Tangents and its Equations Test 27

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Tangents and its Equations Test 27

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Weekly Quiz Competition

Question 1
1 / -0

The equation of normal of $$x^2+y^2-2x+4y-5=0$$ at $$(2,\,1)$$ is

A
$$y=3x-5$$

B
$$2y=3x-4$$

C
$$y=3x+4$$

D
$$y=x+1$$

Solution

Given equation is $$x^2+y^2-2x+4y-5=0$$
On differentiating, we get
$$2x+2y\dfrac{dy}{dx}-2+4\dfrac{dy}{dx}=0$$
$$\Rightarrow\;(y+2)\dfrac{dy}{dx}=1-x$$
$$\Rightarrow\,\begin{pmatrix}\dfrac{dy}{dx}\end{pmatrix}_{(2,\,1)}=\dfrac{1-2}{1+2}=\dfrac{-1}{3}$$
$$\Rightarrow\,-\begin{pmatrix}\dfrac{dx}{dy}\end{pmatrix}_{(2,\,1)}=3$$
Now, equation of normal is
$$(y-1)=3(x-2)$$
$$y-1=3x-6$$
$$\Rightarrow\;y=3x-5$$
Question 2
1 / -0

The equation of tangent to the curve $$y=3{ x }^{ 2 }-x+1$$ at the point $$(1,3)$$ is

A
$$y=5x+2$$

B
$$y=5x-2$$

C
$$y=\cfrac{1}{5}x+2$$

D
$$y=\cfrac{1}{5}x-2$$

Solution

Given, $$y=3x^2-x+1$$ and the given point is $$(1,3)$$
Therefore, $$\cfrac { dy }{ dx } =6x-1\quad $$
$$\therefore { \left( \cfrac { dy }{ dx } \right) }_{ (1,3) }=6-1=5$$
$$\therefore$$ $$m=5$$
Equation of tangent is
$$y=mx+c$$
$$y=5x+c$$ ...(i)
Put $$(1,3)$$ in equation $$(i)$$
$$3=5\times 1+c$$
$$\therefore$$ $$c=-2$$
Equation of tangent $$y=5x-2$$.
Question 3
1 / -0

The equation of tangent to the curve $$y = x^{2} + 4x + 1$$ at $$(-1, -2)$$ is

A
$$2x - y = 0$$

B
$$2x + y - 5 = 0$$

C
$$2x - y - 1 = 0$$

D
$$x + y - 1 = 0$$

Solution

The given equation of the curve is,
$$y = x^{2} + 4x + 1$$
On differentiating w.r.t. $$x$$, we get
$$\dfrac {dy}{dx} = 2x + 4$$
$$\therefore \left (\dfrac {dy}{dx}\right ) = 2(-1) + 4 = -2 + 4 = 2$$.
$$\therefore$$ slope of tangent at $$(-1, -2)$$ is $$2$$.
The equation of the tangent at $$(-1, -2)$$ is,
$$y-(-2) = 2(x - (-1))$$
$$\Rightarrow y + 2 = 2(x + 1)$$
$$\Rightarrow y + 2 = 2x + 2$$
$$\Rightarrow 2x - y = 0$$
Hence, the correct answer from the given alternatives is option A.
Question 4
1 / -0

The slope of the normal to the curve $$y = 3x^2$$ at the point whose $$x$$-coordinate $$2$$ is

A
$$\dfrac{1}{13}$$

B
$$\dfrac{1}{14}$$

C
$$\dfrac{-1}{12}$$

D
$$\dfrac{1}{12}$$

Solution

$$y=3x^{2}$$. Differentiating the equation of the curve with respect to $$x$$.
$$\dfrac{dy}{dx}=6x$$
Now
Slope$$_{x=2}=\dfrac{dy}{dx}|_{x=2}$$
$$=6(2)$$
$$=12$$.
Hence the slope of the tangent at $$x=2$$ is $$12$$. Since the normal is perpendicular to the tangent, therefore, the
Slope$$_\text{normal}\times$$Slope$$_\text{tangent}=-1$$ or Slope$$_\text{normal}=\dfrac{-1}{\text{Slope}_\text{tangent}}=\dfrac{-1}{12}$$
Question 5
1 / -0

The slope of the tangent to the curve $$y=3{ x }^{ 2 }+3\sin { x } $$ at $$x=0$$ is

A
$$3$$

B
$$2$$

C
$$1$$

D
$$-1$$

Solution

$$y=3x^2+3\sin x$$
$$\Rightarrow \dfrac{dy}{dx}=6x+3\cos x$$
Thus slope of tangent to the given curve at $$x=0$$ is
$$=\dfrac{dy}{dx}\bigg|_{x=0}=(6x+3\cos x)\bigg|_{x=0}=6(0)+3\cos 0=0+3\cdot 1=3$$
Question 6
1 / -0

The equation of the normal to the curve $$\theta =\dfrac { 1 }{ t } $$ at the point $$\left( -3,-\dfrac { 1 }{ 3 } \right) $$ is:

A
$$3\theta=27t-80$$

B
$$5\theta=27t-80$$

C
$$3\theta=27t+80$$

D
$$\theta =\cfrac { 1 }{ t } $$

Solution

The slope of the tangent at any point on the curve is given by
$$\dfrac{d\theta}{dt}$$
Therefore, differentiating the equation of the curve with respect to 't',
$$\theta=t^{-1}$$
$$\dfrac{d\theta}{dt}=\dfrac{-1}{t^{2}}$$.
Now
$$\dfrac{d\theta}{dt}_{t=-3}=\dfrac{-1}{9}$$.
Hence the slope of normal will be
$$=\dfrac{-1}{\text{Slope}_\text{tangent}}$$
$$=9$$.
Hence the equation of the normal will be
$$\dfrac{\theta-(\dfrac{-1}{3})}{t-(-3)}=9$$
$$\dfrac{3\theta+1}{t+3}=27$$
$$3\theta+1=27t+81$$
or
$$3\theta=27t+80$$.
Question 7
1 / -0

The tangent to the curve $$y=x^3+1$$ at (1, 2) makes an agnle $$\theta$$ with y axis, then the value of tan $$\theta$$ is.

A
$$3$$

B
$$\frac{1}{3}$$

C
$$-\frac{1}{3}$$

D
$$-3$$

Solution

Given curve is $$y=x^3+1$$
Differentiate it
$$\Rightarrow \dfrac{dy}{dx}=3x^2+0=3x^2$$
Thus slope of tangent to the curve at the point $$(1,2)$$ is
$$=\dfrac{dy}{dx}\bigg|_{x=1}=3(1)^2=3$$
If line makes an angle $$\theta$$ with y-axis then it will make angle $$\pi-\theta$$ with positive x-axis
Thus slope $$=\tan(\pi-\theta)=3$$
$$\Rightarrow \tan\theta=-3$$
Question 8
1 / -0

What is the slope of the tangent to the curve $$ x = t^2 + 3t - 8, y = 2t^2 - 2t - 5$$ at t = 2 ?

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

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

C
1

D
$$\dfrac{5}{6}$$

Solution

Differentiating $$x,y$$ w.r.t $$t,$$ we get
$$\dfrac { dy }{ dt } =\dfrac { d }{ dt } (2t^2-2t-5)=4t-2\\ \dfrac { dx }{ dt } =\dfrac { d }{ dt } (t^2+3t-8)=2t+3\\$$
$$\therefore \dfrac { dy }{ dx } =\dfrac { \left( \dfrac { dy }{ dt } \right) }{ \left( \dfrac { dx }{ dt } \right) } =\dfrac { 4t-2 }{ 2t+3 } $$
At $$t=2$$ derivative will be $$\dfrac {dy}{dx}=\dfrac {4\times 2-2}{2\times 2+3}=\dfrac {6}{7}$$
Thus, the slope of tangent is $$\dfrac{6}{7}$$.
Question 9
1 / -0

If the tangent to the function $$y = f(x)$$ at $$(3, 4)$$ makes an angle of $$\dfrac {3\pi}{4}$$ with the positive direction of x-axis in anticlockwise direction then $$f'(3)$$ is

A
$$-1$$

B
$$1$$

C
$$\dfrac {1}{\sqrt {3}}$$

D
$$\sqrt {3}$$

Solution

Given $$y=f(x)$$
Differentiating w.r.t x, we get
$$y' = f'(x)$$ which is the slope of the tangent

$$f'(3) = \tan \dfrac {3\pi}{4} = -\tan \dfrac {\pi}{4} = -1$$
Hence, $$f'(3)=-1$$.
Question 10
1 / -0

The equation of tangent to the curve $$y = 3x^{2} - x + 1$$ at $$P(1, 3)$$ is ____

A
$$5x - y = 2$$

B
$$x + 5y = 16$$

C
$$5x - y + 2 = 0$$

D
$$5x = y$$

Solution

Given curve is $$y=3x^2-x+1$$
$$\dfrac{dy}{dx}=6x-1$$, differentiate the curve
So slope of tangent at $$(1,3)$$ is, $$m =\dfrac{dy}{dx}\bigg |_{(1,3)} =6(1)-1=5$$
Hence equation of tangent passes through $$(1,3)$$ is given by
$$y-3=5(x-1)$$
$$\Rightarrow y-3=5x-5$$
$$\Rightarrow 5x-y=2$$

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Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 27

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Tangents and its Equations Test 27

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Question 1

$$y=3x-5$$

$$2y=3x-4$$

$$y=3x+4$$

$$y=x+1$$

Solution

Question 2

$$y=5x+2$$

$$y=5x-2$$

$$y=\cfrac{1}{5}x+2$$

$$y=\cfrac{1}{5}x-2$$

Solution

Question 3

$$2x - y = 0$$

$$2x + y - 5 = 0$$

$$2x - y - 1 = 0$$

$$x + y - 1 = 0$$

Solution

Question 4

$$\dfrac{1}{13}$$

$$\dfrac{1}{14}$$

$$\dfrac{-1}{12}$$

$$\dfrac{1}{12}$$

Solution

Question 5

$$3$$

$$2$$

$$1$$

$$-1$$

Solution

Question 6

$$3\theta=27t-80$$

$$5\theta=27t-80$$

$$3\theta=27t+80$$

$$\theta =\cfrac { 1 }{ t } $$

Solution

Question 7

$$3$$

$$\frac{1}{3}$$

$$-\frac{1}{3}$$

$$-3$$

Solution

Question 8

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

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

1

$$\dfrac{5}{6}$$

Solution

Question 9

$$-1$$

$$1$$

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

$$\sqrt {3}$$

Solution

Question 10

$$5x - y = 2$$

$$x + 5y = 16$$

$$5x - y + 2 = 0$$

$$5x = y$$

Solution

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