Tangents and its Equations Test 7

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Tangents and its Equations Test 7

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Weekly Quiz Competition

Question 1
1 / -0

The equation of the tangent to the curve $$y = x + \dfrac{4}{{{x^2}}}$$ , that is parallel to the $$x-$$axis , is

A
$$y=1$$

B
$$y=2$$

C
$$y=3$$

D
$$y=0$$

Solution

$$y=x+\dfrac{4}{x^{2}}$$

$$\dfrac{dy}{dx}=1-\dfrac{8}{x^{3}}$$
For tangent parallel to $$x$$-axis,
$$\dfrac{dy}{dx}=0\Rightarrow 1-\dfrac{8}{x^{3}}=0$$
$$\Rightarrow x^{3}=8$$
$$\Rightarrow x=2$$
Question 2
1 / -0

The coordinates of the feet of the normals drawn from the point (14, 7) to the curve $$y^2 - 16 x - 8y = 0$$ are

A
(0, 0)

B
(3, 4)

C
(3, -4)

D
(8, 16)

Solution
Question 3
1 / -0

The slope of the tangent to the curve $$y=sinx$$ where it crosses the $$x-axis$$ is

A
$$1$$

B
$$-1$$

C
$$ \pm 1$$

D
$$ \pm 2$$

Solution

Consider the given equation of the curve.
$$y=\sin x$$ ……(1)

On differentiating equation (1) with respect to $$x$$, we get
$$\dfrac{dy}{dx}=\cos x$$

When given curve crosses the x-axis then.
$$ y=0 $$
$$\sin x=0$$
$$x=0$$

Therefore,
$$ {{\left( \dfrac{dy}{dx} \right)}_{y=0}}=\cos 0 $$
$$ {{\left( \dfrac{dy}{dx} \right)}_{y=0}}=1 $$

Hence, this is the answer.
Question 4
1 / -0

The equation of normal to the curve $$y=\left| { x }^{ 2 }-\left| x \right| \right| $$ at $$x=-2$$ is

A
$$3y=2x+10$$

B
$$3y=x+8$$

C
$$2y=x+6$$

D
$$2y=3x+10$$

Solution

$$y=\mid x^{2}-\mid x \mid \mid=(\mid x\ \mid )(\mid \mid x\mid-1\mid )$$
At $$x=-2,y=2$$

$$\dfrac{dy}{dx}=\dfrac{\mid x\mid }{x}\mid \mid x\mid-1\mid+\dfrac{\mid \mid x\mid-1\mid}{\mid x\mid -1}\dfrac{\mid x^{2}\mid}{x}$$

$$\bigg(\dfrac{dy}{dx}\bigg)_{x=-2}=-1-2=-3$$

The equation of normal is $$y=\dfrac{x}{3}+\bigg(2+\dfrac{2}{3}\bigg)$$

$$3{y}=x+8$$
Question 5
1 / -0

The equation of the normal at $$t=\dfrac{\pi}{2}$$ to the curve $$x=2\sin t, y=2\cos t$$ is?

A
$$x=0$$

B
$$y=0$$

C
$$y=2x+3$$

D
$$y=3$$

Solution

$$dx = 2\cos t , dy = -2\sin t$$
$$-\cfrac{dx}{dy} = -\cot t$$
$$\cfrac{dx}{dy} = \cot t$$
Slope = $$0$$ as it is x-axis
Hence, $$y = 0$$
Question 6
1 / -0

The equation of the tangents at $$(2, 3)$$ on the curve $$y^{2}=ax^{3}+b$$ is $$y=4x-5$$ find the value of $$a$$ & $$b$$.

A
$$2$$ and $$-2$$

B
$$4$$ and $$-7$$

C
$$2$$ and $$-7$$

D
None of these
Question 7
1 / -0

The equation of the tangent to the curve $$y=2\sin{x}+\sin{2x}$$ at $$x=\cfrac{\pi}{3}$$ on it is

A
$$y-3=0$$

B
$$y+\sqrt{3}=0$$

C
$$2y-3=0$$

D
$$2y-3\sqrt{3}=0$$

Solution

$$x=\dfrac{\pi }{3}\Rightarrow y=2\left(\dfrac{\sqrt{3}}{2}\right)+\dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}$$
$$y^{1}= 2 \cos x+2 \cos 2x$$
at $$x=\dfrac{\pi }{3}\Rightarrow y^{1}=2\left(\dfrac{1}{2}\right)+2\left(\dfrac{-1}{2}\right)$$
$$=0$$
$$\therefore $$ equation of tangent is
$$\dfrac{y-3\sqrt{3}/2}{x=\pi /3}=0$$
$$\Rightarrow y=\dfrac{3\sqrt{3}}{2}$$
Question 8
1 / -0

Tangent drawn to $$y=ax^2+bx+c$$ at$$(5,4)$$ is parallel to $$x-axis.$$ If $$a$$ $$\epsilon $$ $$[2,4]$$. Then maximum value of c.

A
$$54$$

B
$$56$$

C
$$104$$

D
$$106$$

Solution

$$y=a{x^2}+b{x}+c\implies \dfrac{d y}{d x}=2{a}{x}+b$$
At $$x=5,y=4\implies 25{a}+5{b}+c=4$$ and also tangent is parallel to $$x-axis$$
$$\implies 10{a}+b=0\implies b=-10{a}\implies c=4-5{b}-25{a}=4+25{a}$$
$$a\in [2,4]\implies c\in [54,104]$$
Question 9
1 / -0

The Point (s) on the cure $${ y }^{ 3 }+{ 3x }^{ 2 }=12y$$ where the tangent is vertical (parallel to y-axis), is/are.

A
$$\left[ \pm \dfrac { 4 }{ \sqrt { 3 } } ,-2 \right] $$

B
$$\left( \pm \dfrac { \sqrt { 11 } }{ 3 } ,1 \right) $$

C
$$(0,0)$$

D
$$\left( \pm \dfrac { 4 }{ \sqrt { 3 } } ,2 \right) $$

Solution

curve given $$y^3+3x^2=12y$$

Thus differentiating w.r.t $$x$$

$$3y^2\dfrac {dy}{dx}+6x=12\dfrac {dy}{dx}$$
$$\dfrac {dy}{dx}(y^2-4)+2x=0$$
$$\dfrac {dy}{dx} =\dfrac {-2x}{y^2-4}$$

For tangent to be $$\parallel$$ to $$y-$$axis $$\dfrac {dy}{dx}\rightarrow \infty$$

Thus $$y^2-4=0$$

$$y=\pm 2$$
Putting $$y=+2$$ in curve we get $$x=\pm \dfrac {4}{\sqrt 3}$$
Putting $$y=-2$$ in curve $$x$$ does not exit

Thus $$y=-2$$ not in range as $$x^2 < 0$$

Thus points where tangent is $$\parallel$$ to $$y-$$axis is $$\left [\pm \dfrac {4}{\sqrt {3}},\ 2 \right]$$ Answer.
Question 10
1 / -0

The angle made by the tangent line at (1, 3) on the curve $$y=4x-{ x }^{ 2 }$$ with $$\overset { - }{ OX } $$ is

A
$${ tan }^{ -1 }(2)$$

B
$${ tan }^{ -1 }(1/3)$$

C
$${ tan }^{ -1 }(3)$$

D
$$\pi /4$$

Solution

Given,

$$y=4x-x^2$$

$$\dfrac{dy}{dx}=4-2x$$

$$\dfrac{dy}{dx}_{(1,3)}=4-2(1)=2$$

Therefore, angle made by tangent,

$$\tan \theta =2$$

$$\therefore \theta =\tan ^{-1}2$$

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Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 7

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Tangents and its Equations Test 7

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SHARING IS CARING

Question 1

$$y=1$$

$$y=2$$

$$y=3$$

$$y=0$$

Solution

Question 2

(0, 0)

(3, 4)

(3, -4)

(8, 16)

Solution

Question 3

$$1$$

$$-1$$

$$ \pm 1$$

$$ \pm 2$$

Solution

Question 4

$$3y=2x+10$$

$$3y=x+8$$

$$2y=x+6$$

$$2y=3x+10$$

Solution

Question 5

$$x=0$$

$$y=0$$

$$y=2x+3$$

$$y=3$$

Solution

Question 6

$$2$$ and $$-2$$

$$4$$ and $$-7$$

$$2$$ and $$-7$$

None of these

Question 7

$$y-3=0$$

$$y+\sqrt{3}=0$$

$$2y-3=0$$

$$2y-3\sqrt{3}=0$$

Solution

Question 8

$$54$$

$$56$$

$$104$$

$$106$$

Solution

Question 9

$$\left[ \pm \dfrac { 4 }{ \sqrt { 3 } } ,-2 \right] $$

$$\left( \pm \dfrac { \sqrt { 11 } }{ 3 } ,1 \right) $$

$$(0,0)$$

$$\left( \pm \dfrac { 4 }{ \sqrt { 3 } } ,2 \right) $$

Solution

Question 10

$${ tan }^{ -1 }(2)$$

$${ tan }^{ -1 }(1/3)$$

$${ tan }^{ -1 }(3)$$

$$\pi /4$$

Solution

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