Continuity and Differentiability Test

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Continuity and Differentiability Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

If $$x = a \left (\cos t + \log \tan \dfrac {t}{2}\right ), y = a\sin t$$, then $$\dfrac {dy}{dx} =$$

A
$$\tan t$$

B
$$\cot t$$

C
$$-\cot t$$

D
$$-\tan t$$

Solution

Solution:
We have, $$x=a\left(\cos t+\log\tan\cfrac t2\right), y =a\sin t$$
Now,
$$\cfrac{dx}{dt}=-a\sin t+\cfrac{a}{\tan \cfrac t2}\times \sec^2\cfrac t2\times \cfrac12$$
$$=-a\sin t+\cfrac{a}{2\sin \cfrac t2\cos\cfrac t2}$$
$$=-a\sin t+\cfrac{a}{\sin t}$$ $$(\because \ sint=2\sin\cfrac t2\cos\cfrac t2)$$
$$=\cfrac{a(1-\sin^2t)}{\sin t}$$
$$=\cfrac{a\cos^2t}{\sin t}.................(i)$$
And
$$\cfrac{dy}{dt}=a\cos t................(ii)$$
Now,
From eqn.(i) and eqn.(ii), we get
$$\cfrac{dy}{dt}\times\cfrac{dt}{dx}=a\cos t\times\cfrac{\sin t}{a\cos^2 t}$$
$$\Longrightarrow\cfrac{dy}{dx}=\tan t$$
Hence, A is the correct option.
Question 2
1 / -0

If $$f(x)=\cos ^{ -1 }{ \left\{ \cfrac { 1-{ \left( \log _{ e }{ x } \right) }^{ 2 } }{ 1+{ \left( \log _{ e }{ x } \right) }^{ 2 } } \right\} } $$, then $$f'(e)$$

A
Does not exist

B
Is equal to $$\cfrac { 2 }{ e } $$

C
Is equal to $$\cfrac { 1 }{ e } $$

D
Is equal to $$1$$

Solution

We have
$$\quad f(x)=\cos ^{ -1 }{ \left\{ \cfrac { 1-{ \left( \log _{ e }{ x } \right) }^{ 2 } }{ 1+{ \left( \log _{ e }{ x } \right) }^{ 2 } } \right\} } $$
$$\quad =2\tan ^{ -1 }{ \left( \log _{ e }{ x } \right) } \quad \left[ \because \log _{ e }{ x } >0 \right] $$ (in the $$nbd$$ of $$x=e$$)
$$\Rightarrow f'(x)=\cfrac { 2 }{ 1+{ \left( \log _{ e }{ x } \right) }^{ 2 } } .\cfrac { 1 }{ x } $$
$$\Rightarrow f'(e)=\cfrac { 2 }{ 1+{ \left( \log _{ e }{ e } \right) }^{ 2 } } .\cfrac { 1 }{ e } $$
$$=\cfrac { 2 }{ 1+1 } .\cfrac { 1 }{ e } $$
$$=\cfrac { 1 }{ e } $$
Question 3
1 / -0

If $$y=\tan ^{ -1 }{ \left( \cfrac { a\cos { x } -b\sin { x } }{ b\cos { x } +a\sin { x } } \right) } $$, then $$\cfrac { dy }{ dx } $$ is equal to

A
$$2$$

B
$$-1$$

C
$$\cfrac { a }{ b } $$

D
$$0$$

Solution

$$y=\tan ^{ -1 }{ \left( \cfrac { a\cos { x } -b\sin { x } }{ b\cos { x } +a\sin { x } } \right) } $$
Let $$a=r\sin { \theta } ;b=r\cos { \theta } $$
$$\therefore y=\tan ^{ -1 }{ \left\{ \cfrac { r\sin { \left( \theta -x \right) } }{ r\cos { \left( \theta -x \right) } } \right\} } $$
$$\Rightarrow y=\theta -x,y=\tan ^{ -1 }{ \left( \cfrac { a }{ b } \right) } -x$$
$$\therefore \cfrac { dy }{ dx } =-1$$
Question 4
1 / -0

If $$x=\cos { \theta } ,y=\sin { 5\theta } $$, then
$$(1-{ x }^{ 2 })\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -x\cfrac { dy }{ dx } =$$

A
$$-5y$$

B
$$5y$$

C
$$25y$$

D
$$-25y$$

Solution

We have
$$x=\cos { \theta } ,y=\sin { \theta } $$
$$\quad \therefore \cfrac { dy }{ dx } =\cfrac { \cfrac { dy }{ d\theta } }{ \cfrac { dx }{ d\theta } } =-\cfrac { 5\cos { 5\theta } }{ \sin { \theta } } $$
$$\Rightarrow \cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =-5\cfrac { d }{ d\theta } \left( \cfrac { 5\cos { 5\theta } }{ \sin { \theta } } \right) .\cfrac { d\theta }{ dx } \quad $$
$$=\cfrac { -25\sin { \theta } \sin { 5\theta } -5\cos { \theta } \cos { 5\theta } }{ \sin ^{ 3 }{ \theta } } $$
$$\quad \therefore \left( 1-{ x }^{ 2 } \right) \cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -x\cfrac { dy }{ dx } $$
$$=\left( 1-\cos ^{ 2 }{ \theta } \right) \left[ \cfrac { -25\sin { \theta } \sin { 5\theta } -5\cos { \theta } \cos { 5\theta } }{ \sin ^{ 3 }{ \theta } } \right] -\cos { \theta } \left[ \cfrac { -5\cos { 5\theta } }{ \sin { \theta } } \right] $$
$$=-25\sin { 5\theta } =-25y$$
Question 5
1 / -0

If $$y=x\log { \left( \cfrac { x }{ a+bx } \right) } $$, then $${ x }^{ 3 }\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } $$ is equal to

A
$$x\cfrac { dy }{ dx } -y$$

B
$${ \left( x\cfrac { dy }{ dx } -y \right) }^{ 2 }$$

C
$$y\cfrac { dy }{ dx } -x$$

D
$${ \left( y\cfrac { dy }{ dx } -x \right) }^{ 2 }$$

Solution

From the given relation $$y=x\log { \left( \cfrac { x }{ a+bx } \right) } $$
$$\cfrac { y }{ x } =\log { x } -\log { (a+bx) } $$
On differentiating w.r.t $$x$$, we get
$$\cfrac { \left( x\cfrac { dy }{ dx } -y \right) }{ { x }^{ 2 } } =\cfrac { 1 }{ x } -\cfrac { 1 }{ a+bx } b $$
Therefore, $$ x\cfrac { dy }{ dx } -y=\cfrac { ax }{ a+bx } $$ ...(i)
Again differentiating both sides w.r.t $$x$$ we get
$$x\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +\cfrac { dy }{ dx } -\cfrac { dy }{ dx } =\cfrac { (a+bx)a-axb }{ { (a+bx) }^{ 2 } } $$
$$\Rightarrow { x }^{ 3 }\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\cfrac { { a }^{ 2 }{ x }^{ 2 } }{ { (a+bx) }^{ 2 } } ={ \left( x\cfrac { dy }{ dx } -y \right) }^{ 2 }$$ ..... (from Eq(i))
Question 6
1 / -0

If $$ u = 2 (t - \sin t ) $$ and $$ v = 2 (1- \cos t), $$ then $$ \dfrac {dv}{du} $$ at $$ t = \dfrac {2 \pi}{3} $$ is equal to :

A
$$ \sqrt3$$

B
$$ - \sqrt3$$

C
$$ 2 \sqrt3$$

D
$$ \dfrac {2} {\sqrt3} $$

E
$$ \dfrac {1} {\sqrt3} $$

Solution

Given, $$ u = 2 (t - \sin t ) $$ and $$ v = 2 (1- \cos t), $$
On differentiating w.r.t. t, we get
$$ \dfrac {du}{dt} = 2 ( 1- \cos t) $$ and $$ \dfrac {dv}{dt} = 2 ( \sin t ) $$
Therefore, $$ \dfrac {dv}{du} = \dfrac {dv/dt} {du/dt} $$
$$ = \dfrac { 2 ( \sin t)} { 2 (1 - \cos t )} = \dfrac {2 \sin \frac {t}{2} \cos \frac {t}{2}}{2 \sin^2 \frac {t}{2}} $$
$$ \Rightarrow \dfrac {dv}{du} = \cot \dfrac {t}{2} $$
At $$ t = \dfrac { 2 \pi}{3} $$, we have
$$ \dfrac {dv}{du} = \cot \left( \dfrac {2 \pi}{3 \times 2 } \right) = \cot \dfrac {\pi}{3} = \dfrac {1}{\sqrt3} $$
Question 7
1 / -0

Let $$f(x)=2\tan^{-1}x+\sin^{-1}\left(\displaystyle\frac{2x}{1+x^2}\right)$$. Then

A
$$f'(2)=f'(3)$$

B
$$f'(2)=0$$

C
$$f'(1/2)=16/5$$

D
All of these

Solution

We know that,
$$\sin^{-1}\left(\displaystyle\frac{2x}{1+x^2}\right)=\left\{\begin{matrix} 2\tan^{-1}x, & if -1\leq x \leq 1\\ \pi -2\tan^{-1}x, & if x> 1\\ -\pi -2\tan^{-1}x, & if x < -1\end{matrix}\right.$$
Therefore, $$ f(x)=\left\{\begin{matrix} 4\tan^{-1} x, & if -1 \leq x \leq 1\\ \pi & if x>1 \\ -\pi, & if x < -1\end{matrix}\right.$$
$$\Rightarrow f'(x)=\left\{\begin{matrix} \displaystyle\frac{4}{1+x^2}, & if -1 < x <1 \\ 0, & if |x|>1\end{matrix}\right.$$
$$\Rightarrow f'(2)=f'(3)=0$$ and $$f'\left(\displaystyle\frac{1}{2}\right)=\frac{4}{1+\displaystyle \left(\displaystyle \frac{1}{2}\right)^2}=\displaystyle \frac{16}{5}$$
Question 8
1 / -0

If $${ y }^{ x }-{ x }^{ y }=1$$, then the value of $$\cfrac { dy }{ dx } $$ at $$x=1$$ is

A
$$2\left( 1-\log { 2 } \right) $$

B
$$2\left( 1+\log { 2 } \right) $$

C
$$2-\log { 2 } $$

D
$$2+\log { 2 } $$

Solution

We have
$$\quad { y }^{ x }-{ x }^{ y }=1....(i)$$
$$\Rightarrow { e }^{ x\log { y } }-{ e }^{ y\log { x } }=1$$
On differentiating w.r.t. $$x$$ we get
$${ y }^{ x }\left( \cfrac { x }{ y } .\cfrac { dy }{ dx } +\log { y } \right) -{ x }^{ y }\left( \cfrac { dy }{ dx } \log { x } +\cfrac { y }{ x } \right) =0$$
On putting $$x=1$$ and $$y=2$$ we get
$$2\left( \cfrac { 1 }{ 2 } .\cfrac { dy }{ dx } +\log { 2 } \right) -(0+2)=0$$
$$\cfrac { dy }{ dx } =2-2\log { 2 } $$
Question 9
1 / -0

The value of a for which the function $$f(x)=\left\{\begin{matrix} \tan^{-1} a-3x^2, & 0<x<1 \\ -6x, & x\geq 1\end{matrix}\right.$$ has a maximum at $$x=1$$, is?

A
$$0$$

B
$$1$$

C
$$2$$

D
None of these

Solution

We have,
$$\displaystyle f(x)=\left\{\begin {matrix} \tan^{-1} a-3x^2, & 0 < x < 1\\ -6x , & x\geq 1\end{matrix}\right.$$
If $$f(x)$$ attains a maximum at $$x=1$$, then $$f'(1)$$ must exist and should be zero. This means that $$f(x)$$ must be continuous and differentiable at $$x=1$$.
We observe that $$f(x)$$ will be continuous at $$x=1$$, if $$\tan^{-1}a=-3$$.
But, (LHD at $$x=1$$)$$=$$(RHD at $$x=1$$)$$=-6\neq 0$$ for any value of $$a$$.
Hence, there is no value of $$a$$ for which $$f(x)$$ has a minimum at $$x=1$$.
Question 10
1 / -0

If $$ y^x = 2^x , $$ then $$ \dfrac {dy}{dx} $$ is equal to :

A
$$ \dfrac {y}{x} \log \left( \dfrac {2}{y} \right) $$

B
$$ \dfrac {x}{y} \log \left( \dfrac {2}{y} \right) $$

C
$$ \dfrac {y}{x} \log \left( \dfrac {y}{2} \right) $$

D
$$ \dfrac {x}{y} \log \left( \dfrac {y}{2} \right) $$

E
$$ \dfrac {y}{x} \log \left( 2y \right) $$

Solution

Given, $$ y^x = 2^x $$
On taking $$\log$$ on both sides, we get
$$ x \log y = x \log 2 $$
On differentiating w.r.t. $$x,$$ we get
$$ x\times \dfrac {1}{y} \times \dfrac {dy}{dx} + \log y = \log 2 $$
$$ \Rightarrow \dfrac {dy}{dx} = \dfrac {y}{x} [ \log 2 - \log y ] $$
$$ \Rightarrow \dfrac {dy}{dx}= \dfrac {y}{x} \left[ \log \dfrac{2}{y} \right] $$

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 20

Result

Continuity and Differentiability Test - 20

Score

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Rank

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SHARING IS CARING

Question 1

$$\tan t$$

$$\cot t$$

$$-\cot t$$

$$-\tan t$$

Solution

Question 2

Does not exist

Is equal to $$\cfrac { 2 }{ e } $$

Is equal to $$\cfrac { 1 }{ e } $$

Is equal to $$1$$

Solution

Question 3

$$2$$

$$-1$$

$$\cfrac { a }{ b } $$

$$0$$

Solution

Question 4

$$-5y$$

$$5y$$

$$25y$$

$$-25y$$

Solution

Question 5

$$x\cfrac { dy }{ dx } -y$$

$${ \left( x\cfrac { dy }{ dx } -y \right) }^{ 2 }$$

$$y\cfrac { dy }{ dx } -x$$

$${ \left( y\cfrac { dy }{ dx } -x \right) }^{ 2 }$$

Solution

Question 6

$$ \sqrt3$$

$$ - \sqrt3$$

$$ 2 \sqrt3$$

$$ \dfrac {2} {\sqrt3} $$

$$ \dfrac {1} {\sqrt3} $$

Solution

Question 7

$$f'(2)=f'(3)$$

$$f'(2)=0$$

$$f'(1/2)=16/5$$

All of these

Solution

Question 8

$$2\left( 1-\log { 2 } \right) $$

$$2\left( 1+\log { 2 } \right) $$

$$2-\log { 2 } $$

$$2+\log { 2 } $$

Solution

Question 9

$$0$$

$$1$$

$$2$$

None of these

Solution

Question 10

$$ \dfrac {y}{x} \log \left( \dfrac {2}{y} \right) $$

$$ \dfrac {x}{y} \log \left( \dfrac {2}{y} \right) $$

$$ \dfrac {y}{x} \log \left( \dfrac {y}{2} \right) $$

$$ \dfrac {x}{y} \log \left( \dfrac {y}{2} \right) $$

$$ \dfrac {y}{x} \log \left( 2y \right) $$

Solution

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