Application of Derivatives Test

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Application of Derivatives Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

Consider the curve represented parametrically by the equation
$$\displaystyle x = t^3 - 4t^2 - 3t$$ and $$\displaystyle y = 2t^2 + 3t - 5$$ where $$\displaystyle t \: \epsilon \: R$$.
If $$H$$ denotes the number of point on the curve where the tangent is horizontal and $$V$$ the number of point where the tangent is vertical then

A
$$H = 2$$ and $$V = 1$$

B
$$H = 1$$ and $$V = 2$$

C
$$H = 2$$ and $$V = 2$$

D
$$H = 1$$ and $$V = 1$$

Solution

$$\dfrac{dx}{dt} = 3t^2-8t-3$$
$$\dfrac{dy}{dt} = 4t+3$$
$$\Rightarrow \dfrac{dy}{dt} = \dfrac{4t+3}{3t^2-8t-3}$$
clerarly denominator has two roots and numerator has single root
Hence given curve will have two vertical tangent and one horizontal tangent
$$\Rightarrow H =1 , V =2$$
Question 2
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For a $$\displaystyle a\epsilon \left [ \pi , 2\pi \right ],$$ the function $$\displaystyle f\left ( x \right )= \frac{1}{3}\sin a \tan ^{3}x+\left ( \sin a-1 \right )\tan x+ \frac{\sqrt{a-2}}{\sqrt{8-a}}$$

A
$$\displaystyle x= n\pi \left ( n\epsilon I \right )$$ as critical points

B
no critical points

C
$$\displaystyle x= 2n\pi \left ( n\epsilon I \right )$$ as critical points

D
$$\displaystyle x= \left ( 2n+1 \right )\pi \left ( n\epsilon I \right )$$ as critical points.

Solution

$$f\left( x \right) =\dfrac { 1 }{ 3 } \sin a\tan ^{ 3 } x+\left( \sin a-1 \right) \tan x+\dfrac { \sqrt { a-2 } }{ \sqrt { 8-a } } $$
$$f'\left( x \right) =\sin { a } \tan ^{ 2 }{ x } \sec ^{ 2 }{ x } +\left( \sin a-1 \right) \sec ^{ 2 }{ x } =\sec ^{ 2 }{ x } \left( \sin { a } \left( \sec ^{ 2 }{ x } -1 \right) +\left( \sin a-1 \right) \right) $$
$$\Rightarrow f'\left( x \right) =\sec ^{ 2 }{ x } \left( \sin { a } \sec ^{ 2 }{ x } -1 \right) $$
for critical points $$f'(x)=0$$
$$\Rightarrow \sec ^{ 2 }{ x } =0$$
$$\Rightarrow x\in \emptyset $$
or $$\sec ^{ 2 }{ x } =\dfrac { 1 }{ \sin { a } } $$
since, for $$a\epsilon \left[ \pi ,2\pi \right], \quad \sin { a } <0$$
Therefore, $$x\in \emptyset$$
Thus, $$f(x)$$ have no critical points.

Ans: B
Question 3
1 / -0

The point(s) on the curve $$y^{3}+3x^{2}=12y$$ the tangent is vertical is (are)

A
$$\left ( \pm 4/\sqrt{3}\: -2 \right )$$

B
$$\left ( \pm \sqrt{11/3},1 \right )$$

C
$$\left ( 0,0 \right )$$

D
$$\left ( \pm 4/\sqrt{3}\: ,2 \right )$$

Solution

$$3y^{2}.y'+6x=12y'$$
Or
$$y'(3y^{2}-12)=-6x$$
Or
$$y'(y^{2}-4)=-2x$$
Or
$$y'=\dfrac{-2x}{y^{2}-4}$$
Hence for the tangent to be vertical
$$y^{2}-4=0$$
$$y=\pm2$$
Substituting $$y=2$$, we get
$$3x^{2}=12y-y^{3}$$
$$=24-8$$
$$=16$$
Hence
$$x=\dfrac{\pm4}{\sqrt{3}}$$
$$y=-2$$ gives
$$3x^{2}=-16$$
This is not possible.
Hence the required point is
$$(\dfrac{\pm4}{\sqrt{3}},2)$$.
Question 4
1 / -0

The coordinates of the point $$P$$ on the curve $$y^{2}= 2x^{3}$$ the tangent at which is perpendicular to the line $$4x-3y + 2 = 0$$, are given by

A
$$\left ( 2,4 \right )$$

B
$$\left ( 1,\sqrt{2} \right )$$

C
$$\left ( 1/2,-1/2 \right )$$

D
$$\left ( 1/8,-1/16 \right )$$

Solution

$$4x-3y+2=0$$
Or
$$3y=4x+2$$
Or
$$y=\dfrac{4}{3}x+\dfrac{2}{3}$$.
Hence slope of the tangent will be
$$=\dfrac{-3}{4}$$
Now
$$\dfrac{dy}{dx}=\dfrac{-3}{4}$$
$$y=\pm\sqrt{2}.x^{\frac{3}{2}}$$
Hence
$$\dfrac{dy}{dx}$$

$$=\pm\sqrt{2}.\dfrac{3}{2}.x^{\frac{1}{2}}$$

$$=\dfrac{-3}{4}$$

Or
$$-\sqrt{2}.\dfrac{3}{2}.x^{\dfrac{1}{2}}=\dfrac{-3}{4}$$
Or
$$2\sqrt{2}.x^{\dfrac{1}{2}}=1$$
Or
$$x=\dfrac{1}{8}$$.
Hence
$$y=-\sqrt{2}.x^{\dfrac{3}{2}}$$
$$=-\sqrt{2}.\dfrac{1}{16\sqrt{2}}$$
$$=-\dfrac{1}{16}$$
Hence the co-ordinates are $$(\dfrac{1}{8},\dfrac{-1}{16})$$
Question 5
1 / -0

Find the co-ordinates of the point (s) on the curve $$\displaystyle y= \frac{x^{2}-1}{x^{2}+1}, x> 0$$ such that tangent at these point (s)have the greatest slope.

A
$$\displaystyle \left ( \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{2}} \right ).$$

B
$$\displaystyle \left ( \frac{1}{\sqrt{3}},-\frac{1}{{2}} \right ).$$

C
$$\displaystyle \left ( \frac{1}{{3}},-\frac{4}{{5}} \right ).$$

D
$$\displaystyle \left ( {\sqrt{3}},\frac{1}{{2}} \right ).$$

Solution

$$\displaystyle y= \dfrac{x^{2}-1}{x^{2}+1}= \dfrac{x^{2}+1-2}{x^{2}+1}= 1-\dfrac{2}{x^{2}+1}$$

$$\Rightarrow y=1-\dfrac{2}{x^{2}+1}$$

S=slope $$\displaystyle = \dfrac{dy}{dx}= \dfrac{4x}{\left ( x^{2}+1 \right )^{2}}$$

For maximum and minimum of S, $$\displaystyle \dfrac{dS}{dx}= 0$$

$$\displaystyle \dfrac{dS}{dx}= 4\dfrac{\left ( x^{2}+1 \right )^{2}.1-x.2\left ( x^{2}+1 \right ).2x}{\left ( x^{2}+1 \right )^{4}}$$

$$\displaystyle \dfrac{dS}{dx}= 4\dfrac{x^{2}+1-4x^{2}}{\left ( x^{2}+1 \right )^{3}}$$

or $$\displaystyle \dfrac{dS}{dx}= 4\dfrac{1-3x^{2}}{\left ( 1+x^{2} \right )^{3}}=0$$

$$\displaystyle \therefore x= \dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}$$

Now consider change of sign at $$\displaystyle x= \dfrac{1}{\sqrt{3}}, \dfrac{dS}{dx}$$ changes from +ve to -ve.
Hence S has maximum at $$\displaystyle x= \dfrac{1}{\sqrt{3}}$$.
When $$\displaystyle x= \dfrac{1}{\sqrt{3}}$$ the value of $$\displaystyle y= -\dfrac{1}{2}$$
Hence the point is $$\displaystyle \left ( \dfrac{1}{\sqrt{3}},-\dfrac{1}{\sqrt{2}} \right ).$$
Question 6
1 / -0

The tangent to the curve $$x= a\sqrt{\cos 2\theta }\cos \theta $$, $$y= a\sqrt{\cos 2\theta }\sin \theta
$$ at the point corresponding to $$\theta = \pi /6$$ is

A
parallel to the $$x$$-axis

B
parallel to the $$y$$-axis

C
parallel to line $$y = x$$

D
none of these

Solution

$$\displaystyle \dfrac{dx}{d\theta }= -a\sqrt{\cos 2\theta }\sin \theta +\dfrac{-a\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$$

$$\displaystyle = -a\dfrac{\cos 2\theta \sin \theta +\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$$

$$\displaystyle = \dfrac{-a\sin 3\theta }{\sqrt{\cos 2\theta }}$$

$$\displaystyle \dfrac{dy}{d\theta }= a\sqrt{\cos 2\theta }\, \cos \theta -a\dfrac{\sin \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$$

$$
\displaystyle = \dfrac{a\cos 3\theta }{\sqrt{\cos 2\theta }}$$

Hence $$\displaystyle \dfrac{dy}{dx}= -\cot 3\theta \Rightarrow \dfrac{dy}{dx}|_{\theta = \pi /6 }= 0$$
So the tangent to the curve at $$\theta = \pi /6 $$ is parallel to the $$x$$-axis.
Question 7
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The equation of the tangents to $$\displaystyle 4x^{2}-9y^{2}=36$$ which are perpendicular to the straight line $$\displaystyle 2y+5x= 10$$ are

A
$$\displaystyle 5\left ( y-3 \right )=2 \left ( x-\sqrt{\frac{117}{4}} \right )$$

B
$$\displaystyle 5\left ( y-2 \right )=2 \left ( x-\sqrt{-18} \right )$$

C
$$\displaystyle 5\left ( y+2 \right )=2 \left ( x-\sqrt{-18} \right )$$

D
none of these

Solution

Given, $$2y+5x=10$$
$$2y'+5=0$$
$$y'=\dfrac{-5}{2}$$.
Hence the required slope is $$\dfrac{2}{5}$$.
Now
$$4x^{2}-9y^{2}=36$$.
$$4x^{2}-36=9y^{2}$$
$$y=\dfrac{1}{3}.\sqrt{4x^{2}-36}$$.
$$y'=\dfrac{8x}{6\sqrt{4x^{2}-36}}$$ $$=\dfrac{2}{5}$$
$$40x=12\sqrt{4x^{2}-36}$$
$$10x=3\sqrt{4x^{2}-36}$$
$$100x^{2}=36x^{2}-324$$
$$64x^{2}=-324$$
$$x^{2}=\dfrac{-81}{16}$$.
Hence no real values of x.
Hence answer is none of these.
Question 8
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The tangent to the curve $$y=e^{x}$$ drawn at the point $$\left ( c,e^{c} \right )$$ intersects the line joining the points $$(c -1,e^{c-1})$$ and $$(c +1,e^{c+1}) $$

A
on the left of $$x = c$$

B
on the right of $$x = c$$

C
at no paint

D
at all points

Solution

Given equation of curve
$$y=e^x$$
$$\dfrac{dy}{dx}=e^x$$
Slope of tangent at point $$(c,e^c)$$ is $$e^c$$
Equation of tangent at $$(c,e^{c})$$ is
$$y-e^{c}=e^{c}(x-c)$$ ....(1)

Equation of straight line joining $$A\left ( c+1,e^{c+1} \right )$$ and $$B\left ( c-1,e^{c-1} \right )$$ is

$$\displaystyle y-e^{c+1}=\dfrac{e^{c+1}-e^{c-1}}{2}\left ( x-c-1 \right )$$ .....(2)

Subtracting (2) from (1), we get
$$\displaystyle e^{c}\left ( e-1 \right )=e^{c}\left [ \left ( x-c \right )-\dfrac{1}{2} \left ( e-e^{-1} \right )\left ( x-c \right )+\dfrac{1}{2}\left ( e-e^{-1} \right )\right ]$$

$$\displaystyle \Rightarrow \dfrac{1}{2}\left ( e+e^{-1} \right )-1=\left ( x-c \right )\left [ 1-\dfrac{1}{2}\left ( e-e^{-1} \right ) \right ]$$

$$\displaystyle \Rightarrow x-c=\dfrac{e+e^{-1}-2}{2-e+e^{-1}}< 0$$

$$[ \because e+e^{-1}>2$$ and $$2+e^{-1}-e<0 ]$$
$$\Rightarrow x<c$$
Thus the two lines meet to the left of $$x = c$$.
Question 9
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For the curve $${x}^{2}+4xy+8{y}^{2}=64$$ the tangents are parallel to the $$x$$-axis only at the points

A
$$(0,2\sqrt { 2 } )$$ and $$(0,-2\sqrt { 2 } )$$

B
$$(8,-4)$$ and $$(-8,4)$$

C
$$(8\sqrt { 2 } ,-2\sqrt { 2 } )$$ and $$(-8\sqrt { 2 } ,2\sqrt { 2 } )$$

D
$$(9,0)$$ and $$(-8,0)$$

Solution

Given curve is $${x}^{2}+4xy+8{y}^{2}=64.....(i)$$
On differentiating w.r.t $$x$$ ,we get
$$2x+4(y+x\cfrac{dy}{dx})+16y\cfrac{dy}{dx}=0$$
$$\Rightarrow$$ $$2x+4y+(4x+16y)\cfrac{dy}{dx}=0$$
$$\Rightarrow$$ $$\cfrac{dy}{dx}=-\cfrac{(x+2y)}{2(x+4y)}$$
Since, tangent are parallel to $$x$$-axis only
ie., $$\cfrac{dy}{dx}=0$$
$$\Rightarrow$$ $$-\cfrac{(x+2y)}{2(x+4y)}=0$$
$$\Rightarrow$$ $$x+2y=0.....(ii)$$
Now, on putting the values of $$x$$ from eqs. $$(i)$$ in $$(ii)$$ we get
$$4{y}^{2}-8{y}^{2}+8{y}^{2}=64$$
$$\Rightarrow$$ $${y}^{2}=16$$
$$\Rightarrow$$ $$y=\pm 4$$
from eq. $$(ii)$$
When $$y=4,x=-8$$
and when $$y=-4, x=8$$
Hence, required points are $$(-8,4)$$ and $$(8,-4)$$
Question 10
1 / -0

y = $$[x(x-3)]^2$$ is increasing when-

A
$$0 < x < \dfrac{3}{2}$$

B
$$0 < x <$$ $$\infty$$

C
$$- \infty$$ $$< x < 0$$

D
$$1 < x < 3$$

Solution

$$y={ \left[ x\left( x-3 \right) \right] }^{ 2 }$$ increases when For function to be increasing $$\cfrac { dy }{ dx } >0$$
$$\cfrac { dy }{ dx } =2x(x-3)\times (2x-3)>0$$
Now,
$$x(x-3)(2x-3)>0$$
Plotting on number line
(Image)
At a value less than zero, expression is negative
Therefore $$y={ \left[ x\left( x-3 \right) \right] }^{ 2 }$$ increases for
$$x\epsilon \left( 0,\cfrac { 3 }{ 2 } \right) \bigcup { \left( 3,\infty \right) } $$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 58

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Application of Derivatives Test - 58

Score

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SHARING IS CARING

Question 1

$$H = 2$$ and $$V = 1$$

$$H = 1$$ and $$V = 2$$

$$H = 2$$ and $$V = 2$$

$$H = 1$$ and $$V = 1$$

Solution

Question 2

$$\displaystyle x= n\pi \left ( n\epsilon I \right )$$ as critical points

no critical points

$$\displaystyle x= 2n\pi \left ( n\epsilon I \right )$$ as critical points

$$\displaystyle x= \left ( 2n+1 \right )\pi \left ( n\epsilon I \right )$$ as critical points.

Solution

Question 3

$$\left ( \pm 4/\sqrt{3}\: -2 \right )$$

$$\left ( \pm \sqrt{11/3},1 \right )$$

$$\left ( 0,0 \right )$$

$$\left ( \pm 4/\sqrt{3}\: ,2 \right )$$

Solution

Question 4

$$\left ( 2,4 \right )$$

$$\left ( 1,\sqrt{2} \right )$$

$$\left ( 1/2,-1/2 \right )$$

$$\left ( 1/8,-1/16 \right )$$

Solution

Question 5

$$\displaystyle \left ( \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{2}} \right ).$$

$$\displaystyle \left ( \frac{1}{\sqrt{3}},-\frac{1}{{2}} \right ).$$

$$\displaystyle \left ( \frac{1}{{3}},-\frac{4}{{5}} \right ).$$

$$\displaystyle \left ( {\sqrt{3}},\frac{1}{{2}} \right ).$$

Solution

Question 6

parallel to the $$x$$-axis

parallel to the $$y$$-axis

parallel to line $$y = x$$

none of these

Solution

Question 7

$$\displaystyle 5\left ( y-3 \right )=2 \left ( x-\sqrt{\frac{117}{4}} \right )$$

$$\displaystyle 5\left ( y-2 \right )=2 \left ( x-\sqrt{-18} \right )$$

$$\displaystyle 5\left ( y+2 \right )=2 \left ( x-\sqrt{-18} \right )$$

none of these

Solution

Question 8

on the left of $$x = c$$

on the right of $$x = c$$

at no paint

at all points

Solution

Question 9

$$(0,2\sqrt { 2 } )$$ and $$(0,-2\sqrt { 2 } )$$

$$(8,-4)$$ and $$(-8,4)$$

$$(8\sqrt { 2 } ,-2\sqrt { 2 } )$$ and $$(-8\sqrt { 2 } ,2\sqrt { 2 } )$$

$$(9,0)$$ and $$(-8,0)$$

Solution

Question 10

$$0 < x < \dfrac{3}{2}$$

$$0 < x <$$ $$\infty$$

$$- \infty$$ $$< x < 0$$

$$1 < x < 3$$

Solution

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