Tangents and its Equations Test 4

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Tangents and its Equations Test 4

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Weekly Quiz Competition

Question 1
1 / -0

If tangent to curve at a point is perpendicular to $$x$$ - axis then at that point -

A
$$\displaystyle \frac{dy}{dx}=0 $$

B
$$\displaystyle \frac{dx}{dy}=0 $$

C
$$\displaystyle \frac{dy}{dx}=1 $$

D
$$\displaystyle \frac{dy}{dx}=-1 $$

Solution

We know slope of tangent to the any curve $$'y'$$ is $$m=\cfrac{dy}{dx}$$
Now if line is perpendicular to x-axis this means its slope is $$\infty$$
$$\Rightarrow \cfrac{dy}{dx} \to \infty \Rightarrow \cfrac{dx}{dy}=0$$
Hence, option 'C' is correct.
Question 2
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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 3
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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 4
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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 5
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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 6
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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 7
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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
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If a tangent to the curve $$\displaystyle y=6x-{ x }^{ 2 }$$ is parallel to the line $$\displaystyle 4x-2y-1=0$$, then the point of tangency on the curve is:

A
(2, 8)

B
(8, 2)

C
(6, 1)

D
(4, 2)

Solution

$$y=6x-x^2$$
$$\Rightarrow y' =6-2x\ as\ tangent \ is\ parallel \ to \ line 4x-2y-1=0 \ $$ slope $$=2$$
$$\Rightarrow 6-2x=2$$
$$\Rightarrow x=2$$
So $$y=6\cdot 2-2^2=8$$
Hence the point of tangency is $$(2,8)$$
Question 9
1 / -0

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

A
$$\dfrac {1}{4}$$

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

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

D
$$1$$

Solution

$$m=\dfrac {dy}{dx} = \dfrac {1}{1 + x^{3}}$$
$$m = \left (\dfrac {dy}{dx}\right )_ {x = 1} = \dfrac {1}{1 + 1} = \dfrac {1}{2}$$
Question 10
1 / -0

The equation of the tangent to the curve $$y=4xe^x$$ at $$\left(-1, \displaystyle\frac{-4}{e}\right)$$ is

A
$$y=-1$$

B
$$y=-\displaystyle\frac{4}{e}$$

C
$$x=-1$$

D
$$x=\displaystyle\frac{-4}{e}$$

Solution

Given curve is $$y=4xe^x$$
$$\displaystyle\frac{dy}{dx}=4e^x+4xe^x$$
At $$\left(-1, -\displaystyle\frac{4}{e}\right)\left(\displaystyle\frac{dy}{dx}\right)_{(\displaystyle -1, -4/e)}=4e^{-1}+4(-1)e^{-1}$$
$$\therefore$$ Equation of tangent is $$\left(y+\displaystyle\frac{4}{e}\right)=0(x+1)$$
$$\Rightarrow y=-\displaystyle\frac{4}{e}$$

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Tangents and its Equations Test 4

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Question 1

$$\displaystyle \frac{dy}{dx}=0 $$

$$\displaystyle \frac{dx}{dy}=0 $$

$$\displaystyle \frac{dy}{dx}=1 $$

$$\displaystyle \frac{dy}{dx}=-1 $$

Solution

Question 2

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

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

$$\displaystyle x=\pi$$

No where

Solution

Question 3

$$1$$

$$0$$

$$\displaystyle \infty $$

None of these

Solution

Question 4

$$\dfrac 2 3$$

$$-\dfrac 2 3$$

$$\dfrac 3 2$$

$$-\dfrac 3 2$$

Solution

Question 5

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

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

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

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

Solution

Question 6

$$(0, 0)$$

$$(a, a)$$

$$(a, 0)$$

$$(0, a)$$

Solution

Question 7

$$(1, 2)$$

$$(2, 1)$$

$$(1, 4)$$

$$( 2, 2)$$

Solution

Question 8

(2, 8)

(8, 2)

(6, 1)

(4, 2)

Solution

Question 9

$$\dfrac {1}{4}$$

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

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

$$1$$

Solution

Question 10

$$y=-1$$

$$y=-\displaystyle\frac{4}{e}$$

$$x=-1$$

$$x=\displaystyle\frac{-4}{e}$$

Solution

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