Application of Derivatives Test

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Application of Derivatives Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

The point on the curve $$y = \sqrt {x - 1}$$ where the tangent is perpendicular to the line $$2x + y - 5 = 0$$ is

A
$$(2, -1)$$

B
$$(10, 3)$$

C
$$(2, 1)$$

D
$$(5, -2)$$

Solution

$$\dfrac {dy}{dx} = \dfrac {1}{2\sqrt {x - 1}} = m_{1}$$ is the slope of tangent to $$y=\sqrt{x-1}$$
Slope of the line $$2x + y - 5 = 0$$ is $$m_{2} = -2$$
For lines are perpendicular
$$m_{1} m_{2} = -1$$
$$\implies \left (\dfrac {1}{2\sqrt {x - 1}}\right )(-2) = -1$$
$$\implies \dfrac {2}{2\sqrt {x - 1}} = 1$$
$$\implies \sqrt {x - 1} = 1$$
Squaring both sides,
$$\implies x - 1 = 1$$
$$\implies x = 2$$
$$\therefore y = \sqrt {x - 1}$$
$$= \sqrt {2 - 1}$$
$$= \sqrt {1}$$
$$\therefore y = 1$$
$$\therefore (2, 1)$$ is the point on the curve $$y=\sqrt{x-1}$$
Question 2
1 / -0

The slope of the normal to the curve $$x=1-a\sin { \theta } $$, $$y=b\cos ^{ 2 }{ \theta }$$ at $$ \theta =\dfrac { \pi }{ 2 } $$ is

A
$$\dfrac { a }{ 2b } $$

B
$$\dfrac { 2a }{ b } $$

C
$$\dfrac { a }{ b } $$

D
$$\dfrac { -a }{ 2b } $$

Solution

Given, $$x=1-a\sin { \theta } $$ and $$y=b\cos ^{ 2 }{ \theta }$$
On differentiating with respect to $$\theta $$, we get
$$\dfrac { dx }{ d\theta } =-a\cos { \theta }$$
and $$ \dfrac { dy }{ d\theta } =2b\cos { \theta } \left( -\sin { \theta } \right)$$
Then, $$ \dfrac { dy }{ dx } =\dfrac { { dy }/{ d\theta } }{ { dx }/{ d\theta } } =\dfrac { 2b }{ a } \sin { \theta }$$
$$ \therefore$$ Slope of normal at the point $$ \theta =\dfrac { \pi }{ 2 }$$ is
$$ -\dfrac { dx }{ dy } =-\dfrac { 1 }{ { dy }/{ dx } } $$
$$=-\dfrac { 1 }{ \dfrac { 2b }{ a } \sin { \left( \dfrac { \pi }{ 2 } \right) } } =-\dfrac { a }{ 2b } $$
Question 3
1 / -0

If an edge of a cube measure $$2$$ m with a possible error of $$0.5$$ cm. Find the corresponding error in the calculated volume of the cube.

A
$$0.6\ m^{3}$$

B
$$0.06\ m^{3}$$

C
$$0.006\ m^{3}$$

D
$$0.0006\ m^{3}$$

Solution

Let $$x$$ be the length of an edge of a cube and $$V$$ be the volume of that cube.
Thus $$ V = x^{3}$$
On differentiating w.r.t. $$x$$, we get
$$\dfrac {dV}{dx} = 3x^{2}$$
Let $$\delta V$$ be error in $$V$$ and corresponding error $$\delta x$$ in $$x$$.
$$\therefore \delta V = \dfrac {dV}{dx} \delta x = 3x^{2} \delta x$$
Given that, $$x = 2$$ m and $$\delta x = 0.5$$ cm $$= \dfrac {0.5}{100} $$ m
$$\therefore \delta V = 3(2)^{2} \left (\dfrac {0.5}{100}\right )$$
$$= \dfrac {12\times 0.5}{100} = \dfrac {6}{100} = 0.06\ cm^{3}$$
Question 4
1 / -0

The tangents to curve $$y={ x }^{ 3 }-2{ x }^{ 2 }+x-2$$ which are parallel to straight line $$y=x$$, are

A
$$x+y=2$$ and $$x-y=\dfrac { 86 }{ 27 } $$

B
$$x-y=2$$ and $$x-y=\dfrac { 86 }{ 27 } $$

C
$$x-y=2$$ and $$x+y=\dfrac { 86 }{ 27 } $$

D
$$x+y=2$$ and $$x+y=\dfrac { 86 }{ 27 } $$

Solution

Given,
$$y={ x }^{ 3 }-2{ x }^{ 2 }+x-2$$

On differentiating both sides with respect to $$x$$, we get

$$\dfrac { dy }{ dx } =3{ x }^{ 2 }-4x+1$$

and $$y=x$$

$$\Rightarrow \dfrac { dy }{ dx } =1$$

$$\therefore$$ Slope of tangent will be $$ 3{ x }^{ 2 }-4x+1$$.

Since, the tangent is parallel to line $$y=x$$.

$$\therefore 3{ x }^{ 2 }-4x+1=1$$

$$\Rightarrow 3{ x }^{ 2 }-4x=0$$

$$\Rightarrow x\left( 3x-4 \right) =0$$

$$\Rightarrow x=0,\dfrac { 4 }{ 3 }$$

When $$x=0$$, then $$y=-2$$

When $$x=\dfrac { 4 }{ 3 } $$, then $$y=\dfrac { -50 }{ 27 }$$

Now, equation of tangents at point $$ \left( 0,-2 \right)$$ is

$$ y-{ y }_{ 1 }=\dfrac { dy }{ dx } \left( x-{ x }_{ 1 } \right)$$

$$ \Rightarrow y+2=1\left( x-0 \right)$$

$$ \Rightarrow y+2=x$$

$$\Rightarrow x-y=2$$ ...(i)

and equation of tangents at point $$\left( \dfrac { 4 }{ 3 } ,-\dfrac { 50 }{ 27 } \right) $$ is

$$y-{ y }_{ 1 }=\dfrac { dy }{ dx } \left( x-{ x }_{ 1 } \right) $$

$$y+\dfrac { 50 }{ 27 } =x-\dfrac { 4 }{ 3 }$$

$$ \Rightarrow x-y=\dfrac { 50 }{ 27 } +\dfrac { 4 }{ 3 }$$

$$ \Rightarrow x-y=\dfrac { 50+36 }{ 27 } $$

$$\Rightarrow x-y=\dfrac { 86 }{ 27 } $$ ....(ii)

Hence, equations (i) and (ii) are required equations of the tangents.
Question 5
1 / -0

The slope of tangent to the curve $$ x=t^2 + 3t - 8, y = 2t^2 - 2t - 5 $$ at the point $$(2, -1)$$ is :

A
$$ \dfrac {22}{7} $$

B
$$ \dfrac {6}{7} $$

C
$$-6$$

D
None of these

Solution

Given curves are $$ x=t^{2} + 3t - 8$$ ... $$(i)$$ and $$y = 2t^{2} - 2t - 5 $$ ... $$(ii)$$
At $$(2,-1),$$ From $$(i)$$
$$t^{2}+3t-10=0\implies t=2$$ or $$t=-5$$
From $$(ii)$$
$$2t^{2}-2t-4=0\implies t^{2}-t-2=0\implies t=2$$ or $$t=-1$$
From both the solutions, we get $$t=2$$

Differentiating both the equations w.r.t. $$t$$, we get
$$\dfrac{dx}{dt}=2t+3$$ ........ $$(iii)$$
$$\dfrac{dy}{dt}=4t-2$$ ....... $$(iv)$$

Now, $$ \dfrac{dy}{dx} = \dfrac {\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$$

$$ = \dfrac {4t-2}{2t+3}$$ .... From $$(iii)$$ and $$(iv)$$
$$\therefore \dfrac{dy}{dx} = \dfrac {4t-2}{2t+3}$$ is the slope of tangent to the given curve
$$\therefore \left|\dfrac{dy}{dx}\right|_{(2,-1)} = \left|\dfrac {4t-2}{2t+3}\right|_{t=2}=\dfrac{8-2}{4+3}=\dfrac{6}{7}$$ is the slope of tangent to the given curve at $$(2,-1)$$
Question 6
1 / -0

The points at which the tangent to the curve $$y = x^3 - 3x^2 - 9x + 7$$ is parallel to the x-axis are

A
$$(3, - 20)$$ and $$(- 1, 12)$$

B
$$(3, 20)$$ and $$(1, 12)$$

C
$$(1, -10)$$ and $$(2, 6)$$

D
None of these

Solution

Tangent to the curve is parallel to the axis is when slope of the tangent is 0.
$$\therefore$$ Equation of the curve is
$$y=x^3-3x^2-9x+7=0$$ ...... $$(i)$$
$$\therefore \dfrac{dy}{dx}=3x^2-6x-9$$

Now, the tangent is parallel to x-axis, then slope of the tangent is zero or we can say that $$\dfrac{dy}{dx}=0$$.
$$\Rightarrow 3x^2-6x-9=0$$
$$\Rightarrow 3(x^2-2x-3)=0$$
$$\Rightarrow (x-3)(x+1)=0$$
$$\Rightarrow x=3, -1$$

When $$x=3$$, then from Eq. $$(i),$$ we get
$$y=(3)^3-(3)\cdot (3)^2-9\cdot 3+7$$
$$=27-27-27+7=-20$$

When $$x =-1,$$ then from Eq. $$(i),$$ we get
$$y=(-1)^3-3(-1)^2-9(-1)+7$$
$$=-1-3+9+7=12$$
Hence, the points at which the tangent is parallel to x-axis are $$(3, -20)$$ and $$(-1, 12)$$.
Question 7
1 / -0

If $$y=8{ x }^{ 3 }-60{ x }^{ 2 }+144x+27$$ is a strictly decreasing function in the interval

A
$$(-5,6)$$

B
$$\left( -\infty ,2 \right) $$

C
$$(5,6)$$

D
$$\left( 3,\infty \right) $$

E
$$(2,3)$$

Solution

Given, $$y=8{ x }^{ 3 }-60{ x }^{ 2 }+144x+27$$

$$\cfrac { dy }{ dx } =24{ x }^{ 2 }-120x+144\quad $$

For function to be strictly decreasing

$$\cfrac { dy }{ dx } <0$$

$$\Rightarrow 24{ x }^{ 2 }-120x+144<0$$

$$\Rightarrow \left( { x }^{ 2 }-5x+6 \right) <0$$

$$\Rightarrow \left( { x }^{ 2 }-2x-3x+6 \right) <0$$

$$\Rightarrow (x-3)(x-2)<0$$

$$\therefore x\in (2,3)$$
Question 8
1 / -0

If the tangent at $$(1,1)$$ on $${ y }^{ 2 }=x{ (2-x) }^{ 2 }$$ meets the curve again at $$P$$, then $$P$$ is

A
$$(4,4)$$

B
$$(-1,2)$$

C
$$\left( \cfrac { 9 }{ 4 } ,\cfrac { 3 }{ 8 } \right) $$

D
$$(1,2)$$

Solution

We have
$$\Rightarrow { y }^{ 2 }={ x }^{ 3 }-4{ x }^{ 2 }+4x$$ .....(i)

$$\Rightarrow 2y\cfrac { dy }{ dx } =3{ x }^{ 2 }-8x+4$$

$$\Rightarrow \cfrac { dy }{ dx } =\cfrac { 3{ x }^{ 2 }-8x+4 }{ 2y } $$

$$\Rightarrow { \left[ \cfrac { dy }{ dx } \right] }_{ (1,1) }=\cfrac { 3-8+4 }{ 2 } =-\cfrac { 1 }{ 2 } $$

The equation of the tangent at $$(1,1)$$ is

$$-1=-\cfrac { 1 }{ 2 } (x-1)$$

$$\Rightarrow x+2y-3=0$$ .....(ii)

On solving Eqs. (i) and (ii) we get

$$x=9/4$$ and $$y=3/8$$

Hence,the coordinates of $$P$$ are $$\left( \cfrac { 9 }{ 4 } ,\cfrac { 3 }{ 8 } \right) $$.
Question 9
1 / -0

The tangent to the curve $$y=a{ x }^{ 2 }+bx$$ at $$\left( 2,-8 \right) $$ is parallel to $$X$$-axis. Then,

A
$$a=2, b=-2$$

B
$$a=2, b=-4$$

C
$$a=2, b=-8$$

D
$$a=4, b=-4$$

Solution

Given, $$y=a{ x }^{ 2 }+bx$$
On differentiating with respect to $$x$$, we get
$$\dfrac { dy }{ dx } =2ax+b$$
At $$\left( 2,-8 \right) ,$$ $${ \left( \dfrac { dy }{ dx } \right) }_{ \left( 2,-8 \right) }=4a+b$$
Since tangent is parallel to $$X$$-axis.
Thus $$ \dfrac { dy }{ dx } =0\Rightarrow b=-4a$$ ....(i)
Now, point $$\left( 2,-8 \right)$$ is on the curve $$ y=a{ x }^{ 2 }+bx$$
Therefore, $$ -8=4a+2b$$ ....(ii)
On solving equations (i) and (ii), we get
$$a=2,b=-8$$
Question 10
1 / -0

Find the critical points of the function $$f (x)= (x - 2)^{2/3} (2x + 1)$$

A
$$-1$$ and $$2$$

B
$$1$$

C
$$1$$ and $$- 2$$

D
$$1$$ and $$2$$

Solution

We have,$$f(x)=(x-2)^{2/3}(2x+1)$$

$$\Rightarrow f'(x)=\frac {2}{3}(x-2)^{-1/3}(2x+1)+(x-2)^{2/3}\cdot 2$$

$$=\dfrac {2(2x+1)}{3(x-2)^{1/3}}+2 (x-2)^{2/3}$$

$$= \dfrac {2(2x+1)+6(x-2)}{3(x-2)^{1/3}}$$

$$=\dfrac {4x+2+6x-12}{3(x-2)^{1/3}}$$

$$=\dfrac {10(x-1)}{3(x-2)^{1/3}}$$

For critical points,

$$f'(x)=0$$

$$\Rightarrow x=1$$

and $$f'(x)$$ is not defined at x= 2.

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 37

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Application of Derivatives Test - 37

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SHARING IS CARING

Question 1

$$(2, -1)$$

$$(10, 3)$$

$$(2, 1)$$

$$(5, -2)$$

Solution

Question 2

$$\dfrac { a }{ 2b } $$

$$\dfrac { 2a }{ b } $$

$$\dfrac { a }{ b } $$

$$\dfrac { -a }{ 2b } $$

Solution

Question 3

$$0.6\ m^{3}$$

$$0.06\ m^{3}$$

$$0.006\ m^{3}$$

$$0.0006\ m^{3}$$

Solution

Question 4

$$x+y=2$$ and $$x-y=\dfrac { 86 }{ 27 } $$

$$x-y=2$$ and $$x-y=\dfrac { 86 }{ 27 } $$

$$x-y=2$$ and $$x+y=\dfrac { 86 }{ 27 } $$

$$x+y=2$$ and $$x+y=\dfrac { 86 }{ 27 } $$

Solution

Question 5

$$ \dfrac {22}{7} $$

$$ \dfrac {6}{7} $$

$$-6$$

None of these

Solution

Question 6

$$(3, - 20)$$ and $$(- 1, 12)$$

$$(3, 20)$$ and $$(1, 12)$$

$$(1, -10)$$ and $$(2, 6)$$

None of these

Solution

Question 7

$$(-5,6)$$

$$\left( -\infty ,2 \right) $$

$$(5,6)$$

$$\left( 3,\infty \right) $$

$$(2,3)$$

Solution

Question 8

$$(4,4)$$

$$(-1,2)$$

$$\left( \cfrac { 9 }{ 4 } ,\cfrac { 3 }{ 8 } \right) $$

$$(1,2)$$

Solution

Question 9

$$a=2, b=-2$$

$$a=2, b=-4$$

$$a=2, b=-8$$

$$a=4, b=-4$$

Solution

Question 10

$$-1$$ and $$2$$

$$1$$

$$1$$ and $$- 2$$

$$1$$ and $$2$$

Solution

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