Application of Derivatives Test

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Application of Derivatives Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

Two lines drawn through the point $$A ( 4,0 )$$ divide the area bounded by the curve $$y = \sqrt { 2 } \sin ( \pi x / 4 )$$ and $$x$$ - axis between the lines $$x = 2$$ and $$x = 4$$ into three equal parts. Sum of the slopes of the drawn lines is:

A
$$- 4 \sqrt { 2 } / \pi$$

B
$$- \sqrt { 2 } / \pi$$

C
$$- 2 \sqrt { 2 } / \pi$$

D
None

Solution

Area bounded by $$y=\sqrt{2}\sin\left(\dfrac{\pi x}{4}\right)$$ and $$x-axis$$ between the lines $$x=2$$ and $$x=4,$$
$$\Delta=\sqrt{2}\displaystyle\int_2^4\sin\dfrac{\pi x}{4}dx$$

$$=\left[-\dfrac{4\sqrt{2}}{\pi}.\cos\dfrac{\pi x}{4}\right]^4_2$$

$$=\dfrac{4\sqrt{2}}{\pi}\,unit^2$$

Let the drawn lines are $$L_1:y-m_1(x-4)=0$$ and $$L_2:y-m_2(x-4)=0,$$ meeting the line $$x=2$$ at the points $$A$$ and $$B,$$ respectively.
Clearly, $$A=(2,-2m_1);\,B=(2,-2m_2)$$ [ Fig ]
Now,
$$\Delta_{ACD}=\dfrac{\Delta}{3}\Rightarrow \dfrac{4\sqrt{2}}{3\pi}=\dfrac{1}{2}.2.(-2m_1)$$
$$\Rightarrow$$ $$m_1=-\dfrac{2\sqrt{2}}{3\pi}$$
Also, $$\Delta_{BCD}=\dfrac{2\Delta}{3}$$
$$\Rightarrow$$ $$\dfrac{8\sqrt{2}}{3\pi}=\dfrac{1}{2}\times 2-2m_2$$
$$\Rightarrow$$ $$m_2=\dfrac{-4\sqrt{2}}{3\pi}$$
$$\Rightarrow$$ Required sum $$=-\dfrac{2\sqrt{2}}{3\pi}+\left(\dfrac{-4\sqrt{2}}{3\pi}\right)$$

$$=\dfrac{-6\sqrt{2}}{3\pi}$$

$$=\dfrac{-2\sqrt{2}}{\pi}$$
Question 2
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If $$- 4 \leq x \leq 4$$ then critical points of $$f ( x ) = x ^ { 2 } - 6 | x | + 4$$ are

A
$$3 , - 2$$

B
$$6 , - 6$$

C
$$3 , - 3$$

D
$$0,1$$
Question 3
1 / -0

The number of critical points of the function $$f(x)=|x-1||x-2|$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
none of these

Solution
Question 4
1 / -0

$$f(x)=\dfrac{x}{5}+\dfrac{4}{x}(x\neq 0)$$ in increasing in

A
$$(-5,\ 0)$$

B
$$(0,\ 5)$$

C
$$(-\infty ,\ -5)\cup (5,\ \infty )$$

D
$$(-5,\ 5)$$

Solution
Question 5
1 / -0

The tangent to the curve $$2a^2y=x^3-3ax^2$$ is parallel to the x-axis at the points

A
$$(0 , 0) , (2a, - 2a)$$

B
$$(0 , 0) , (-2a, 2a)$$

C
$$(0 , 0) , (-2a, - 2a)$$

D
$$(2 , 2) , (0 , 0)$$

Solution

Let $$(x_1,y_1)$$ represent the required points.

The slope of the $$x-$$axis is $$0$$

Here $$2a^2y=x^3-3ax^2$$, since the points lies on the curve we get

$$2a^2y_1=x_1^3-3ax_1^2 \quad ...(1)$$

Consider $$2a^2y=x^3-3ax^2$$

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

$$2a^2\dfrac{dy}{dx}=3x^2-6ax$$

$$\Rightarrow \dfrac{dy}{dx}=\dfrac{3x^2-6ax}{2a^2}$$

Slope of the tangent at $$(x_1,y_1)=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=\dfrac{3x_1^2-6ax_1}{2a^2}$$

It is given that slope of the tangent at $$(x_1,y_1)=$$ slope of the $$x-$$axis.

$$\Rightarrow \dfrac{3x_1^2-6ax_1}{2a^2}=0$$

$$\Rightarrow 3x_1^2-6ax_1=0$$

$$\Rightarrow x_1(3x_1-6a)=0$$

$$\Rightarrow x_1=0$$ or $$x_1=2a$$

Also, from $$(1)$$

$$2a^2y_1=0$$ or $$2a^2y_1=8a^3-12a^3$$

$$\Rightarrow y_1=0$$ or $$y_1=-2a$$

Thus, the required points are $$(0,0)$$ and $$(2a,-2a)$$
Question 6
1 / -0

The line $$3x-4y=0$$

A
is a tangent to the circle $${x}^{2}+{y}^{2}=25$$

B
is a normal to the circle $${x}^{2}+{y}^{2}=25$$

C
does not meet the circle $${x}^{2}+{y}^{2}=25$$

D
does not pass thro' the origin

Solution

$$3x=4y$$
$$x=\dfrac { 4y }{ 3 } $$
$${ x }^{ 2 }+{ y }^{ 2 }=25$$
$${ x }^{ 2 }-\left( 1+\dfrac { 16 }{ 9 } \right) =25$$
$${ x }^{ 2 }=9$$
$$x=\pm 3$$
$$y=\pm 9/4$$
Now, $${ x }^{ 2 }+{ y }^{ 2 }=25$$
$$2x+2y\dfrac { dy }{ dx } =0$$
$$\dfrac { dy }{ dx } =\dfrac { -x }{ y } =\dfrac { -\left( 3 \right) }{ \left( 9/4 \right) } =-4/3$$
$$\dfrac { -dx }{ dy } =3/4\Rightarrow $$ slope of normal at circle and $$3/4\Rightarrow $$ solpe of $$3x-4y=0$$
So, Line is normal at circle.
Question 7
1 / -0

If $$x-2y+k=0$$ is a common tangent to $$\displaystyle{ y }^{ 2 }=4x\quad \& \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { 3 } } =1\left( a>\sqrt { 3 } \right) $$, then the value of a, k and other common tangent are given by

A
$$a = 2$$

B
$$a = -2$$

C
$$x+2y+4 = 0$$

D
$$k=4$$

Solution

Given:$$x-2y+k=0$$
$$\Rightarrow\,2y=x+k$$
$$\Rightarrow\,y=\dfrac{x}{2}+\dfrac{k}{2}$$
Comparing with $$y=mx+c$$ we get $$m=\dfrac{1}{2}$$ and $$c=\dfrac{k}{2}$$
$$\therefore\,y=mx+c$$ is tangent to $${y}^{2}=4ax$$ where $$c=\dfrac{a}{m}$$
$$\Rightarrow\,{y}^{2}=4x,\,a=1$$
$$\Rightarrow\,c=\dfrac{1}{m}$$
$$\Rightarrow\,\dfrac{k}{2}=\dfrac{1}{\dfrac{1}{2}}$$
$$\Rightarrow\,\dfrac{k}{2}=2$$
$$\therefore\,k=4$$
Substituting $$m=\dfrac{1}{2}$$ and $$c=\dfrac{1}{m}=\dfrac{1}{\dfrac{1}{2}}=2$$ in $$y=mx+c$$ we get
$$\Rightarrow\,y=\dfrac{x}{2}+\dfrac{4}{2}$$
$$\Rightarrow\,y=\dfrac{x}{2}+2$$
Given:$$y=mx+c$$ is tangent to $$\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$$
$$\therefore\,$$ Required condition is $${c}^{2}={a}^{2}{m}^{2}+{b}^{2}$$
$$\Rightarrow\,4={a}^{2}{\left(\dfrac{1}{2}\right)}^{2}+3$$
$$\Rightarrow\,\dfrac{{a}^{2}}{4}=4-3=1$$
$$\Rightarrow\,{a}^{2}=4$$
$$\Rightarrow\,a=2$$ since $$a>0$$
Question 8
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The tangent at any point of the curve $$x={ at }^{ 3 },y={ at }^{ 4 }$$ divides the abscissa of the point of contact in the ratio

A
$$1:4$$

B
$$3:2$$

C
$$1:3$$

D
$$3:1$$

Solution

Given:
$$x = at^2$$
$$y = at^4$$

Then,
$$\Rightarrow t^3 = \dfrac{x}{a}, t^4= \dfrac{y}{a}$$

$$\Rightarrow t = \left(\dfrac{x}{a}\right)^{1/3}\, t = \left(\dfrac{y}{a}\right)^{1/4}$$

$$\Rightarrow \left(\dfrac{x}{a}\right)^{1/3} = \left(\dfrac{y}{a}\right)^{1/4}$$

$$\Rightarrow \left(\dfrac{x}{a}\right)^{\frac{1}{3}\times \frac{4}{12}} = \left(\dfrac{y}{a} \right)^{\frac{1}{4}\times 12}$$

$$\Rightarrow \dfrac{x^4}{a^4} = \dfrac{y^3}{a^2}$$

$$\Rightarrow y^3 = \dfrac{x^4}{a}$$ at $$P(h,k)$$

Also relation $$k^3 = \dfrac{h^4}{a}$$

Now by differentiation of after equation we get

$$3y^2 \dfrac{dy}{dx} = \dfrac{4x^2}{a}$$

$$\dfrac{dy}{dx} = \dfrac{4x^3}{3ay^2}$$

Now $$M_T = \dfrac{dy}{dx} = \dfrac{4x^3}{3ay^2}$$

Now we will find slope of tangent

$$M_T / P(h,x) = \dfrac{4h^3}{3ax^2}$$

equation of tangent $$P(h,k)$$

$$y - k = \dfrac{4h^3}{3ak^2}(x-h)$$

$$\Rightarrow $$ Let $$y = 0$$

Then, $$-k = \dfrac{4h^3}{3ak^2}(x-h)$$

$$\Rightarrow \dfrac{-3ak^3}{4h^3} = x - h$$

Earlier we found that $$k^3 = \dfrac{h^4}{a}$$

Then $$ax^3 = h^4$$

$$\dfrac{-3h^4}{4h^2} = x-h$$

$$\Rightarrow \dfrac{-3h}{4}+h = x$$

$$\Rightarrow x = \dfrac{h}{4}$$

$$\Rightarrow \dfrac{x}{h} = \dfrac{1}{4}$$

Hence option (a) $$1 : 4$$ is the correct choice
Question 9
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If the slope of one of the lines represented $${a^3}{x^2} + 2hxy + {b^3}{y^2} = 0$$ be the square of the other, then $$ab(a+b)$$ is equal to:

A
$$2h$$

B
$$-2h$$

C
$$8h$$

D
$$-8h$$

Solution
Question 10
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Slope of tangent to the circle $$( x - r ) ^ { 2 } + y ^ { 2 } = r ^ { 2 }$$ at the point $$( x , y )$$ lying on the circle is

A
$$\frac { x } { y - r }$$

B
$$\frac { r - x } { y }$$

C
$$\frac { y ^ { 2 } - x ^ { 2 } } { 2 x y }$$

D
$$\frac { y ^ { 2 } + x ^ { 2 } } { 2 x y }$$

Solution

$$(x-r)^2+y^2=r^2$$

$$\dfrac{d}{dx}(x-r)^2+\dfrac{d}{dx}(y^2)=\dfrac{d}{dx}(r^2)$$

$$2(x-r)+2y\dfrac{dy}{dx}=0$$

$$x-r+y\dfrac{d}{dx}=0$$

Therefore the slope of the tangent is,

$$\dfrac{dy}{dx}=\dfrac{r-x}{y}$$