Application of Derivatives Test

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Application of Derivatives Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

The angle between the curves $$x^{2} + y^{2} = 25$$ and $$x^{2} + y^{2} - 2x + 3y - 43 = 0$$ at $$(-3, 4)$$ is

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

B
$$\tan^{-1}\left (\dfrac {1}{68}\right )$$

C
$$\dfrac {\pi}{2}$$

D
$$\tan^{-1}\left (\dfrac {3}{4}\right )$$

Solution

slope of first curve at $$(-3,4)$$
$$m_1 = dy/dx = \dfrac{-x}{y} = \dfrac{3}{4}$$

slope of second curve at $$(-3,4)$$
$$m_2 = \dfrac{-2x+2}{2y+3} = \dfrac{8}{11}$$

angle between them $$= tan^{-1} \dfrac{m_1 - m_2}{1+m_1 +m_2}$$
$$=tan^{-1} \dfrac{1/44}{17/11} =tan^{-1} \dfrac{1}{68}$$
Question 2
1 / -0

If the tangent at each point of the curve $$y=\cfrac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$$ makes an acute angle with positive direction of X-axis then

A
$$a\ge 1$$

B
$$-1\le a\le 1$$

C
$$a\le -1$$

D
None of these

Solution

On differentiating, We get
$$\dfrac{dy}{dx}=2x^2-4ax+2$$
If Tangent makes acute angle with +ve x axis, then
$$\dfrac{dy}{dx}\ge0$$ For all x
$$x^2-2ax+1\ge0\\ \Rightarrow 4a^2-4\le0\Rightarrow a^2\le1\\ -1\le a\le1$$
Question 3
1 / -0

The slope of the tangent at each point of the curve is equal to the sum of the coordinate of the point. Then, the curve that passes through the origin is

A
$$x + y = e^{x} - 1$$

B
$$e^{x} = x + y$$

C
$$y = e^{x}$$

D
$$y = e^{x} + 1$$

Solution

Given, $$\dfrac {dy}{dx}=x+y\Rightarrow dy=xdx+ydx\Rightarrow dy+dx=xdx+ydx+dx$$

So, $$d(x+y)=(x+y+1)dx\Rightarrow dx=\dfrac {d(x+y+1)}{x+y+1}$$
Integrating on both sides,
$$x=\ln(x+y+1)$$
On taking antilog,
$$e^x=x+y+1$$
Or. $$e^x-1=x+y$$
So, Option A is correct answer.
Question 4
1 / -0

If $$g\left( x \right) =2f\left( 2{ x }^{ 3 }-3{ x }^{ 2 } \right) +f\left( 6{ x }^{ 2 }-4{ x }^{ 3 }-3 \right)$$ $$\forall \ x\ \in \ R$$ and $$f^{''}\left( x \right) > 0$$, $$\forall \ x \in \ R$$, then $$g\left ( x \right)$$ is increasing in the interval

A
$$\left( \infty ,-\dfrac { 1 }{ 2 } \right)$$

B
$$\left( -\dfrac { 1 }{ 2 } ,\ 0 \right) \bigcup \left( 1,\infty \right)$$

C
$$\left( 0,\infty \right)$$

D
None of these

Solution
Question 5
1 / -0

The slope of the tangent at the point $$(h, h)$$ of the circle $$x^{2} + y^{2} = a^{2}$$ is :

A
$$0$$

B
$$1$$

C
$$-1$$

D
Depends on $$h$$

Solution

Slope of the tangent for the cicle $$x^{2}+y^{2}=a^{2}$$ can be calculated by differentiating the circle's equation.
On differentiating w.r.t x, $$2x + 2y\dfrac{dy}{dx}=0$$
Slope $$m$$$$=\space \dfrac{dy}{dx}=-\dfrac{x}{y}$$
Point given is $$(h,h)$$
Putting point in the slope equation $$m=\dfrac{dy}{dx}=-\dfrac{h}{h}$$
Slope $$m=-1$$
Hence, $$(C)$$
Question 6
1 / -0

The angle at which the curve $$y={ x }^{ 2 }$$ and the curve $$x=\cfrac { 5 }{ 3 } \cos { t } ,y=\cfrac { 5 }{ 4 } \sin { t } $$ intersect is

A
$$\tan ^{ -1 }{ \cfrac { 2 }{ 41 } } $$

B
$$\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } $$

C
$$-\tan ^{ -1 }{ \cfrac { 2 }{ 41 } } $$

D
$$2\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } $$

Solution

Given
$$y={ x }^{ 2 }...(i)$$
$$\quad x=\cfrac { 5 }{ 3 } \cos { t } ;y=\cfrac { 5 }{ 4 } \sin { t } ...(ii)$$
Which is parametric equation, we change this equation is caresian equation as follows
$$\cos { t } =\cfrac { 3 }{ 5 } x;\sin { t } =\cfrac { 4 }{ 5 } y$$
On Squaring and adding both i.e., $$\cos{t}$$ and $$\sin {t}$$ we get
$$\cfrac { 9 }{ 25 } { x }^{ 2 }+\cfrac { 16 }{ 25 } { y }^{ 2 }=\cos ^{ 2 }{ t } +\sin ^{ 2 }{ t } $$
$$\Rightarrow 9{ x }^{ 2 }+16{ y }^{ 2 }=25....(iii)\quad \quad \left[ \because \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1 \right] $$
$$\therefore$$ The intersection points at Eq.(i) and (iii) are $$(1,1)$$ and $$(-1,1)$$
Now, slope of tangent of Eq.(i) at point $$(1,1)$$ is
$${ m }_{ 1 }=\cfrac { dy }{ dx } =2x\quad \therefore { m }_{ 1 }={ \left| \cfrac { dy }{ dx } \right| }_{ (1,1) }=2$$
And slope of tangent of Eq. (iii) at point $$(1,1)$$ is
$${ m }_{ 2 }=\cfrac { dy }{ dx } =-\cfrac { 9 }{ 16 } $$
$$\therefore$$ Angle at point of intersection of Eqs. (i) and (iii) we get
$${ \theta }_{ 1 }=\tan ^{ -1 }{ \left| \cfrac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| } =\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } $$
similarly, slope of tangent of Eq. (i) at point $$(-1,1)$$
$${ m }_{ 1 }={ \left| \cfrac { dy }{ dx } \right| }_{ (-1,1) }=-2$$
And slope of tangent of Eq. (iii) at point $$(-1,1)$$
$${ m }_{ 2 }=\cfrac { dy }{ dx } =\cfrac { 9 }{ 16 } $$
$$\therefore$$ Angle at point of intersection of Eqs. (i) and (iii) we get
$${ \theta }_{ 2 }=\tan ^{ -1 }{ \left| \cfrac {-2- \cfrac { 9 }{ 16 } }{ 1-\cfrac { 18 }{ 16 } } \right| } =\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } \quad \quad $$
Question 7
1 / -0

The slope of the tangent to the curve $$y=\int _{ 0 }^{ x }{ \frac { dt }{ 1+{ t }^{ 3 } } } $$ at the point where $$x=1$$ is

A
$$\frac { 1 }{ 4 } $$

B
$$\frac { 1 }{ 3 } $$

C
$$\frac { 1 }{ 2 } $$

D
$$1$$

Solution

$$\frac { dy }{ dx } =\frac { 1 }{ 1+{ x }^{ 3 } } $$
$$m=\left( \frac { dy }{ dx } \right) x=1=\frac { 1 }{ 1+1 } =\frac { 1 }{ 2 } $$
Question 8
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The equation of the curve satisfying the differential equation $$y_{2}(x^{2} + 1) = 2xy_{1}$$ passing through the point $$(0, 1)$$ and having slope of tangent at $$x = 0$$ as $$3$$ is

A
$$y = x^{2} + 3x + 2$$

B
$$y = x^{2} + 3x + 1$$

C
$$y = x^{3} + 3x + 1$$

D
None of these

Solution

Given $$\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \left( { x }^{ 2 }+1 \right) =2x\cfrac { dy }{ dx } $$
$$y={ x }^{ 2 }+3x+1$$
Slope, $${ \left( \cfrac { dy }{ dx } \right) }_{ x=0 }={ (2x+3) }_{ x=0 }\Rightarrow 2\times 0+3\Rightarrow 3$$
At point $$x=0$$
$$y={ (0) }^{ 2 }+3\times 0+1$$
Thus this curve passes through point $$(0,1)$$ and have slope of tangent at $$x=0$$ as $$3$$
$$y={ x }^{ 2 }+3x+1$$ is the answer
Question 9
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The area of the triangle formed by the positive x-axis, the tangent and normal to the curve $${ x }^{ 2 }+{ y }^{ 2 }=16{ a }^{ 2 }$$ at the point $$\left( 2\sqrt { 2 } a,2\sqrt { 2 } a \right) $$ is

A
$${a}^{2}$$

B
$$16{a}^{2}$$

C
$$4{a}^{2}$$

D
$$8{a}^{2}$$

Solution

Point on the curve is $$(2\sqrt{2}a,2\sqrt{2}a)$$

Equation of tangent at this point is $$x_1x+y_1y=16a^2$$

here $$x_1 = 2\sqrt{2}a$$ and $$y_1=2\sqrt{2}a$$

$$Area = \dfrac{1}{2} \times{base}\times{height}$$

Point of intersection of tangent to the x axis is $$\dfrac{16a^2}{2\sqrt{2}a}$$
So,
Base = $$4\sqrt{2}a$$

Height =$$2\sqrt{2}a $$

$$Area = \dfrac{1}{2}\times{4\sqrt{2}a}\times{2\sqrt{2}a}$$

$$Area = 8a^2$$
Question 10
1 / -0

A tangent PT is drawn to the circle $$x^2+y^2=4$$ at the point $$P(\sqrt{3}, 1)$$. A straight line L, perpendicular to PT is a tangent to the circle $$(x-3)^2+y^2=1$$. $$(1)$$ A possible equation of L is?

A
$$x-\sqrt{3}y=1$$

B
$$x+\sqrt{3}y=1$$

C
$$x-\sqrt{3}y=-1$$

D
$$x+\sqrt{3}y=5$$

Solution

Slope of tangent to $$x^{2}+y^{2}=4$$ at $$P(\sqrt{3},1)$$ is given by differentiating given equation w.r.t. x
$$\dfrac{dy}{dx}=\dfrac{-x}{y}=-\sqrt{3}$$
So slope of line L $$=\dfrac{-1}{-\sqrt{3}}=\dfrac{1}{\sqrt{3}}$$
Let equation of L be $$x-\sqrt3y+c=0$$
As L is tangent to $$(x-3)^{2}+y^{2}=1$$
So perpendicular distance of L from its center should be equal to its radius i.e. $$1$$.
Centre $$(3,0)$$
$$\dfrac{|3+c|}{\sqrt4}=1$$
$$\Rightarrow c=-1 or -5$$
So equation of L is $$x-\sqrt3y=1$$
or
$$x-\sqrt3y=5$$
Hence option A is correct.

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 39

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Application of Derivatives Test - 39

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SHARING IS CARING

Question 1

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

$$\tan^{-1}\left (\dfrac {1}{68}\right )$$

$$\dfrac {\pi}{2}$$

$$\tan^{-1}\left (\dfrac {3}{4}\right )$$

Solution

Question 2

$$a\ge 1$$

$$-1\le a\le 1$$

$$a\le -1$$

None of these

Solution

Question 3

$$x + y = e^{x} - 1$$

$$e^{x} = x + y$$

$$y = e^{x}$$

$$y = e^{x} + 1$$

Solution

Question 4

$$\left( \infty ,-\dfrac { 1 }{ 2 } \right)$$

$$\left( -\dfrac { 1 }{ 2 } ,\ 0 \right) \bigcup \left( 1,\infty \right)$$

$$\left( 0,\infty \right)$$

None of these

Solution

Question 5

$$0$$

$$1$$

$$-1$$

Depends on $$h$$

Solution

Question 6

$$\tan ^{ -1 }{ \cfrac { 2 }{ 41 } } $$

$$\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } $$

$$-\tan ^{ -1 }{ \cfrac { 2 }{ 41 } } $$

$$2\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } $$

Solution

Question 7

$$\frac { 1 }{ 4 } $$

$$\frac { 1 }{ 3 } $$

$$\frac { 1 }{ 2 } $$

$$1$$

Solution

Question 8

$$y = x^{2} + 3x + 2$$

$$y = x^{2} + 3x + 1$$

$$y = x^{3} + 3x + 1$$

None of these

Solution

Question 9

$${a}^{2}$$

$$16{a}^{2}$$

$$4{a}^{2}$$

$$8{a}^{2}$$

Solution

Question 10

$$x-\sqrt{3}y=1$$

$$x+\sqrt{3}y=1$$

$$x-\sqrt{3}y=-1$$

$$x+\sqrt{3}y=5$$

Solution

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