-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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