Application of Derivatives Test

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Application of Derivatives Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

$$f(x)=\cfrac{x}{5}+\cfrac{5}{x}$$ is increaing in

A
$$(-5,0)$$

B
$$(0,5)$$

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

D
$$(-5,5)$$

Solution

Given,

$$f(x)=\dfrac{x}{5}+\dfrac{5}{x}$$

$$\Rightarrow x=\dfrac{y}{5}+\dfrac{5}{y}$$

$$x\cdot \:5y=y^2+25$$

$$y^2-x\cdot \:5y+25=0$$

$$y=\dfrac{-\left(-x5\right)\pm \sqrt{\left(-x5\right)^2-4\cdot \:1\cdot \:25}}{2\cdot \:1}$$

$$y=\dfrac{5x \pm 5\sqrt{x^2-4}}{2}$$

$$x\le \:-2\quad \mathrm{or}\quad \:x\ge \:2$$

$$\therefore f\left(x\right)\le \:-2\quad \mathrm{or}\quad \:f\left(x\right)\ge \:2$$

and

$$\mathrm{Increasing}:-\infty \:<x<-5,\:\mathrm{Decreasing}:-5<x<0,\:\mathrm{Decreasing}:0<x<5,\:\mathrm{Increasing}:5<x<\infty \:$$

$$\therefore (-\infty ,-5)\cup (5,\infty)$$
Question 2
1 / -0

The complex number $$\frac{(-\sqrt 3 + 3i)(1-i)}{(3 + \sqrt 3i) (i) (\sqrt 3 + \sqrt 3i)}$$ when represented in the Argand diagram is

A
in the second quadrant

B
in the first quadrant

C
on the y-axis (imaginary axis)

D
on the x-axis (real axis)

Solution
Question 3
1 / -0

If the line $$ax+y=c$$, touches both the curves $${x}^{2}+{y}^{2}=1$$ and $${y}^{2}=4\sqrt{2}x$$, then $$\left| c \right| $$ is equal to:

A
$$1/2$$

B
$$2$$

C
$$\sqrt{2}$$

D
$$\cfrac { 1 }{ \sqrt { 2 } } $$

Solution

Tangent to $${y}^{2}=4\sqrt{2}x$$ is $$y=mx+\cfrac { \sqrt { 2 } }{ m } $$ is also tangent to $${x}^{2}+{y}^{2}=1$$

$$\Rightarrow \left| \cfrac { \sqrt { 2 } /m }{ \sqrt { 1+{ m }^{ 2 } } } \right| =1\Rightarrow m=\pm 1$$

Tangent will be $$y=x+\sqrt{2}$$ or $$y=-x-\sqrt{2}$$

compare with $$y=-ax+c$$

$$\Rightarrow a=\pm 1;c=\pm \sqrt { 2 } $$

$$\therefore |c|=\sqrt 2$$
Question 4
1 / -0

Let $$f"(x) > 0$$ and $$\phi (x) = f(x) + f(2 - x) , x \in (0,2)$$ be a function, then the function $$\phi (x)$$ is

A
increasing in $$(0, 1)$$ and decreasing $$(1, 2)$$

B
decreasing in $$(0,1)$$ and increasing $$(1, 2)$$

C
increasing in $$(0, 2)$$

D
decreasing in $$(0, 2)$$

Solution

$$f"(x) > 0, y = f(x) ; x \in (0, 2)$$
$$\phi (x) = f(x) + f(2 - x)$$
$$\phi ' (x) = f' (x) - f' (2 - x)$$
for $$\phi (x) $$ to be increasing
$$\phi' (x) > 0 \Rightarrow f'(x) > f'(2 - x)$$
$$\Rightarrow x > 2 - x $$ ($$f'(x)$$ is increasing in $$(0, 2)$$)
$$\Rightarrow x > 1$$
$$\Rightarrow x \in (1, 2)$$
For $$\phi (x)$$ to be decreasing
$$\phi ' (x) = 0 \Rightarrow f'(x) < f'(2 - x)$$
$$\therefore x \in (0, 1)$$
Question 5
1 / -0

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

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

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

C
$$-6$$

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

Solution

$$x=t^2+3t-8$$ ........$$(1)$$
$$y=2t^2-2t-5$$ ..........$$(2)$$

Differentiate $$(1)$$, we get
$$\dfrac{dx}{dt}=2t+3$$

Differentiate $$(2)$$, we get
$$\dfrac{dy}{dx}=4t-2$$
$$m=\dfrac{dy}{dx}=\dfrac{4t-2}{2t+3}$$

Given point is $$(2, -1)$$
Put the point in original x and y, we get
$$\Rightarrow x=t^2+3t-8$$
$$2=t^2+3t-8$$
$$t^2+3t-10=0$$
$$(t-2)(t+5)=0$$
$$t=2, t=-5$$

$$\Rightarrow y=2t^2-2t-5$$
$$(-1)=2t^2-2t-5$$
$$2t^2-2t-4=0$$
$$(t+1)(t-2)=0$$
$$t=-1, t=2$$

Since, $$t=2$$ common in both parts, so we take
$$\dfrac{dy}{dx}=\dfrac{4t-2}{2t-3}$$ at $$t=2$$

At $$t=2$$
$$\dfrac{dy}{dx}=\dfrac{4(2)-2}{2(2)-3}=\dfrac{8-2}{4+3}=\dfrac{6}{7}$$

$$m=\dfrac{dy}{dx}=\dfrac{6}{7}$$.
Question 6
1 / -0

At what points the slope of the tangent to the curve $$x^2+y^2-2x-3=0$$ is zero?

A
$$(3, 0), (-1, 0)$$

B
$$(3, 0), (1, 2)$$

C
$$(-1, 0), (1, 2)$$

D
$$(1, 2), (1, -2)$$

Solution

$$x^2+y^2-2x-3=0$$ is zero.

Differentiate w.r.t. x
$$2x+2y\dfrac{dy}{dx}-2=0$$

$$2y\cdot \dfrac{dy}{dx}=2-2x$$

$$\dfrac{dy}{dx}=\dfrac{2(1-x)}{2y}$$

$$\dfrac{dy}{dx}=\dfrac{1-x}{y}$$ ........$$(1)$$

If line is parallel to x-axis
Angle with x-axis $$=\theta =0$$
Slope of x-axis$$=\tan\theta =\tan 0^o=0$$
Slope of tangent $$=$$ Slope of x-axis

$$\dfrac{dy}{dx}=0$$

$$\dfrac{1-x}{y}=0$$

$$x=1$$

Finding y when $$x=1$$
$$x^2+y^2-2x-3=0\\$$
$$(1)^2+y^2-2(1)-3=0\\$$
$$1+y^2-2-3=0\\$$
$$y=\pm 2$$

Hence, the points are $$(1, 2)$$ and $$(1, -2)$$.
Question 7
1 / -0

The point on the curve $$y=12x-x^2$$, where the slope of the tangent is zero will be

A
$$(0, 0)$$

B
$$(2, 16)$$

C
$$(3, 9)$$

D
$$(6, 36)$$

Solution

Let the point be $$P(x, y)$$
$$y=12x-x^2$$
$$\dfrac{dy}{dx}=12-2x$$
$$\left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=12-2x_1$$

since slope of tangent is zero
so $$\left(\dfrac{dy}{dx}\right)_{x_1, y_1}=0$$
$$12-2x_1=0$$
$$2x_1=12$$
$$x_1=6$$

Also curve passing through tangent
$$y_1=12x_1-x^2_1$$
$$y_1=12\times 6-36$$
$$y_1=72-36$$
$$y_1=36$$
The points are $$(6, 36)$$.
Question 8
1 / -0

The slope of the tangent to the curve $$x=3t^2+1, y=t^3-1$$ at $$x=1$$ is

A
$$\dfrac {1}{2}$$

B
$$0$$

C
$$-2$$

D
$$\infty$$

Solution

Given curves are,
$$x=3t^2+1$$ ---- ( 1 )
$$y=t^3-1$$ ---- ( 2 )
Substituting $$x=1$$ in ( 1 ) we get,
$$\Rightarrow$$ $$3t^2+1=1$$
$$\Rightarrow$$ $$3t^2=0$$
$$\Rightarrow$$ $$t=0$$
Differentiate ( 1 ) w.r.t. $$t,$$ we get
$$\Rightarrow$$ $$\dfrac{dx}{dt}=6t$$
Differentiate ( 2 ) w.r.t. $$t,$ we get
$$\Rightarrow$$ $$\dfrac{dy}{dt}=3t^2$$

$$\Rightarrow$$ $$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{3t^2}{6t}$$

$$\therefore$$ $$\dfrac{dy}{dx}=\dfrac{t}{2}$$

Slope of the tangent $$=\left(\dfrac{dy}{dx}\right)_{t=0}=\dfrac{0}{2}=0$$
Question 9
1 / -0

Mark the correct alternative of the following.
The point on the curve $$9y^2=x^3$$, where the normal to the curve makes equal intercepts with the axes is?

A
$$(4, \pm 8/3)$$

B
$$(-4, 8/3)$$

C
$$(-4, -8/3)$$

D
$$(8/3, 4)$$

Solution

Let the required point be $$(x_1, y_1)$$

equation of the curve is $$9y^2=x^2$$

since $$(x_1, y_1)$$ lies on the curve, therefore
$$9y^2_1=x^3_1$$ .......$$(1)$$

Now $$9y^2=x^3$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{x^2}{6y}$$

$$\Rightarrow \left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=\dfrac{x^2}{6y^1}$$

since normal to the curve at $$(x_1, y_1)$$ makes equal intersepts with the coordinate axis, therefore slope of the normal $$=\pm 1$$

$$\Rightarrow \dfrac{1}{-(dy/dx)_{(x_1, y_1)}}=\pm 1$$

$$\Rightarrow (dy/dx)_{(x_1, y_1)}=\pm 1$$

$$\rightarrow \dfrac{x^2_1}{6y^1}=\pm 1$$

$$\Rightarrow x^4_1=36y^2_1=36\left(\dfrac{x^3_1}{9}\right)$$ (using $$1$$)

$$\Rightarrow x^4_1=4x^3_1\Rightarrow x^3_1(x_1-4)=0$$

$$\Rightarrow x_1=0, 4$$

Putting $$x_1=0$$ in $$(1)$$ we get
$$9y^2_1=0\Rightarrow y_1=0$$

Putting $$x_1=4$$ in $$(1)$$ we get
$$9y^2_1=(4)^3\Rightarrow y_1=\pm \dfrac{8}{3}$$

But the line making equal intersepts with the coordinate axes cannot pass through origin.

Hence, the required points are $$\left(4, \dfrac{8}{3}\right)$$ and $$\left(4, \dfrac{-8}{3}\right)$$.
Question 10
1 / -0

Mark the correct alternative of the following.
The line $$y=mx+1$$ is a tangent to the curve $$y^2=4x$$, if the value of m is?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$1/2$$

Solution

Given equation of the tangent to the given curve
$$y=mx+1$$
Now substituting the value of y in
$$y^2=4x$$, we get
$$\Rightarrow (mx+1)^2=4x$$
$$\Rightarrow m^2x^2+1+2mx-4x=0$$
$$\Rightarrow m^2x^2+x(2m-4)+1=0$$ ..........$$(1)$$
Since, a tangent touches the curve at one point, the root of equation $$(1)$$ must be equal.
Thus, we get
Discriminant, $$D=b^2-4ac=0$$
$$(2m-4)^2-4(m^2)(1)=0$$
$$\Rightarrow 4m^2-16m+16-4m^2=0$$
$$\Rightarrow -16m+16=0$$
$$\Rightarrow m=1$$.

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 48

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Application of Derivatives Test - 48

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SHARING IS CARING

Question 1

$$(-5,0)$$

$$(0,5)$$

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

$$(-5,5)$$

Solution

Question 2

in the second quadrant

in the first quadrant

on the y-axis (imaginary axis)

on the x-axis (real axis)

Solution

Question 3

$$1/2$$

$$2$$

$$\sqrt{2}$$

$$\cfrac { 1 }{ \sqrt { 2 } } $$

Solution

Question 4

increasing in $$(0, 1)$$ and decreasing $$(1, 2)$$

decreasing in $$(0,1)$$ and increasing $$(1, 2)$$

increasing in $$(0, 2)$$

decreasing in $$(0, 2)$$

Solution

Question 5

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

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

$$-6$$

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

Solution

Question 6

$$(3, 0), (-1, 0)$$

$$(3, 0), (1, 2)$$

$$(-1, 0), (1, 2)$$

$$(1, 2), (1, -2)$$

Solution

Question 7

$$(0, 0)$$

$$(2, 16)$$

$$(3, 9)$$

$$(6, 36)$$

Solution

Question 8

$$\dfrac {1}{2}$$

$$0$$

$$-2$$

$$\infty$$

Solution

Question 9

$$(4, \pm 8/3)$$

$$(-4, 8/3)$$

$$(-4, -8/3)$$

$$(8/3, 4)$$

Solution

Question 10

$$1$$

$$2$$

$$3$$

$$1/2$$

Solution

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