Tangents and its Equations Test 22

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Tangents and its Equations Test 22

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Weekly Quiz Competition

Question 1
1 / -0

The points of contact of the vertical tangents $$x= 2-3\sin \theta $$, $$y= 3+2\cos \theta $$ are

A
$$\left ( 2,5 \right ),\left ( 2,1 \right )$$

B
$$\left ( -1,3 \right ),\left ( 5,3 \right )$$

C
$$\left ( 2,5 \right ),\left ( 5,3 \right )$$

D
$$\left ( -1,3 \right ),\left ( 2,1 \right )$$

Solution

For the tangents to be vertical,
$$\dfrac{dy}{dx}=\infty$$
Or
$$\dfrac{dx}{dy}=0$$
Or
$$\dfrac{dx}{d\theta}=0$$
Or
$$-3\cos\theta=0$$
Or
$$\theta=\dfrac{2n-1}{2}\pi$$
Hence
$$\theta=\dfrac{\pi}{2},\dfrac{3\pi}{2}$$.
Now
$$x_{\tfrac{\pi}{2}}$$
$$=-1$$
$$y_{\tfrac{\pi}{2}}$$
$$=3$$.
Similarly
$$x_{\tfrac{3\pi}{2}}=5$$
$$y_{\tfrac{3\pi}{2}}=3$$.
Hence the points are
$$(-1,3)$$ and $$(3,5)$$.
Question 2
1 / -0

The lines tangent to the curves $$\displaystyle y^{3}-x^{2}y+5y-2x=0$$ and $$\displaystyle x^{4}-x^{3}y^{2}+5x+2y=0$$ at the origin intersect at an angle $$\displaystyle \theta $$ equal to

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

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

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

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

Solution

Given curves are, $$\displaystyle y^{3}-x^{2}y+5y-2x=0 ..(1)$$ and $$\displaystyle x^{4}-x^{3}y^{2}+5x+2y=0 ..(2)$$
differentiating both w.r.t $$x$$
$$3y^2\cfrac{dy}{dx}-x^2\cfrac{dy}{dx}-2xy+5\cfrac{dy}{dx}-2=0$$
and
$$4x^3-3x^2y^2-2x^3y^2\cfrac{dy}{dx}+5+2\cfrac{dy}{dx}=0$$
Thus, by putting coordinates $$(0,0)$$ of origin for $$(x,y)$$ in above equation $$(1)$$, slope of tangent at origin to the first curve is $$m_1=\cfrac{2}{5}$$
and, by putting coordinates $$(0,0)$$ of origin for $$(x,y)$$ in equation $$(2)$$ above, slope of tangent at origin to the second curve is $$m_2=-\cfrac{5}{2}$$
Clearly $$m_1.m_2=-1$$
Hence both the lines are perpendicular.
Question 3
1 / -0

Equation of the line through the point $$(1/2,2)$$ and tangent to the parabola $$\displaystyle y=\frac{-x^{2}}{2}+2$$ and secant to the curve $$\displaystyle y=\sqrt{4-x^{2}}$$ is

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

B
$$2x + 2y - 9 = 0$$

C
$$y - 2 = 0$$

D
none

Solution

Equation of tangent is $$\displaystyle y-2=m\left ( x-\dfrac{1}{2} \right )$$
$$\displaystyle y=mx+2-\dfrac{m}{2}$$
Put it in the parabolas $$\displaystyle mx+2-\dfrac{m}{2}=-\dfrac{x^{2}}{2}+2$$
$$\displaystyle \dfrac{x^{2}}{2}+mx-\dfrac{m}{2}=0$$
since D = 0
$$\displaystyle \Rightarrow m^{2}+m=0$$
m = 0, -1 Two tangents are there
(i) y = 2
(ii)$$\displaystyle y=-x+2+\dfrac{1}{2}$$
$$\displaystyle \Rightarrow y=-x+\dfrac{5}{2}$$
$$\displaystyle y-=-x+\dfrac{5}{2}$$
The line y = 2 is tangent but $$\displaystyle y=-x+\dfrac{5}{2}$$ is secant for the curve
Question 4
1 / -0

If the curve $$\displaystyle { \left( \frac { x }{ a } \right) }^{ n }+{ \left( \frac { y }{ b } \right) }^{ n }=2$$ touches the straight line $$\displaystyle \frac { x }{ a } +\frac { y }{ b } =2$$, then find the value of $$n$$.

A
$$2$$

B
$$3$$

C
$$4$$

D
any real number

Solution

Given $$\displaystyle { \left( \frac { x }{ a } \right) }^{ n }+{ \left( \frac { y }{ b } \right) }^{ n }=2$$
Differentiating both sides w.r.t $$x,$$ we get
$$\displaystyle \frac { n }{ a } { \left( \frac { x }{ a } \right) }^{ n-1 }+\frac { b }{ b } { \left( \frac { y }{ b } \right) }^{ n-1 }\times \frac { dy }{ dx } =0$$
$$\displaystyle \Rightarrow \frac { dy }{ dx } =-\frac { n }{ a } { \left( \frac { x }{ a } \right) }^{ n-1 }\times \frac { b }{ a } { \left( \frac { b }{ y } \right) }^{ n-1 }$$
$$\displaystyle \therefore \frac { dy }{ dx } $$ at $$\displaystyle \left( a,b \right) =\frac { b }{ a } $$
$$\therefore$$ Tangent is $$\displaystyle y-b=-\frac { b }{ a } \left( x-a \right) \Rightarrow bx+ay=2ab\Rightarrow \frac { x }{ a } +\frac { y }{ b } =2$$
for all values of $$n$$ $$(\because\displaystyle\frac{dy}{dx}$$ is independent of $$n)$$
Question 5
1 / -0

Find all the tangents to the curve $$\displaystyle y=\cos \left ( x+y \right ),-2\pi \leq \times \leq 2\pi $$ that are parallel to the line $$x + 2y = 0$$

A
$$\displaystyle x + 2\: y=\dfrac {\pi }2$$ & $$\displaystyle x +2y=-\dfrac {3\pi }2$$

B
$$\displaystyle x + 2\: y=\dfrac {\pi }2$$ & $$\displaystyle x +2y=-\dfrac {\pi }2$$

C
$$\displaystyle x + 2\: y=-\dfrac {3\pi }2$$ & $$\displaystyle x +2y=-\dfrac {3\pi }2$$

D
$$\displaystyle x + 2\: y=\dfrac {3\pi }2$$ & $$\displaystyle x +2y=-\dfrac {3\pi }2$$

Solution

Given equation is $$x+2y=0$$
Slope $$\displaystyle = - \frac{1}{2}$$
$$\displaystyle = - \frac{1}{2}=-sin(x+y)(1+y')$$
$$\displaystyle \Rightarrow -\frac{1}{2}=-\sin \left ( x+y \right )\left ( 1-\frac{1}{2} \right )$$
$$\displaystyle \Rightarrow \sin \left ( x+y \right )=1$$
$$\displaystyle \Rightarrow \cos \left ( x+y \right )=0\Rightarrow y=0\cos x=0$$
$$\displaystyle \therefore x=\frac{\pi }{2},\frac{3\pi }{2},-\frac{\pi }{2}.-\frac{3\pi }{2}$$
$$\displaystyle \sin x=1$$ is possible for x = $$\displaystyle\frac{\pi }{2}$$ or $$\displaystyle -\frac{3\pi }{2}$$
Equation are : $$\displaystyle y-0=-\frac{1}{2}\left ( x-\frac{\pi }{2} \right )$$ and $$\displaystyle y-0=-\frac{1}{2}\left ( x+\frac{3\pi }{2} \right )$$
Question 6
1 / -0

A function is defined parametrically by the equations
x= $$\displaystyle 2t+t^{2}\sin \frac{1}{t}$$ if $$\displaystyle t\neq 0$$; $$ 0 $$, otherwise
and
y = $$\displaystyle \frac{1}{t}\sin t^{2}$$ if $$\displaystyle t\neq 0$$; $$ 0 $$, otherwise
Find the equation of the tangent and normal at the point for t = 0 if they exist

A
Tangent ;$$ 2y - x = 0;$$ Normal: $$2x + y = 0$$

B
Tangent ;$$ 3y - x = 0;$$ Normal: $$2x + y = 0$$

C
Tangent ;$$ 2y - x = 0;$$ Normal: $$3x + y = 0$$

D
Tangent ;$$ 3y - x = 0;$$ Normal: $$3x + y = 0$$

Solution

Given, $$x= 2t+ t^2 \sin \dfrac {1}{t}$$ and $$y=\dfrac {1}{t} \sin t^2$$
At $$t = 0$$ the point is origin
$$\displaystyle \frac{dx}{dt}=\lim_{t\rightarrow 0}\frac{2t+t^{2}\sin \dfrac 1t-0}{t}=2$$
$$\displaystyle \frac{dy}{dt}=\lim_{t-0}\frac{\dfrac{1}{t}\sin t^{2}}{t}=1$$
$$\displaystyle \frac{dy}{dx}=\dfrac{1}{2}$$
equation of tangent is $$\displaystyle y-0=\frac{1}{2}\left ( x-0 \right )\Rightarrow 2y-x=0$$
equation of normal is $$y - 0 = -2(x - 0)\Rightarrow y+2x=0$$
Question 7
1 / -0

The curve $$\displaystyle y=ax^{3}+bx^{2}+cx+5$$ touches the $$x$$ - axis at $$P(-2, 0)$$ and cuts the $$y$$-axis at a point $$Q$$, where its gradient is $$3$$. Find $$a, b, c$$.

A
$$\displaystyle a=-\frac{1}{5}, b=1,c=3$$

B
$$\displaystyle a=-\frac{1}{4}, b=-1,c=4$$

C
$$\displaystyle a=-\frac{1}{4}, b=0,c=3$$

D
$$\displaystyle a=-\frac{1}{3}, b=1,c=-3$$

Solution

Given, $$y= ax^3+bx^2+cx+5$$

$$\therefore \displaystyle \frac{dy}{dx}= 3ax^{2}+2bx+c$$

Since the curve touches $$x$$-axis at $$\left ( -2,0 \right )$$ so

$$\displaystyle \frac{dy}{dx}|_{\left ( -2,0 \right )}= 0\Rightarrow 12a-4b+c= 0$$ .... $$\left ( i \right )$$

The curve cut the $$y$$-axis at $$\left ( 0,8 \right )$$, so

$$\displaystyle \frac{dy}{dx}|_{\left ( 0,8 \right )}= 3\Rightarrow c= 3$$

Also the curve passes through $$\left ( -2,0 \right )$$, so

$$0= -8a+4b-2c+8\Rightarrow -8a+4b-2= 0$$ ..... $$ \left ( ii \right )$$

Solving $$ \left ( i \right )$$ and $$ \left ( ii \right )$$ $$a = -\dfrac {1}{4}$$, $$b =0$$
Question 8
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Let $$f$$ be a continuous, differentiable and bijective function. If the tangent to $$y=f\left( x \right) $$ at $$x=b$$, then there exists at least one $$c\in \left( a,b \right) $$ such that

A
$$f'\left( c \right) =0$$

B
$$f'\left( c \right) >0$$

C
$$f'\left( c \right) <0$$

D
none of these

Solution

Since the same line is tangent at one point $$x=a$$ and normal at other point $$x=b$$
$$\Rightarrow$$ Tangent at $$x=b$$ will be perpendicular to tangent at $$x=a$$
$$\Rightarrow$$ Slope of tangent changes from positive to negative or negative to positive. Therefore, it takes the value zero somewhere. Thus, there exists a point $$c\in \left( a,b \right) $$ where $$f'\left( c \right) =0$$.
Question 9
1 / -0

Find the equation of the normal to the curve $$\displaystyle y = \left ( 1+x \right )^{y}+\sin ^{-1}\left ( \sin ^{2}x \right )$$ at $$x = 0$$

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

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

C
$$x + y - 3 = 0$$

D
$$x + y - 4 = 0$$

Solution

Clearly at $$x=0 , y=1$$

Thus curve is, $$\displaystyle y = \left ( 1+x \right )^{y}+\sin ^{-1}\left ( \sin ^{2}x \right )$$

Differentiating w.r.t $$x$$

$$\cfrac{dy}{dx}=(1+x)^y[\cfrac{y}{1+x}+\log(1+x).\cfrac{dy}{dx}]+\cfrac{2\sin x.\cos x}{\sqrt{1+\sin^4x}}$$

Putting $$x=0 , y=1$$, we get, $$\cfrac{dy}{dx}=1=m$$ (say)

Thus slope of the normal is $$=-\cfrac{1}{m}=-1$$

Hence require equation is given by,

$$(y-1)=-1(x-0)$$

$$\Rightarrow x+y=1$$
Question 10
1 / -0

Find the equation of normal to the curve $$\displaystyle x^{2}=4y$$ passing through the point $$(1, 2)$$

A
$$x + y = 3$$

B
$$x-y=3$$

C
$$2x-y=4$$

D
$$2x-3y=1$$

Solution

$$x^2=4y$$

$$\Rightarrow \cfrac{dy}{dx}=\cfrac{x}{2}=m$$ (say)
Thus slope of normal at $$(1,2)$$ is $$m'=-\cfrac{1}{m}=\dfrac{-2}{x}=\dfrac{-2}{2}=-1$$

Hence, equation of normal at $$(1,2)$$ to the parabola is,

$$(y-2)=-1(x-1)$$

$$\Rightarrow x+y=3$$

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Tangents and its Equations Test 22

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Tangents and its Equations Test 22

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Question 1

$$\left ( 2,5 \right ),\left ( 2,1 \right )$$

$$\left ( -1,3 \right ),\left ( 5,3 \right )$$

$$\left ( 2,5 \right ),\left ( 5,3 \right )$$

$$\left ( -1,3 \right ),\left ( 2,1 \right )$$

Solution

Question 2

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

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

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

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

Solution

Question 3

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

$$2x + 2y - 9 = 0$$

$$y - 2 = 0$$

none

Solution

Question 4

$$2$$

$$3$$

$$4$$

any real number

Solution

Question 5

$$\displaystyle x + 2\: y=\dfrac {\pi }2$$ & $$\displaystyle x +2y=-\dfrac {3\pi }2$$

$$\displaystyle x + 2\: y=\dfrac {\pi }2$$ & $$\displaystyle x +2y=-\dfrac {\pi }2$$

$$\displaystyle x + 2\: y=-\dfrac {3\pi }2$$ & $$\displaystyle x +2y=-\dfrac {3\pi }2$$

$$\displaystyle x + 2\: y=\dfrac {3\pi }2$$ & $$\displaystyle x +2y=-\dfrac {3\pi }2$$

Solution

Question 6

Tangent ;$$ 2y - x = 0;$$ Normal: $$2x + y = 0$$

Tangent ;$$ 3y - x = 0;$$ Normal: $$2x + y = 0$$

Tangent ;$$ 2y - x = 0;$$ Normal: $$3x + y = 0$$

Tangent ;$$ 3y - x = 0;$$ Normal: $$3x + y = 0$$

Solution

Question 7

$$\displaystyle a=-\frac{1}{5}, b=1,c=3$$

$$\displaystyle a=-\frac{1}{4}, b=-1,c=4$$

$$\displaystyle a=-\frac{1}{4}, b=0,c=3$$

$$\displaystyle a=-\frac{1}{3}, b=1,c=-3$$

Solution

Question 8

$$f'\left( c \right) =0$$

$$f'\left( c \right) >0$$

$$f'\left( c \right) <0$$

none of these

Solution

Question 9

$$x + y - 1 = 0$$

$$x + y - 2 = 0$$

$$x + y - 3 = 0$$

$$x + y - 4 = 0$$

Solution

Question 10

$$x + y = 3$$

$$x-y=3$$

$$2x-y=4$$

$$2x-3y=1$$

Solution

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