Continuity and Differentiability Test

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Continuity and Differentiability Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

If $$x=t^{2},y=t^{3} \ then \ \ y_{2}=$$

A
$$\dfrac{3}{2}$$

B
$$\dfrac{3}{4t}$$

C
$$\dfrac{3}{2t}$$

D
$$\dfrac{3t}{2}$$

Solution

Given $$x=t^{2},y=t^{3}$$
$$\displaystyle \frac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\frac{3t^{2}}{2t}=\frac{3t}{2}$$

$$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d}{dt}\left ( \frac{dy}{dx} \right )\frac{dt}{dx}=\left ( \frac{3}{2} \right )\frac{1}{2t}=\frac{3}{4t}$$
Question 2
1 / -0

Suppose $$f(x)=x+\cot x$$ then $$(\sin^{2}x)f''(x)+2x=$$

A
$$-2f(x)$$

B
$$f(x)$$

C
$$2f(x)$$

D
$$-f(x)$$

Solution

Given, $$f(x)=x+\cot x$$

Then $$f^{'}(x)=1-cosec^{2}x=\cot^{2}x$$

$$f^{''}(x)=+2\cot n\ cosec ^{2}x$$

Therefore, $$\sin^{2}xf''(x)=2 \cot x$$

$$\sin^{2}xf''(x)+2x=2cotx+2x$$

$$\sin^{2}xf^{''}(x)+2x=2(cotx+x)$$
$$\therefore \sin^{2}xf^{''}(x)+2x=2f(x)$$
Question 3
1 / -0

lf $$y=e^{-2x}\cos 3x$$ and $$\displaystyle \frac{d^{2}y}{dx^{2}}+a(\frac{dy}{dx})+by=0$$, then $$a=?,b=?$$

A
$$4, 13$$

B
$$13, 4$$

C
$$-4, 13$$

D
$$-4, -13$$

Solution

$$y'=-2e^{-2x}\cos3x-3e^{-2x}\sin3x$$
$$y''=-2y'+6e^{-2x}\sin3x-9e^{-2x}\cos3x$$
$$y''=-2y'+2(-y^{1}-2y)-9y$$
$$y''=-4y'-13y$$
$$y''+4y'+13y=0$$
Then, $$a=4, b=13$$.
Question 4
1 / -0

$$y=a\cos ec(b-x)$$, then $$yy_{2}-2y_{1}^{2}=$$

A
$$-{y}^{2}$$

B
$$-y$$

C
$${y}^{2}$$

D
$$y$$

Solution

$$y^{‘}=-a cosec(b-x)cot(b-x)(-1)$$
$$y^{‘}=y cot(b-x)$$
$$y^{‘‘}=y^{‘}cot(b-x)+y \ cosec^{2}(b-x)$$
$$y^{‘‘}=\frac{y^{12}}{y}+y(1+\frac{y^{2}}{y^{2}})$$

$$y y^{‘‘}=2(y')^{2}+y^{2}$$
Question 5
1 / -0

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

A
$$\displaystyle \frac{1}{\sqrt{2}}$$

B
$$\sqrt{2}$$

C
$$1$$

D
none of these

Solution

$$y = f(\tan x)$$ and $$z = g (\sec x)$$

$$\dfrac{dy}{dx}= f'(\tan x) . \sec^{2}x$$ and $$\dfrac{dz}{dx}= g'(\sec x) . \sec x \tan x$$

$$\therefore \dfrac{dy}{dz}= \dfrac{f'(\tan x).\sec^{2} x}{g'(\sec x).\sec x \tan x}$$

$$\left.\begin{matrix}\dfrac{dy}{dz}\end{matrix}\right|_{x=\tfrac{\pi }{4}}=\dfrac{f'(1).(\sqrt{2})^{2}}{g'(\sqrt{2}).(\sqrt{2})}=\dfrac{1}{\sqrt{2}}$$
Question 6
1 / -0

If $$y=\displaystyle \frac{1}{1+x+x^{2}+x^{3}}$$, then $$y_{2}(0)=$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

Given, $$y=\dfrac{1}{(1+x)(1+x^{2})}$$
$$\Rightarrow (1+x^{2})y=\dfrac{1}{1+x}$$
$$\Rightarrow 2xy+(1+x^{2})y^{'}=\dfrac{-1}{(1+x)^{2}}$$
$$\Rightarrow 2y+2xy^{'}+2xy^{'}(1+x^{2})y^{''}=\dfrac{2}{(1+x^{2})^{3}}$$
$$\Rightarrow 2+y^{''}=2$$
$$\Rightarrow y^{''}=0$$
Question 7
1 / -0

If $$y^x=x^y$$, then find $$\displaystyle\frac{dy}{dx}$$.

A
$$\displaystyle\frac{x(y\log{y}-y)}{y(x\log{x}-x)}$$

B
$$\displaystyle\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$

C
$${y(x\log{y}-y)}$$

D
$$\displaystyle\frac{y(x\log{x}-y)}{x(y\log{y}-x)}$$

Solution

$$y^x=x^y$$

Taking log on both sides

$$\Rightarrow$$ $$\log{y^x}=\log{x^y}$$

$$\Rightarrow$$ $$x\log{y}=y\log{x}$$

$$\Rightarrow$$ $$\displaystyle x\frac{1}{y}\frac{dy}{dx}+\log{y}\times 1=\frac{y}{x}+\log{x}\frac{dy}{dx}$$

$$\Rightarrow$$ $$\displaystyle\left(\frac{x}{y}-\log{x}\right)\frac{dy}{dx}=\frac{y}{x}-\log{y}$$

$$\Rightarrow$$ $$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{y}{x}-\log{y}}{\displaystyle\frac{x}{y}-\log{x}}=\frac{y(y-x\log{y})}{x(x-y\log{x})}=\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$
Question 8
1 / -0

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

A
$$\displaystyle\frac{4}{3}$$

B
$$\displaystyle -\frac{4}{3}$$

C
$$\displaystyle\frac{3}{4}$$

D
$$\displaystyle -\frac{3}{4}$$

Solution

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 $$\displaystyle \frac { dy }{ dt } =\frac { \left( \left( 1+{ t }^{ 2 } \right) \displaystyle \frac { d }{ dt } \left( 1-{ t }^{ 2 } \right) -\left( 1-{ t }^{ 2 } \right) \displaystyle \frac { d }{ dt } \left( 1+{ t }^{ 2 } \right) \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } } $$
$$\displaystyle\frac{dx}{dt}=\frac{(1+t^2)2-2t\times 2t}{{(1+t^2)}^2}=\frac{2-2t^2}{{(1+t^2)}^2}$$

$$\displaystyle\frac{dy}{dt}=\frac{(1+t^2)(-2t)-(1-t^2)2t}{{(1+t^2)}^2}=\frac{-4t}{{(1+t^2)}^2}$$
$$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}}=\frac{-4t}{2-2t^2}=\frac{2t}{t^2-1}$$
$$\therefore { \left( \displaystyle \frac { dy }{ dx } \right) }_{ t=2 }=\displaystyle \frac { 4 }{ 3 } $$
Question 9
1 / -0

$$\displaystyle\frac{dy}{dx}$$ for $$y=x^x$$ is

A
$$x^x(1-\log{x})$$

B
$$x^x(1-\log{y})$$

C
$$x^x(1+\log{y})$$

D
$$x^x(1+\log{x})$$

Solution

$$y=x^x$$
Taking $$\log $$ on both the sides
$$\log { y } =\log { \left( { x }^{ x } \right) } $$
$$\log { y } =x\log { x } $$
Differentiate w.r to $$x$$
$$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =x\frac { d }{ dx } \left( \log { x } \right) +\left( \log { x } \right) \frac { d }{ dx } \left( x \right) $$
$$\displaystyle \frac { dy }{ dx } =y\left[ \frac { x }{ x } +\log { x } \right] $$
$$\displaystyle \frac { dy }{ dx } ={ x }^{ x }\left[ 1+\log { x } \right] $$
Question 10
1 / -0

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 ?

A
$$\displaystyle \frac{5}{2}$$

B
$$\displaystyle \frac{2}{5}$$

C
$$\displaystyle \frac{7}{5}$$

D
none of these

Solution

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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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 15

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Continuity and Differentiability Test - 15

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SHARING IS CARING

Question 1

$$\dfrac{3}{2}$$

$$\dfrac{3}{4t}$$

$$\dfrac{3}{2t}$$

$$\dfrac{3t}{2}$$

Solution

Question 2

$$-2f(x)$$

$$f(x)$$

$$2f(x)$$

$$-f(x)$$

Solution

Question 3

$$4, 13$$

$$13, 4$$

$$-4, 13$$

$$-4, -13$$

Solution

Question 4

$$-{y}^{2}$$

$$-y$$

$${y}^{2}$$

$$y$$

Solution

Question 5

$$\displaystyle \frac{1}{\sqrt{2}}$$

$$\sqrt{2}$$

$$1$$

none of these

Solution

Question 6

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 7

$$\displaystyle\frac{x(y\log{y}-y)}{y(x\log{x}-x)}$$

$$\displaystyle\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$

$${y(x\log{y}-y)}$$

$$\displaystyle\frac{y(x\log{x}-y)}{x(y\log{y}-x)}$$

Solution

Question 8

$$\displaystyle\frac{4}{3}$$

$$\displaystyle -\frac{4}{3}$$

$$\displaystyle\frac{3}{4}$$

$$\displaystyle -\frac{3}{4}$$

Solution

Question 9

$$x^x(1-\log{x})$$

$$x^x(1-\log{y})$$

$$x^x(1+\log{y})$$

$$x^x(1+\log{x})$$

Solution

Question 10

$$\displaystyle \frac{5}{2}$$

$$\displaystyle \frac{2}{5}$$

$$\displaystyle \frac{7}{5}$$

none of these

Solution

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