Application of Derivatives Test

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Application of Derivatives Test - 55

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Weekly Quiz Competition

Question 1
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The slope of the normal to the curve $$ y = 2x ^{2} + 3 \sin x $$ at $$ x = 0 $$ is

A
$$3$$

B
$$1/3$$

C
$$-3$$

D
$$-1 /3$$

Solution

Given, $$ y = 2x ^{2} + 3 \sin x $$
slope $$ = \dfrac{dy}{dx} = 4x + 3 \cos x $$
$$\dfrac{dy}{dy}|_{x=0} = 4 ( 0 ) = 3 $$
slope of normal = $$ -1/ \dfrac{dy}{dx} = -1/3 $$
Question 2
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The line $$ y = x + 1 $$ is a tangent to the curve $$y^{2} = 4 x $$ at the point

A
$$ ( 1 , 2 ) $$

B
$$ ( 2 , 1 ) $$

C
$$ ( 1 , -2 ) $$

D
$$ ( -1 , 2 )$$

Solution

Given,
$$y^{2} = 4 x \\ \Rightarrow 2y \dfrac{dx}{dx} = 4 \\ \Rightarrow \dfrac{dx}{dy} = 2/y $$
but given that $$ y = x + 1 $$ is a tangent to the curve its slope = $$ 1 ( y = mx +c ) $$
$$\dfrac 2y =1\implies y=2$$
$$x=2-1=1$$
The line $$ y = x + 1 $$ is a tangent to the curve $$y^{2} = 4 x $$ at the point $$(1,2)$$.
Question 3
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The points on the curve $$9 y^{2} = x^{3}$$, where the normal to the curve makes equal intercepts with the axes are ...........

A
$$\left ( 4, \pm \dfrac{8}{3} \right )$$

B
$$\left ( 4, \dfrac{-8}{3} \right )$$

C
$$\left ( 4, + \dfrac{8}{3} \right )$$

D
$$\left (\pm 4, \dfrac{8}{3} \right )$$
Question 4
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For $$a\in[\pi,2\pi]$$ and $$n\in I$$, the critical points of $$\displaystyle f(x)=\frac{1}{3}\sin a\tan^{3}x+(\sin a - 1 ) \tan x +\sqrt{\frac{a-2}{8-a}}$$ is

A
$$x=n\pi$$

B
$$x=2n\pi$$

C
$$x=(2n+1)\pi$$

D
no critical points

Solution

$$\displaystyle f(x)=\frac{1}{3}\sin a \tan^{3}(x)+(\sin a - 1 ) \tan x +\sqrt{\dfrac{a-2}{8-a}}$$
$$f'(x)=\sec^{2}x (\sin a \tan^{2}x+\sin a -1)$$
For critical points, $$\>f'(x)=0$$
$$\Rightarrow \sec^{2}x (\sin a \tan^{2}x+\sin a -1)=0$$
$$\Rightarrow \displaystyle \tan^{2}x=\frac{1-\sin a}{\sin a}$$ [$$\because \sec^{2}x=0$$ (not possible)]
Since, $$a\in [\pi,2\pi]$$
$$\Rightarrow \displaystyle \frac{1-\sin a}{\sin a} <0$$
$$\Rightarrow \tan^{2}x <0$$ which is false.
Hence, no critical points.
Question 5
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The value of a for which the function $$\displaystyle \mathrm{f}(\mathrm{x})=(4\mathrm{a}-3)(\mathrm{x}+\log 5)+2(\mathrm{a}-7)\cot\frac{\mathrm{x}}{2}\sin^{2}\frac{\mathrm{x}}{2}$$ does not possess critical points is

A
$$(-\displaystyle \infty,-\frac{4}{3})\mathrm{\cup}(2,\infty)$$

B
$$(-\infty, -1)$$

C
$$[1, \infty)$$

D
$$(-2,\infty)$$

Solution

$$f\left( x \right) =\left( 4a-3 \right) \left( x+\log { 5 } \right) +2\left( a-7 \right) \cot { \cfrac { x }{ 2 } } \sin ^{ 2 }{ \cfrac { x }{ 2 } } $$
$$f'\left( x \right) =\left( 4a-3 \right) \left( 1+0 \right) +2\left( a-7 \right) \left[ \cot { \cfrac { x }{ 2 } } \times 2\sin { \cfrac { x }{ 2 } } \cos { \cfrac { x }{ 2 } } \times \cfrac { 1 }{ 2 } +\sin ^{ 2 }{ \cfrac { x }{ 2 } } \times \left( -\csc ^{ 2 }{ \cfrac { x }{ 2 } } \right) \times \cfrac { 1 }{ 2 } \right] $$
$$f'\left( x \right) =\left( 4a-3 \right) +2\left( a-7 \right) \left[ \cos ^{ 2 }{ \cfrac { x }{ 2 } } -\cfrac { 1 }{ 2 } \right] $$
$$f'\left( x \right) =\left( 4a-3 \right) +2\left( a-7 \right) \times \cfrac { \left( \cos { x } \right) }{ 2 } \quad \left[ \because 1+\cos { x } =2\cos ^{ 2 }{ \cfrac { x }{ 2 } } \right] $$
$$f'\left( x \right) =\left( 4a-3 \right) +\left( a-7 \right) \cos { x } $$
Now, for no critical points
$$f'\left( x \right) \neq 0$$
$$\left( 4a-3 \right) \neq \left( 7-a \right) \cos { x } $$
$$\cos { x } \neq \cfrac { 4a-3 }{ 7-a } $$
$$\therefore \cfrac { 4a-3 }{ 7-a } >1$$$$4a-3>7-a$$
$$5a>10$$
$$a>2$$
and
$$\cfrac { 4a-3 }{ 7-a } <-1$$
$$4a-3<a-7$$
$$3a<-4$$
$$a<-{ 4 }/{ 3 }$$
$$\therefore a\epsilon \left( -\infty ,{ -4 }/{ 3 } \right) \bigcup { (2,\infty ) } $$ is correct.
Question 6
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Directions For Questions

$$\displaystyle \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1$$ is the equation ot an ellipse. Tangents are drawn to the ellipse and it's auxillary circle at the points where a common ordinate cuts them.

...view full instructions

The greatest inclination between the tangents is

A
$$\displaystyle \tan^{-1} \left ( \dfrac{\mathrm{a}+\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$$

B
$$\displaystyle \tan^{-1} \left (\dfrac{\mathrm{a}-\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$$

C
$$\tan^{-1}\sqrt{\dfrac{\mathrm{a}}{\mathrm{b}}}$$

D
$$\tan^{-1}\sqrt{\dfrac{\mathrm{b}}{\mathrm{a}}}$$

Solution

The slope of the tangent to the ellipse $$=-\displaystyle \dfrac{\mathrm{b}}{\mathrm{a}}\cot\theta$$
The slope of tangent to the circle is $$-\cot\theta$$
$$\therefore$$ the angle $$\phi$$ between the tangents is given by
$$\tan\phi=\dfrac{(-\dfrac{\mathrm{b}}{\mathrm{a}}+1)\cot\theta}{1+\dfrac{\mathrm{b}}{\mathrm{a}}\cot^{2}\theta}=\dfrac{(\mathrm{a}-\mathrm{b})}{\mathrm{a}\tan\theta+\mathrm{b}\cot\theta}=\dfrac{\mathrm{a}-\mathrm{b}}{(\sqrt{\mathrm{a}\tan\theta}-\sqrt{\mathrm{b}\cot\theta})^{2}+2\sqrt{\mathrm{a}\mathrm{b}}}$$
This is maximum when $$\sqrt{\mathrm{a}\tan\theta}-\sqrt{\mathrm{b}\cot\theta}=0$$
$$\therefore$$ the maximum angle $$=\displaystyle \tan^{-1}(\dfrac{\mathrm{a}-\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}})$$
Question 7
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A function $$y=f(x)$$ has a second order derivative $$f''(x)=6(x-1)$$ .
If its graph passes through the point $$(2,1)$$ and at that point the tangent to the graph is $$y=3x-5$$, then the function is

A
$$(x-1)^{2}$$

B
$$(x+1)^{2}$$

C
$$(x+1)^{3}$$

D
$$(x-1)^{3}$$

Solution

$$f''(x)=6(x-1)$$
$$ \Rightarrow f'(x) = 3x^2 -6x+k$$...(1)
Slope of tangent at $$(2,1)$$ is $$3$$.
$$ \Rightarrow 3=3 \times 2^2 -6 \times 2 +k$$
$$ \Rightarrow k=3$$...(2)
From (1) and (2)
$$f(x)=x^3-3x^2+3x+c$$
Using $$f(x)=y=1$$ when $$x=2$$, we get
$$1=2+c \Rightarrow c=-1$$
Hence, the function is
$$f(x)=x^3-3x^2+3x-1$$
$$ \Rightarrow f(x)=(x-1)^3$$
Question 8
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If $$ f(x) = \displaystyle \frac{x}{{\sin x}}$$ and $$g(x) = \displaystyle \frac{x}{{\tan x}}$$ where $$0 < x \leq 1$$ then in the interval

A
Both f(x) and g(x) are increasing functions

B
Both f(x) and g(x) are decreasing functions

C
f(x) is an increasing function

D
g(x) is an increasing function

Solution

We know that if $$f'(x)>0$$ then $$f$$ is increasing on that interval and if $$f'(x)<0$$ then $$f$$ is decreasing on the interval.

Given $$f(x)=\dfrac{x}{sinx}$$

$$\implies f'(x)=\dfrac{sinx-xcosx}{sin^2x}=0$$

$$\implies sinx-xcosx=0$$

$$\implies x=\dfrac{sinx}{cosx}=tanx$$

In the interval $$(0,1)$$ there is no solution for $$x=tanx$$

Therefore, $$f'(x)>0$$

Hence, $$f(x)$$ is an increasing function.

Given $$g(x)=\dfrac{x}{tanx}$$

$$\implies g'(x)=\dfrac{tanx-xsec^2x}{tan^2x}=0$$

$$\implies tanx-xsec^2x=0$$

$$\implies x=\dfrac{sec^2x}{tanx}=\dfrac{1}{sinxcosx}$$

In the interval $$(0,1)$$, $$g(x)$$ is not an increasing function.
Question 9
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A curve passes through $$(2, 0)$$ and the slope of the tangent at any point $$(x, y)$$ is $$x^2 -2x$$ for all values of $$x$$. The point of minimum ordinate on the curve where $$x > 0$$ is $$(a, b)$$'
Then find the value of $$a + 6b$$.

A
$$2$$

B
$$4$$

C
$$-2$$

D
$$-4$$

Solution

$$\dfrac {dy}{dx} = x^2 - 2x$$
$$y = \dfrac {x^3}{3} - x^2 + C$$
Passing through (2, 0)
$$\Rightarrow C = \dfrac {4}{3}$$
$$\therefore y = \dfrac {x^3}{3} - x^2 + \dfrac {4}{3}$$
$$y' = x^2 - 2x = x(x - 2)$$
For maxima or minima, $$y'=0$$
$$ \Rightarrow x= 0,2$$
$$y''>0$$ at $$x=2$$
At x = 2, y takes the minimum value.
$$\therefore $$ minimum value of y is $$= \dfrac {8}{3} - 4 +\dfrac {4}{3} = 0$$
$$\therefore a = 2, b = 0$$
$$a + 6b = 2$$
Question 10
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Suppose $$a,b,c$$ are such that the curve $$y = ax^2 + bx + c$$ is tangent to $$y = 3x -3$$ at $$(1, 0)$$ and is also tangent to $$y = x + 1$$ at $$(3, 4)$$ then the value of $$(2a -b -4c)$$ equals

A
$$7$$

B
$$8$$

C
$$9$$

D
$$10$$

Solution

$$y = ax^2 + bx + c $$
Since $$(1,0)$$ lies on the curve
$$ \Rightarrow a + b + c = 0 $$ ...... (i)
Slope of curve $$\displaystyle =\frac {dy}{dx} = 2ax + b$$
Slope of curve at (1,0)$$=2a+b$$
Slope of curve at (3,4)$$=6a+b$$
Slope of tangent $$y=3x-3$$ is 3
Slope of tangent $$y=x+1$$ is 1
$$(\dfrac {dy}{dx})_{x=1} = 3$$ and $$(\dfrac {dy}{dx})_{x=3} = 1$$
$$2a + b = 3$$ .... (ii)
$$6a + b = 1$$ .... (iii)
on solving we get a $$= -\dfrac {1}{2}, b = 4, c = -\dfrac {7}{2}$$
$$\therefore 2a - b - 4c = -1 -4 + 14 = 9$$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 55

Result

Application of Derivatives Test - 55

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SHARING IS CARING

Question 1

$$3$$

$$1/3$$

$$-3$$

$$-1 /3$$

Solution

Question 2

$$ ( 1 , 2 ) $$

$$ ( 2 , 1 ) $$

$$ ( 1 , -2 ) $$

$$ ( -1 , 2 )$$

Solution

Question 3

$$\left ( 4, \pm \dfrac{8}{3} \right )$$

$$\left ( 4, \dfrac{-8}{3} \right )$$

$$\left ( 4, + \dfrac{8}{3} \right )$$

$$\left (\pm 4, \dfrac{8}{3} \right )$$

Question 4

$$x=n\pi$$

$$x=2n\pi$$

$$x=(2n+1)\pi$$

no critical points

Solution

Question 5

$$(-\displaystyle \infty,-\frac{4}{3})\mathrm{\cup}(2,\infty)$$

$$(-\infty, -1)$$

$$[1, \infty)$$

$$(-2,\infty)$$

Solution

Question 6

$$\displaystyle \tan^{-1} \left ( \dfrac{\mathrm{a}+\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$$

$$\displaystyle \tan^{-1} \left (\dfrac{\mathrm{a}-\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$$

$$\tan^{-1}\sqrt{\dfrac{\mathrm{a}}{\mathrm{b}}}$$

$$\tan^{-1}\sqrt{\dfrac{\mathrm{b}}{\mathrm{a}}}$$

Solution

Question 7

$$(x-1)^{2}$$

$$(x+1)^{2}$$

$$(x+1)^{3}$$

$$(x-1)^{3}$$

Solution

Question 8

Both f(x) and g(x) are increasing functions

Both f(x) and g(x) are decreasing functions

f(x) is an increasing function

g(x) is an increasing function

Solution

Question 9

$$2$$

$$4$$

$$-2$$

$$-4$$

Solution

Question 10

$$7$$

$$8$$

$$9$$

$$10$$

Solution

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