Application of Derivatives Test

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Application of Derivatives Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

The slope of the curve $$\displaystyle y=\sin x+\cos ^{2}x $$ is zero at the point where -

A
$$\displaystyle x=\frac{\pi }{4}$$

B
$$\displaystyle x=\frac{\pi }{2}$$

C
$$\displaystyle x=\pi$$

D
No where

Solution

$$y=\sin x+\cos^2x$$
$$\cfrac{dy}{dx}=\cos x-2\cos x .\sin x$$
For slope to be zero, $$\cfrac{dy}{dx}=0$$
$$\Rightarrow \cos x-2\cos x .\sin x=0\Rightarrow \cos x(1-2\sin x)=0$$
$$\Rightarrow \cos x=0$$ or $$\sin x=\cfrac{1}{2}$$
$$\Rightarrow x=\cfrac{\pi}{2}, \cfrac{\pi}{6}$$ in first quadrant.
Hence, option 'B' is correct.
Question 2
1 / -0

The slope of the tangent to the curve $$\displaystyle y=\sin x$$ at point $$(0, 0)$$ is

A
$$1$$

B
$$0$$

C
$$\displaystyle \infty $$

D
None of these

Solution

Given $$y=\sin x$$
$$\Rightarrow \cfrac{dy}{dx}=\cos x$$
Thus, slope of tangent at $$(0,0)$$ is,
$$m=\left (\cfrac{dy}{dx}\right )_{(0.0)}=\cos (0)=1$$
Hence, option 'A' is correct.
Question 3
1 / -0

If tangent at a point of the curve $$y = f(x)$$ is perpendicular to $$2x - 3y = 5$$ then at that point $$\displaystyle \dfrac{dy}{dx}$$ equals

A
$$\dfrac 2 3$$

B
$$-\dfrac 2 3$$

C
$$\dfrac 3 2$$

D
$$-\dfrac 3 2$$

Solution

We know slope of tangent to the curve is $$\displaystyle \frac{dy}{dx}$$
Now slope of line $$2x - 3y = 5$$ is $$\displaystyle \frac{2}{3}$$
and tangent to $$y=f(x)$$ is perpendicular to this line.
$$\displaystyle \therefore $$ $$\displaystyle \frac{dy}{dx}=-\frac{3}{2} $$
Hence, option 'D' is correct.
Question 4
1 / -0

The inclination of the tangent w.r.t. $$x$$ - axis to the curve $$\displaystyle x^{2}+2y=8x-7$$ at the point $$x = 5$$ is

A
$$\displaystyle\dfrac{ \pi }4$$

B
$$\displaystyle\dfrac{ \pi }3$$

C
$$\displaystyle\dfrac{3 \pi }4$$

D
$$\displaystyle\dfrac{ \pi }2$$

Solution

Given, $$\displaystyle x^{2}+2y=8x-7$$
$$\Rightarrow 2y=8x-x^2-7$$
$$\therefore 2\cfrac{dy}{dx}=8-2x\Rightarrow \cfrac{dy}{dx}=4-x$$
Thus slope of tangent to the given curve at $$x=5$$ is, $$m=-1$$
If $$'\theta '$$ is the inclination of the tangent w.r.t $$x-axis$$ then,
$$\tan\theta=-1\Rightarrow \theta=\cfrac{3\pi}{4}$$
Hence, option 'C' is correct.
Question 5
1 / -0

At what point the tangent to the curve $$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$$ is perpendicular to the $$x$$ - axis

A
$$(0, 0)$$

B
$$(a, a)$$

C
$$(a, 0)$$

D
$$(0, a)$$

Solution

Let the point be $$P\displaystyle \left ( x_{1},y_{1} \right )$$
Now $$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$$
Differentiating w.r.t $$x$$
$$\displaystyle \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\times \frac{dy}{dx}=0 $$
$$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{y}{x}}$$
Thus slope of tangent at P is $$=\displaystyle \left ( \frac{dy}{dx} \right )_{\left ( x_{1},y_{1} \right )}=-\sqrt{\frac{y_{1}}{x_{1}}}$$
if tangent $$\displaystyle \perp $$ to x axis then $$\displaystyle \frac{dy}{dx}=\frac{1}{0} $$
$$\displaystyle \therefore x_1 = 0\Rightarrow y_1 = a$$
Therefore, required point is $$(0, a)$$
Hence, option 'D' is correct.
Question 6
1 / -0

The slope of the tangent to the curve $$\displaystyle y=-x^{3}+3x^{2}+9x-27$$ is maximum when x equals.

A
$$1$$

B
$$3$$

C
$$\dfrac 12$$

D
$$-\dfrac 12$$

Solution

Given, $$\displaystyle y=-x^{3}+3x^{2}+9x-27$$
Slope at any point on thegiven curve is $$m= \cfrac{dy}{dx}=-3x^2+6x+9$$
Now for maximum value of slope we must have $$\cfrac{dm}{dx}=0=-6x+6\Rightarrow x = 1$$
Question 7
1 / -0

The line $$y = x + 1$$ is a tangent to the curve $$ y^2 = 4x$$ at the point.

A
$$(1, 2)$$

B
$$(2, 1)$$

C
$$(1, 4)$$

D
$$( 2, 2)$$

Solution

We have $$y=x+1, y^2=4x$$
Solving these, $$\Rightarrow (x+1)^2=4x\Rightarrow x^2-2x+1=0$$
$$\Rightarrow (x-1)^2=0\Rightarrow x=1\therefore y = (x+1)_{x=1}=2$$
Thus required point of contact is $$(1,2)$$
Question 8
1 / -0

The rate of change of the area of a circle with respect to its radius $$r$$ at $$r = 6 cm$$ is.

A
$$10 \pi$$

B
$$12 \pi$$

C
$$8 \pi$$

D
$$11 \pi$$

Solution

The area $$(A)$$ of circle with radius $$r$$ is, $$A =\pi r^2$$
Thus rate of change of area of circle with radius is,
$$\displaystyle\Rightarrow \frac{dA}{dr}=2\pi r$$.
Hence at $$r=6$$ cm $$,\displaystyle\frac{dA}{dr}=12\pi$$ cm
Question 9
1 / -0

The side of a square sheet is increasing at the rate of $$4 cm$$ per minute. The rate by which the area increasing when the side is $$8 cm$$ long is.

A
$$60$$ $$ cm^2/minute$$

B
$$66$$ $$ cm^2/minute$$

C
$$62$$ $$ cm^2/minute$$

D
$$64$$ $$ cm^2/minute$$

Solution

Given $$\dfrac{dx}{dt}=4 cm/min$$, $$x=8 cm$$
Area of square $$A = x^2$$

$$\dfrac {dA}{dt}=2x\dfrac {dx}{dt}$$

$$\Rightarrow \dfrac{dA}{dt}= 2 \times 8 \times 4 = 64 cm^2/minute$$
Question 10
1 / -0

The slope of the normal to the curve $$y = 2x^2+ 3 \sin x$$ at $$x = 0$$ is.

A
$$3$$

B
$$\dfrac{1}{3}$$

C
$$-3$$

D
$$-\dfrac{1}{3}$$

Solution

$$y = 2x^2+ 3 \sin x$$
$$\cfrac{dy}{dx} = 4x+3\cos x$$
Thus slope of tangent at $$x=0$$ is $$4(0)+3\cos 0=3$$
Hence slope of normal at the same point is $$-\cfrac{1}{m}=-\cfrac{1}{3}$$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 14

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Application of Derivatives Test - 14

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SHARING IS CARING

Question 1

$$\displaystyle x=\frac{\pi }{4}$$

$$\displaystyle x=\frac{\pi }{2}$$

$$\displaystyle x=\pi$$

No where

Solution

Question 2

$$1$$

$$0$$

$$\displaystyle \infty $$

None of these

Solution

Question 3

$$\dfrac 2 3$$

$$-\dfrac 2 3$$

$$\dfrac 3 2$$

$$-\dfrac 3 2$$

Solution

Question 4

$$\displaystyle\dfrac{ \pi }4$$

$$\displaystyle\dfrac{ \pi }3$$

$$\displaystyle\dfrac{3 \pi }4$$

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

Solution

Question 5

$$(0, 0)$$

$$(a, a)$$

$$(a, 0)$$

$$(0, a)$$

Solution

Question 6

$$1$$

$$3$$

$$\dfrac 12$$

$$-\dfrac 12$$

Solution

Question 7

$$(1, 2)$$

$$(2, 1)$$

$$(1, 4)$$

$$( 2, 2)$$

Solution

Question 8

$$10 \pi$$

$$12 \pi$$

$$8 \pi$$

$$11 \pi$$

Solution

Question 9

$$60$$ $$ cm^2/minute$$

$$66$$ $$ cm^2/minute$$

$$62$$ $$ cm^2/minute$$

$$64$$ $$ cm^2/minute$$

Solution

Question 10

$$3$$

$$\dfrac{1}{3}$$

$$-3$$

$$-\dfrac{1}{3}$$

Solution

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