Application of Derivatives Test

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Application of Derivatives Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

The line $$y=x$$ is a tangent to the parabola $$\displaystyle y= ax^{2}+bx+c$$ at the point $$x=1$$.If the parabola passes through the point $$(-1,0)$$, then determine $$a, b, c.$$

A
$$\displaystyle a= \frac{1}{2}, b= \frac{1}{4}, c= \frac{1}{3}.$$

B
$$\displaystyle a= \frac{1}{4}, b= \frac{1}{2}, c= \frac{1}{4}.$$

C
$$\displaystyle a= 2, b= 1, c= 4.$$

D
$$\displaystyle a= 4, b= 2, c= 4.$$

Solution

Given equation of parabola is
$$\displaystyle y= ax^{2}+bx+c$$
$$\displaystyle \frac{dy}{dx}= 2ax+b$$
Slope of tangent to the curve at $$x=1$$ is $$2a+b$$
Given tangent is $$y=x$$ . Slope of this tangent is $$1.$$
So, $$2a+b=1$$ ...(1)

Since, the parabola passes through $$(-1,0)$$
$$\displaystyle \therefore a-b+c= = 0$$ ...(2)

Given $$y=x$$ is a tangent at $$x=1$$
$$\displaystyle \therefore y= 1.$$
Hence $$(1,1)$$ lies both on tangent and parabola
$$\displaystyle \therefore a+b+c= 1$$ ...(3)

Solving (1), (2) and (3), we get
$$\displaystyle a= \frac{1}{4}, b= \frac{1}{2}, c= \frac{1}{4}.$$
Question 2
1 / -0

Find the points on the curve $$y=x/(1-x^{2})$$ where the tangents makes an angle of $$\pi /4$$ with x-axis

A
$$(\sqrt { 3 } ,-\sqrt { \dfrac { 2 }{ 3 } } ),(-\sqrt { 2 } ,\sqrt { \dfrac { 2 }{ 3 } } )$$

B
$$(\sqrt { 3 } ,-\sqrt { \dfrac { 3 }{ 4 } } ),(-\sqrt { 3 } ,\sqrt { \dfrac { 3 }{ 4 } } )$$

C
$$(\sqrt { 3 } ,-\sqrt { \dfrac { 3 }{ 2 } } ),(-\sqrt { 3 } ,\sqrt { \dfrac { 3 }{ 2 } } )$$

D
none of these

Solution

$$\dfrac{dy}{dx}=tan45^{0}$$
Or
$$\dfrac{dy}{dx}=1$$
Or
$$\dfrac{(1-x^{2})-x(-2x)}{(1-x^{2})^{2}}=1$$
Or
$$1-x^{2}+2x^{2}=(1-x^{2})^{2}$$
Or
$$1+x^{2}=1-2x^{2}+x^{4}$$
Or
$$x^{4}-3x^{2}=0$$
Or
$$x^{2}[x^{2}-3]=0$$
$$x=0$$ or
$$x=\pm\sqrt{3}$$
Hence
$$y=\dfrac{\sqrt{3}}{1-3}=-\dfrac{\sqrt{3}}{2}$$

$$y=\dfrac{-\sqrt{3}}{1-3}=\dfrac{\sqrt{3}}{2}$$

Hence
$$\left(\sqrt{3},-\dfrac{\sqrt{3}}{2}\right),\left(-\sqrt{3},\dfrac{\sqrt{3}}{2}\right)$$.
Question 3
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The line $$\dfrac xa+\dfrac yb=1$$ touches the curve $$\displaystyle y=be^{-x/a}$$ at the point

A
$$(a,b/a)$$

B
$$(-a,b/a)$$

C
$$(a,a/b)$$

D
None of these

Solution

Simplifying the equation of the line we get
$$bx+ay=ab$$
$$ay=-bx+ab$$
Or
$$y=\dfrac{-b}{a}.x+b$$
Hence
$$\dfrac{dy}{dx}=\dfrac{-b}{a}$$
Or
$$-\dfrac{b}{a}.e^{-x/a}.=\dfrac{-b}{a}$$
Or
$$e^{-x/a}=1$$
Or
$$x=0$$
Hence
$$y=b$$.
Therefore the point is $$(0,b)$$
Question 4
1 / -0

The curve $$\displaystyle y-e^{xy}+x=0$$ has a vertical tangent at the point

A
$$(1,\ 1)$$

B
$$no\ point$$

C
$$(0,\ 1)$$

D
$$(1,\ 0)$$

Solution

$$\dfrac{dy}{dx}-e^{xy}[x\dfrac{dy}{dx}+y]+1=0$$
Or
$$\dfrac{dy}{dx}[1-x.e^{xy}]+1-y.e^{xy}=0$$
Or
$$\dfrac{dy}{dx}=\dfrac{-(1-y.e^{xy})}{1-x.e^{xy}}$$
Now
$$1-xe^{xy}=0$$ since the slope of the tangent is $$90^{0}$$.
Hence
$$xe^{xy}=1$$
Or
$$x(x+y)=1$$
Or
$$x^{2}+xy=1$$
Or
$$y=\dfrac{1-x^{2}}{x}$$
Or
$$y=\dfrac{1}{x}-x$$ ...(i)
Considering $$y=0$$,
$$x^{2}=1$$
$$x=\pm1$$.
Hence we get 2 points $$(1,0)$$ and $$(-1,0)$$.
Out of these 2, only $$(1,0)$$ lies on the curve.
Hence the required point is $$(1,0)$$.
Question 5
1 / -0

If $$\displaystyle x\cos \alpha +y\sin \alpha =p$$ touches $$\displaystyle x^{2}+a^{2}y^{2}=a^{2},$$ then

A
$$\displaystyle p^{2}=a^{2}\sin^{2}\alpha +\cos^{2}\alpha $$

B
$$\displaystyle p^{2}=a^{2}\cos^{2}\alpha +\sin^{2}\alpha $$

C
$$\displaystyle 1/p^{2}=\sin^{2}\alpha +\alpha^{2}\cos^{2}\alpha $$

D
$$\displaystyle 1/p^{2}=\cos^{2}\alpha +\alpha^{2}\sin^{2}\alpha $$

Solution

Solving for y, we have
$$\displaystyle y=\dfrac { p }{ \sin { \alpha } } -x\cot { \alpha } $$
Putting this value in $${ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }={ a }^{ 2 }$$, we have
$$\displaystyle { x }^{ 2 }+{ a }^{ 2 }{ \left( \dfrac { p }{ \sin { \alpha } } -x\cot { \alpha } \right) }^{ 2 }={ a }^{ 2 }$$

$$\displaystyle \Rightarrow { x }^{ 2 }\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) -\dfrac { 2a^{ 2 }xp\cot { \alpha } }{ \sin { \alpha } } +\dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha } } -{ a }^{ 2 }=0$$

The discriminant of this equation must be zero. So
$$\displaystyle { a }^{ 4 }\dfrac { { p }^{ 2 }\cot ^{ 2 }{ \alpha } }{ \sin ^{ 2 }{ \alpha } } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) \left( \dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha } } -{ a }^{ 2 } \right) \\ \Rightarrow { a }^{ 2 }{ p }^{ 2 }\cot ^{ 2 }{ \alpha } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) \left( { p }^{ 2 }-\sin ^{ 2 }{ \alpha } \right) \\ \Rightarrow { p }^{ 2 }\left( { a }^{ 2 }\cot ^{ 2 }{ \alpha } -1-{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) =\sin ^{ 2 }{ \alpha } -{ a }^{ 2 }\cos ^{ 2 }{ \alpha } \\ \Rightarrow { p }^{ 2 }={ a }^{ 2 }\cos ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \alpha } $$
Question 6
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The critical points of the function $$\displaystyle f\left ( x\right )=\left ( x-2 \right )^{2/3}\left ( 2x+1\right )$$ are

A
$$-1,2$$

B
$$1$$

C
$$\displaystyle 1,-\frac{1}{2}$$

D
$$1,2$$

Solution

Given, $$f(x)=(x-2)^{2/3}(3x+1)$$
$$\displaystyle f'\left( x \right) =\dfrac { 2 }{ 3 } { \left( x-2 \right) }^{ \frac { -1 }{ 3 } }\left( 2x+1 \right) +2{ \left( x-2 \right) }^{ \frac { 2 }{ 3 } }=10\left( x-1 \right) { \left( x-2 \right) }^{ \frac { -1 }{ 3 } }$$
$$f'\left( x \right) =0$$
$$\Rightarrow x=1$$
Also $$f'\left( x \right) $$ does not exits at $$x=2$$
Hence the critical points are $$x=1,2$$
Question 7
1 / -0

If $$\displaystyle f\left ( 0 \right )=0$$ and $$\displaystyle f''\left ( x \right )>0$$ for all $$x > 0$$, then $$\displaystyle \frac{f(x)}{x}$$

A
decreases on $$\displaystyle \left ( 0, \infty \right )$$

B
increases on $$\displaystyle \left ( 0, \infty \right )$$

C
decreases on $$\displaystyle \left ( 1, \infty \right )$$

D
neither increases nor decreases on $$\displaystyle \left ( 0, \infty \right )$$

Solution

Given, $$f\left( 0 \right) =0$$
$$f''(x)>0$$, it means $$f'(x)$$ is also increasing
Let $$g(x)=\cfrac { f\left( x \right) }{ x } $$
$$g'\left( x \right) =\cfrac { xf'\left( x \right) -f\left( x \right) }{ { x }^{ 2 } } $$
$$f'\left( x \right) $$ is increasing and $$x\epsilon \left( 0,\infty \right) $$ thus
$$f'\left( x \right) =$$positive
$$f(x)=$$positive
$${ x }^{ 2 }=$$positive
Therefore, $$g\left( x \right) =$$positive $$>0$$
Thus, $$\cfrac { f\left( x \right) }{ x } $$ increases for $$x\epsilon \left( 0,\infty \right) $$
Question 8
1 / -0

The number of critical points of the fuction $$\displaystyle f'\left ( x \right ),$$ where $$\displaystyle f'\left ( x \right )= \frac{\left | x-2 \right |}{x^{3}}$$ are

A
$$0$$

B
$$1$$

C
$$3$$

D
$$4$$

Solution

$$\displaystyle f\left( x \right) =\begin{cases} \dfrac { x-2 }{ { x }^{ 3 } } ,\quad \quad x>1 \\ \dfrac { 2-x }{ { x }^{ 3 } } ,\quad \quad x<1,x\neq 0 \end{cases}$$

$$\displaystyle f'\left( x \right) =\begin{cases} \dfrac { 2\left( 3-x \right) }{ { x }^{ 4 } } ,\quad x\in \left( -\infty ,0 \right) \cup \left( 0,1 \right) \\ \dfrac { 2\left( x-3 \right) }{ { x }^{ 4 } } ,\quad \quad x\in \left( 1,\infty \right) \end{cases}$$

$$\displaystyle f''\left( x \right) =\begin{cases} \dfrac { 6\left( x-4 \right) }{ { x }^{ 5 } } ,\quad x\in \left( -\infty ,0 \right) \cup \left( 0,1 \right) \\ \dfrac { 6\left( 4-x \right) }{ { x }^{ 5 } } ,\quad \quad x\in \left( 1,3 \right) \cup \left( 3,\infty \right) \end{cases}$$

$$f''\left( x \right) $$ doesn't exits at $$x=3$$
Thus critical point of $$f'\left( x \right) $$ is 3.
Question 9
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The angle at which the curve $$y=ke^{kx}$$ intersects the $$y$$ -axis is

A
$$\tan ^{-1}(k^{2})$$

B
$$\cot ^{-1}(k^{2})$$

C
$$\sin ^{-1}\left ( 1/\sqrt{1+k^{4}} \right )$$

D
$$

\sec ^{-1}\left ( 1/\sqrt{1+k^{4}} \right )$$

Solution

$$\displaystyle \dfrac{dy}{dx}=k^{2}e^{kx}$$.

The curve intersects $$y$$-axis at $$\left ( 0,k \right )$$

So, $$\displaystyle \dfrac{dy}{dx}|_{\left ( 0,k \right )}=k^{2}$$.

If $$\theta $$ is the angle at which the given
curve intersects the $$y$$-axis then $$\displaystyle \tan \left ( \pi /2-\theta \right )=\dfrac{k^{2}-0}{1+0.k^{2}}=k^{2}$$.

Hence $$\theta =\cot ^{-1}k^{2}$$
Question 10
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Let $$\displaystyle f\left ( x \right )=x^{3}+ax+b$$ with $$\displaystyle a\neq b$$ and suppose the tangent lines to the graph of $$f$$ at $$x = a$$ and $$x = b$$ have the same gradient Then the value of $$f (1)$$ is equal to

A
$$0$$

B
$$1$$

C
$$\displaystyle -\frac{1}{3}$$

D
$$\displaystyle \frac{2}{3}$$

Solution

$$f(x)=x^3+ax+b$$
$$f'(x)=3x^2+a$$
Given gradient at $$x=a$$ and at $$x=b$$ are same
$$\Rightarrow 3a^2 +a=3b^2+a\Rightarrow b^2=a^2$$
But given $$a\neq b$$
$$\Rightarrow a+b=0 ..(1)$$
Hence $$ f(1)=1+a+b=1$$ using (1)

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 28

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Application of Derivatives Test - 28

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Question 1

$$\displaystyle a= \frac{1}{2}, b= \frac{1}{4}, c= \frac{1}{3}.$$

$$\displaystyle a= \frac{1}{4}, b= \frac{1}{2}, c= \frac{1}{4}.$$

$$\displaystyle a= 2, b= 1, c= 4.$$

$$\displaystyle a= 4, b= 2, c= 4.$$

Solution

Question 2

$$(\sqrt { 3 } ,-\sqrt { \dfrac { 2 }{ 3 } } ),(-\sqrt { 2 } ,\sqrt { \dfrac { 2 }{ 3 } } )$$

$$(\sqrt { 3 } ,-\sqrt { \dfrac { 3 }{ 4 } } ),(-\sqrt { 3 } ,\sqrt { \dfrac { 3 }{ 4 } } )$$

$$(\sqrt { 3 } ,-\sqrt { \dfrac { 3 }{ 2 } } ),(-\sqrt { 3 } ,\sqrt { \dfrac { 3 }{ 2 } } )$$

none of these

Solution

Question 3

$$(a,b/a)$$

$$(-a,b/a)$$

$$(a,a/b)$$

None of these

Solution

Question 4

$$(1,\ 1)$$

$$no\ point$$

$$(0,\ 1)$$

$$(1,\ 0)$$

Solution

Question 5

$$\displaystyle p^{2}=a^{2}\sin^{2}\alpha +\cos^{2}\alpha $$

$$\displaystyle p^{2}=a^{2}\cos^{2}\alpha +\sin^{2}\alpha $$

$$\displaystyle 1/p^{2}=\sin^{2}\alpha +\alpha^{2}\cos^{2}\alpha $$

$$\displaystyle 1/p^{2}=\cos^{2}\alpha +\alpha^{2}\sin^{2}\alpha $$

Solution

Question 6

$$-1,2$$

$$1$$

$$\displaystyle 1,-\frac{1}{2}$$

$$1,2$$

Solution

Question 7

decreases on $$\displaystyle \left ( 0, \infty \right )$$

increases on $$\displaystyle \left ( 0, \infty \right )$$

decreases on $$\displaystyle \left ( 1, \infty \right )$$

neither increases nor decreases on $$\displaystyle \left ( 0, \infty \right )$$

Solution

Question 8

$$0$$

$$1$$

$$3$$

$$4$$

Solution

Question 9

$$\tan ^{-1}(k^{2})$$

$$\cot ^{-1}(k^{2})$$

$$\sin ^{-1}\left ( 1/\sqrt{1+k^{4}} \right )$$

$$\sec ^{-1}\left ( 1/\sqrt{1+k^{4}} \right )$$

Solution

Question 10

$$0$$

$$1$$

$$\displaystyle -\frac{1}{3}$$

$$\displaystyle \frac{2}{3}$$

Solution

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$$

\sec ^{-1}\left ( 1/\sqrt{1+k^{4}} \right )$$