Tangents and its Equations Test 25

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Tangents and its Equations Test 25

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Weekly Quiz Competition

Question 1
1 / -0

The minimum value of the polynomial.
$$p(x)=3{ x }^{ 2 }-5x+2$$

A
$$-\frac { 1 }{ 6 }$$

B
$$\frac { 1 }{ 6 }$$

C
$$\frac { 1 }{ 12 }$$

D
$$-\frac { 1 }{ 12 }$$

Solution

$$p(x)=3{ x }^{ 2 }-5x+2$$
$$=3\left( x2-\frac { 5 }{ 3 } x+\frac { 2 }{ 3 } \right) $$
$$=3\left[ { x }^{ 2 }-\frac { 5 }{ 3 } x+\left( { \frac { 5 }{ 6 } } \right) ^{ 2 }-\left( { \frac { 5 }{ 6 } } \right) ^{ 2 }+\frac { 2 }{ 3 } \right] $$
$$=3\left[ \left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }-\frac { 25 }{ 36 } +\frac { 2 }{ 3 } \right] $$
$$=3\left[ \left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }+\frac { -25+24 }{ 36 } \right] $$
$$=3\left[ \left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }-\frac { 1 }{ 36 } \right] =3\left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }-\frac { 1 }{ 12 } $$
Question 2
1 / -0

If the tangent to the curve $$x = a(8 + sin \theta), y = a(1 + cos \theta )$$ at $$\theta = \displaystyle \frac{\pi}{3}$$ makes an angle $$\alpha$$ with x-axis, then $$\alpha$$ is equal to

A
$$\displaystyle \frac{\pi}{3}$$

B
$$\displaystyle \frac{2\pi}{3}$$

C
$$\displaystyle \frac{\pi}{6}$$

D
$$\displaystyle \frac{5\pi}{6}$$

Solution

$$x=a(8+\sin \theta)$$

$$\frac{d x}{d \theta}=a \cos \theta$$

$$y=a(1+\cos \theta)$$
$$\frac{d y}{d \theta}=-a \sin \theta$$

Now,
$$\frac{d y}{d \theta}=-a \sin \theta$$
$$\frac{d x}{d \theta}=a \cos \theta$$
$$\frac{d y}{d x}=-\tan \theta$$
$$\frac{d y}{d x}\left(\operatorname{at} \theta=\frac{\pi}{3}\right)=-\sqrt{3}=m$$

Now,
$$m=\tan \alpha$$
$$\Rightarrow \tan \alpha=-\sqrt{3}$$

Hence,
$$\alpha=\frac{2 \pi}{3}$$
Question 3
1 / -0

The curve which passes through $$(1, 2)$$ and whose tangent at any point has a slope that is half of slope of the line joining origin to the point of contact, is -

A
A rectangle hyperbola

B
A circle

C
A parabola

D
A straight line through origin

E
Answer required

Solution

Let the arbitrary point be $$x,y.$$
Now the slope of the line joining the origin and the point is
$$=\dfrac{y-0}{x-0}$$
$$=\dfrac{y}{x}$$.
The equation of the curve
$$y=f(x)$$.
Now slope of the tangent
$$=\dfrac{dy}{dx}$$
$$=\dfrac{y}{2x}$$ ... as per the given condition.
Hence
$$2\dfrac{dy}{y}=\dfrac{dx}{x}$$
Integrating both sides we get
$$2\ln y=\ln x+\ln c$$
Now $$y_{x=1}=2$$
Hence
$$2\ln (2)=\ln (1)+\ln (c)$$
Or
$$c=2^{2}=4$$
Hence
$$2\ln y=\ln x+\ln 4$$
Or
$$\ln (y^{2})=\ln 4x$$
Or
$$y^{2}=4x$$ is the required equation.
This is an equation of parabola.
Question 4
1 / -0

A curve $$\displaystyle y=f\left( x \right) ;\left( y>0 \right) $$ passes thorugh $$(1,1)$$ and at point $$\displaystyle P(x,y)$$ tangents cuts x-axis and y-axis at A and B respectively. If P divides AB internally in the ratio $$3 : 2$$, then the value of $$\displaystyle f\left( \frac { 1 }{ 8 } \right) $$ is

A
$$4$$

B
$$\displaystyle \frac { 1 }{ 4 } $$

C
$$\displaystyle 16\sqrt { 2 } $$

D
$$\displaystyle \frac { 1 }{ 16\sqrt { 2 } } $$

Solution

$$\displaystyle \frac { dy }{ dx } =-\frac { k }{ h } =-\frac { 2y }{ 3x } $$
$$\displaystyle \Rightarrow \quad 3\frac { dy }{ y } +2\frac { dx }{ x } =0$$
$$\displaystyle \Rightarrow \quad 3\ln { y }^{ 3 }{ x }^{ 3 }=C$$
$$\displaystyle \because \quad $$passing through $$(1,1)$$
$$\displaystyle \Rightarrow \quad c=0$$
$$\displaystyle \Rightarrow \quad { y }^{ 3 }{ x }^{ 2 }=1$$
at $$\displaystyle x=\frac { 1 }{ 8 }$$
$$ \Rightarrow \quad y=4$$
Question 5
1 / -0

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

A
$$\cfrac{22}{7}$$

B
$$\cfrac{6}{7}$$

C
$$\cfrac{7}{6}$$

D
$$\cfrac{-6}{7}$$

E
answer required

Solution

Slope to any curve is given by $$\dfrac{dy}{dx}$$

Here, $$\cfrac{dx}{dt}=2t+3$$ ...(1)
And $$\cfrac{dy}{dt}=4t-2$$ ...(2)
Dividing (2) by (1), we get $$\cfrac {dy}{dx} = \cfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{4t-2}{2t+3}$$
At point $$(2,-1)$$, $$t=2$$
So slope is given by,
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{4\times 2-2}{2\times 2+3}=\dfrac{6}{7}$$
Question 6
1 / -0

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

E
answer required

Solution

Lets put $$y=mx+1$$ in $${y}^{2}=4x$$ we get
$${(mx+1)}^{2}=4x$$
$$\Rightarrow {m}^{2}{x}^{2}+1+2mx-4x=0$$
Tangent touches a curve at one point so descriminant of the equation should be zero.
$$\Rightarrow {(2m-4)}^{2}-4\times 1\times {m}^{2}=0$$
$$\Rightarrow 4{m}^{2}+16-16m-4{m}^{2}=0$$
$$\Rightarrow m=1$$
Question 7
1 / -0

Angle between $${ y }^{ 2 }=x$$ and $${ x }^{ 2 }=y$$ at the origin is

A
$$2\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $$

B
$$\tan ^{ -1 }{ \left( \dfrac { 4 }{ 3 } \right) } $$

C
$$\dfrac { \pi }{ 2 } $$

D
$$\dfrac { \pi }{ 4 } $$

Solution

It is clear from the graph that both the curves have a tangent at the coordinate axes, so the angle between the curves is $$\dfrac { \pi }{ 2 } $$.

Alternative
Given curves are $${ y }^{ 2 }=x$$ and $${ x }^{ 2 }=y$$

On differentiating with respect to $$x$$, we get

$$2y\dfrac { dy }{ dx } =1$$ and $$2x=\dfrac { dy }{ dx } $$

$$\Rightarrow \dfrac { dy }{ dx } =\dfrac { 1 }{ 2y } $$ and $$\dfrac { dy }{ dx } =2x$$

At $$\left( 0,0 \right) $$

$${ m }_{ 1 }=\dfrac { dy }{ dx } =\infty $$ and $${ m }_{ 2 }=\dfrac { dy }{ dx } =0$$

$$\therefore \tan { \theta } =\dfrac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } =\infty $$

$$\Rightarrow \theta =\dfrac { \pi }{ 2 } $$
Question 8
1 / -0

If the line $$\alpha\,x+by+c=0$$ is a tangent to the curve $$xy=4$$, then

A
$$a < 0,\,b > 0$$

B
$$a \le o,\,b > 0$$

C
$$a < 0,\,b < 0$$

D
$$a \le 0,\,b < 0$$

Solution

Slope of line $$ax+by+c=0$$ is
$$\Rightarrow\;-\dfrac{a}{b}$$
$$y=\dfrac{4}{x}=1,\,\dfrac{dy}{dx}=-\dfrac{4}{x^2}$$,
$$-\dfrac{a}{b}=-\dfrac{4}{x^2}\Rightarrow\,\dfrac{a}{b}=\dfrac{4}{x^2} > 0$$
$$a < 0,\,b < 0$$
Question 9
1 / -0

Let $$y=e^{x^2}$$ and $$y=e^{x^2}\sin\, x$$ be two given curves. Then, angle between the tangents to the curves at any point their intersection is

A
$$0$$

B
$$\pi$$

C
$$\dfrac{\pi}{2}$$

D
$$\dfrac{\pi}{4}$$

Solution

For intersecting points, $$e^{x^2}=e^{x^2}\sin\,x$$
$$\Rightarrow e^{x^2}(\sin x-1)=0$$
$$\Rightarrow e^{x^2}=0$$ or $$\sin\,x=1$$

But $$e^{x^2}\neq 0\Rightarrow \sin x=1$$
$$\Rightarrow x=\dfrac{\pi}{2}$$

Now,
$$y=e^{x^2}$$

$$\therefore \dfrac{dy}{dx}=e^{x^2}.2x=2xe^{x^2}$$ $$[\because \sin\,x=1, \cos \, x = 0]$$

Since, both the curves has equal slope.
Hence, angle between the tangents at intersecting point is $$0$$.
Question 10
1 / -0

The slope at any point of a curve $$y=f\left( x \right) $$ is given by $$\dfrac { dy }{ dx } =3{ x }^{ 2 }$$ and it passes through $$\left( -1,1 \right) $$. The equation of the curve is

A
$$y={ x }^{ 3 }+2$$

B
$$y=-{ x }^{ 3 }-2$$

C
$$y=3{ x }^{ 3 }+4$$

D
$$y=-{ x }^{ 3 }+2$$

Solution

Given, $$\dfrac { dy }{ dx } =3{ x }^{ 2 }$$
$$\Rightarrow dy=3{ x }^{ 2 }dx$$
On integrating, we get
$$y=\dfrac { 3{ x }^{ 3 } }{ 3 } +c$$
$$\Rightarrow y={ x }^{ 3 }+c$$
It passes through $$\left( -1,1 \right) $$.
$$\therefore 1={ \left( -1 \right) }^{ 3 }+c$$
$$\Rightarrow c=2$$
$$\therefore y={ x }^{ 3 }+2$$

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Tangents and its Equations Test 25

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Tangents and its Equations Test 25

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Question 1

$$-\frac { 1 }{ 6 }$$

$$\frac { 1 }{ 6 }$$

$$\frac { 1 }{ 12 }$$

$$-\frac { 1 }{ 12 }$$

Solution

Question 2

$$\displaystyle \frac{\pi}{3}$$

$$\displaystyle \frac{2\pi}{3}$$

$$\displaystyle \frac{\pi}{6}$$

$$\displaystyle \frac{5\pi}{6}$$

Solution

Question 3

A rectangle hyperbola

A circle

A parabola

A straight line through origin

Answer required

Solution

Question 4

$$4$$

$$\displaystyle \frac { 1 }{ 4 } $$

$$\displaystyle 16\sqrt { 2 } $$

$$\displaystyle \frac { 1 }{ 16\sqrt { 2 } } $$

Solution

Question 5

$$\cfrac{22}{7}$$

$$\cfrac{6}{7}$$

$$\cfrac{7}{6}$$

$$\cfrac{-6}{7}$$

answer required

Solution

Question 6

$$1$$

$$2$$

$$3$$

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

answer required

Solution

Question 7

$$2\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $$

$$\tan ^{ -1 }{ \left( \dfrac { 4 }{ 3 } \right) } $$

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

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

Solution

Question 8

$$a < 0,\,b > 0$$

$$a \le o,\,b > 0$$

$$a < 0,\,b < 0$$

$$a \le 0,\,b < 0$$

Solution

Question 9

$$0$$

$$\pi$$

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

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

Solution

Question 10

$$y={ x }^{ 3 }+2$$

$$y=-{ x }^{ 3 }-2$$

$$y=3{ x }^{ 3 }+4$$

$$y=-{ x }^{ 3 }+2$$

Solution

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