Application of Derivatives Test

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Application of Derivatives Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

If the curves $${y}^{2}=6x,9{x}^{2}+b{y}^{2}=16$$ intersect each other at right angles, then the values of $$b$$ is

A
$$6$$

B
$$\cfrac{7}{2}$$

C
$$4$$

D
$$\cfrac{9}{2}$$

Solution

given the curves intersect at right angles so the tangents to curves at intersection points are perpendicular

consider curve 1 $$y^2=6x$$

$$2y.y_{}{'}=6=>y_{}{'}=\dfrac{6}{2y}$$

now slope of tangent to second curve is $$18x+2by.y_{}{'}=0=>y_{}{'}=\dfrac{-9x}{by}$$

now product of slopes is $$-1$$

$$\dfrac{6}{2y}\times \dfrac{-9x}{by}=-1$$

but $$6x=y^2$$

$$\dfrac{-9y^2}{2by^2}=-1$$

$$b=\dfrac{9}{2}$$
Question 2
1 / -0

The equation of normal to the curve $$y=\left| { x }^{ 2 }-\left| x \right| \right| $$ at $$x=-2$$ is

A
$$3y=2x+10$$

B
$$3y=x+8$$

C
$$2y=x+6$$

D
$$2y=3x+10$$

Solution

$$y=\mid x^{2}-\mid x \mid \mid=(\mid x\ \mid )(\mid \mid x\mid-1\mid )$$
At $$x=-2,y=2$$

$$\dfrac{dy}{dx}=\dfrac{\mid x\mid }{x}\mid \mid x\mid-1\mid+\dfrac{\mid \mid x\mid-1\mid}{\mid x\mid -1}\dfrac{\mid x^{2}\mid}{x}$$

$$\bigg(\dfrac{dy}{dx}\bigg)_{x=-2}=-1-2=-3$$

The equation of normal is $$y=\dfrac{x}{3}+\bigg(2+\dfrac{2}{3}\bigg)$$

$$3{y}=x+8$$
Question 3
1 / -0

If the tangent to the curve $$y=x\log { x } $$ at $$\left( c,f\left( x \right) \right) $$ is parallel to the line-segment joining $$A\left(1,0\right)$$ and $$B\left(e,e\right)$$, then c=...... .

A
$$\dfrac {e-1}{e}$$

B
$$\log { \dfrac { e-1 }{ e } } $$

C
$${ e }^{ \dfrac { 1 }{ 1-e } }$$

D
$${ e }^{ \dfrac { 1 }{ e-1 } }$$

Solution

slope of AB $$\displaystyle =\frac{e}{e-1}$$
$$\displaystyle \frac{dy}{dx}=1+\log x$$
slope at $$(c,f\left(x\right ))$$
$$\displaystyle 1+\log c=\frac{e}{e-1}\implies=e^{\frac{1}{e-1}}$$
Question 4
1 / -0

The intercept on x-axis made by tangent to the curve, $$\displaystyle y=\int _{ 0 }^{ x }{ \left| t \right| } dt,x\in R$$, which are parallel to the line $$y=2x$$, are equal to

A
$$\pm 1$$

B
$$\pm 2$$

C
$$\pm 3$$

D
$$\pm 4$$

Solution

$$y = \int_{0}^{x}{\left| t \right| dt}$$
$$\Rightarrow \cfrac{dy}{dx} = \left| x \right|$$
Since tangent to the curve is parallel to the line $$y = 2x$$.
$$\therefore \cfrac{dy}{dx} = 2$$
$$\Rightarrow x = \pm 2$$
Therefore,
$$y = \int_{0}^{\pm 2}{\left| t \right| dt}$$
$$\Rightarrow y = \pm 2$$
Therefore,
Equation of tangents are-
$$y - 2 = 2 \left( x - 2 \right)$$
$$y + 2 = 2 \left( x + 2 \right)$$
For $$x$$-intercept, substitute $$y = 0$$, we have
$$0 - 2 = 2 \left( x - 2 \right) \Rightarrow x = +1$$
$$0 + 2 = 2 \left( x + 2 \right) \Rightarrow x = -1$$
Hence the intercept on $$x$$-axis are $$\pm 1$$.
Question 5
1 / -0

The slope of the tangent to the curve $$y=sinx$$ where it crosses the $$x-axis$$ is

A
$$1$$

B
$$-1$$

C
$$ \pm 1$$

D
$$ \pm 2$$

Solution

Consider the given equation of the curve.
$$y=\sin x$$ ……(1)

On differentiating equation (1) with respect to $$x$$, we get
$$\dfrac{dy}{dx}=\cos x$$

When given curve crosses the x-axis then.
$$ y=0 $$
$$\sin x=0$$
$$x=0$$

Therefore,
$$ {{\left( \dfrac{dy}{dx} \right)}_{y=0}}=\cos 0 $$
$$ {{\left( \dfrac{dy}{dx} \right)}_{y=0}}=1 $$

Hence, this is the answer.
Question 6
1 / -0

The Point (s) on the cure $${ y }^{ 3 }+{ 3x }^{ 2 }=12y$$ where the tangent is vertical (parallel to y-axis), is/are.

A
$$\left[ \pm \dfrac { 4 }{ \sqrt { 3 } } ,-2 \right] $$

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

C
$$(0,0)$$

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

Solution

curve given $$y^3+3x^2=12y$$

Thus differentiating w.r.t $$x$$

$$3y^2\dfrac {dy}{dx}+6x=12\dfrac {dy}{dx}$$
$$\dfrac {dy}{dx}(y^2-4)+2x=0$$
$$\dfrac {dy}{dx} =\dfrac {-2x}{y^2-4}$$

For tangent to be $$\parallel$$ to $$y-$$axis $$\dfrac {dy}{dx}\rightarrow \infty$$

Thus $$y^2-4=0$$

$$y=\pm 2$$
Putting $$y=+2$$ in curve we get $$x=\pm \dfrac {4}{\sqrt 3}$$
Putting $$y=-2$$ in curve $$x$$ does not exit

Thus $$y=-2$$ not in range as $$x^2 < 0$$

Thus points where tangent is $$\parallel$$ to $$y-$$axis is $$\left [\pm \dfrac {4}{\sqrt {3}},\ 2 \right]$$ Answer.
Question 7
1 / -0

The angle made by the tangent line at (1, 3) on the curve $$y=4x-{ x }^{ 2 }$$ with $$\overset { - }{ OX } $$ is

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

B
$${ tan }^{ -1 }(1/3)$$

C
$${ tan }^{ -1 }(3)$$

D
$$\pi /4$$

Solution

Given,

$$y=4x-x^2$$

$$\dfrac{dy}{dx}=4-2x$$

$$\dfrac{dy}{dx}_{(1,3)}=4-2(1)=2$$

Therefore, angle made by tangent,

$$\tan \theta =2$$

$$\therefore \theta =\tan ^{-1}2$$
Question 8
1 / -0

equation of tangent at $$(0,0)$$ for the equation $$y^2=16x$$

A
$$y=0 $$

B
$$x=0$$

C
$$x+y=0$$

D
$$x-y=0$$

Solution

The equation is $$y^2=16x$$
The point is $$(0,0)$$
The slope of tangent $$2y \dfrac{dy}{dx}=16\\\dfrac{dy}{dx}=\dfrac 8y\\\left.\dfrac{dy}{dx}\right|_{(0,0)}=\infty$$
Equation of tangent is $$y-0=\dfrac{8}{0}(x-0)\\x=0$$
Question 9
1 / -0

The area of triangle formed by tangent and normal at point $$(\sqrt{3}, 1)$$ of the curve $$x^2+y^2=4$$ and x-axis is?

A
$$\dfrac{4}{\sqrt{3}}$$

B
$$\dfrac{2}{\sqrt{3}}$$

C
$$\dfrac{8}{\sqrt{3}}$$

D
$$\dfrac{5}{\sqrt{3}}$$

Solution

Slope of OP$$=\dfrac{1}{\sqrt{3}}$$, slope of PQ$$=-\sqrt{3}$$
$$y-1=-\sqrt{3}(x-\sqrt{3})=-\sqrt{3}x+3$$
$$\Rightarrow \sqrt{3}x+y=4$$ and $$Q\left(\dfrac{4}{\sqrt{3}}, 0\right)$$
$$\Delta OPQ=\dfrac{2}{\sqrt{3}}$$.
Question 10
1 / -0

The function $$f(x)=x^3-6x^2+9x+3$$ is decreasing for

A
$$1 < x < 3$$

B
$$x > 1$$

C
$$x < 1$$

D
$$x < 1$$ or $$x > 3$$

Solution

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 17

Result

Application of Derivatives Test - 17

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SHARING IS CARING

Question 1

$$6$$

$$\cfrac{7}{2}$$

$$4$$

$$\cfrac{9}{2}$$

Solution

Question 2

$$3y=2x+10$$

$$3y=x+8$$

$$2y=x+6$$

$$2y=3x+10$$

Solution

Question 3

$$\dfrac {e-1}{e}$$

$$\log { \dfrac { e-1 }{ e } } $$

$${ e }^{ \dfrac { 1 }{ 1-e } }$$

$${ e }^{ \dfrac { 1 }{ e-1 } }$$

Solution

Question 4

$$\pm 1$$

$$\pm 2$$

$$\pm 3$$

$$\pm 4$$

Solution

Question 5

$$1$$

$$-1$$

$$ \pm 1$$

$$ \pm 2$$

Solution

Question 6

$$\left[ \pm \dfrac { 4 }{ \sqrt { 3 } } ,-2 \right] $$

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

$$(0,0)$$

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

Solution

Question 7

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

$${ tan }^{ -1 }(1/3)$$

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

$$\pi /4$$

Solution

Question 8

$$y=0 $$

$$x=0$$

$$x+y=0$$

$$x-y=0$$

Solution

Question 9

$$\dfrac{4}{\sqrt{3}}$$

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

$$\dfrac{8}{\sqrt{3}}$$

$$\dfrac{5}{\sqrt{3}}$$

Solution

Question 10

$$1 < x < 3$$

$$x > 1$$

$$x < 1$$

$$x < 1$$ or $$x > 3$$

Solution

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