MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 1

The equation of normal to the curve

$3x^2-y^2 =8$ which is parallel to the line

$x+ 3y=8$ is

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$3x-y=8$
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$3x+y+8=0$
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$x+3y \pm 8 = 0$
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$x+ 3y=0$

Let N be the set of positive integers. For all

$n \in N$ , let

$f_n = (n + 1)^{1/3} - n^{1/3}$ and

$A = \left\{n \in N : f_{n +1} < \dfrac{1}{3(n + 1)^{2/3}} < f_n \right\}$
Then

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A = N
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A is a finite set
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the complement of A in N is nonempty, but finite
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A and its complement in N are both infinite

for

$f(x)=\displaystyle \int _{ 0 }^{ x }2 \left| t \right| dt$ , the tangent lines which are parallel to the bisector of the first co-ordinate angle is

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$y=x-\frac{1}{4}$
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$y=x+\frac{1}{4}$
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$y=x-\frac{3}{2}$
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$y=x+\frac{3}{2}$

At any point on the curve

$2x^{2}y^{2}-x^{4}=c$ , the mean proportional between the abscissa and the difference between the abscissa and the sub-normal drawn to the curve at the same point is equal to

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ordinate
0%
radius vector
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x-intercept of tangent
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sub-tangent

Given

$g(x)= \dfrac{x+2}{x-1}$ and the line 3x + y -10 =0, then the line is

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tangent to g(x)
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normal to g(x)
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chord of g(x)
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none of these

The angle formed bt the positive y-axis and the tangent to

$y = x^{2}+4x-17$ at

$(5/2, -3/4)$ is

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$\tan ^{-1}(9)$
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$\dfrac{\pi }{2}\tan ^{-1}(9)$
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$\dfrac{\pi }{2}\tan ^{-1}(9)$
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None of these

The abscissa of a point on the curve

$xy = (a+y)^{2}$ , the normal which cuts off numerically equal intercept from the coordinate axes, is

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

The co-ordinates of the point (s) on the graph of the function

$f(x)= \dfrac{x^{3}}{3} - \dfrac{5x^{2}}{2} + 7x - 4$ , where the tangent drawn cuts off intercept from the co-ordinate axes which

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(2, 8/3)
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(3, 7/2)
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(1, 5/6)
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None of these

The equation of the curve

$y = be^{-x/a}$ at the point where it crosses the y-axis is

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$\dfrac{x}{a}-\dfrac{y}{b}=1$
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$ax = by = 1$
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$ax-by=1$
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$\dfrac{x}{a}+\dfrac{y}{b}=1$

A curve is represented by the equations

$x=sec^{2}t$ and

$y=\cot t,$ where t is a parameter. If the tangent at the point P on the curve, where

$t=\pi /4$ , meets the curve again at the point Q, then

$\left | PQ \right |$ is equal to

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$\dfrac{5\sqrt{3}}{2}$
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$\dfrac{5\sqrt{5}}{2}$
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$\dfrac{2\sqrt{5}}{3}$
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$\dfrac{3\sqrt{5}}{2}$

The angle between the tangents at ant point P and the line joining P to the original, where P is a point on the curve in

$(x^{2}+y^{2})=c \tan ^{-1}\dfrac{y}{x},c$ is a constnt, is

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independent of x
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independent of y
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independent of x but dependent on y
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independent of y but dependent on x

Let f be a continuous, differetiable and bijective function. If the tangent to y= f (x) at x = a is also the normal to y = f (x) at x = b then there exists at least one

$c \epsilon (a, b)$ such that

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f'(c) = 0
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$f (c)> 0$
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$f (c)< 0$
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None of these

The slope of the tangent to the curve

$y=\int _{ 0 }^{ x }{ \frac { dt }{ 1+{ t }^{ 3 } } }$ at the point where

$x=1$ is

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$\frac { 1 }{ 4 }$
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$\frac { 1 }{ 3 }$
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$\frac { 1 }{ 2 }$
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$1$

Explanation

$\frac { dy }{ dx } =\frac { 1 }{ 1+{ x }^{ 3 } }$

$m=\left( \frac { dy }{ dx } \right) x=1=\frac { 1 }{ 1+1 } =\frac { 1 }{ 2 }$

$\left | f(x_{1})-f(x_{2}) \right |< (x_{1}-x_{2})^{2}$ for all

$x_{1}$

$x_{2}$

$\in$ R. Find the equation of tangent to the curve y = f(x) at the point (1, 2).

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$x=2$
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$y=2$
0%
$y=1$
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$x=1$

Explanation

$\left| f\left( { x }_{ 1 } \right) -f\left( { x }_{ 2 } \right) \right| \le { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 },\forall { x }_{ 1 },{ x }_{ 2 }\in R$

$\displaystyle\Rightarrow \left| f\left( { x }_{ 1 } \right) -f\left( { x }_{ 2 } \right) \right| \le { \left| { x }_{ 1 }-{ x }_{ 2 } \right| }^{ 2 }$

$\left( { x }^{ 2 }={ \left| x \right| }^{ 2 } \right)$

$\displaystyle \therefore \left| \frac { f\left( { x }_{ 1 } \right) -f\left( { x }_{ 2 } \right) }{ { x }_{ 1 }-{ x }_{ 2 } } \right| \le { \left| { x }_{ 1 }-{ x }_{ 2 } \right| }\Rightarrow \lim _{ { x }_{ 1 }\rightarrow { x }_{ 2 } }{ \left| \frac { f\left( { x }_{ 1 } \right) -f\left( { x }_{ 2 } \right) }{ { x }_{ 1 }-{ x }_{ 2 } } \right| } \le \lim _{ { x }_{ 1 }\rightarrow { x }_{ 2 } }{ \left| { x }_{ 1 }-{ x }_{ 2 } \right| }$

$\displaystyle\Rightarrow \left| f'\left( { x }_{ 1 } \right) \right| \le 0,\forall { x }_{ 1 }\in R$

$\therefore \left| f'\left( x \right) \right| \le0$ , which showsw

$\left| f'\left( x \right) \right| =0$ (as modulus is non negative or

$\left| f'\left( x \right) \right| \ge 0$ )

$\therefore f'\left( x \right)=0$ or

$f\left( x \right)$ is a constant function.

$\Rightarrow$ Equation of tangent at

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

$\displaystyle \frac { y-2 }{ x-1 } =f'\left( x \right)$ or

$y-2=0$ [

$\because$ as

$f'(x)=0$ ]

$\Rightarrow y-2=0$ is required equation of tangent.

The distance, from the origin, of the normal to the curve,

$x = 2\cos t + 2t\sin t, y = 2\sin t - 2t\cos t$ at

$t = \dfrac {\pi}{4}$ , is

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$2$
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$4$
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$\sqrt {2}$
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$2\sqrt {2}$

Explanation

$x = 2 \cos t + 2t \sin t$
$\cfrac { dx }{ dt } =2t\cos t+2\sin t-2\sin t=2t\cos t$
$y = 2 \sin t - 2t \cos t$
$\cfrac { dy }{ dt } =2\cos t+2t\sin t-2\cos t=2t\sin t$
$\cfrac { dy }{ dx } =\cfrac { 2t\sin t }{ 2t\cos t } (t=\dfrac{\pi }4)$
$\cfrac { dy }{ dx } =1$

$\cfrac { dx }{ dy } =-1$ for slope of normal.
$x\left( \cfrac { \pi }{ 4 } \right) =2\cos \left( \cfrac { \pi }{ 4 } \right) +2.\cfrac { \pi }{ 4 } \sin \left( \cfrac { \pi }{ 4 } \right) =\sqrt { 2 } \left( 1+\cfrac { \pi }{ 4 } \right)$
$y\left( \cfrac { \pi }{ 4 } \right) =2\sin \left( \cfrac { \pi }{ 4 } \right) -2.\cfrac { \pi }{ 4 } \cos \left( \cfrac { \pi }{ 4 } \right) =\sqrt { 2 } \left( 1-\cfrac { \pi }{ 4 } \right)$
Equation of normal with given point
$x+y-2\sqrt { 2 } =0$
$d=\cfrac { (2\sqrt { 2) } }{ \sqrt { 2 } } =2$

The equation of a normal to the curve,

$\sin y = x \sin \displaystyle \left ( \frac{\pi}{3} + y \right )$ at

$x = 0$ , is

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$2x - \sqrt 3 y = 0$
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$2y - \sqrt 3 x = 0$
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$2y + \sqrt 3 x = 0$
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$2x + \sqrt 3 y = 0$

Explanation

$\sin y = x \sin \displaystyle \left ( \frac{\pi}{3} + y \right ) ..(1)$

Putting

$x =0\Rightarrow \sin y = 0\Rightarrow y = 0$

Now differentiating eqn (1) w.r.t

$x$

$\displaystyle \cos y\frac{dy}{dx}=\sin \displaystyle \left ( \frac{\pi}{3} + y \right )+x\cos \displaystyle \left ( \frac{\pi}{3} + y \right )\frac{dy}{dx}$

Putting

$x=0$ and

$y=0$ to get the slope of tangent at these points,

$\cfrac{dy}{dx} = \sin \displaystyle \left ( \frac{\pi}{3} \right )=\cfrac{\sqrt{3}}{2}=m$ (say)

Thus slope of normal at

$(0,0)$ is

$=-\cfrac{1}{m}=-\cfrac{2}{\sqrt{3}}$

Hence required equation of normal is,

$y-0=-\dfrac{2}{\sqrt3}(x-0)$

$2x + \sqrt 3 y = 0$

The tangent at the point

$(2, -2)$ to the curve,

$x^2y^2-2x=4(1-y)$ does not pass through the point.

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$(8, 5)$
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$\left(4, \displaystyle\frac{1}{3}\right)$
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$(-2, -7)$
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$(-4, -9)$

Explanation

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

$2xy^2+2y\cdot x^2\cdot\displaystyle\frac{dy}{dx}-2=-4\cdot\frac{dy}{dx}$

$\displaystyle\frac{dy}{dx}(2y\cdot x^2+4)=2-2x\cdot y^2$

$\left.\displaystyle\frac{dy}{dx}\right|_{2, -2}=\displaystyle\frac{2-2\times 2\times 4}{2(-2)\times 4+4}=\frac{+14}{+12}=\frac{7}{6}$
eq of tangent :

$(y+2)=\displaystyle\frac{7}{6}(x-2)\Rightarrow 7x-6y=26$ .

$(-2,-7)$ does not satisfy above eq.

If tangent to the curve

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

$x$ -axis, then its point of contact is:

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$(a, a)$
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$(0, a)$
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$(0, 0)$
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$(a, 0)$

Explanation

We have,

$x=at^2\Rightarrow \dfrac{dx}{dt}=2at$

And

$y=2at\Rightarrow \dfrac{dy}{dt}=2a$

$\therefore \dfrac{dy}{dx}=\dfrac{1}{t}=\infty$ , for tangent to the curve perpendicular to

$x-$ axis

$\Rightarrow t = 0$

so the point of contact is

$(at^2,2at)=(0,0)$

The normal to the curve

$x = a (\cos\theta +\theta \sin \theta ), y = a (\sin \theta -\theta \cos\theta )$ at any point

$\theta$ is such that

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it passes through the origin
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it makes angle $\frac{\pi }{2}+\theta$ with the x-axis
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it passes through $\left ( a\frac{\pi }{2} ,-a\right )$
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it is at a constant distance from the origin

Explanation

Clearly

$\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\tan\theta =$ slope of normal

$=-\cot\theta$
Equation of normal at

$\theta$ ' is

$\mathrm{y}-\mathrm{a}(\sin\theta -\theta \cos\theta)=-\cot\theta(\mathrm{x}-\mathrm{a}(\cos\theta +\theta \sin\theta)$

$=$

$y\sin{\theta}-a\sin^{2}\theta +a\theta \cos\theta \sin\theta =-\mathrm{x}\cos\theta +a\cos^{2}\theta +a\theta \sin\theta \cos\theta$

$= x\cos{\theta} + y\sin{\theta} =a$
Clearly this is an equation of straight line which is at a constant distance

$a$ from origin.

The normal to the curve,

$x^2+2xy-3y^2=0$ , at (1, 1)

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does not meet the curve again
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meets the curve again in the second quadrant
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meets the curve again in the third quadrant
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meets the curve again in the fourth quadrant

Explanation

$x^2 +2xy -3y^2 =0$

This is the equation of a pair of straight lines.

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

$2x+2y+2xy'-6yy'=0$

$\Rightarrow \dfrac{x+y}{3y-x}=y'$

$(1,1)$

the slope of the tangent is

$y'=1.$

The slope of the normal is

$-1$ .

Hence,

The normal has the equation :

$x+y = 2$

We need to find its intersection with

$x+3y =0$

On solving we get,

$2-y +3y =0 \Rightarrow 2y=-2$

$y =-1 , x=3$

Hence, it meets it in the fourth quadrant.

The equation of the tangent to the curve

$\displaystyle \mathrm{y}=\mathrm{x}+\frac{4}{\mathrm{x}^{2}}$ , that is parallel to the

$x-axis$ , is

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$\mathrm{y}=1$
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$\mathrm{y}=2$
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$\mathrm{y}=3$
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$\mathrm{y}=0$

Explanation

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

$\Rightarrow \displaystyle \frac{dy}{dx}=1-\frac{8}{x^{3}}$

Since, tangent is parallel to x-axis i.e.

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

$0=1-\dfrac{8}{x^3}$

$x^3=8$

$\Rightarrow x=2$

substitute in the equation of the curve

$y=2+\dfrac{4}{2^2}=2+\dfrac{4}{4}=2+1$

$\Rightarrow y=3$

.
Equation of tangent is given as

${y}-3=0({x}-2)$ [slope-point form]

$\Rightarrow y-3=0$

Hence, option 'C' is correct.

The tangent to the curve

$\mathrm{y}=\mathrm{e}^{\mathrm{x}}$ drawn at the point

$(\mathrm{c}, \mathrm{e}^{\mathrm{c}})$ intersects the line joining the points

$(\mathrm{c}- \mathrm{l}, \mathrm{e}^{c-1})$ and

$(\mathrm{c}+1, \mathrm{e}^{c-1})$

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on the left of $x = c$
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on the right of $x = c$
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at no point
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at all points

Explanation

$\dfrac{dy}{dx}_{x=c}$

$=e^{x}|_{x=c}$

$=e^{c}$

Therefore equation of the tangent will be

$\dfrac{y-e^{c}}{x-c}=e^{c}$

$y-e^{c}=e^{c}x-ce^{c}$

$y-e^{c}x=e^{c}(1-c)$ ...(i)

Now the slope of the line joining

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

$(c+1,e^{c+1})$

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

$=\dfrac{e^{c+1}-e^{c-1}}{2}$

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

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

Hence

$y-(e^{c-1})=(x-(c-1))\dfrac{e^{c-1}(e^{2}-1)}{2}$ ...(ii)

From i, we know

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

Substituting in the given equation, we get

$2(e^{c}x+e^{c}-e^{c}c-e^{c-1})=x(e^{c+1}-e^{c-1})-c(e^{c+1}-e^{c-1})+(e^{c+1}-e^{c-1})$

$2e^{c}-2e^{c}c-2e^{c-1}=x(e^{c+1}-e^{c-1}-2e^{c})-ce^{c+1}+ce^{c-1}+e^{c+1}-e^{c-1}$

$x(e^{c+1}-e^{c-1}-2e^{c})+(1-c)(e^{c+1}-e^{c-1})=2e^{c}(1-c)-2e^{c-1}$

$x(e^{c+1}-e^{c-1}-2e^{c})=(1-c)(2e^{c}-e^{c+1}+e^{c-1})-2e^{c-1}$

$x(e^{c+1}-e^{c-1}-2e^{c})=-(1-c)(e^{c+1}-e^{c-1}-2e^{c})-2e^{c-1}$

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

$=c-1-\dfrac{2e^{c-1}}{e^{c+1}-e^{c-1}-2e^{c}}$

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

$=c-k$ ...(i)

Where k is constant.

Now

$c-k<c$

Hence the intersection is towards the left of

$x=c$ .

If the tangent to the conic,

$y - 6 = x^2$ at (2, 10) touches the circle,

$x^2 + y^2 + 8x - 2y = k$ (for some fixed k) at a point

$(\alpha, \beta)$ ; then

$(\alpha, \beta)$ is;

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$\displaystyle \left( -\frac{4}{17}, \frac{1}{17} \right)$
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$\displaystyle \left( -\frac{7}{17}, \frac{6}{17} \right)$
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$\displaystyle \left( -\frac{6}{17}, \frac{10}{17} \right)$
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$\displaystyle \left( -\frac{8}{17}, \frac{2}{17} \right)$

Explanation

Given equation of conic is

$y - 6 = x^2$

$\dfrac{dy}{dx}=2x$
Slope of tangent at

$(2,10)$ is

$4.$

Equation of tangent to conic is

$y-10=4(x-2)$

$\Rightarrow y-10=4x-8$

$\Rightarrow y=4x+2$ .....(1)

Given equation of circle is

$x^{ 2 }+y^{ 2 }+8x-2y=k$

$\Rightarrow (x+4)^2+(y-1)^2=k+17$

Given

$(\alpha, \beta)$ is a point of tangency for fixed

$k$
Radius = Length of tangent from center

$(-4,1)$

$\Rightarrow \sqrt{k+17}=\dfrac{15}{\sqrt{17}}$

$\Rightarrow k+17=\dfrac{225}{17}$

So, equation of circle at

$(\alpha, \beta)$ is

$(\alpha+4)^2+(\beta-1)^2=\dfrac{225}{17}$

Now, eqn (1) is tangent to the circle for fixed

$k$ at

$(\alpha,\beta)$

$\Rightarrow (\alpha+4)^2+(4\alpha+1)^2=\dfrac{225}{17}$

$\Rightarrow 17\alpha^2+16\alpha+17=\dfrac{225}{17}$

$\Rightarrow 289\alpha^2+272\alpha+64=0$

$\Rightarrow \alpha =\dfrac{-8}{17}$ (Here

$D=0$ )

So, by (1),

$\beta =\dfrac{2}{17}$

Let

$C$ be a curve given by

$y(x) = 1 + \sqrt {4x - 3}, x > \dfrac {3}{4}$ . If

$P$ is a point on

$C$ , such that the tangent at

$P$ has slope

$\dfrac {2}{3}$ , then a point through which the normal at

$P$ passes, is:

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$(1, 7)$
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$(3, -4)$
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$(4, -3)$
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$(2, 3)$

Explanation

$\dfrac {dy}{dx} = \dfrac {1}{2\sqrt {4x - 3}} \times 4 = \dfrac {2}{3}$

$\Rightarrow 4x - 3 = 9$

$\Rightarrow x = 3$
So,

$y = 4$
Equation of normal at

$P(3, 4)$ is

$y - 4 = -\dfrac {3}{2} (x - 3)$
i.e.

$2y - 8 = -3x + 9$

$\Rightarrow 3x + 2y - 17 = 0$

The equation of the normal to the circle

$\displaystyle { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ at point

$(x', y')$ will be:

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$\displaystyle x'y-xy'=0$
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$\displaystyle xx'-yy'=0$
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$\displaystyle x'y+xy=0$
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$\displaystyle xx'+yy'=0$

Explanation

Given,

$\displaystyle { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$
On differentiating w.r.t.

$x$ , we get

$\displaystyle 2x+2y\dfrac { dy }{ dx } =0$

$\displaystyle \Rightarrow \quad \frac { dy }{ dx } =-\frac { x }{ y }$

$\displaystyle \Rightarrow \quad { \left( \frac { dy }{ dx } \right) }_{ \left( x',y' \right) }=-\frac { x' }{ y' }$
Slope of tangent is

$-\dfrac { x' }{ y' }$
So, Slope of normal will be

$-\dfrac{ dx }{ dy }$ i.e.

$\dfrac{y'}{x'}$

$\displaystyle \therefore$ Equation of normal is

$\displaystyle y-y'=\frac { y' }{ x' } \left( x-x' \right)$

$\displaystyle \Rightarrow \quad x'y-y'x'=xy'-y'x'$

$\displaystyle \Rightarrow \quad x'y-xy'=0$

What is the

$x$ -coordinate of the point on the curve

$f(x) = \sqrt {x}(7x - 6)$ , where the tangent is parallel to

$x$ -axis?

0%
$-\dfrac {1}{3}$
0%
$\dfrac {2}{7}$
0%
$\dfrac {6}{7}$
0%
$\dfrac {1}{2}$

Explanation

Given,

$f(x) = \sqrt {x}(7x - 6) = 7x^{3/2} - 6x^{1/2}$
Thus

$f'(x) = 7\times \dfrac {3}{2}x^{1/2} - 6\times \dfrac {1}{2} x^{-1/2}$
When tangent is parallel to

$x$ axis

$f'(x) = 0$ .
Therefore,

$\dfrac {21}{2}x^{1/2} - 3x^{-1/2} = 0$

$\Rightarrow \dfrac {21}{2}\sqrt {x} = \dfrac {3}{\sqrt {x}}$

$\Rightarrow 7x = 2$

$\Rightarrow x = \dfrac {2}{7}$

Determine the equation of tangent at vertex of the parabola

$\displaystyle (x+4)^{2}=-4(y-2)$ .

0%
$y=0$
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$y=2$
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$x=0$
0%
$x+4=0$

Explanation

Given hyperbola is,

$(x+4)^2=-4(y-2)$

Differentiating w.r.t

$x$

$2(x+4)=-4\cfrac{dy}{dx}\Rightarrow \cfrac{dy}{dx}=-\cfrac{x+4}{2}$

Clearly the vertex of the parabola is,

$(-4,2)$

Hence slope of the tangent at the vertex is,

$m=\left(\cfrac{dy}{dx}\right)_{(-4,2)}=-\cfrac{-4+4}{2}=0$

Ergo required tangent is,

$(y-2)=0(x+4)=0$

$\Rightarrow y=2$

What is/are the tangents to

$\displaystyle y=(x^{3}-1)(x-2)$ at the points where the curve cuts the x-axis

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$y+3x=3$
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$y+2x=3$
0%
$y-7x+14=0$
0%
$y -5x-14=0$

Explanation

Given curve is,

$\displaystyle y=(x^{3}-1)(x-2)$
For intersection of this curve with x-axis put

$y=0$

$\Rightarrow (x^{3}-1)(x-2)=0\Rightarrow x = 1,2$
Thus points are

$(1,0)$ and

$(2,0)$
Now

$\cfrac{dy}{dx}=(x^3-1)+(x-2).3x^2=4x^3-6x^2-1$
So slope of the tangent are

$m_1=\left(\cfrac{dy}{dx}\right)_{x=1}=-3$ and

$m_2=\left(\cfrac{dy}{dx}\right)_{x=2}=7$
Hence required tangents are,

$y+3x=3,\, y-7x+14=0$

Normal to the parabola

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

$m$ is the slope of the normal is

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$\displaystyle y=mx+2am-am^{3}$
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$\displaystyle y=mx-2am-am^{3}$
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$\displaystyle y=mx-2am+am^{3}$
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none of these

Explanation

Given,

$y^2=4ax$

$\Rightarrow 2ydy=4a$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{y}$
Thus slope of normal to the given parabola at any point is,

$\displaystyle =-\left(\frac{1}{\frac{dy}{dx}}\right)=-\frac{y}{2a}=m$ (given)

$\Rightarrow y=-2am$
Thus from

$y^2=4ax,$

$x=\dfrac{y^2}{4a}=am^2$
Therefore the point is

$(am^2,-2am)$
Hence the required line passing through

$(am^2,-2am)$ and having slope

$m$ is,

$(y+2am)=m(x-am^2)$

$\Rightarrow y= mx-2am-am^3$

Tangent to parabola

$\displaystyle y^{2}=4x+5$ which is parallel to

$y=2x+7$

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

Explanation

Given,

$y^2=4x+5$

$\therefore \displaystyle 2y\frac{dy}{dx}=4$

$\therefore \dfrac{dy}{dx}=\dfrac{2}{y}=2,$ (given)

$\therefore y=1 \Rightarrow x=-1$

$\therefore$ required tangent is given by,

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

$y=2x+3$

The slope of tangent to the curve

$y=\int_{0}^{x}\displaystyle \frac{dx}{1+x^{3}}$ at the point where

$x=1$ is

0%
$\displaystyle \frac{1}{2}$
0%
$1$
0%
$\displaystyle \frac{1}{4}$
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none of these

Explanation

$y=\int _{ 0 }^{ x }{ \cfrac { dx }{ 1+{ x }^{ 3 } } } \\ \cfrac { dy }{ dx } =\cfrac { 1 }{ 1+{ x }^{ 3 } } =\cfrac { 1 }{ 2 }$

The values of '

$a$ ' for which

$y=x^{2}+ ax+25$ touches

$x$ -axis are

0%
$\pm 10$
0%
$\pm 2$
0%
$\pm 1$
0%
0

Explanation

${y}'=2x+a=0$

For any quadratic equation to touch x-axis it must have equal roots at that point, so the equation must be in form of perfect square.

$y=(x+5)^{2}$ at

$a=10$

The equation of the straight line which is tangent at one point and normal at another point to the curve

$y=8{ t }^{ 3 }-1,x=4{ t }^{ 2 }+3$ is

0%
$\displaystyle \sqrt { 2 } x-y=\frac { 89\sqrt { 2 } }{ 27 } -1$
0%
$\displaystyle \sqrt { 2 } x-y=\frac { 89\sqrt { 2 } }{ 27 } +1$
0%
$\displaystyle \sqrt { 2 } x+y=\frac { 89\sqrt { 2 } }{ 27 } -1$
0%
$\displaystyle \sqrt { 2 } x+y=\frac { 89\sqrt { 2 } }{ 27 } +1$

Explanation

We have,

$\displaystyle y=8{ t }^{ 3 }-1$ and

$\displaystyle x=4{ t }^{ 2 }+3$

$\displaystyle \Rightarrow \dfrac { dy }{ dt } =24{ t }^{ 2 }$ and

$\displaystyle \dfrac { dx }{ dt } =8t$

$\displaystyle \therefore \dfrac { dy }{ dx } =\dfrac { 24{ t }^{ 2 } }{ 8t } =3t$

$\therefore$ tangent at

$'t'$ is

$\displaystyle y-\left( 8{ t }^{ 3 }-1 \right) =3t\left( x-4{ t }^{ 2 }-3 \right)$ ...(1)

Let the tangent meet the curve again at the point

$t'$ (say)

$\displaystyle \therefore \left( 8t'-1 \right) -\left( 8{ t }^{ 3 }-1 \right) =3t\left( 4{ t' }^{ 2 }+3-4{ t }^{ 2 }-3 \right)$

$\displaystyle \Rightarrow 8\left( { t' }^{ 3 }-{ t }^{ 3 } \right) =3t.4\left( t'-t \right) \left( t'+t \right)$

$\displaystyle \Rightarrow 2\left( { t' }^{ 2 }+{ t' }^{ 2 }+tr \right) =3t\left( t'+t \right)$

$\displaystyle \Rightarrow 2{ t' }^{ 2 }-tt'-{ t }^{ 2 }=0$

$\displaystyle \Rightarrow \left( t'-t \right) \left( 2t'+t \right) =0\Rightarrow t'=t$ or

$\displaystyle \dfrac { -t }{ 2 }$

$\therefore$ the line (1), which is tangent at the point

$'t'$ is normal at the point

$\displaystyle \dfrac { -t }{ 2 }$

$\displaystyle \therefore 3t=\dfrac { 2 }{ 3t } \Rightarrow t=\pm \dfrac { \sqrt { 2 } }{ 3 }$ Hence, the required lines are

$\displaystyle \sqrt { 2 } x-y=\dfrac { 89\sqrt { 2 } }{ 27 } +1$ and

$\displaystyle \sqrt { 2 } x+y=\dfrac { 89\sqrt { 2 } }{ 27 } -1$

The equation of the normal to the curve

$\displaystyle 2y=3-x^{2}$ at the point

$(1,1)$

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

Explanation

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

$\cfrac{dy}{dx}=-x =m$

Thus slope of normal at

$(1,1)$ is

$=-\left(\cfrac{1}{m}\right)_{(1,1)} = 1$

Hence required normal is,

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

The equation of the normal to the curve

$\displaystyle x^{3}+y^{3}=6xy$ at the point

$(3,3).$

0%
$x+y-6=0$
0%
$-x-y+6=0$
0%
$x-y=0$
0%
$x+y=0$

Explanation

$\displaystyle \dfrac{dy}{dx}=-\dfrac{fx}{fy}=-\dfrac{3x^{2}-6y}{3y^{2}-6x}=-1$ at

$(3,3)$

$\therefore$ Required normal is

$y-3=1(x-3)$ or

$x-y=0$

Normal to the curve

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

$(1,2)$

0%
$x+y=3$
0%
$x-y=3$
0%
$2x+y=4$
0%
$x+2y=5$

Explanation

From the equation of the curve

$\displaystyle x^{2}=4y$ ,
we get

$dy/dx=x/2.$ If

$m$ be the slope of the normal to

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

then

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

$\displaystyle \therefore x=\dfrac{-2}m$ and

$y=\dfrac{x^{2}}4=1/m^{2}$
Thus normal is

$\displaystyle y-\frac{1}{m^{2}}=m\left ( x+\frac{2}{m} \right )$ .

If it passes through the point (1,2),

then

$\displaystyle \therefore 2-\frac{1}{m^{2}}=m\left ( 1+\frac{2}{m} \right )=m+2 \therefore m^{3}=-1$

$\displaystyle m=-1$

$\therefore$ Required normal is,

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

$\Rightarrow x+y=3$

Find the equation of a line passing through

$(-2,3)$ and parallel to tangent at origin for the circle

$\displaystyle x^{2}+y^{2}+x-y=0$

0%
$x -2 y + 5 = 0$
0%
$x -4 y + 3 = 0$
0%
$x - y + 5 = 0$
0%
$2x - y + 6 = 0$

Explanation

Given,

$x^{2}+y^{2}+x-y=0$
Differentiating w.r.t

$x$ , we get

$2x+2y\cfrac{dy}{dx}+1-\cfrac{dy}{dx}=0$

$\Rightarrow \cfrac{dy}{dx}=\cfrac{1+2x}{1-2y}$
Thus slope of the tangent at origin is,

$m=\left(\cfrac{dy}{dx}\right)_{(0,0)}=1$
Hence required line through

$(-2,3)$ is,

$(y-3)=1(x+2)\Rightarrow x-y+5=0$

Find the equations of tangents to parabola

$\displaystyle y^{2}= 4ax$ which are drawn from the point (2a,3a).

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

Explanation

Any tangent to parabola is given by,

$\displaystyle y= mx+\frac{a}{m}$
Now given it passes through

$(2a, 3a)$

$\Rightarrow \displaystyle 3a= 2am+\dfrac{a}{m}\Rightarrow 2m^2-3m+1=0\Rightarrow m =1,\dfrac{1}{2}$
Hence equation of tangents are,

$(y-3a)=1(x-2a)$ and

$(y-3a)=\cfrac{1}{2}(x-2a)$

$\Rightarrow \displaystyle x-y+a= 0$ and

$x-2y+4a= 0$

$y = f(x)$ be the equation of a parabola which is touched by the line

$y = x$ at the point where

$x = 1$ Then

0%
$f'(1) = 1$
0%
$f'(0) = f'(1)$
0%
$2f(0) = 1 - f'(0)$
0%
$f(0) + f'(0) + f"(0) = 1$

Explanation

Let,

$y=ax^2+bx+c$

$\therefore y'=2ax+b$
Given line

$y=x$ touches this parabola at

$x=1$

$\Rightarrow y(1)=1\Rightarrow a+b+c=1 ..(1)$
and

$y"(1)=1\Rightarrow 2a+b=1 ..(2)$
Solving (1) and (2) we get

$2c=1-b \Rightarrow 2f(0)=1-f'(0)$

The slope of the normal to the curve

$y = 2x^2+ 3 \sin x$ at

$x = 0$ is.

0%
$3$
0%
$\dfrac{1}{3}$
0%
$-3$
0%
$-\dfrac{1}{3}$

Explanation

$y = 2x^2+ 3 \sin x$

$\cfrac{dy}{dx} = 4x+3\cos x$
Thus slope of tangent at

$x=0$ is

$4(0)+3\cos 0=3$
Hence slope of normal at the same point is

$-\cfrac{1}{m}=-\cfrac{1}{3}$

The normal drawn at the point

$\displaystyle P\left ( at_{1}^{2},2at_{1} \right )$ on the parabola meets the curve again at

$\displaystyle Q\left ( at_{2}^{2},2at_{2} \right ).$ then

$\displaystyle t_{2} =?$

0%
$\displaystyle t_{2}= -t_{1}-\dfrac{2}{t_{1}}$
0%
$\displaystyle t_{2}= -t_{1}+\dfrac{2}{t_{1}}$
0%
$\displaystyle t_{2}= +t_{1}-\dfrac{2}{t_{1}}$
0%
$\displaystyle t_{2}= +t_{1}+\dfrac{2}{t_{1}}$

Explanation

The normal at

$\displaystyle t_{1}$ is easily found to be

$\displaystyle y= -t_{1}x+2at_{1}+at_{1}^{3}$ It passes through the point

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

$\displaystyle \therefore 2at_{2}= -t_{1}at_{2}^{2}+2at_{1}+at_{1}^{3}$ or

$\displaystyle 2a\left ( t_{2}-t_{1} \right )= -at_{1}\left ( t_{2}^{2}-t_{1}^{2} \right )$

$\displaystyle \therefore 2= -t_{1}\left ( t_{2}+t_{1} \right ) \because t_{2}\neq t_{1}$

$\displaystyle \therefore t_{2}= -t_{1}-\frac{2}{t_{1}}$

Ans: A

The slope of the tangent to the curve

$\displaystyle y=-x^{3}+3x^{2}+9x-27$ is maximum when x equals.

0%
$1$
0%
$3$
0%
$\dfrac 12$
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$-\dfrac 12$

Explanation

Given,

$\displaystyle y=-x^{3}+3x^{2}+9x-27$
Slope at any point on thegiven curve is

$m= \cfrac{dy}{dx}=-3x^2+6x+9$
Now for maximum value of slope we must have

$\cfrac{dm}{dx}=0=-6x+6\Rightarrow x = 1$

Find the tangents and normal to the curve

$y(x-2)(x-3)-x+7=0,$ at point (7,0) are

0%
$x-20y-7=0,$ $20x+y-140=0$ .
0%
$x+20y-7=0,$ $20x-y-140=0$ .
0%
$7x-20y-1=0,$ $20x+7y-100=0$ .
0%
$7x+20y-1=0,$ $20x-7y-100=0$ .

Explanation

Given curve is,

$y(x-2)(x-3)-x+7=0\Rightarrow y(x^2-5x+6)-x+7=0$
Differentiating w.r.t

$x$

$\cfrac{dy}{dx}(x^2-5x+6)+y(2x-5)-1=0$
putting (7,0) to get slope of the tangent

$\cfrac{dy}{dx}(7^2-5.7+6)+0(2x-5)-1=0\Rightarrow \cfrac{dy}{dx}=\cfrac{1}{20}$

$\Rightarrow$ slope of tangent is

$m = \cfrac{1}{20}$ and slope of normal is,

$m'=-\cfrac{1}{m}=-20$
Hence required lines are

$x-20y-7=0,$ and

$20x+y-140=0$ .

Find the distance between the point

$(1,1)$ and the tangent to the curve

$\displaystyle y=e^{2x}+x^{2}$ drawn from the point where the curve cuts

$y$ -axis

0%
$\displaystyle \frac{\sqrt3}{\sqrt{5}}$
0%
$\displaystyle \frac{3}{\sqrt{5}}$
0%
$\displaystyle \frac{2}{\sqrt{5}}$
0%
$\displaystyle \frac{\sqrt2}{\sqrt{5}}$

Explanation

Clearly the the point on the

$y$ -axis is

$(0,1)\equiv P$ (say)
Now

$\cfrac{dy}{dx}=2e^{2x}+2x$
Thus slope at

$P$ is

$m=\left(\cfrac{dy}{dx}\right )_{(0,1)}=2$
Thus required tangent is

$(y-1)=2(x-0)\Rightarrow 2x-y+1=0$
Hence, distance of

$(1,1)$ from this line is

$=\left |\cfrac{2\times 1-1+1}{\sqrt{2^2+1^2}}\right |=\cfrac{2}{\sqrt{5}}$

$\displaystyle \frac{x}{a}+\frac{y}{b}=1$ is a tangent to the curve

$\displaystyle x=Kt,y=\frac{K}{t},K> 0$ than

0%
$a>0, b>0$
0%
$a>0, b<0$
0%
$a<0, b>0$
0%
$a<0, b<0$

Explanation

$\displaystyle x=Kt,y=\frac{K}{t},K> 0$

$\cfrac{dx}{dt}=K, \cfrac{dy}{dt}=-\cfrac{K}{t^2}$
Hence slope of tangent at point 't' is,

$m=\cfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=-\cfrac{1}{t^2}< 0 \forall t \epsilon R$
Hence for line

$\cfrac{x}{a}+\cfrac{y}{b}=1$ to be tangent to the given curve
it slope should be negative

$\cfrac{-b}{a}<0\Rightarrow \cfrac{b}{a}>0$

$a>0, b> 0$ or

$a<0, b<0$

The curve

$\displaystyle y-e^{xy}+x=0$ has a vertical tangent at

0%
$(1, 1)$
0%
$(0, 1)$
0%
$(1, 0)$
0%
$(0,0)$

Explanation

Given,

$\displaystyle y-e^{xy}+x=0$
Differentiating w.r.t

$x$

$\cfrac{dy}{dx}-e^{xy}(y+x\cfrac{dy}{dx})+1=0$

$\Rightarrow \cfrac{dy}{dx}=\cfrac{1-ye^{xy}}{xe^{xy}-1}$
Thus for vertical tangent,

$\cfrac{dy}{dx}=\infty$

$\Rightarrow xe^{xy}-1=0$
Hence required point is

$(1,0)$

Find the equation of the tangent to the curve

$\displaystyle y=x^{2}+1$ at the point

$(1,2).$

0%
$2y=x$
0%
$y=2x$
0%
$y+x=2$
0%
$y+2x=0$

Explanation

Given

$y=x^2+1$

$y'=2x$ at point

$(1,2)$

$\Rightarrow y'=2$

The equation of tangent

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

$\Rightarrow y=2x$

Ans: B

The equation of normal to the curve

$\displaystyle y=e^{x}$ at the point

$(0,1)$ is -

0%
$x + y = 1$
0%
$x - y = 1$
0%
$ey - x = e$
0%
$e(y - 1) + x = 0$

Explanation

$\displaystyle y=e^x$

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

$\therefore \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( 0,1 \right )}=1 =m$ (say)

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

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

$(y - 1) = -1 (x -0\times a)$

$\Rightarrow x + y = 1$
Hence, option 'A' is correct.

If tangent to curve at a point is perpendicular to

$x$ - axis then at that point -

0%
$\displaystyle \frac{dy}{dx}=0$
0%
$\displaystyle \frac{dx}{dy}=0$
0%
$\displaystyle \frac{dy}{dx}=1$
0%
$\displaystyle \frac{dy}{dx}=-1$

Explanation

We know slope of tangent to the any curve

$'y'$ is

$m=\cfrac{dy}{dx}$
Now if line is perpendicular to x-axis this means its slope is

$\infty$

$\Rightarrow \cfrac{dy}{dx} \to \infty \Rightarrow \cfrac{dx}{dy}=0$
Hence, option 'C' is correct.

The equation of normal to the curve

$\displaystyle y^{2}=16x$ at the point

$(1, 4)$ is

0%
$2x + y = 6$
0%
$2x - y + 2 = 0$
0%
$x + 2y = 9$
0%
None of these

Explanation

$\displaystyle y^{2}=16x$

$\Rightarrow \displaystyle 2y\frac{dy}{dx}=16$

$\Rightarrow \therefore \displaystyle \frac{dy}{dx}=\frac{8}{y}$
Thus slope of tangent at

$(1, 4)$ is

$m=2$

$\therefore$ Slope is normal is

$=\displaystyle -\frac{1}{2}$
Therefore, the equation of normal at

$(1, 4)$ is given by,

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

$\displaystyle \Rightarrow x+2y=9$
Hence, option 'C' is correct.

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 1 - 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 1 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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