Continuity and Differentiability Test

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Continuity and Differentiability Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

lf $$y=\sqrt{2x-x^{2}}$$ then $$y^{3}.y''=$$

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

$$y^{2}=2x-x^{2}$$

$$2yy'=2-2x$$

$$yy'=1-x$$

$$(y')^{2}+yy''=-1$$

$$y^{3}y''=-y^{2}-(yy')^{2}$$

$$y^{3}y''=-(2x-x^{2})-(1-2x+x^{2})$$

$$y^{3}y''=-1$$
Question 2
1 / -0

$${A} :$$ If $$x = ct, y = \dfrac{c}{t}$$, then at $$t=1, \dfrac{dy}{dx} =$$
$$B:$$ If $$x=3\cos\theta -\cos^{3}\theta, y=3\sin\theta-\sin^{3}\theta$$, then at $$\theta = \dfrac{\pi}{3}, \dfrac{dy}{dx} = $$
$$C:$$ If $$x = a\left(t+ \dfrac{1}{t}\right), y = a\left(t-\dfrac{1}{t}\right)$$, then at $$t =2, \dfrac{dy}{dx} = $$
$$D:$$ Derivative of $$\log(\sec x)$$ with respect to $$\tan x$$ at $$x = \dfrac{\pi}{4}$$ is$$\\$$
Arrangement of the above values in the increasing order is

A
$$C, D, A, B$$

B
$$C, A, D, B$$

C
$$A, B, D, C$$

D
$$B, D, A, C$$

Solution

For $$A$$
$$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times\dfrac{dt}{dx} =-\dfrac{c}{t^2}\times\dfrac{1}{c} = -\dfrac{1}{t^2}$$

At $$t=1, \dfrac{dy}{dx} = -1$$

For $$B$$

$$\dfrac{dy}{dx}=\dfrac{dy}{d\theta}\times\dfrac{d\theta}{dx}$$

$$= 3\cos\theta-3\sin^2\theta\times \cos\theta \times \dfrac{1}{-3\sin\theta+3\sin\theta\times \cos^2\theta}$$

$$=-\dfrac{3\cos\theta(1-\sin^2\theta)}{3\sin\theta(1-\cos^2\theta)}= -\dfrac{\cos^3\theta}{\sin^3\theta}$$

At $$\theta = \dfrac{\pi}{3}, \dfrac{dy}{dx} = -\dfrac{1}{3\sqrt3}$$

For $$C$$
$$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times\dfrac{dt}{dx} =a\left(1+\dfrac{1}{t^2}\right)\times\dfrac{1}{a\left(1-\dfrac{1}{t^2}\right)}=\dfrac{t^2+1}{t^2-1}$$

At $$t = 2, \dfrac{dy}{dx} = \dfrac{5}{3}$$

For $$D$$,
$$\dfrac{dy}{dx}= \dfrac{\sec x\tan x}{\sec x}\times\dfrac{1}{\sec^2x} = \sin x\cos x$$

At $$x = \dfrac{\pi}{4}, \dfrac{dy}{dx} = \dfrac{1}{2}$$

So, the arrangement would be $$A, B, D, C$$
Question 3
1 / -0

Let $$f(x)=\sin x,g(x)=x^{2},\ h(x)=\log x$$,If $$F(x)=h\{f(g(x))\}$$, then $$\displaystyle \frac{d^{2}F}{dx^{2}}=$$

A
$$2\cos^{3}x$$

B
$$2\cot x^{2}-4x^{2}\cos ec^{2}x^{2}$$

C
$$2x\cot x^{2}$$

D
$$-2\cos ec^{2}x$$

Solution

$$F(x)=log \sin x^{2}$$
$$ \dfrac {dF(x)}{dx}= \dfrac {2x \cos(x^{2})}{\sin x^{2}}$$

$$ \dfrac {d^{2}F(x)}{dx^{2}}= \dfrac {(2 \cos x^{2}-4x^{2}\sin x^{2})\sin x^{2}-(2x \cos x^{2})^{2}}{\sin^{2}x^{2}}$$

$$ \dfrac {d^{2}F(x)}{dx^{2}}=2\cot x^{2}-4x^{2} cosec {2x^{2}}$$
Question 4
1 / -0

If $$t\left( 1+{ x }^{ 2 } \right) =x$$ and $${ x }^{ 2 }+{ t }^{ 2 }=y,$$ then at $$x=2,$$ the value of $$\displaystyle\frac{dy}{dx}$$

A
$$\displaystyle\frac{488}{125}$$

B
$$\displaystyle\frac{88}{125}$$

C
$$\displaystyle\frac{101}{125}$$

D
None of these

Solution

$$\displaystyle { x }^{ 2 }+{ t }^{ 2 }=y\Rightarrow \frac { dy }{ dx } =2x+2t.\frac { dt }{ dx } $$ ...(1)
As $$\displaystyle t=\frac { x }{ 1+{ x }^{ 2 } } \Rightarrow \frac { dt }{ dx } =\frac { 1-{ x }^{ 2 } }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } } $$
Substitute these value of $$t$$ and $$\displaystyle\frac{dt}{dx}$$ in (1), we get
$$\displaystyle \frac { dy }{ dx } =2x+\frac { 2x }{ 1+{ x }^{ 2 } } .\frac { 1-{ x }^{ 2 } }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } } $$
On ptiing $$x=2$$ in $$\displaystyle\frac{dy}{dx}$$, we get
$$\displaystyle \frac { dy }{ dx } =\frac { 488 }{ 125 } $$
Question 5
1 / -0

If $$\mathrm{x}=\mathrm{a}\mathrm{t}^{2},\ \mathrm{y}=2\mathrm{a}\mathrm{t}$$, then $$\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}$$ is

A
$$-\dfrac{1}{t^{2}}$$

B
$$-\dfrac{1}{2at^{2}}$$

C
$$-\dfrac{1}{t^{3}}$$

D
$$-\dfrac{1}{2at^{3}}$$

Solution

We have $$x=at^2, y=2at$$
$$\Rightarrow \dfrac{dx}{dt}=2at=y, \dfrac{dy}{dt}=2a$$
Thus $$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{2a}{y}$$
$$\Rightarrow \dfrac{d^2y}{dx^2}=2a\left(-\dfrac{1}{y^2}\right)\dfrac{dy}{dx}=2a\left(-\dfrac{1}{y^2}\right)\dfrac{2a}{y}=-\dfrac{(2a)^2}{y^3}=-\dfrac{(2a)^2}{(2at)^3}=-\dfrac{1}{2at^3}$$
Question 6
1 / -0

$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[l \mathrm{o}\mathrm{g}(\mathrm{a}\mathrm{x})^{\mathrm{x}}]$$, where $$a$$ is a constant, is equal to

A
$$1$$

B
$$\log {ax}$$

C
$$1/a$$

D
$$\log {(ax)+1}$$

Solution

$$\frac{d}{dx}\left ( x \log ax \right )= \log (ax) +\dfrac{ax}{ax}=1+\log(ax)$$
Question 7
1 / -0

If $$r=\sqrt{x^{2}+y^{2}+z^{2}}$$ and $$x=2\sin 3t,y=2\cos 3t,z=8t$$ then $$\displaystyle \frac{dr}{dt}=$$

A
$$\displaystyle \frac{32t}{\sqrt{1+16t^{2}}}$$

B
$$\displaystyle \frac{16t}{\sqrt{1+16t^{2}}}$$

C
$$\displaystyle \frac{t}{\sqrt{1+16t^{2}}}$$

D
$$\displaystyle \frac{4t}{\sqrt{1+16t^{2}}}$$

Solution

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
$$2r\dfrac{dr}{dt}=2\times 64\times t$$
$$\implies \dfrac{dr}{dt}=\dfrac{64\times t}{r}$$

$$\implies \dfrac{dr}{dt}=\dfrac{64\times t}{\sqrt{4+64t^{2}}}=\dfrac{64t}{2\sqrt{1+16t^{2}}}=\dfrac{32t}{\sqrt{1+16t^{2}}}$$
Question 8
1 / -0

If $$ x=\cos\theta+\theta\sin\theta$$, $$ y=\sin\theta-\theta\cos\theta$$, then $${y}_{2}$$ is

A
$$\displaystyle \frac{\sec^{3}\theta}{\theta}$$

B
$$\displaystyle \frac{\theta}{\sec^{3}\theta}$$

C
$$\displaystyle \frac{\theta}{\cos^{3}\theta}$$

D
$$\sec^{2}\theta$$

Solution

The given information in the question is:
$$x = \cos \theta + \theta \sin \theta $$ and $$y = \sin \theta -\theta \cos \theta $$
Taking their first derivatives with respect to $$\theta$$ we get,
$$\Rightarrow \cfrac{dx}{d\theta }= -\sin \theta + \sin \theta +\theta \cos \theta $$
$$\Rightarrow \cfrac{dx}{d\theta }= \theta \cos \theta $$ and
$$\Rightarrow \cfrac{dy}{d\theta }= \cos \theta -\cos \theta +\theta \sin \theta $$
$$\Rightarrow \cfrac{dy}{d\theta }= \theta \sin \theta $$
so $$\cfrac{dy}{dx} = \cfrac{dy}{d\theta} \times \cfrac{d\theta}{dx}$$
$$\Rightarrow \cfrac{dy}{dx} = \cfrac{\theta \sin \theta}{\theta \cos \theta}$$
$$\Rightarrow \cfrac{dy}{dx}= \tan \theta$$
Now $$\cfrac{d^2y}{dx^2}=\cfrac{d}{dx}\left ( \cfrac{dy}{dx} \right )$$
$$\Rightarrow \cfrac{d^2y}{dx^2}=\cfrac{d}{dx}\left ( \tan \theta \right )$$
$$\Rightarrow \cfrac{d^2y}{dx^2}=\cfrac{d}{d\theta}\left ( \tan \theta \right )\times \cfrac{d\theta}{dx}$$
$$\Rightarrow \cfrac{d^2y}{dx^2}=\sec ^2\theta\times \cfrac{1}{\theta \cos \theta}$$
$$\Rightarrow \cfrac{d^2y}{dx^2}=\cfrac{\sec ^3 \theta}{\theta}$$
Question 9
1 / -0

If $$y=\sin^{-1}x\ then \ (1-x^{2})\displaystyle \frac{d^{2}y}{dx^{2}}=$$

A
$$-x\displaystyle \frac{dy}{dx}$$

B
$$0$$

C
$$x\displaystyle \frac{dy}{dx}$$

D
$$x(\displaystyle \frac{dy}{dx})^{2}$$

Solution

Given $$y=\sin^{-1}x$$

Differentiating w.r.t. x, we get,

$$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^{2}}}$$

Differentiating w.r.t. $$x$$, we get,

$$\dfrac{d^{2}y}{dx^{2}}=\dfrac{x}{(1-x^{2})^{3/2}}$$

$$(1-x^{2})\dfrac{d^{2}y}{dx^{2}}=\dfrac{x}{\sqrt{1-x^2}}=x\dfrac{dy}{dx}$$
Question 10
1 / -0

lf $$y=\sqrt{\cos 2x}$$, then $$y\displaystyle \frac{d^{2}y}{dx^{2}}+2y^{2}=$$

A
$$0$$

B
$$-(\displaystyle \frac{dy}{dx})^{2}$$

C
$$(\displaystyle \frac{dy}{dx})^{2}$$

D
$$y\displaystyle \frac{dy}{dx}$$

Solution

$$y^{'}=\dfrac{-sin2x}{\sqrt{cos2x}}$$

$$yy^{'}=-sin2x$$

$$(y^{'})^2+yy^{''}=-2 cos2x$$

$$(y^{'})^2+yy^{''}=-2y^{2}$$

$$yy^{''}+2y^{2}=-(y^{'})^2$$

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 14

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Continuity and Differentiability Test - 14

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SHARING IS CARING

Question 1

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 2

$$C, D, A, B$$

$$C, A, D, B$$

$$A, B, D, C$$

$$B, D, A, C$$

Solution

Question 3

$$2\cos^{3}x$$

$$2\cot x^{2}-4x^{2}\cos ec^{2}x^{2}$$

$$2x\cot x^{2}$$

$$-2\cos ec^{2}x$$

Solution

Question 4

$$\displaystyle\frac{488}{125}$$

$$\displaystyle\frac{88}{125}$$

$$\displaystyle\frac{101}{125}$$

None of these

Solution

Question 5

$$-\dfrac{1}{t^{2}}$$

$$-\dfrac{1}{2at^{2}}$$

$$-\dfrac{1}{t^{3}}$$

$$-\dfrac{1}{2at^{3}}$$

Solution

Question 6

$$1$$

$$\log {ax}$$

$$1/a$$

$$\log {(ax)+1}$$

Solution

Question 7

$$\displaystyle \frac{32t}{\sqrt{1+16t^{2}}}$$

$$\displaystyle \frac{16t}{\sqrt{1+16t^{2}}}$$

$$\displaystyle \frac{t}{\sqrt{1+16t^{2}}}$$

$$\displaystyle \frac{4t}{\sqrt{1+16t^{2}}}$$

Solution

Question 8

$$\displaystyle \frac{\sec^{3}\theta}{\theta}$$

$$\displaystyle \frac{\theta}{\sec^{3}\theta}$$

$$\displaystyle \frac{\theta}{\cos^{3}\theta}$$

$$\sec^{2}\theta$$

Solution

Question 9

$$-x\displaystyle \frac{dy}{dx}$$

$$0$$

$$x\displaystyle \frac{dy}{dx}$$

$$x(\displaystyle \frac{dy}{dx})^{2}$$

Solution

Question 10

$$0$$

$$-(\displaystyle \frac{dy}{dx})^{2}$$

$$(\displaystyle \frac{dy}{dx})^{2}$$

$$y\displaystyle \frac{dy}{dx}$$

Solution

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