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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 2 - MCQExams.com
CBSE
Class 12 Commerce Maths
Continuity And Differentiability
Quiz 2
Find
d
y
d
x
if
x
=
a
(
θ
−
sin
θ
)
and
y
=
a
(
1
−
cos
θ
)
.
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cot
(
θ
2
)
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tan
(
θ
2
)
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−
cos
(
θ
2
)
0%
None of these
Explanation
We have
x
=
a
(
θ
−
sin
θ
)
and
y
=
a
(
1
−
cos
θ
)
Differentiate w.r.t
θ
∴
and
\displaystyle\frac{dy}{d\theta}=a\sin{\theta}
or
\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{d\theta}}{\displaystyle\frac{dx}{d\theta}}
=\displaystyle\frac{a\sin{\theta}}{a(1-\cos{\theta})}=\frac{2\sin{\left(\displaystyle\frac{\theta}{2}\right)}\cos{\left(\displaystyle\frac{\theta}{2}\right)}}{2\sin^{2}{\left(\displaystyle\frac{\theta}{2}\right)}}=\cot{\left(\displaystyle\frac{\theta}{2}\right)}
The derivative of
f(\tan x)
w.r.t.
g (\sec x)
at
x=\displaystyle \frac{\pi }{4},
where
{f}'(1)=2
and
{g}'(\sqrt{2})=4,
is
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\displaystyle \frac{1}{\sqrt{2}}
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\sqrt{2}
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1
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none of these
Explanation
y = f(\tan x)
and
z = g (\sec x)
\dfrac{dy}{dx}= f'(\tan x) . \sec^{2}x
and
\dfrac{dz}{dx}= g'(\sec x) . \sec x \tan x
\therefore \dfrac{dy}{dz}= \dfrac{f'(\tan x).\sec^{2} x}{g'(\sec x).\sec x \tan x}
\left.\begin{matrix}\dfrac{dy}{dz}\end{matrix}\right|_{x=\tfrac{\pi }{4}}=\dfrac{f'(1).(\sqrt{2})^{2}}{g'(\sqrt{2}).(\sqrt{2})}=\dfrac{1}{\sqrt{2}}
If
r=\sqrt{x^{2}+y^{2}+z^{2}}
and
x=2\sin 3t,y=2\cos 3t,z=8t
then
\displaystyle \frac{dr}{dt}=
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\displaystyle \frac{32t}{\sqrt{1+16t^{2}}}
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\displaystyle \frac{16t}{\sqrt{1+16t^{2}}}
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\displaystyle \frac{t}{\sqrt{1+16t^{2}}}
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\displaystyle \frac{4t}{\sqrt{1+16t^{2}}}
Explanation
Given :
x=2\sin 3t,y=2\cos 3t,z=8t
r=\sqrt{x^{2}+y^{2}+z^{2}}
=\sqrt{(2\sin 3t)^{2}+(2 \cos 3t)^{2}+(8t)^{2}}
=\sqrt{4\sin^{2}3t + 4\cos^{2}3t+64t^{2}}
=\sqrt{4+64t^{2}}
\implies r^{2}=4+64t^{2}
Differentiating above equation we get
2r\dfrac{dr}{dt}=2\times 64\times t
\implies \dfrac{dr}{dt}=\dfrac{64\times t}{r}
\implies \dfrac{dr}{dt}=\dfrac{64\times t}{\sqrt{4+64t^{2}}}=\dfrac{64t}{2\sqrt{1+16t^{2}}}=\dfrac{32t}{\sqrt{1+16t^{2}}}
If
y^x=x^y
, then find
\displaystyle\frac{dy}{dx}
.
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\displaystyle\frac{x(y\log{y}-y)}{y(x\log{x}-x)}
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\displaystyle\frac{y(x\log{y}-y)}{x(y\log{x}-x)}
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{y(x\log{y}-y)}
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\displaystyle\frac{y(x\log{x}-y)}{x(y\log{y}-x)}
Explanation
y^x=x^y
Taking log on both sides
\Rightarrow
\log{y^x}=\log{x^y}
\Rightarrow
x\log{y}=y\log{x}
\Rightarrow
\displaystyle x\frac{1}{y}\frac{dy}{dx}+\log{y}\times 1=\frac{y}{x}+\log{x}\frac{dy}{dx}
\Rightarrow
\displaystyle\left(\frac{x}{y}-\log{x}\right)\frac{dy}{dx}=\frac{y}{x}-\log{y}
\Rightarrow
\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{y}{x}-\log{y}}{\displaystyle\frac{x}{y}-\log{x}}=\frac{y(y-x\log{y})}{x(x-y\log{x})}=\frac{y(x\log{y}-y)}{x(y\log{x}-x)}
If
\displaystyle x=\frac{2t}{1+t^2}
,
\displaystyle y=\frac{1-t^2}{1+t^2}
, then
\displaystyle\frac{dy}{dx}
at
t=2
is
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\displaystyle\frac{4}{3}
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\displaystyle -\frac{4}{3}
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\displaystyle\frac{3}{4}
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\displaystyle -\frac{3}{4}
Explanation
Given that
\displaystyle x=\frac{2t}{1+t^2}
,
\displaystyle y=\frac{1-t^2}{1+t^2}
By differentiating w.r. to
t
, we get
\displaystyle \frac { dx }{ dt } =\frac { \left( \left( 1+{ t }^{ 2 } \right)\displaystyle \frac { d }{ dt } \left( 2t \right) -2t\displaystyle \frac { d }{ dt } \left( 1+{ t }^{ 2 } \right) \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } }
And
\displaystyle \frac { dy }{ dt } =\frac { \left( \left( 1+{ t }^{ 2 } \right) \displaystyle \frac { d }{ dt } \left( 1-{ t }^{ 2 } \right) -\left( 1-{ t }^{ 2 } \right) \displaystyle \frac { d }{ dt } \left( 1+{ t }^{ 2 } \right) \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } }
\displaystyle\frac{dx}{dt}=\frac{(1+t^2)2-2t\times 2t}{{(1+t^2)}^2}=\frac{2-2t^2}{{(1+t^2)}^2}
\displaystyle\frac{dy}{dt}=\frac{(1+t^2)(-2t)-(1-t^2)2t}{{(1+t^2)}^2}=\frac{-4t}{{(1+t^2)}^2}
\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}}=\frac{-4t}{2-2t^2}=\frac{2t}{t^2-1}
\therefore { \left( \displaystyle \frac { dy }{ dx } \right) }_{ t=2 }=\displaystyle \frac { 4 }{ 3 }
\displaystyle\frac{dy}{dx}
for
y=x^x
is
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x^x(1-\log{x})
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x^x(1-\log{y})
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x^x(1+\log{y})
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x^x(1+\log{x})
Explanation
y=x^x
Taking
\log
on both the sides
\log { y } =\log { \left( { x }^{ x } \right) }
\log { y } =x\log { x }
Differentiate w.r to
x
\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =x\frac { d }{ dx } \left( \log { x } \right) +\left( \log { x } \right) \frac { d }{ dx } \left( x \right)
\displaystyle \frac { dy }{ dx } =y\left[ \frac { x }{ x } +\log { x } \right]
\displaystyle \frac { dy }{ dx } ={ x }^{ x }\left[ 1+\log { x } \right]
Let the function
y=f(x)
be given by
x=t^{5}-5t^{3}-20t+7
and
y=4t^{3}-3t^{2}-18t+3
, where
t\epsilon \left ( -2, 2 \right )
. Then
f^{'}(x)
at
t=1
is ?
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\displaystyle \frac{5}{2}
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\displaystyle \frac{2}{5}
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\displaystyle \frac{7}{5}
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none of these
Explanation
Given,
x=t^{5}-5t^{3}-20t+7
\displaystyle \frac{dx}{dt}=5t^{4}-15t^{2}-20
\displaystyle \left (\frac{dx}{dt}\right)_{t=1}=-30
Also, given
y=4t^{3}-3t^{2}-18t+3
\displaystyle \frac{dy}{dt}=12t^{2}-6t-18
\displaystyle \left (\frac{dy}{dt}\right)_{t=1}=-12
So,
\displaystyle \left (\frac{dy}{dx}\right)_{t=1}=\frac{2}{5}
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect
Explanation
As
\displaystyle f\left ( x \right )= g\left ( x \right ).\sin x.
f'\left( x \right)=g\left( x \right)\cos { x } +g'\left( x \right)\sin { x }
Putting
x=0
f'\left( 0 \right)=g\left( 0 \right)=0
Now,
\displaystyle f'\left( 0 \right)=\lim _{ x\rightarrow 0 }{ \dfrac { f'\left( x \right)-f'\left( 0 \right) }{ x } }
\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } +g'\left( x \right)\sin { x } -g\left( 0 \right) }{ x } }
\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } -g\left( 0 \right) }{ x } } +\lim _{ x\rightarrow 0 }{ \dfrac { g'\left( x \right)\sin { x } }{ x } }
\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } -g\left( 0 \right) }{ x } }
\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } -g\left( 0 \right) }{ \sin { x } } }
=\lim _{ x\rightarrow 0 }{ \left( g\left( x \right)\cot { x-g\left( 0 \right)\csc { x } } \right) }
\displaystyle\frac{dy}{dx}
at
\displaystyle t=\frac{\pi}{4}
for
\displaystyle x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]
and
y=a\sin{t}
is
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1
0%
0
0%
-1
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\displaystyle \frac { 1 }{ 2 }
Explanation
\displaystyle x=\frac{\pi}{4}
for
\displaystyle x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]
and
y=a\sin{t}
Differentiating w.r.t
t
, we get
\displaystyle \frac { dx }{ dt } =a\left[ \frac { d }{ dt } \cos { t } +\frac { 1 }{ 2 } \frac { d }{ dt } \left( \log { \tan ^{ 2 }{ \frac { t }{ 2 } } } \right) \right]
\displaystyle\frac{dx}{dt}=a\left[-\sin{t}+\frac{1}{\displaystyle\tan{\frac{t}{2}}}\sec^2{\frac{t}{2}}\times\frac{1}{2}\right]
\displaystyle \frac { dx }{ dt } =a\left[ -\sin { t } +\frac { 1 }{ 2\displaystyle \frac { \sin { { t }/{ 2 } } }{ \cos { { t }/{ 2 } } } .\cos ^{ 2 }{ { t }/{ 2 } } } \right]
\displaystyle =a\left[-\sin{t}+\frac{1}{\displaystyle 2\sin{\frac{t}{2}}\cos{\frac{t}{2}}}\right]
\displaystyle =a\left[-\sin{t}+\frac{1}{\sin{t}}\right]
=a\left[ \displaystyle \frac { 1-\sin ^{ 2 }{ t } }{ \sin { t } } \right] =a\left[ \displaystyle \frac { \cos ^{ 2 }{ t } }{ \sin { t } } \right]
\displaystyle\frac{dy}{dt}=a\cos{t}
\displaystyle\therefore\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}}=\frac{a\cos{t}}{\displaystyle\frac{a\cos^2{t}}{\sin{t}}}=\tan{t}
At
\displaystyle t=\frac{\pi}{4}
,
\displaystyle\frac{dy}{dx}=1
The relation between the parameter '
t
' and the angle
\alpha
between the tangent to the given curve and the
x-
axis is given by, '
t
' equals to
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\displaystyle \dfrac {\pi}{2}-\alpha
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\displaystyle \dfrac {\pi}{4}+\alpha
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\alpha-\displaystyle \dfrac {\pi}{4}
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\displaystyle \dfrac {\pi}{4}-\alpha
Explanation
x=e^t \cos t
and
y=e^t \sin t
\displaystyle \dfrac { dy }{ dt } =\displaystyle \dfrac { d }{ dt } \left( { e }^{ t }\sin { t } \right)
\displaystyle \dfrac { dy }{ dt } ={ e }^{ t }\left( \cos { t } +\sin { t } \right)
\displaystyle \dfrac { dx }{ dt } =\displaystyle \dfrac { d }{ dt } \left( { e }^{ t }\cos { t } \right)
\displaystyle \dfrac { dx }{ dt } ={ e }^{ t }\left( \cos { t } -\sin { t } \right)
\therefore \displaystyle \dfrac { dy }{ dx } =\displaystyle \dfrac { \cos { t } +\sin { t } }{ \cos { t } -\sin { t } }
\therefore \tan { \left( \displaystyle \dfrac { \pi }{ 4 } +t \right) =\tan { \alpha } }
\displaystyle \dfrac { \pi }{ 4 } +t=\alpha
t=\alpha -\displaystyle \dfrac { \pi }{ 4 }
If
y={(\tan{x})}^{\displaystyle{(\tan{x})}^{\displaystyle\tan{x}}}
, then find
\displaystyle\frac{dy}{dx}
at
\displaystyle x=\frac{\pi}{4}
.
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0
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1
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-1
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2
Explanation
Taking
\log
on both sides, we get
\log{y}={(\tan{x})}^{\displaystyle\tan{x}}\log{\tan{x}}
Again taking
\log
on both the sides
\log{\log{y}}=[\tan{x}\log{\tan{x}}]+\log{\log{\tan{x}}}
Differentiating w.r.t.
x
, we get
\displaystyle\frac{1}{y\log{y}}\frac{dy}{dx}=\log { \tan { x } \frac { d }{ dx } \left( \tan { x } \right) +\tan { x } \frac { d }{ dx } \left( \log { \tan { x } } \right) +\frac { d }{ dx } \left( \log { \log { \tan { x } } } \right) }
\displaystyle \frac{dy}{dx}=\log { \tan { x } .\sec ^{ 2 }{ x } +\tan { x } .\frac { \sec ^{ 2 }{ x } }{ \tan { x } } +\frac { \sec ^{ 2 }{ x } }{ \tan { x } .\log { \tan { x } } } }
\displaystyle \frac { dy }{ dx } =y\log { y.\sec ^{ 2 }{ x } \left[ \log { \tan { x } +1+\frac { 1 }{ \tan { x } .\log { \tan { x } } } } \right] }
At
\displaystyle x=\frac{\pi}{4}
,
y=1
and
\log{y}=0
So, putting this values in the equation of
\displaystyle \frac { dy }{ dx }
, we get
\displaystyle\therefore{\left(\frac{dy}{dx}\right)}_{\displaystyle x=\frac{\pi}{4}}=0
If
x=a\sec^{3}{\theta}
and
y=a\tan^{3}{\theta}
, then
\displaystyle\frac{dy}{dx}
at
\theta=\displaystyle\frac{\pi}{3}
is
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\displaystyle\frac{\sqrt{3}}{2}
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\displaystyle\frac{\sqrt{5}}{2}
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\displaystyle\frac{\sqrt{2}}{3}
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\displaystyle\frac{\sqrt{4}}{3}
Explanation
We have
x=a\sec^{3}{\theta}
and
y=a\tan^{3}{\theta}
Differentiate w. r to
\theta
\displaystyle\frac{dx}{d\theta}=3a\sec^{2}{\theta}\displaystyle\frac{d}{d\theta}(\sec{\theta})=3a\sec^{3}{\theta}\tan{\theta}
\displaystyle\frac{dy}{d\theta}=3a\tan^{2}{\theta}\displaystyle\frac{d}{d\theta}(\tan{\theta})=3a\tan^{2}{\theta}\sec^{2}{\theta}
\therefore \displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{d\theta}}{\displaystyle\frac{dx}{d\theta}}=\frac{3a\tan^{2}{\theta}\sec^{2}{\theta}}{3a\sec^{3}{\theta}\tan{\theta}}=\frac{\tan{\theta}}{\sec{\theta}}=\sin{\theta}
\displaystyle{\left(\frac{dy}{dx}\right)}_{\displaystyle\theta=\frac{\pi}{3}}=\sin{\frac{\pi}{3}}=\frac{\sqrt{3}}{2}
If
\displaystyle y^{x}=x^{\sin y}
, find
\cfrac{dy}{dx}
.
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\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]
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\displaystyle \frac{y}{x}\left [ \frac{x \log y+\sin y}{y \:\log x\: \cos y+x} \right ]
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\displaystyle \frac{-y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]
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\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y+x} \right ]
Explanation
Given
\displaystyle y^{x}\:=x^{\sin \:y}
.
Take log both sides.
\displaystyle x \log y = \sin y \: \log \:x.
Differentiate w.r.t. x
\displaystyle \log y+x.\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \sin y+ (\log x)\cos y .\frac{dy}{dx}
\displaystyle \left ( \log y - \frac{\sin y}{x} \right )=\frac{dy}{dx}\left [ \cos y \log x-\frac{x}{y} \right ]
\displaystyle \therefore \frac{dy}{dx}=\frac{y}{x}\left [ \frac{x \log y-\sin y}{y \log x \cos y-x}\right ]
Discuss the applicability of Rolle's theorem to
\displaystyle f(x)=\log \left[\frac{x^{2}+ab}{(a+b)x}\right],
in the interval
[a,b].
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Yes Rolle's theorem is applicable and the stationary point is
x=\sqrt { ab }
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No Rolle's theorem is not applicable due to the discontinuity in the given interval
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Yes Rolle's theorem is applicable and the stationary point is
x= ab
0%
none of these
Explanation
We have
\displaystyle f(a)=\log \left [ \frac{a^{2}+ab}{(a+b)a} \right ]=\log 1=0
and
\displaystyle f(b)=\log \left [ \frac{b^{2}+ab}{(a+b)b} \right ]=\log1=0
\Rightarrow f(a)=f(b)=0.
Also, it
can be easily seen that
f(x)
is continuous on
[a,b]
and differentiable on
[a,b]
.
Thus all the three conditions of Rolle's
theorem are satisfied. Hence
\displaystyle f^{'}(x)=0
for at last one value of x in
[a,b]
Now
\displaystyle f^{'}(x)=0 = \frac{2x}{x^{2}+ab}-\frac{1}{x}=0\Rightarrow \displaystyle 2x^{2}-(x^{2}+ab)=0=x^{2}=ab
or
\displaystyle x=\sqrt{ab}
which is also known as stationary point.
Verify the Rolle's theorem for the function
\displaystyle f(x)=x^{2}-3x+2
on the interval[1,2]
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No Rolle's theorem is not applicable in the given interval
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Yes Rolle's theorem is applicable in the given interval and the stationary point
x=\frac { 5 }{ 4 }
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Yes Rolle's theorem is applicable in the given interval and the stationary point
x=\frac { 3 }{ 2 }
0%
nnone of these
Explanation
It can be easily seen that
f(x)=x^{2}-3x+2
is continuous as differentiable on R (being a polynomial)
\Rightarrow f(x)
is continous in (1,2) and differentiable in [1,2]. Also, we have
f(1)=f(2)=0
.
Thus,
f(x)
satisfies all the conditions of Rolle's theorem in
[1,2]
\Rightarrow \displaystyle \exists
at least one number, say
x
in
[1,2]
such that
\displaystyle f^{'}(c)=0.
Now,
\displaystyle f^{'}(x)=2x-3=0\Rightarrow x=\frac{3}{2}
Since, the root (stationary point)
\displaystyle x=\frac{3}{2}
lies in the interval(1,2).
Hence Rolle's theorem is verified.
Differentiate
\displaystyle x^{\sin^{-1}x}
w.r.t.
\displaystyle \sin ^{-1}x.
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\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]
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-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]
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\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]
0%
-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]
Explanation
Let
\displaystyle y = x^{\sin^{-1}x}
and
z = \displaystyle \sin ^{-1}x\Rightarrow \sin z = x
we have to find
\cfrac{dy}{dz}
\Rightarrow y = (\sin z)^z
taking
\log
both side
\log y = z\log \sin z
Differentiating w.r.t
z
\displaystyle \frac{1}{y}\frac{dy}{dz}=\log\sin z+z\frac{\cos z}{\sin z}
\therefore \displaystyle \frac{dy}{dz}=y\left(\log\sin z+z\frac{\cos z}{\sin z}\right)=\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]
\displaystyle y=(\cot x)^{\sin x}+(\tan x)^{\cos x}
.Find dy/dx
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\sin x(\cot x)^{ \sin x-1 }(-cosec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(-sinx) }
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\sin x(\cot x)^{ \sin x-1 }(-co\sec ^{ 2 } x)+\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x }
0%
\sin x(\cot x)^{ \sin x-1 }(-sec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x+1 } }co\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(sinx) }
0%
None of these
Explanation
Given
\displaystyle y=(\cot x)^{\sin x}+(\tan x)^{\cos x}
Let
u=(\cot x)^{\sin x}
Taking log on both sides
\log u=\sin x \log(\cot x)
Differentiating w.r.t. x, we get,
\dfrac{1}{u}\dfrac{du}{dx}=\dfrac{\sin x}{\cot x}(-cosec^2 x)+\cos x \log \cot x
\Rightarrow \dfrac{du}{dx}=\sin x(\cot x)^{\sin x -1}(-cosec^2 x)+(\cot x)^{\sin x} \cos x \log \cot x
Now,
v=(\tan x)^{\cos x}
Taking log on both sides
\log v=\cos x \log(\tan x)
Differentiating w.r.t. x, we get,
\dfrac{1}{v}\dfrac{dv}{dx}=\dfrac{\cos x}{\tan x}(\sec^2 x)-\sin x \log \tan x
\Rightarrow \dfrac{dv}{dx}=\cos x(\tan x)^{\cos x -1}(\sec^2 x)+(\tan x)^{\cos x}(-\sin x) \log \tan x
Since,
y=u+v
\Rightarrow \dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}
\Rightarrow \dfrac{dy}{dx}=\sin x(\cot x)^{\sin x -1}(-cosec^2 x)+(\cot x)^{\sin x} \cos x \log \cot x+
\cos x(\tan x)^{\cos x -1}(\sec^2 x)+(\tan x)^{\cos x}(-\sin x) \log \tan x
Verify Rolle's theorem for
\displaystyle f(x)=x(x+3)e^{-x/2}
in
(-3,0)
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Yes Rolle's theorem is applicable and the stationary point is
x=-2
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Yes Rolle's theorem is applicable and the stationary point is
x=-1
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No Rolle's theorem is not applicable in the given interval
0%
Both A and B
Explanation
We have
\displaystyle f(x)=x(x+3)e^{-x/2}
\displaystyle \therefore f^{'}(x)=(2x+3)e^{-x/2}+(x^{2}+3x)e^{-x/2}\left ( -\frac{1}{2} \right )
\displaystyle =e^{-x/2}\left [ 2x+3-\frac{1}{2}[x^{2}+3x] \right ]=-\frac{1}{2}[x^{2}-x-6]e^{-x/2}
f^{'}(x)
exist for every value of x in the interval[-3,0]
Hence,
f(x)
is differentiable and hence, continous in the interval
[-3,0]
Also, we have
\displaystyle f(-3)=f(0)=0\Rightarrow
All the three conditions of Rolle's theorem are satisfied.
So
\displaystyle f^{'}(x)=0\Rightarrow \frac{1}{2}(x^{2}-x-6)e^{-x/6}=0\Rightarrow x^{2}-x-6=0 \Rightarrow x=3,-2
Since, the value
x=-2
lies in the open interval
[-3,0]
the Rolle's theorem is verified.
Verify Rolle's theorem the function
\displaystyle f(x)=x^{3}-4x
on
[-2,2].
If you think it is applicable in the given interval then find the stationary point ?
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Yes Rolle's theorem is applicable and stationary point is
x=\pm \dfrac{2}{\sqrt{3}}
0%
No Rolle's theorem is not applicable
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Yes Rolle's theorem is applicable and
x=2\ or\ -2
0%
none of these
Explanation
The function
\displaystyle f(x)=x^{3}-4x
is a polynomial and so it is continuous and differentiable at all x
\displaystyle \epsilon
R.
In particular it is continuous in the closed interval
[-2,2]
.
Also
f(-2)]=0=f(2)
. Thus ,
f(x)
satisfies all three conditions of Rolle's theorem in
(-2,2)
.
Therefore , there must exist at least one real number
'x'
in the
open interval
(-2,2)
for which
f'(x)=0
Also
\displaystyle f'(x)= 3x^{2}-4
Now
\displaystyle f^{'}(x)=0
gives
3x^{2}-4=0
or
x=\pm \dfrac{2}{\sqrt{3}}
which is also known as stationary point.
Both these value lie in the open interval
(-2,2)
and thus the conclusion of Rolle's theorem is verified.
Verify the Rolle's theorem for the function
\displaystyle f(x)=x^{2}
in
(-1,1)
Report Question
0%
Yes Rolle's theorem is applicable and the stationary point is
x=\frac { 1 }{ 2 }
0%
Yes Rolle's theorem is applicable and the stationary point is
x=0
0%
No Rolle's theorem is not applicable in the given interval
0%
Both A and B
Explanation
f\left( x \right) ={ x }^{ 2 }
{ x }^{ 2 }
is continuous in
[-1,1]
since it is quadratic equation
f'\left( x \right) =2x
2x
is defined in
(-1,1)
\Rightarrow f\left( x \right)
is differentiable on
(-1,1)
f\left( -1 \right) =f\left( 1 \right) =1
There exists a
c,a\le c\le b
such that
f'\left( c \right) =0
2c=0
\Rightarrow c=0
which lies in
(-1,1)
Verify Rolle's theorem for the function
\displaystyle f(x)=10x-x^{2}
in the interval [0,10]
Report Question
0%
Yes Rolle's theorem is applicable and the stationary point is
x=5
0%
Yes Rolle's theorem is applicable and the stationary point is
x=4
0%
No Rolle's theorem is not applicable in the given interval
0%
none of these
Explanation
f\left( x \right) =10x-{ x }^{ 2 }\quad \quad \left[ 0,10 \right]
10x-{ x }^{ 2 }
is continuous in
\left[ 0,10 \right]
since it is polynomial function.
f'\left( x \right) =10-2x
is defined for all values of x in
(0,10)
\Rightarrow f\left( x \right)
is differentiable on
(0,10)
f\left( 0 \right) =f\left( 10 \right) =0
\therefore
There exists a
c,a\le c\le b
0\le c\le 10
Such that
f'\left( c \right) =0
10-2x=0
x=5
5
lies in
\left[ 0,10 \right]
If
y = \displaystyle (\tan x)^{\log x}
, then
\cfrac{dy}{dx} =
Report Question
0%
(\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]
0%
\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } logx
0%
\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x }
0%
none of these
Explanation
Let
y = \displaystyle (\tan x)^{\log x}
\log y = \log x .\log \tan x
Differentiating both side w.r.t
x
\cfrac{1}{y}.\cfrac{dy}{dx} = \cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x)
\Rightarrow \cfrac{dy}{dx} = (\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]
Verify Lagrange's mean value for the function
f(x)=\displaystyle \sin x
in
\displaystyle \left [ 0,\frac{\pi }{2} \right ]
Report Question
0%
No Lagrange's theorem is not applicable
0%
Yes Lagrange's theorem is applicable and
\displaystyle c =\cos ^{-1}\left ( \frac{2}{\pi } \right )
0%
Yes Lagrange's theorem is applicable and
\displaystyle c =\sin ^{-1}\left ( \frac{4}{\pi } \right )
0%
none of these
Explanation
The function
f\left( x \right)=\sin { x }
is continuous and differentiable for all
x\in R
.
In particular it is continuous in the closed interval
\displaystyle \left[ 0,\frac { \pi }{ 2 } \right]
and differentiable in the open interval
\displaystyle \left( 0,\frac { \pi }{ 2 } \right)
as is required for the application of lagrange's mean value theorem for it.
By lagrange's mean value theorem, there must exist at least one value of
'c'
of
x
lying in the open interval
\displaystyle \left( 0,\frac { \pi }{ 2 } \right)
such that
\displaystyle \frac { f\left( \frac { \pi }{ 2 } \right) -f\left( 0 \right) }{ \left( \frac { \pi }{ 2 } \right) -0 } =f'\left( c \right)
...(1)
Let us verify it, we have
\displaystyle f\left( 0 \right) =\sin { \left( 0 \right) } =0;f\left( \frac { \pi }{ 2 } \right) =\sin { \left( \frac { \pi }{ 2 } \right) } =1
Also
f'\left( x \right) =\cos { x }
gives
f'\left( c \right) =\cos { c }
.
Putting these values in (1), we have
\displaystyle \frac { 1-0 }{ \frac { \pi }{ 2 } } =\cos { \left( c \right) \Rightarrow \cos { \left( c \right) } } =\frac { 2 }{ \pi } \Rightarrow c=\cos ^{ -1 }{ \left( \frac { 2 }{ \pi } \right) }
Since
\displaystyle 0<\frac { 2 }{ \pi } <1,
therefore the principal value of
\displaystyle c=\cos ^{ -1 }{ \left( \frac { 2 }{ \pi } \right) }
lies in the open interval
\displaystyle \left( 0,\frac { \pi }{ 2 } \right)
and so is the required value of
c
This verifies lagrange's mean value theorem
Find 'c' of the mean value theorem, if
\displaystyle f(x)=x(x-1)(x-2);a=0, b=1/2
Report Question
0%
\displaystyle c=\frac{1+\sqrt{21}}{6}
0%
\displaystyle c=\frac{1-\sqrt{21}}{6}
0%
\displaystyle c=\frac{1+\sqrt{6}}{3}
0%
\displaystyle c=\frac{1-\sqrt{6}}{3}
Explanation
We have
f\left( a \right) =f\left( 0 \right) =0
and
\displaystyle f\left( b \right) =f\left( \frac { 1 }{ 2 } \right) =\frac { 3 }{ 8 }
\displaystyle \therefore \frac { f\left( b \right) -f\left( a \right) }{ b-a } =\frac { \frac { 3 }{ 8 } -0 }{ \frac { 1 }{ 2 } -0 } =\frac { 3 }{ 4 }
Now
f\left( x \right) ={ x }^{ 3 }-3{ x }^{ 2 }+2x\\ \Rightarrow f'\left( x \right) ={ 3x }^{ 2 }-6x+2\\ \Rightarrow f'\left( c \right) ={ 3c }^{ 2 }-6c+2
Putting all these value in lagrange's mean value theorem
\displaystyle \frac { f\left( b \right) -f\left( a \right) }{ b-a } =f'\left( c \right) ,\left( a<c,b \right)
We get
\displaystyle \frac { 3 }{ 4 } ={ 3c }^{ 2 }-6c+2\Rightarrow c=1\pm \frac { \sqrt { 21 } }{ 6 }
Hence
\displaystyle c=\frac { 1-\sqrt { 21 } }{ 6 }
lies in the open interval
\displaystyle \left( 0,\frac { 1 }{ 2 } \right)
Therefore it is the required value
Discuss the applicapibility of Rolle's Theorem to the function
\displaystyle f(x)=x^{2/3}
in (-1,1).
Report Question
0%
Yes Rolle's theorem is applicable and the stationary point
x=0
0%
Yes Rolle's theorem is applicable and the stationary point
x=\frac { 1 }{ 2 }
0%
No Rolle's theorem is not applicable in the given interval
0%
none of these
Explanation
\displaystyle f\left( x \right)={ x }^{ \frac { 2 }{ 3 } }\Rightarrow f'\left( x \right) =\frac { 2 }{ x } { x }^{ -\frac { 1 }{ 3 } }
\displaystyle\therefore \lim _{ x\rightarrow 0 }{ f'\left( x \right) } =\lim _{ x\rightarrow 0 }{ \frac { 2 }{ 3 } { \left( 0+h \right) }^{ -\frac { 1 }{ 3 } } } =\infty
\displaystyle Rf'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ h } \right\} } =\lim _{ h\rightarrow 0 }{ \left\{ \frac { { h }^{ \frac { 2 }{ 3 } }-1 }{ h } \right\} } =+\infty
\displaystyle Lf'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ -h } \right\} } =f'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { { \left( -h \right) }^{ \frac { 2 }{ 3 } } }{ h } \right\} } =-\infty
\displaystyle \therefore Lf'\left( 0 \right) \neq Rf'\left( 0 \right)
\displaystyle\therefore f'\left( 0 \right)
does not exist showing that
f'(x)
does not exists in the open interval
(-1,1)
Hence, Rolle's Theorem is not applicable
although
f(-1)=f(1)=1
and
f(x)
is continuous in the closed interval
(-1,1)
If
\displaystyle x=2 \cos t-cos 2t, y=2\sin t-\sin 2t,
find the value of
dy/dx
.
Report Question
0%
\tan\displaystyle \frac{3t}{2}
0%
\tan\displaystyle \frac{-3t}{2}
0%
\tan\displaystyle \frac{3t}{4}
0%
\tan\displaystyle \frac{-3t}{4}
Explanation
\dfrac{dx}{dt}=\dfrac{d}{dt}\left(2\cos \left(t\right)-\cos \left(2t\right)\right)
=\dfrac{d}{dt}\left(2\cos \left(t\right)\right)-\dfrac{d}{dt}\left(\cos \left(2t\right)\right)
=-2\sin \left(t\right)-\left(-2\sin \left(2t\right)\right)
=-2\sin \left(t\right)+2\sin \left(2t\right)
\dfrac{dy}{dt}=\dfrac{d}{dt}\left(2\sin \left(t\right)-\sin \left(2t\right)\right)
=\dfrac{d}{dt}\left(2\sin \left(t\right)\right)-\frac{d}{dt}\left(\sin \left(2t\right)\right)
=2\cos \left(t\right)-\cos \left(2t\right)\cdot \:2
\dfrac{dy}{dx}=\dfrac{dy}{dt}\times\dfrac{dt}{dx}=\dfrac{cos(t)-cos(2t)}{-sin(t)+sin(2t)}=\dfrac{2sin(\dfrac{3t}{2})sint}{2cos(\dfrac{3t}{2})sin(t)}=tan(\dfrac{3t}{2})
The function
f
is defined, where
x=2t-\left| t \right|, t\in R
. Draw the graph of
f
for the interval
-1\le x\le 1
. Also discuss its continuity and differentiability at
x=0
Report Question
0%
continuous and differentiable at
x=0
0%
not continuous but differentiable at
x=0
0%
neither continuous nor differentiable at
x=0
0%
differentiable but not continuous at
x=0
Explanation
For
t\ge 0
x=2t-t=t
y={t}^{2}+{t}^{2}=2{t}^{2}
For
t<0
x=3t, y=0
0\le x< 1 \Rightarrow 0\le t<1
[
\therefore x=t
]
-1\le x<0 \Rightarrow \cfrac{-1}{3}\le t<0
[
\therefore x=3t
]
f
is continuous and differentiable at
x=0
The function
f(x) = \displaystyle \sin ^{-1}(\cos x)is :-
Report Question
0%
discontinuous at
x = 0
0%
continuous at
x = 0
0%
differentiable at
x = 0
0%
none of these
Explanation
f(0-) = \displaystyle \lim_{h\to 0}\sin ^{-1}(\cos -h)=\lim_{h\to 0}\sin ^{-1}(\cos h) = \sin^{-1}1
, Since
\cos(-x) = \cos x
and
f(0+) = \displaystyle \lim_{h\to 0}\sin ^{-1}(\cos h) = \sin^{-1}1
Clearly at
x=0
, L.H.L
=
R.H.L
=f(0) \Rightarrow f(x)
is continuous at
x =0
Also
\displaystyle {y}'=-\frac{\sin x}{\sqrt{1-\cos ^{2}x}}=-\frac{\sin x}{\sqrt{\sin ^{2}x}}=-\frac{\sin x}{\left | \sin x \right |}
So the function is not differentiable at the points where
\sin x=0
,
that is, for
x=k\pi \left ( k\in I \right )
. In particular,
x=0
.
Examine the origin for continuity and derivability in case of the function
f
defined by
f(x)=x\tan ^{ -1 }{ (1/x) }
,
x\ne 0
and
f(0)=0
.
Report Question
0%
continuous but not differentiable at
x=0
0%
continuous and differentiable at
x=0
0%
not continuous and not differentiable at
x=0
0%
none of these
Explanation
f(0^+) =\displaystyle \lim_{h \rightarrow 0} (htan^{-1}(\cfrac{1}{h})) = 0 \times \cfrac{\pi}{2} = 0
f(0^-) = \displaystyle \lim_{h \rightarrow 0 }(-h tan^{-1}(\cfrac{1}{-h})) = 0 \times \cfrac{- \pi}{2} = 0
Hence the function is continuous at
x = 0.
f'(0^+) = \displaystyle \lim_{h \rightarrow 0} \cfrac{htan^{-1}(\cfrac{1}{h})}{h} = \cfrac{\pi}{2}
f(0^+) = \displaystyle \lim_{h \rightarrow 0} (- h tan^{-1}(\cfrac{1}{-h})) = \cfrac{-\pi}{2}
Hence the function is not differentiable at
x = 0.
State true or false:
If
x=t^2+3t-8, y=2t^2-2t-5
, then
\dfrac {dy}{dx}
at
(2, -1)
is
\dfrac {6}{7}
.
Report Question
0%
True
0%
False
Explanation
\displaystyle x={ t }^{ 2 }+3t-8\Rightarrow \frac { dx }{ dt } =2t+3
\displaystyle y={ 2t }^{ 2 }-2t-5\Rightarrow \frac { dy }{ dt } =4t-2
\displaystyle \therefore \frac { dy }{ dx } =\frac { 4t-2 }{ 2t+3 }
At
\left( 2,-1 \right)
2={ t }^{ 2 }+3t-8\Rightarrow { t }^{ 2 }+3t-10=0\\ \Rightarrow { t }^{ 2 }+5t-2t-10=0\\ \Rightarrow \left( t+5 \right) \left( t-2 \right) =0\Rightarrow t=-5,t=2
and
-1={ 2t }^{ 2 }-2t-5\Rightarrow { 2t }^{ 2 }-2t-4=0\\ \Rightarrow { t }^{ 2 }-t-2=0\Rightarrow { t }^{ 2 }-2t+t-2=0\\ \Rightarrow t=2,t=-1
\therefore t=2
Hence
\displaystyle \frac { dy }{ dx } =\frac { 4.2-2 }{ 2.2+3 } =\frac { 6 }{ 7 }
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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