MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 6

The points of contact of the vertical tangents

$x= 2-3\sin \theta$ ,

$y= 3+2\cos \theta$ are

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$\left ( 2,5 \right ),\left ( 2,1 \right )$
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$\left ( -1,3 \right ),\left ( 5,3 \right )$
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$\left ( 2,5 \right ),\left ( 5,3 \right )$
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$\left ( -1,3 \right ),\left ( 2,1 \right )$

Explanation

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

$(3,5)$ .

Tangent is drawn to ellipse

$\displaystyle \frac{x^{2}}{27}+y^{2}=1$ at

$\left ( 3\sqrt{3}\cos \theta ,\sin \theta \right )$ (where

$\theta \in \left ( 0,\dfrac{\pi}2 \right )).$ Then the value of

$\theta$ such that sum of intercepts on axes made by this tangent is least is

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$\dfrac {\pi}3$
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$\dfrac {\pi}6$
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$\dfrac {\pi}8$
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$\dfrac {\pi}4$

Explanation

Let

$x=3\sqrt{3}\cos\theta$
And

$y=\sin\theta$
Hence,

$\dfrac{dy}{dx}$

$=-\dfrac{\cos\theta}{3\sqrt{3}\sin\theta}$ ...(i)
Hence equation of the tangent will be

$y-\sin\theta=-\dfrac{\cos\theta}{3\sqrt{3}\sin\theta}(x-3\sqrt{3}.\cos\theta)$

$3\sqrt{3}\sin\theta.y-3\sqrt{3}\sin^{2}\theta=-x\cos\theta+3\sqrt{3}\cos^{2}\theta$

$3\sqrt{3}\sin\theta.y+x\cos\theta=3\sqrt{3}$ .
Hence the intercepts are

$cosec\theta$ and

$3\sqrt{3}.\sec\theta$ .
Hence,

$k=cosec\theta+3\sqrt{3}.\sec\theta$
Now

$\dfrac{dk}{d\theta}$

$=-cosec\theta.\cot\theta+3\sqrt{3}.\sec\theta.\tan\theta$

$=0$

$3\sqrt{3}.\sec\theta.\tan\theta=cosec\theta.\cot\theta$

$3\sqrt{3}.\tan^{2}\theta.\dfrac{1}{\cos\theta}=\dfrac{1}{\sin\theta}$

$3\sqrt{3}.\tan^{2}\theta=\dfrac{\cos\theta}{\sin\theta}$

$3\sqrt{3}.\tan^{3}\theta=1$

$\tan^{3}\theta=\dfrac{1}{3\sqrt{3}}$

$\tan\theta=\dfrac{1}{\sqrt{3}}$

$\theta=\tan^{-1}(\dfrac{1}{\sqrt{3}})$

$\theta=30^{0}$

$=\dfrac{\pi}{6}$ .

The tangent to the curve

$\displaystyle 3xy^{2}-2x^{2}y=1$ at

$(1, 1)$ meets the curve again at the point

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$\displaystyle \left ( \frac{16}{5},\frac{1}{20} \right )$
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$\displaystyle \left (- \frac{16}{5},-\frac{1}{20} \right )$
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$\displaystyle \left ( \frac{1}{20},\frac{16}{5} \right )$
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$\displaystyle \left ( -\frac{1}{20},\frac{16}{5} \right )$

Explanation

Given,

$3xy^{2}-2x^{2}y=1$

Differentiating with respect to x, gives us

$3y^{2}+6xyy'-4xy-2x^{2}y'=0$

At (1,1)

$3+6y'-4-2y'=0$

$4y'=1$

$y'=\dfrac{1}{4}$

Hence the equation of the tangent is

$\dfrac{y-1}{x-1}=\dfrac{1}{4}$

$4y-4=x-1$

$x-4y=-3$

$x=4y-3$

Substituting in the equation of the curve, give us

$3(4y-3)y^{2}-2(4y-3)^{2}y=1$

$(4y-3)y[3y-2(4y-3)]=1$

$(4y-3)y(6-5y)=1$

$y(4y-3)(5y-6)=-1$

$y(20y^{2}-24y-15y+18)=-1$

$y(20y^{2}-39y+18)=-1$

$20y^{3}-39y^{2}+18y+1=0$

$y=1,\dfrac{-1}{20}$

$\Rightarrow x=1,\dfrac{-16}{5}$

Hence the tangent again cuts the curve at

$(\dfrac{-16}{5},\dfrac{-1}{20})$

The angle at which the curve

$y=ke^{kx}$ intersects the

$y$ -axis is

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$\tan ^{-1}(k^{2})$
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$\cot ^{-1}(k^{2})$
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$\sin ^{-1}\left ( 1/\sqrt{1+k^{4}} \right )$
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$\sec ^{-1}\left ( 1/\sqrt{1+k^{4}} \right )$

Explanation

$\displaystyle \dfrac{dy}{dx}=k^{2}e^{kx}$ .

The curve intersects

$y$ -axis at

$\left ( 0,k \right )$

So,

$\displaystyle \dfrac{dy}{dx}|_{\left ( 0,k \right )}=k^{2}$ .

$\theta$ is the angle at which the given

curve intersects the

$y$ -axis then

$\displaystyle \tan \left ( \pi /2-\theta \right )=\dfrac{k^{2}-0}{1+0.k^{2}}=k^{2}$ .

Hence

$\theta =\cot ^{-1}k^{2}$

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

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$\displaystyle \frac{\pi }{6}$
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$\displaystyle \frac{\pi }{4}$
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$\displaystyle \frac{\pi }{3}$
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$\displaystyle \frac{\pi }{2}$

Let

$\displaystyle f\left ( x \right )=x^{3}+ax+b$ with

$\displaystyle a\neq b$ and suppose the tangent lines to the graph of

$f$ at

$x = a$ and

$x = b$ have the same gradient Then the value of

$f (1)$ is equal to

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$0$
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$1$
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$\displaystyle -\frac{1}{3}$
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$\displaystyle \frac{2}{3}$

Explanation

$f(x)=x^3+ax+b$

$f'(x)=3x^2+a$
Given gradient at

$x=a$ and at

$x=b$ are same

$\Rightarrow 3a^2 +a=3b^2+a\Rightarrow b^2=a^2$
But given

$a\neq b$

$\Rightarrow a+b=0 ..(1)$
Hence

$f(1)=1+a+b=1$ using (1)

A curve with equation of the form

$\displaystyle y=ax^{4}+bx^{3}+cx+d$ has zero gradient at the point (0, 1) and also touches the x-axis at the point (-1, 0) then the values of x for which the curve has a negative gradient are

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x > -1
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x < 1
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x < -1
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$\displaystyle -1\leq \times \leq 1$

Explanation

Given,

$\displaystyle y=ax^{4}+bx^{3}+cx+d$

$\Rightarrow y'=4ax^3+3bx^2+c$
Using given conditions,

$y(0)=1\Rightarrow d=1$

$y'(0)=0\Rightarrow c=0$

$y(-1)=0\Rightarrow a-b=-1 ..(1)$
and

$y'(-1)=0\Rightarrow 4a-3b=0 ..(2)$
Solving equation (1) and (2) we get,

$a=3,b=4$
Hence the polynomial is,

$y=3x^4+4x^3+1$

$y'=12x^2(1+x)$
Now for negative gradient

$y' < 0\Rightarrow 12x^2(1+x)< 0$

$\Rightarrow x< -1$

For the curve represented parametrically by the equations

$\displaystyle x=2\ln\cot t+1$ and

$\displaystyle y=\tan t+\cot t$

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normal at $t = \displaystyle \dfrac {\pi }4$ is parallel to y=axis
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tangent at $t = \displaystyle \dfrac {\pi }4$ is parallel to x-axis
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tangent at is parallel to the line $y = x$
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normal at is parallel to the line $y = x$

Explanation

Given,

$x= 2 \ln \cot t+1$ and

$y=\tan t+\cot t$

$\therefore \displaystyle \frac{dx}{dt}=\frac{2\left ( -\mathrm{cosec} ^{2t} \right )}{\cot t}$
at

$\displaystyle t = \frac{\pi }{4},\frac{dx}{dt}=-4$
and

$\displaystyle \frac{dy}{dt}=\sec ^{2}t-\mathrm{cosec} ^{2}t$

$\displaystyle at t =\frac{\pi }{4}\:\:\;\frac{dy}{dt}=0$

$\displaystyle \frac{dy}{dx}=0$ for tangent & hence it is parallel to x-axis & its normal is parallel to y axis

If a variable tangent to the curve

$\displaystyle x^{2}y=c^{3}$ makes intercepts

$a, b$ on

$x$ and

$y$ axis respectively then the value of

$\displaystyle x^{2}$ is

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$\displaystyle 27c^{3}$
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$\displaystyle \frac4{27}c^{3}$
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$\displaystyle \frac{27}{4}c^{3}$
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$\displaystyle \frac4{9}c^{3}$

Explanation

Let any point on the curve is

$\displaystyle \left ( \frac{c}{t},ct^{2} \right )$

$\displaystyle y=\frac{c^{3}}{x^{2}}\Rightarrow \frac{dy}{dx}=-\frac{2c^{3}}{x^{3}}=-\frac{2c^{3}}{\dfrac {c^{3}}{t^{3}}}$

$\displaystyle \frac{dy}{dx}=-2t^{3}$
Equation of tangent is

$\displaystyle y-ct^{2}=-2t^{3}\left ( x-\frac{c}{t} \right )$
for x intercept

$\displaystyle 0-ct^{2}=-2t^{3}\left ( a-\frac{c}{t} \right )\Rightarrow \frac{c}{2t}=a-\frac{c}{t}$

$\displaystyle a=\frac{3}{2}\frac{c}{t}$
for y intercept

$\displaystyle b-ct^{2}=-2t^{3}\left ( 0-\frac{c}{t} \right )$

$\displaystyle \Rightarrow b-ct^{2}=2t^{2}c \:\:\Rightarrow b=3ct^{2}$

$\displaystyle a^{2}b=\frac{9}{4},\frac{c^{2}}{t^{2}}\times 3ct^{2}=\frac{27c^{3}}{4}$

The coordinates of the point(s) on the graph of the function

$\displaystyle f(x)=\frac{x^{3}}3{-\frac{5x^{2}}{2}}+7x-4$ where the tangent drawn cut off intercepts from the coordinate axes which are equal in magnitude but opposite in sign is

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$(2,\dfrac 83)$
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$(3, \dfrac 72)$
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$(1,\dfrac 56)$
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none

Explanation

Hence

$\dfrac{dy}{dx}=1$ at that point.
Hence

$x^{2}-5x+7=1$
Or

$x^{2}-5x+6=0$
Or

$(x-2)(x-3)=0$
Or

$x=2$ and

$x=3$

$f(2)=\dfrac{8}{3}$
And

$f(3)=\dfrac{7}{2}$
Hence the points are

$(2,\dfrac{8}{3})$ and

$(3,\dfrac{7}{2})$ .

The number of values of c such that the straight line

$3x + 4y = c$ touches the curve

$\displaystyle \frac{x^{4}}{2}=x+y$ is

0%
$0$
0%
$1$
0%
$2$
0%
$4$

Explanation

$3x+4y=c$
Or

$3+4y'=0$
Or

$y'=\dfrac{-3}{4}$ .
This is the slope of the tangent.
Now
Differentiating the equation of the curve with respect to

$x$ , we get

$2x^{3}=1+y'$
Now

$y'=\dfrac{-3}{4}$
Hence

$2x^{3}=\dfrac{1}{4}$
Or

$x^{3}=\dfrac{1}{8}$
Hence

$x=0.5$
Substituting in the equation of the curve we get

$\dfrac{1}{32}=\dfrac{1}{2}+y$
Or

$y=\dfrac{-15}{32}$ .
Hence substituting the above points on the equation of the line, we get

$\dfrac{3}{2}+\dfrac{-15}{8}=c$
Or

$c=\dfrac{-3}{8}$
Hence there is only

$1$ value of

$c$ .

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

$x + 2y = 0$

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

Explanation

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

$\displaystyle y-0=-\frac{1}{2}\left ( x+\frac{3\pi }{2} \right )$

The angle at which the curve

$\displaystyle y=ke^{kx}$ intersects the y - axis is

0%
$\displaystyle \tan ^{-1}k^{2}$
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$\displaystyle \cot ^{-1}\left ( k^{2} \right )$
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$\displaystyle \sin ^{-1\left ( \dfrac{1}{\sqrt{1+k^{4}}} \right )}$
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$\displaystyle \sec {-1\left ( \sqrt{1+k^{4}} \right )}$

Explanation

$\displaystyle \dfrac{dy}{dx}=K^{2}e^{kx}$

$\displaystyle \dfrac{dy}{dx}|_{x=0}=K^{2}=\tan \theta$

(where

$\displaystyle \phi$ is angle made by x-axis)

Let

$\displaystyle \phi$ be the gngle made by y-axis

$\displaystyle \tan \theta =\tan \left ( \dfrac{3\pi }{2}-\phi \right )=\cot \phi$

$\displaystyle \cot \phi =K^{2}$

$\displaystyle \phi =\cot ^{-1}\left ( K^{2} \right )$

$\displaystyle \Rightarrow \phi =\sin ^{-1}\left ( \dfrac{1}{\sqrt{1+K^{4}}} \right )$

The abscissa of the point on the curve

$\displaystyle \sqrt{xy}=a+x$ the tangent at which cuts off equal intercepts from the co-ordinate axes is (a > 0)

0%
$\displaystyle \dfrac{a}{\sqrt{2}}$
0%
$\displaystyle- \dfrac{a}{\sqrt{2}}$
0%
$\displaystyle a\sqrt{2}$
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$\displaystyle -a\sqrt{2}$

Explanation

$\displaystyle y=\dfrac{\left (a+x \right )^{2}}{x}\Rightarrow y=\dfrac{a^{2}}{x}+2a+x$

$\displaystyle \dfrac{dy}{dx}=\dfrac{-a^{2}}{x^{2}}+1=-1$ ( for equal intercepts)

$\displaystyle x^{2}=\dfrac{a^{2}}{2}\Rightarrow x=\pm \dfrac{a}{\sqrt{2}}$

Consider the curve

$\displaystyle f(x)=x^{1/3}$ then

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the equation of tangent at (0, 0) is x = 0
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the equation of normal at (0, 0) is y = 0
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normal to the curve does not exist at (0, 0)
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f(x) and its inverse meet at exactly 3 points

Explanation

$f(x)={x}^{1/3}\Rightarrow\displaystyle f'\left ( x \right )=\frac{1}{3x^{2/3}}$

$\displaystyle f'\left ( 0 \right )\rightarrow \infty$ tangent is vertical at

$x = 0$
Equation of tangent at

$(0,0)$ is

$x = 0$
Equation of normal is

$y = 0$

$\displaystyle f\left ( x \right )=f^{-1}\left ( x \right )$

$\implies \displaystyle x^{\frac{1}{3}}=x^{3}\implies x^{9}=x$

$\displaystyle \implies x = 0, 1, -1$

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

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$2x + 2y - 5 = 0$
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$2x + 2y - 9 = 0$
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$y - 2 = 0$
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none

Explanation

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

Equation of a trangent to the curve

$\displaystyle y\cot x=y^{3}\tan x$ at the point where the abscissa is

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

0%
$\displaystyle 4x+2y=\pi +2$
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$\displaystyle 4x-2y=\pi +2$
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x = 0
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y = 0

Explanation

Put

$x=\cfrac{\pi}{4}$ in the given equation,

$y=y^3\Rightarrow y(y^2-1)=0\Rightarrow y=-1,0,1$
So the points are

$(\cfrac{\pi}{4}, -1),(\cfrac{\pi}{4}, 0)$ and

$(\cfrac{\pi}{4}, 1)$
Now given equation may be written as,

$y=\tan^2x.y^3$
Differentiating w.r.t

$x$

$\cfrac{dy}{dx}=\tan^2x.3y^2\cfrac{dy}{dx}+y^3.2\tan x.\sec^2 x$

$\Rightarrow \cfrac{dy}{dx}=\cfrac{2y^3\sec^2 x.\tan x}{1-3\tan^2x.y^2}$
Thus slope of tangents are,

$m_1=\left(\cfrac{dy}{dx}\right)_{(\cfrac{\pi}{4}, -1)}=\cfrac{2(-1).2.1}{1-3.1.1}=2$

$m_2=\left(\cfrac{dy}{dx}\right)_{(\cfrac{\pi}{4},0)}=\cfrac{2(0).2.1}{1-3.1.0}=0$

$m_3=\left(\cfrac{dy}{dx}\right)_{(\cfrac{\pi}{4}, 1)}=\cfrac{2(1).2.1}{1-3.1.1}=-2$
Hence equation of tangents are,

$(y+1)=2(x-\cfrac{\pi}{4})\Rightarrow 4x-2y=\pi +2$

$(y-0)=0(x-\cfrac{\pi}{4})\Rightarrow y=0$

$(y-1)=2(x-\cfrac{\pi}{4})\Rightarrow 4x+2y=\pi +2$

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

$n$ .

0%
$2$
0%
$3$
0%
$4$
0%
any real number

Explanation

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

$n)$

Find the equation of the normal to the curve

$\displaystyle y = \left ( 1+x \right )^{y}+\sin ^{-1}\left ( \sin ^{2}x \right )$ at

$x = 0$

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$x + y - 1 = 0$
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$x + y - 2 = 0$
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$x + y - 3 = 0$
0%
$x + y - 4 = 0$

Explanation

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,

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

$\Rightarrow x+y=1$

Find the equation of normal to the curve

$\displaystyle x^{2}=4y$ passing through the point

$(1, 2)$

0%
$x + y = 3$
0%
$x-y=3$
0%
$2x-y=4$
0%
$2x-3y=1$

Explanation

$x^2=4y$

$\Rightarrow \cfrac{dy}{dx}=\cfrac{x}{2}=m$ (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,

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

$\Rightarrow x+y=3$

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

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

Explanation

Given,

$x= 2t+ t^2 \sin \dfrac {1}{t}$ and

$y=\dfrac {1}{t} \sin t^2$

$t = 0$ 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

$y - 0 = -2(x - 0)\Rightarrow y+2x=0$

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

$a, b, c$ .

0%
$\displaystyle a=-\frac{1}{5}, b=1,c=3$
0%
$\displaystyle a=-\frac{1}{4}, b=-1,c=4$
0%
$\displaystyle a=-\frac{1}{4}, b=0,c=3$
0%
$\displaystyle a=-\frac{1}{3}, b=1,c=-3$

Explanation

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

$\left ( ii \right )$

$a = -\dfrac {1}{4}$ ,

$b =0$

The slope of the normal to the curve

$\displaystyle x=a\left ( \theta -\sin \theta \right ),\: \: y=a\left ( 1-\cos \theta \right )$ at point

$\displaystyle \theta =\dfrac{\pi }2$ is

0%
$0$
0%
$1$
0%
$-1$
0%
$\dfrac 1{\displaystyle \sqrt{2}}$

Explanation

The slope of normal

$\displaystyle x=a\left ( \theta -\sin \theta \right );\: \: \: \: y=a\left ( 1-\cos \theta \right )\: \: \: at\: \: \: \theta =\dfrac{\pi }{2}$

$\displaystyle \left ( \dfrac{dx}{d\theta } \right )_{\theta =\dfrac{\pi }{2}}=a\left ( 1-\cos \theta \right )=a$

$\displaystyle \left ( \dfrac{dy}{d\theta } \right )_{\theta =\dfrac{\pi }{2}}=a\sin \theta =a$

$\therefore \left ( \dfrac{dy}{dx} \right )=1$
Therefore, slope of normal at the given point is,

$m= \left ( -\dfrac{dx}{dy} \right )=-1$
Hence, option 'C' is correct.

The equation of the normal to the curve

$\displaystyle y^{2}=4ax$ at point

$(a, 2a)$ is

0%
$x - y + a = 0$
0%
$x + y - 3a = 0$
0%
$x + 2y + 4a = 0$
0%
$x + y + 4a = 0$

Explanation

$\displaystyle y^{2}=4ax$

$\displaystyle 2y\frac{dy}{dx}=4a\Rightarrow \cfrac{dy}{dx}=\cfrac{2a}{y}$

$\therefore \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( a,2a \right )}=\left ( \frac{2a}{y} \right )_{\left ( a,2a \right )}=1 =m$ (say)

$\displaystyle \therefore$ Slope of normal at the given point is

$=-\cfrac{1}{m}=-1$
Therefore, the equation of normal is,

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

$\Rightarrow x + y = 3a$
Hence, option 'B' is correct.

The tangent at a point

$P$ of a curve meets the

$y-$ axis at

$A$ and the line parallel to

$y-$ axis at

$A$ , and the line parallel to

$y-$ axis through

$P$ meets the

$x-$ axis at

$B$ . If area of

$\Delta OAB$ is constant (

$O$ being the origin). Then the curve is

0%
$cx^2-xy+k=0$
0%
$x^{2}+y^{2}=cx$
0%
$3x^{2}+4y^{2}=k$
0%
$xy-x^{2}y^{2}+kx=0$

A tangent to the hyperbola

$\displaystyle y=\frac { x+9 }{ x+5 }$ passing though the origin is

0%
$x+25y=0$
0%
$5x+y=0$
0%
$5x-y=0$
0%
$x-25y=0$

Explanation

$\displaystyle y=\frac { x+9 }{ x+5 } =1+\frac { 4 }{ x+5 }$

$\displaystyle \frac { dy }{ dx }$ at

$\displaystyle \left( { x }_{ 1 },{ y }_{ 1 } \right) =-\frac { 4 }{ { \left( { x }_{ 1 }+5 \right) }^{ 2 } }$

Equation of tangent is

$\displaystyle y-{ y }_{ 1 }=-\frac { 4 }{ { \left( { x }_{ 1 }+5 \right) }^{ 2 } } \left( x-{ x }_{ 1 } \right) \Rightarrow y-1-\frac { 4 }{ { x }_{ 1 }+5 } =-\frac { 4 }{ { \left( { x }_{ 1 }+5 \right) }^{ 2 } } \left( x-{ x }_{ 1 } \right)$

Since it passes through origin

$(0,0)$

$\displaystyle -1-\frac { 4 }{ { x }_{ 1 }+5 } =-\frac { 4 }{ { \left( { x }_{ 1 }+5 \right) }^{ 2 } } \Rightarrow { \left( { x }_{ 1 }+5 \right) }^{ 2 }+4\left( { x }_{ 1 }+5 \right) +4{ x }_{ 1 }=0$

$\Rightarrow { { x }_{ 1 } }^{ 2 }+18{ x }_{ 1 }+45=0\Rightarrow \left( { x }_{ 1 }+15 \right) \left( { x }_{ 1 }+3 \right) =0\Rightarrow { x }_{ 1 }=15$ or

${ x }_{ 1 }=-3$

So equation of tangent is

$\displaystyle y-1-\frac { 4 }{ \left( -15+5 \right) } =-\frac { 4 }{ { \left( -15+5 \right) }^{ 2 } } \left( x+15 \right)$

$\displaystyle \Rightarrow y-1+\frac { 2 }{ 5 } =-\frac { 1 }{ 25 } \left( x+15 \right) \Rightarrow y-\frac { 3 }{ 5 } =-\frac { x }{ 25 } -\frac { 3 }{ 5 } \Rightarrow x+25y=0$

asymptotes of the graph

0%
$\displaystyle x=\frac{3\pi }{2}$
0%
$\displaystyle x=-\frac{\pi }{2}$
0%
$\displaystyle x=\frac{\pi }{2}$
0%
$\displaystyle x=-\frac{3\pi }{2}$

Explanation

$f'(x)=\dfrac{\cos(x)}{1+\sin(x)}$
Or

$\dfrac{dy}{dx}=\dfrac{\cos x}{1+\sin x}$
For asymptotes,

$\dfrac{dy}{dx}=\infty$
Or

$dx=0$
Or

$\sin(x)=-1$
Or

$x=\dfrac{3\pi}{2},\dfrac{-\pi}{2}$ .

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

0%
$f'\left( c \right) =0$
0%
$f'\left( c \right) >0$
0%
$f'\left( c \right) <0$
0%
none of these

Explanation

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

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

The slope of normal to the curve

$\displaystyle y^{2}=4ax$ at a point

$\displaystyle \left ( at^{2},2at \right )$ is

0%
$\dfrac 1 t$
0%
$t$
0%
$-t$
0%
$-\dfrac 1 t$

Explanation

$y^2=4ax$

$\Rightarrow 2y\cfrac{dy}{dx}=4a\Rightarrow \cfrac{dy}{dx}=\cfrac{2a}{y}$
Therefore, slope of normal to the given curve is,

$=\left(-\cfrac{dx}{dy}\right)_{(at^2,2at)}=-\cfrac{2at}{2a}=-t$
Hence, option 'C' is correct.

The coordinates of the point P on the graph of the function

$\displaystyle y=e^-{\left | x \right |},$ where area of triangle made by tangent and the coordinate axis has the greatest area, is

0%
$\displaystyle \left ( 1,\frac{1}{e} \right )$
0%
$\displaystyle \left ( -1,\frac{1}{e} \right )$
0%
$\displaystyle \left ( e,e^{-e} \right )$
0%
none

Explanation

The curve of f(x) is symmetrical about y axis.
Let the point of contact of the tangent be

$(a,e^{-a})$
Therefore

$\dfrac{dy}{dx}_{(a,e^{-a})}=-e^{-a}$
Hence the equation of the tangent will be

$y-e^{-a}=-e^{-a}(x-a)$

$y-e^{-a}=-e^{-a}x+ae^{-a}$

$y+e^{-a}x=e^{-a}(1+a)$

$\dfrac{y}{e^{-a}(1+a)}+\dfrac{x}{1+a}=1$
Hence the area formed by the tangent and the coordinate axes will be

$=\dfrac{(x-intercept)(y-intercept)}{2}$

$A=\dfrac{e^{-a}(1+a)^{2}}{2}$

Differentiating with respect to a, we get

$\dfrac{dA}{da}=\dfrac{1}{2}[2(1+a)e^{-a}-(1+a)^{2}e^{-a}]$

$=\dfrac{(1+a)e^{-a}}{2}[2-(1+a)]$

$=\dfrac{(1+a)e^{-a}}{2}(1-a)$

$=0$
Hence

$a=\pm1$
Thus we get the points as

$(1,e^{-1})$ and

$(-1,e^{-1})$ .

On the ellipse,

$4x^2\, +\, 9y^2\, =\, 1$ , the points at which the tangents are parallel to the line

$8x = 9y$ are

0%
$\left ( \displaystyle \frac{2}{5},\,\frac{1}{5} \right )$
0%
$\left ( -\displaystyle \frac{2}{5},\,\frac{1}{5} \right )$
0%
$\left ( -\displaystyle \frac{2}{5},\,-\frac{1}{5} \right )$
0%
$\left ( \displaystyle \frac{2}{5},\,-\frac{1}{5} \right )$

Explanation

We have

${ 4x }^{ 2 }+9{ y }^{ 2 }=1$ ...(1)
Differentiating w.r.t x, we get

$\displaystyle 8x+18y\frac { dy }{ dx } =0\Rightarrow \frac { dy }{ dx } =-\frac { 4x }{ 9y }$
The tangent at point

$\left( x,y \right)$ will be parallel to the line

$8x=9y$ if

$\displaystyle \frac { -4x }{ 9y } =\frac { 8 }{ 9 } \Rightarrow x=-2y$
Substituting this in (1), we get

$\displaystyle 4{ \left( -2y \right) }^{ 2 }+{ 9y }^{ 2 }=1\Rightarrow { 25y }^{ 2 }=1\Rightarrow y=\pm \frac { 1 }{ 5 }$
Thus the point where the tangent are parallel to

$8x=9y$ are

$\displaystyle \left( \frac { -2 }{ 5 } ,\frac { 1 }{ 5 } \right)$ and

$\displaystyle \left( \frac { 2 }{ 5 } ,\frac { -1 }{ 5 } \right)$

The slope of the tangent to the curve

$xy + ax - by = 0$ at the point

$(1, 1)$ is

$2$ then values of

$a$ and

$b$ are respectively -

0%
$1, 2$
0%
$2, 1$
0%
$3, 5$
0%
None of these

Explanation

Given curve is

$xy + ax - by = 0$
Slope of tangent at

$(1, 1)$ is

$\displaystyle x.\frac{dy}{dx}+y+a-b.\frac{dy}{dx}=0$

$\displaystyle \left ( \frac{dy}{dx} \right )_{\left ( 1,1 \right )}=-\left [ \frac{\left ( a+y \right )}{x-b} \right ]_{\left ( 1,1 \right )}=-\frac{\left ( a+1 \right )}{1-b}$

$\displaystyle \therefore -\frac{\left ( a+1 \right )}{1-b}=2$
So,

$2b - a = 3$ ..........(1)

$\displaystyle \because (1, 1)$ lies on curve

$xy + ax - by = 0$

$\displaystyle \therefore a-b=-1$ .........(2)
From (1) and (2), we get

$a = 1, b = 2$
Hence, option 'A' is correct.

$\displaystyle x=t^{2}$ and

$y = 2t$ then equation of normal at

$t = 1$ is -

0%
$x + y + 3 = 0$
0%
$x + y + 1 = 0$
0%
$x + y - 1 = 0$
0%
$x + y - 3 = 0$

Explanation

$\displaystyle x=t^{2},\: \: y=2t$

$\Rightarrow \displaystyle \frac{dx}{dt}=2t\: \: \: \: \: \:, \frac{dy}{dt}=2$

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

$\therefore$ Slope of normal is

$=\displaystyle \left ( \frac{-dx}{dy} \right )_{t=1}=-t=-1$
Therefore, equation of normal is,

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

$\Rightarrow x + y = 3$
Hence, option 'D' is correct.

The equation of the tangent to the curve

$\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{2}{\sqrt{a}}$ at point

$(a, a)$ is

0%
$\displaystyle \frac{a}{\sqrt{x}}+\frac{a}{\sqrt{y}}={2}\sqrt a$
0%
$x + y = 2a$
0%
$\displaystyle \sqrt{x}+\sqrt{y}=2\sqrt{a}$
0%
None of these

Explanation

Given curve is,

$\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{2}{\sqrt{a}}$

$\Rightarrow x^{-1/2}+y^{-1/2}=2a^{-1/2}$

Differentiating w.r.t

$x$

$\dfrac{-1}{2}x^{-3/2}-\dfrac{1}{2}y^{-3/2}\cfrac{dy}{dx}=0$

$\Rightarrow \cfrac{dy}{dx}=-\cfrac{y\sqrt{y}}{x\sqrt{x}}$

Thus slope of tangent at

$(a,a)$ is

$m=-1$

The equation of required tangent is given by,

$(y-a)=-1(x-a)\Rightarrow x+y=2a$

Hence, option 'B' is correct.

Consider the curved mirror

$y=f(x)$ passing through

$(0, 6)$ having the property that all light rays emerging from origin, after getting reflected from the mirror becomes parallel to

$x-$ axis, then the equation of curve is

0%
$y^2=4(x-y)$ or $y^2=36(9+x)$
0%
$y^2=4(1-x)$ or $y^2=36(9-x)$
0%
$y^2=4(1+x)$ or $y^2=36(9-x)$
0%
$None\ of\ these$

The point where the tangent line to the curve

$\displaystyle y=e^{2x}$ at

$(0, 1)$ meets

$x$ - axis is -

0%
$(1, 0)$
0%
$(-1, 0)$
0%
$\displaystyle \left ( -\dfrac12,0 \right )$
0%
None of these

Explanation

$\displaystyle y=e^{2x}$

$\displaystyle \frac{dy}{dx}=e^{2x}\times 2$

$\therefore \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( 0,1 \right )}=2$
Thus equation of tangent at point

$(0,1)$ is,

$y - 1 = 2(x - 0)\Rightarrow y=2x+1$
Now substitute

$y = 0$ to get point of intersection of tangent with

$x-axis$

$\displaystyle x=-\frac{1}{2}$
Therefore, required point is

$\displaystyle \left ( -\frac{1}{2},0 \right )$
Hence, option 'C' is correct.

The slope of the tangents to the curve

$y = (x + 1) (x - 3)$ at the points where it crosses x - axis are

0%
$\displaystyle \pm 2$
0%
$\displaystyle \pm 3$
0%
$\displaystyle \pm 4$
0%
None of these

Explanation

$y = (x + 1)(x - 3) ..(1)$

Substitute

$y = 0$ to get the point of intersection of this curve with

$x-axis$

$0 = (x + 1) (x - 3)$

$\Rightarrow x = -1, 3$

So the points are

$(-1, 0)$ and

$(3, 0)$

Now differentiating eq. (1), we get

$\displaystyle \dfrac{dy}{dx}=\left ( x-3 \right )+\left ( x+1 \right )=2x-2$

$\Rightarrow \displaystyle \left ( \dfrac{dy}{dx} \right )=2\left ( x-1 \right )$

Thus slope at

$(-1,0)$ is

$=\left (\dfrac{dy}{dx} \right )_{\left ( -1,0 \right )}=-4$

and at

$(3,0)$ is

$=\displaystyle \left ( \dfrac{dy}{dx} \right )_{\left ( 3,0 \right )}=4$

Hence, option 'C' is correct.

The equation of tangent at the point

$\displaystyle \left ( at^{2},at^{3} \right )$ on the curve

$\displaystyle ay^{2}=x^{3}$ is

0%
$\displaystyle 3tx-2y=at^{3}$
0%
$\displaystyle tx-3y=at^{3}$
0%
$\displaystyle 3tx+2y=at^{3}$
0%
None of these

The equation of the normal to the curve

$\left (\displaystyle x=at^{2} \right ), \left (y=2at \right )$ at '

$t$ ' point is -

0%
$\displaystyle ty=x+at^{2}$
0%
$\displaystyle y+tx-2at-at^{3}=0$
0%
$\displaystyle y=tx-2at-at^{3}$
0%
None of these

Explanation

Given

$\displaystyle x=at^{2},y=2at$

$\Rightarrow \cfrac{dx}{dt}=2at, \cfrac{dy}{dt}=2a$
Thus slope of normal at point 't' is

$=-\cfrac{dx}{dy}=\dfrac{dx/dt}{dy/dt}=-t$

Therefore, the equation of normal at

$'t'$ is given by,

$(y-2at)=-t(x-at^2)$

$\Rightarrow y+tx-2at-at^3=0$

Hence, option 'B' is correct.

The equation of the tangent to the curve is

$\displaystyle y=2\sin x+\sin 2x$ at the point

$\displaystyle x=\dfrac {\pi }3$ is -

0%
$\displaystyle 2y=\sqrt{3}$
0%
$\displaystyle 3y=\sqrt{2}$
0%
$\displaystyle 2y=3\sqrt{3}$
0%
$2y = 3$

Explanation

$\displaystyle y=2\sin x+\sin 2x$

$\displaystyle \frac{dy}{dx}=2\cos x+2\cos 2x$
The slope of tangent at

$x=\cfrac{\pi}{3}$ is,

$m=\displaystyle \left ( \dfrac{dy}{dx} \right )_{x=\dfrac{\pi }{3}}=2\times \dfrac{1}{2}+2\times \left ( -\dfrac{1}{2} \right )=1-1=0$
Now at

$\displaystyle x=\dfrac{\pi }{3};\: \: \: y=2\times \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}$

$\displaystyle y=\dfrac{\sqrt{3}}{2}\times 3$
So the point is

$\displaystyle \left ( \frac{\pi }{3},\frac{3\sqrt{3}}{2} \right )$
Therefore, required tangent is,

$\displaystyle \left ( y-\dfrac{3\sqrt{3}}{2} \right )=0$

$\Rightarrow y=\dfrac{3\sqrt{3}}{2}$

$\Rightarrow 2y=3\sqrt{3}$
Hence, option 'C' is correct.

The equation of the tangent to the curve

$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$ at the point

$\displaystyle \left ( x_{1},y_{1} \right )$ is -

0%
$\displaystyle \frac{x}{\sqrt{x_{1}}}+\frac{y}{\sqrt{y_{1}}}=\frac{1}{\sqrt{a}}$
0%
$\displaystyle \frac{x}{\sqrt{x_{1}}}+\frac{y}{\sqrt{y_{1}}}=\sqrt{a}$
0%
$\displaystyle x\sqrt{x_{1}}+y\sqrt{y_{1}}=\sqrt{a}$
0%
None of these

Explanation

$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$ at point

$\displaystyle \left ( x_{1},y_{1} \right )$

$\displaystyle \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$

$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{y}{x}}\: \: \Rightarrow \: \: \left ( \frac{dy}{dx} \right )_{\left ( x_{1},y_{1} \right )}=-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}$
Thus equation of tangent is given by,

$\displaystyle y-y_{1}=-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}\left ( x-x_{1} \right )$

$\displaystyle y\sqrt{x_{1}}-y_{1}\sqrt{x_{1}}=-x\sqrt{y_{1}}+x_{1}\sqrt{y_{1}}$

$\displaystyle x\sqrt{y_{1}}+y\sqrt{x_{1}}=\sqrt{x_{1}}\sqrt{y_{1}}\left ( \sqrt{a} \right )$

$\displaystyle \frac{x}{\sqrt{x_{1}}}+\frac{y}{\sqrt{y_{1}}}=\sqrt{a}$
Hence, option 'B' is correct.

The coordinates of the point on the curve

$\displaystyle y=x^{2}+3x+4$ the tangent at which passes through the origin are -

0%
$(-2, 2), (2, 14)$
0%
$(1, -1), (3, 4)$
0%
$(2, 14), (2, 2)$
0%
$(1, 2), (14, 3)$

Explanation

Let the point is

$P(a,b)$
Now the given curve is

$y=x^2+3x+4$
Differentiating w.r.t

$x$

$\cfrac{dy}{dx}=2x+3$
Thus slope of tangent at P is

$=\left(\cfrac{dy}{dx}\right)_{(a,b)}=2a+3$
Hence equation of tangent at P is given by,

$(y-b)=(2a+3)(x-a)$
But given that tangent passes through origin

$\Rightarrow (0-b)=(2a+3)(0-a)\Rightarrow 2a^2+3a=b ..(1)$
Also the point P lies on the given curve,

$b=a^2+3a+4 ...(2)$
Solving (1) and (2) we get the required point P coordinate
which are

$(-2,2) \mbox{or} (2,14)$
Hence, option 'A' is correct.

At what point the tangent line to the curve

$\displaystyle y=\cos \left ( x+y \right ),\left ( -2\pi \leq x\leq 2\pi \right )$ is parallel to

$x + 2y = 0$

0%
$\displaystyle \left ( \dfrac {\pi }2, 0 \right )$
0%
$\displaystyle \left ( -\dfrac {\pi }2, 0 \right )$
0%
$\displaystyle \left (\dfrac{ 3\pi }2, 0 \right )$
0%
$\displaystyle \left (-\dfrac{ 3\pi }2,\dfrac { \pi }2 \right )$

Explanation

$\displaystyle y=\cos \left ( x+y \right )$

$\displaystyle \left ( -2\pi \leq x\leq 2\pi \right )$

$\Rightarrow \displaystyle \frac{dy}{dx}=-\sin \left ( x+y \right )\left ( 1+\frac{dy}{dx} \right )$

$\Rightarrow \displaystyle \frac{dy}{dx}\left ( 1+\sin \left ( x+y \right ) \right )=-\sin \left ( x+y \right )$

$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{\sin \left ( x+y \right )}{1+\sin \left ( x+y \right )}$
Given tangent is parallel to the line,

$x + 2y = 0$

$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{\sin \left ( x+y \right )}{1+\sin \left ( x+y \right )}=-\frac{1}{2}$

$\displaystyle 2\sin \left ( x+y \right )=1+\sin \left ( x+y \right )$

$\displaystyle \sin \left ( x+y \right )=1\Rightarrow \cos(x+y)=0\Rightarrow x+y=\cfrac{\pi}{2}$
Also the point lies on the given curve,

$y=0\Rightarrow x= \dfrac{\pi }{2}$
Thus the point is,

$\displaystyle \left ( \frac{\pi }{2},0 \right )$
Hence, option 'A' is correct.

The point at which the tangent to the curve

$\displaystyle y=x^{3}+5$ is perpendicular to the line

$x + 3y = 2$ are

0%
$(6, 1), (-1, 4)$
0%
$(6, 1) (4, -1)$
0%
$(1, 6), (1, 4)$
0%
$(1, 6), (-1, 4)$

Explanation

Let point be

$\displaystyle \left ( x_{1},y_{1} \right )$

$y = \displaystyle x^{3}+5$

$\Rightarrow \displaystyle \dfrac{dy}{dx}=3x^{2}$
Now, the slope of tangent at this point is

$=\displaystyle \left ( \frac{dy}{dx} \right )_{x_{1}y_{1}}=3x_{1}^{2}$
It is

$\displaystyle \perp$ to

$x + 3y - 2 = 0$
So

$\displaystyle 3x_{1}^{2}\times -\frac{1}{3}=-1\Rightarrow x_1=\pm 1$
Thus,
at

$\displaystyle x_{1}=1$ ,

$\Rightarrow$

$\displaystyle y_{1}=6$
and at

$\displaystyle x_{1}=-1$

$\Rightarrow$

$\displaystyle y_{1}=4$
Thus, the points are

$(1, 6)$ and

$(-1, 4)$
Hence, option 'D' is correct.

A normal

$P(x,y)$ on a curve meets the

$X-$ axis at

$Q$ and

$N$ is the ordinate at

$P$ .

$NQ=\dfrac {x(1+y^2)}{1+x^2}$ .
Then the equation of curves passing through

$(3,1)$ is

0%
$5(1+y^2)=(1+x^2)$
0%
$5(1+y^2)=5(1+x^2)$
0%
$5(1+x^2)=(1+y^2)$
0%
None of these

At what point of the curve

$\displaystyle y=2x^{2}-x+1$ tangent is parallel to

$y = 3x + 4$

0%
$(0, 1)$
0%
$(1, 2)$
0%
$(-1, 4)$
0%
$(2, 7)$

Explanation

Let the point is

$P(a,b)$
Now the given curve is

$y=2x^2-x+1$
Differentiating w.r.t

$x$

$\cfrac{dy}{dx}=4x-1$
Thus slope of tangent at P is

$=\left(\cfrac{dy}{dx}\right)_{(a,b)}=4a-1$
But given the tangent is parallel to line

$y=3x+4$

$\Rightarrow 4a-1=3\Rightarrow a=1$
Also the point P lies on the given curve,

$b=2a^2-a+1=2$
Therefore, the point P is

$(1,2)$
Hence, option 'B' is correct.

Tangents are drawn from origin to the curve

$\displaystyle y=\sin x$ then point of contect lies on -

0%
$\displaystyle x^{2}=y^{2}$
0%
$\displaystyle x^{2}y^{2}=0$
0%
$\displaystyle x^{2}y^{2}=x^{2}-y^{2}$
0%
None of these

Explanation

$y=\sin x$

$\Rightarrow \cfrac{dy}{dx}=\cos x$
Let the point of contact of tangent passing through origin to the given curve is

$P(h,k)$
So equation of tangent to this curve passes through

$(0,0)$ is,

$y=(\cos h) x$
For point of contact this line will pass through point P,

$\Rightarrow k=h\cos h\Rightarrow \cos h=\cfrac{k}{h} ...(1)$
Also point P lies on the given curve,

$\sin h=k ..(1)$
Now using

$(1)^2+(2)^2=1$

$\cfrac{k^2}{h^2}+k^2=1\Rightarrow k^2h^2=h^2-k^2$
Thus locus of

$P(h,k)$ is,

$x^2y^2=x^2-y^2$
Hence, option 'C' is correct.

A tangent to the curve

$\displaystyle y=x^{2}+3x$ passes through a point

$(0, -9)$ if it is drawn at the point -

0%
$(-3, 0)$
0%
$(1, 4)$
0%
$(0, 0)$
0%
$(-4, 4)$

Explanation

Let the point is

$\displaystyle P\left ( x_{1},y_{1} \right )$ on the curve
So,

$\displaystyle \frac{dy}{dx}=2x+3$

$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( x_{1},y_{1} \right )}=2x_{1}+3$
Therefore, the tangent is

$\displaystyle \frac{y-y_{1}}{x-x_{1}}=2x_{1}+3$

$\displaystyle \left ( y-y_{1} \right )=\left ( x-x_{1} \right )\left ( 2x_{1}+3 \right )$

$\displaystyle y-y_{1}=2xx_{1}+3x-2x_{1}^{2}-3x_{1}$

$\displaystyle x\left ( 2x_{1}+3 \right )-y+y_{1}-2x_{1}^{2}-3x_{1}=0$
It passes

$(0, -9)$ then

$\displaystyle 0\times \left ( 2x_{1}+3 \right )+9+y_{1}-2x_{1}^{2}-3x_{1}=0$

$\displaystyle \left ( x_{1},y_{1} \right )$ also lies on the curve

$=\displaystyle y_{1}-2x_{1}^{2}-3x_{1}+9=0$ ..........(1)
So

$\displaystyle y_{1}=x_{1}^{2}+3x_{1}$ ..........(2)
From eq (1) and (2), we get

$\displaystyle y_{1}-2x_{1}^{2}-3x_{1}+9-y_{1}+x_{1}^{2}+3x_{1}=0$

$\displaystyle -x_{1}^{2}+9=0$

$\displaystyle x_{1}^{2}=9$

$\displaystyle x_{1}=\pm 3$
when

$\displaystyle x_{1}= 3$

$\Rightarrow$

$\displaystyle y_{1}= 9+9=18$

$\displaystyle x_{1}= -3$

$\Rightarrow$

$\displaystyle y_{1}= 9-9=0$
Thus the required points are

$(-3, 0)$ and

$(3, 18)$
Hence, option 'A' is correct.

The equation of the normal to the curve

$\displaystyle y^{2}=x^{3}$ at the point whose abscissa is

$8$ is -

0%
$\displaystyle x\pm \sqrt{2}y=104$
0%
$\displaystyle x\pm 3\sqrt{2}y=104$
0%
$\displaystyle 3\sqrt{2}x\pm y=104$
0%
None of these

Explanation

Solving the curve at

$x=8=2^3$

$\displaystyle y^{2}=x^{3}=2^{9}$

$\Rightarrow \displaystyle y=\pm 2^{4}\sqrt{2}=\pm 16\sqrt{2}$
So the points are

$\displaystyle \left ( 8,\pm 16\sqrt{2} \right )$
Now the curve is,

$\displaystyle y^{2}=x^{3}$

$\Rightarrow \displaystyle 2y\times \frac{dy}{dx}=3x^{2}$

$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )=\frac{3x^{2}}{2y}$

$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( 8,\pm 16\sqrt{2} \right )}=\frac{3\times 64}{\pm 2\times 16\sqrt{2}}=\pm 3\sqrt{2}$
Thus the slope of normal is

$=\displaystyle \left ( \frac{dy}{dx} \right )=\mp \frac{1}{3\sqrt{2}}$
Therefore, equation of normal is,

$\displaystyle \left ( y\mp 16\sqrt{2} \right )=\mp \frac{1}{3\sqrt{2}}\left ( x-8 \right )$

$\Rightarrow \displaystyle \mp 3\sqrt{2}y\pm 96=(x-8)$

$\Rightarrow \displaystyle x\pm 3\sqrt{2}y=104$
Hence, option 'B' is correct.

The normal to the curve

$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$ is perpendicular to

$x$ axis at the point

0%
$(0, a)$
0%
$(a, 0)$
0%
$(\dfrac a 4, \dfrac a 4)$
0%
No where

Explanation

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 normal at P is

$=\displaystyle -( \frac{dx}{dy} )_{( x_{1},y_{1} )}=-\sqrt{\frac{x_{1}}{y_{1}}}$
If normal

$\displaystyle \perp$ to x axis then

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

$\displaystyle \therefore y_1 = a\Rightarrow x_1 = 0$
Therefore, required point is

$(a, 0)$
Hence, option 'B' is correct.

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 6 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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