Application of Derivatives Test

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Application of Derivatives Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=min(|x|^{2}-5|x|,1)$$ then $$f(x)$$ is non differentiable at $$\lambda$$ points, then $$\lambda+13$$ equals

A
$$16$$

B
$$14$$

C
$$13$$

D
$$15$$
Question 2
1 / -0

Find the angle between tangent of the curve $$y = (x + 1) (x - 3)$$ at the point where it cuts the axis of $$x$$.

A
$$\tan^{-1} \left(\dfrac{8}{15}\right)$$

B
$$\tan^{-1} \left(\dfrac{15}{8}\right)$$

C
$$\tan^{-1} 4$$

D
$$None\ of\ these$$

Solution

The curve $$y=(x+1)(x-3)=x^2-2x-3$$.........(1) cuts the axis of $$x$$ at $$(-1,0)$$ and $$(3.0)$$. These points are obtained by putting $$y=0$$ in the equation (1).
Now the slope of the tangent to the curve (1) at $$(-1,0)$$ be $$(m_1)=\left.\dfrac{dy}{dx}\right|_{(-1,0)}=[2x-2]_{(-1,0)}=-4$$.
Again the slope of the tangent to the curve (1) at $$(-1,0)$$ be $$(m_2)=\left.\dfrac{dy}{dx}\right|_{(3,0)}=[2x-2]_{(3,0)}=4$$.
Now the angle between the tangents to (1) at those points be
$$\tan^{-1}\left(\dfrac{m_1-m_2}{1+m_1.m_2}\right)$$

$$=\tan^{-1}\left(\dfrac{8}{15}\right)$$.
So the required angle be $$\tan^{-1}\left(\dfrac{8}{15}\right)$$.
Question 3
1 / -0

If the curves $$y^2 = 4ax$$ and $$xy = c^2$$ cut orthogonally then $$\dfrac{c^4}{a^4} =$$

A
$$4$$

B
$$8$$

C
$$16$$

D
$$32$$

Solution

$$y^2=4ax, xy=c^2$$ cut orthogonally
Let they intersect at $$(x_1, y_1)$$
$$y^2=4ax$$
$$2y\dfrac{dy}{dx}=4a$$
$$\dfrac{dy}{dx}=\dfrac{2a}{y}$$
$$\dfrac{dy}{dx}|_{(x_1, y_1)}=\dfrac{2a}{y_1}$$ …..$$(1)$$
$$xy=c^2$$
$$y+x\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{dx}|_{(x_1, y_1)}=\dfrac{-y_1}{x_1}$$ ……$$(2)$$
From $$(1), (2)$$ $$(\because m_1m_2=-1)$$
$$\dfrac{2a}{\not{y_1}}\times \dfrac{\not{-}\not{y_1}}{x_1}=\not{-}1$$
$$x_1=2a$$
From $$y^2=4ax$$
$$y_1=\sqrt{8a^2}=2\sqrt{2}a$$
$$xy=c^2$$
$$x_1y_1=c^2$$
$$2a(2\sqrt{2}a)=c^2$$
$$\dfrac{c^2}{a^2}=4\sqrt{2}$$
$$\dfrac{c^4}{a^4}=32$$.
Question 4
1 / -0

Number of critical points of the function $$\displaystyle f\left( x \right) =\dfrac { 2 }{ 3 } \sqrt { { x }^{ 3 } } -\dfrac { x }{ 2 } +\int _{ 1 }^{ x }{ \left( \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \cos { 2t } -\sqrt { t } \right) } dt$$ which lie in the interval $$\left[ -2\pi ,2\pi \right] $$ is:

A
$$2$$

B
$$6$$

C
$$4$$

D
$$8$$

Solution
Question 5
1 / -0

The equation of the tangent to the curve $$y = b{e^{ -\dfrac{x}{a}}}$$ at a point , where $$x=0$$ is

A
$$\dfrac{x}{a} - \dfrac{y}{b} = 1$$

B
$$\dfrac{y}{b} - \dfrac{x}{a} = 1$$

C
$$\dfrac{x}{a} + \dfrac{y}{b} = 1$$

D
$$\dfrac{x}{b} + \dfrac{y}{a} = 1$$

Solution

$$y=b{ e }^{ \cfrac { -x }{ a } }$$ when $$x=0\, y=b$$
$$\dfrac { dy }{ dx } =\cfrac { -b }{ a } { e }^{ \cfrac { -x }{ a } }$$ at $$x=0$$,
$$\dfrac { dy }{ dx } =\cfrac { -b }{ a } $$
$$y=\cfrac { -bx }{ a } +c$$
$$y=\cfrac { -bx }{ a } +b\\ \cfrac { y }{ b } =\cfrac { -x }{ a } +1\\ \cfrac { x }{ a } +\cfrac { y }{ b } =1$$
Question 6
1 / -0

The function $$y = \dfrac{2x^2 - 1}{x^4}$$ is

A
Always increasing

B
Always decreasing

C
Neither increasing nor decreasing

D
None of these

Solution

$$y = \dfrac{{2{x^2} - 1}}{{{x^4}}}$$
$$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4x \times {x^4} - 4{x^3}\left( {2{x^2} - 1} \right)}}{{{x^8}}}$$
$$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4{x^3} - 4{x^5}}}{{{x^8}}}$$
$$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4{x^3}\left( {1 - {x^2}} \right)}}{{{x^8}}}$$
$$\dfrac{{dy}}{{dx}} \geqslant 0\forall x \in \left[ { - 1,1} \right]$$
So,this function is increasing for $$x \in \left[ { - 1,1} \right]$$ and decreasing for other values of $$x$$
Hence this function is neither increasing nor decreasing
Question 7
1 / -0

The equation of the curve passing through $$(1,3)$$ whose slope at any point $$(x,y)$$ on it is $$\dfrac { y }{ { x }^{ 2 } }$$ is given by

A
$$y={ 3e }^{ -1/x }$$

B
$$y={ 3e }^{ 1-1/x }$$

C
$$y={ ce }^{ 1/x }$$

D
$$y={ 3e }^{ 1/x }$$

Solution

$$\dfrac{{dy}}{{dx}} = \dfrac{y}{{{x^2}}}\,\,\,\,\,\,\,\left( {given} \right)$$

$$\int {\dfrac{{dy}}{{dx}} = \int {\dfrac{{dx}}{{{x^2}}}} } $$

$$ \Rightarrow \log y = - \dfrac{1}{x} + c$$

$$ \Rightarrow y = {e^{ - \frac{1}{x} + c}}$$

$$ \Rightarrow y = {e^{ - \frac{1}{x} + c}}$$

$$ \Rightarrow y = {e^{ - \frac{1}{x}}},c$$

since this curve is passing through point $$\left( {1,3} \right)$$

so, $$3 = {e^{ - \dfrac{1}{t}}}.c$$

$$ \Rightarrow c = 3e$$

therefore , eqn of the curve is

$$y = {e^{ - \dfrac{1}{x}}}.3e$$

$$ \Rightarrow y = 3{e^{ - \dfrac{1}{x} + 1}}$$

$$ \Rightarrow y = 3{e^{1 - \dfrac{1}{x}}}$$
Question 8
1 / -0

An equation of the tangent to the curve $$y=x^{4}$$ from the point $$(2,0)$$ not on the curve is:

A
$$y=0$$

B
$$x=0$$

C
$$x+y=0$$

D
$$none\ of\ these$$

Solution

let a,b be point on curve (x, y)

$$ \displaystyle b = a^4 $$

$$ \displaystyle \frac{dy}{dx} (x^4) = 4x^3 = 4a^3 $$

$$ \displaystyle (y- y_1) = m (x - x_1) \rightarrow$$ be tangent equation passes through (2, 0)

$$ \displaystyle (b-0) = 4a^3 (a-2) $$

$$ \displaystyle a^4 = 4a^3 (a-2) $$

$$ \displaystyle 3a^4 - 8a^3 = 0 $$

$$ \displaystyle a^3 (3a - 8) = 0 $$

$$ \displaystyle a= 0 \quad or \quad a= \frac{8}{3} \quad (0,0)$$ & $$\left[\dfrac{8}{3},\left(\dfrac{8}{3} \right)^4\right] $$ points on curve

$$ \displaystyle b=0 \quad or \quad b= \left(\frac{8}{3}\right)^4 $$

$$ \displaystyle (y - 0) = \frac{0-0}{2-0} (x-0)\Rightarrow y = 0 \rightarrow $$tangent (1)

$$ \displaystyle \left[y-\left(\frac{8}{3}\right)^4\right] = \frac{(\frac{8}{3})^4_{-0}}{(\frac{8}{3})_{-2}} \quad (x - \frac{8}{3}) \rightarrow $$tangent (2)
Question 9
1 / -0

Let $$f:R\rightarrow R$$ be a function defined by $$f\left(x\right)=Min \left\{x+1, \left|x\right|+1\right\}$$. Then which of the following is true?

A
$$f(x)\ge 1$$ for all $$x\in R$$

B
$$f(x)\ge 1$$ is not differentiable at $$x=1$$

C
$$f(x)\ge 1$$ is differentiable at $$x=1$$

D
$$f(x)\ge 1$$ is not differentiable at $$x=0$$

Solution

Function has the minimum value of $$x+1$$ and $$\left|x\right|+1$$
Hence $$f\left(x\right)$$ is differentiable everywhere.
Question 10
1 / -0

Find the points on the ellipse $$\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{9}=1$$ , on which the normals are parallel to the line $$3x-y=1$$.

A
$$(\pm\dfrac{6}{\sqrt {10}},\pm\dfrac{3}{\sqrt {10}})$$

B
$$(\pm\dfrac{3}{\sqrt {10}},\pm\dfrac{1}{\sqrt {10}})$$

C
$$(\pm\dfrac{1}{\sqrt {10}},\pm\dfrac{2}{\sqrt {10}})$$

D
None of these

Solution

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 41

Result

Application of Derivatives Test - 41

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SHARING IS CARING

Question 1

$$16$$

$$14$$

$$13$$

$$15$$

Question 2

$$\tan^{-1} \left(\dfrac{8}{15}\right)$$

$$\tan^{-1} \left(\dfrac{15}{8}\right)$$

$$\tan^{-1} 4$$

$$None\ of\ these$$

Solution

Question 3

$$4$$

$$8$$

$$16$$

$$32$$

Solution

Question 4

$$2$$

$$6$$

$$4$$

$$8$$

Solution

Question 5

$$\dfrac{x}{a} - \dfrac{y}{b} = 1$$

$$\dfrac{y}{b} - \dfrac{x}{a} = 1$$

$$\dfrac{x}{a} + \dfrac{y}{b} = 1$$

$$\dfrac{x}{b} + \dfrac{y}{a} = 1$$

Solution

Question 6

Always increasing

Always decreasing

Neither increasing nor decreasing

None of these

Solution

Question 7

$$y={ 3e }^{ -1/x }$$

$$y={ 3e }^{ 1-1/x }$$

$$y={ ce }^{ 1/x }$$

$$y={ 3e }^{ 1/x }$$

Solution

Question 8

$$y=0$$

$$x=0$$

$$x+y=0$$

$$none\ of\ these$$

Solution

Question 9

$$f(x)\ge 1$$ for all $$x\in R$$

$$f(x)\ge 1$$ is not differentiable at $$x=1$$

$$f(x)\ge 1$$ is differentiable at $$x=1$$

$$f(x)\ge 1$$ is not differentiable at $$x=0$$

Solution

Question 10

$$(\pm\dfrac{6}{\sqrt {10}},\pm\dfrac{3}{\sqrt {10}})$$

$$(\pm\dfrac{3}{\sqrt {10}},\pm\dfrac{1}{\sqrt {10}})$$

$$(\pm\dfrac{1}{\sqrt {10}},\pm\dfrac{2}{\sqrt {10}})$$

None of these

Solution

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