MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 12

Equation of the line through the point

$\left(\dfrac{1}{2}, 2 \right)$ and tangent to the parabola

$\displaystyle y = \frac {-x^2}{2}+2$ and secant to the curve

$\displaystyle y = \sqrt {4 - x^2}$ is :

0%
$2x + 2y - 5 = 0$
0%
$2x + 2y - 3 = 0$
0%
$y - 2 = 0$
0%
$none$

Explanation

We have,

$y = \dfrac{-x^2}{2} +2$

$y' = -x$
At

$(x_1,y_1)$ ,

$y' = -x_1$
The equation of a tangent at

$(x_1, y_1)$ can be written as

$y - y_1 = -x_1 (x- x_1)$
The tangent passes through

$\left(\dfrac12,2 \right)$ .
Hence,

$y_1 -2 = -x_1 \left(x_1 - \dfrac12 \right)$

$-\dfrac{{x_1}^2}{2} = -{x_1}^2 + \dfrac{x_1}{2}$

$x_1 = x_1^2$

$x_1 = 0,1$
The equation at

$x_1= 1, y_1 = \dfrac32$ ,

$y -\dfrac32 = -1(x-1)$

$2x+2y -5 = 0$ . The distance of the line from the origin is

$\dfrac{5}{2\sqrt{2}} <2$ . Hence, this line is a secant to the circle.
The tangent at

$(2,0)$ is given by

$x=2$ . However this is also tangent to the circle.
Hence, option A is correct.

What is the minimum intercept made by the axes on the tangent to the ellipse

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ ?

0%
$a+b$
0%
$2(a+b)$
0%
$\sqrt{ab}$
0%
none of these

Explanation

Equation of ellipse is

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
Let

$P (a\cos\theta, b\sin\theta)$ be a point on the ellipse.Equation of tangent to ellipse at P is

$\dfrac{x}{a}{\cos\theta}+\dfrac{y}{b}{\sin\theta}=1$
So, the tangent intersects X-axis at

$(a\sec\theta,0)$ and Y-axis at

$(0,bcosec\theta)$
Intercepted portion of tangent between axes is

$S=\sqrt{a^2\sec^{2}\theta+b^2cosec^{2}\theta}$
For maximum or minimum ,

$\dfrac{dS}{d\theta}=0$

$\Rightarrow 2a^2\sec^{2}\theta\tan\theta-2b^2cosec^{2}\theta\cot\theta=0$

$\Rightarrow \tan^{4}\theta=\dfrac{b^2}{a^2}$

$\Rightarrow \tan^{2}\theta=\dfrac{b}{a}$
Also,

$\dfrac{d^2S}{d\theta^2}>0$ for

$\tan^{2}\theta=\dfrac{b}{a}$
Hence, S attains minimum at

$\tan^{2}\theta=\dfrac{b}{a}$
Minimum value of

$S=\sqrt{a^2(1+\tan^{2}\theta)+b^2(1+\cot^{2}\theta)}$

$\Rightarrow S_{min}=\sqrt{a(a+b)+b(a+b)}$
Minimum value of S

$=a+b$

Let

$C$ be the curve

$\displaystyle y = x^3$ (where

$x$ takes all real values.) The tangent at

$A$ meets the curve again at

$B$ . If the gradient of the curve at

$B$ is

$K$ times the gradient at

$A$ then

$K$ is equal to

0%
$4$
0%
$2$
0%
$-2$
0%
$\dfrac{1}{4}$

Explanation

Given equation of curve is

$y=x^{3}$
Let the coordinates of A be

$(t,t^{3})$ and coordinates of B be

$(T,T^{3})$

$\displaystyle \dfrac{dy}{dx}=3x^{2}$
Slope of tangent to curve at A

$= 3t^{2}$ .

Now , slope of AB

$=\displaystyle \dfrac{T^3-t^3}{T-t}$

$\Rightarrow \displaystyle \dfrac{T^3-t^3}{T-t} =3t^{2}$

$\Rightarrow {T^3-t^3}=3t^{2}(T-t)$

$\Rightarrow (T-t)(T^2+tT+t^2)-3t^{2}(T-t)=0$

$\Rightarrow (T-t)(T^2+tT-2t^2)=0$

$\Rightarrow T=t, T=-2t$

$T=t$ is not possible. Hence,

$T=-2t$

Given,

$m_B =K\cdot m_A$

$\Rightarrow T^2=Kt^{2}$

$\Rightarrow 4t^{2}=Kt^{2}$

$\Rightarrow K=4$

The graphs

$y=2x^3-4x+2$ and

$y=x^3+2x-1$ intersect at exactly 3 distinct points. The slope of the line passing through two of these points

0%
is equal to 4
0%
is equal to 6
0%
is equal to 8
0%
is not unique

Explanation

Given equation of curves

$y=2x^3-4x+2$

$y=x^3+2x-1$
Let

$(x_{1},y_{1})$ be a point of intersection of the curves.

$\Rightarrow y_{1}=2x_{1}^3-4x_{1}+2$
And

$y_{1}=x_{1}^3+2x_{1}-1$

$\Rightarrow y_1=8x_{1}-4$
Let

$(x_{2},y_{2})$ be another point of intersection of the curves.

$\Rightarrow y_{2}=2x_{2}^3-4x_{2}+2$
And

$y_{2}=x_{2}^3+2x_{2}-1$

$\Rightarrow y_2=8x_{2}-4$
Now, slope

$= \displaystyle \dfrac{y_2-y_1}{x_2-x_1}$

$\Rightarrow m =\displaystyle \dfrac{8(x_2-x_1)}{x_2-x_1}$

$\Rightarrow m=8$

Consider the curve represented parametrically by the equation

$\displaystyle x = t^3 - 4t^2 - 3t$ and

$\displaystyle y = 2t^2 + 3t - 5$ where

$\displaystyle t \: \epsilon \: R$ .
If

$H$ denotes the number of point on the curve where the tangent is horizontal and

$V$ the number of point where the tangent is vertical then

0%
$H = 2$ and $V = 1$
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$H = 1$ and $V = 2$
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$H = 2$ and $V = 2$
0%
$H = 1$ and $V = 1$

The number of points on the curve

$x^{3/2}+y^{3/2}=a^{3/2}$ , where the tangents are equally inclined to the axes, is

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

Explanation

If the tangents are equally inclined to the axes we get

$y'_{x,y}=1$ .

Therefore

$\dfrac{3}{2}\sqrt{x}+\dfrac{3}{2}\sqrt{y}.y'=0$

$y'=1$ implies

$\sqrt{x}+\sqrt{y}=0$

$\sqrt{x}=-\sqrt{y}$

$x=y$ .

$x_{1}=y_{1}$ .

Substituting in the equation we get

$2x^{\frac{3}{2}}=a^{\frac{3}{2}}$

$x=\dfrac{a}{2^{\frac{2}{3}}}$

Therefore

$(x_{1},y_{1})=\left(\dfrac{a}{2^{\frac{2}{3}}},\dfrac{a}{2^{\frac{2}{3}}}\right)$

Therefore there exists only one point.

Let

$\displaystyle f(x) = ln \: mx (m > 0)$ and

$g(x) = px$ . Then the equation

$\displaystyle |f(x)| = g(x)$ has only one solution for

0%
$\displaystyle 0 < p < \frac {m}{e}$
0%
$\displaystyle p < \frac {e}{m}$
0%
$\displaystyle 0 < p < \frac {e}{m}$
0%
$\displaystyle p > \frac {m}{e}$

Explanation

Let

$P(h,k)$ be a point on

$f(x)=\ln mx$
then,

$k=\ln mh$

$\dfrac { dy }{ dx }$ at point P

$=\dfrac { 1 }{ h }$
Equation of tangent at point P

$L_{1}: y-k=\dfrac { \left( x-h \right) }{ h }$
Since, it passes through origin
Therefore,

$k=1$ &

$h=e/m$
and

$L_{1}: y=x/h$

$\Rightarrow L_{1}: y=mx/e$

From the figure:

$|f(x)|=g(x)$ have one solution when slope of

$g(x)$ is greater than slope of

$L_{1}$
Therefore,

$p>m/e$

Ans: D

The equation of the normal to the curve

$\displaystyle (\frac {x}{a})^n + (\frac {y}{b})^n = 2 (n \epsilon N)$ at the point with abscissa equal to 'a' can be:

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$\displaystyle ax + by = a^2 - b^2$
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$\displaystyle ax + by = a^2 + b^2$
0%
$\displaystyle ax - by = a^2 - b^2$
0%
$\displaystyle dx - ay = a^2 - b^2$

Explanation

$\displaystyle (\frac {x}{a})^n + (\frac {y}{b})^n = 2$ .....(i)
Given

$x=a$

$\Rightarrow \displaystyle (\frac{y}{b})^{n}=1$

$\Rightarrow y=\pm b$
So the points are

$(a,b)$ and

$(a,-b)$

Differentiating (i) w.r.t

$x$ , we get

$\displaystyle \frac{dy}{dx}=-\frac{b^{n}}{a^{n}}(\frac{x}{y})^{n-1}$
Slope of tangent at

$(a,b) \displaystyle=-\frac{b}{a}$
Slope of normal at

$(a,b) \displaystyle =\frac{a}{b}$
Equation of normal through

$(a,b)$ is

$y-b=\dfrac {a}{b}(x-a)$

$\Rightarrow ax-by=a^{2}-b^{2}$

Slope of tangent at

$(a,-b) =\dfrac{b}{a}$
Slope of normal at

$(a,-b) \displaystyle =-\frac{a}{b}$
Equation of normal at

$(a,-b)$ is

$\displaystyle y+b=-\frac{a}{b} (x-a)$

$\Rightarrow ax+by=a^{2}-b^{2}$

The co-ordinates of the point P on the graph of the function

$\displaystyle y = e^{-|x|}$ where the portion of the tangent intercepted between the co-ordinate axes has the greatest area, is

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

Explanation

$y={ e }^{ -|x| }$

$y= { e }^{ -x }, x>0$ ,

${ e }^{ x }, x<0$
For

$y={ e }^{ -x }$ let point

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ lies on it
So, slope is

$\cfrac { dy }{ dx } =-{ e }^{ -{ x }_{ 1 } }$
Therefore equation of tangent is

$\cfrac { dy }{ dx } =-{ e }^{ -{ x }_{ 1 } }$
Hence X intercept is

$\left( 1+{ x }_{ 1 } \right)$ and Y intercepts is

$\quad { e }^{ -{ x }_{ 1 } }\left( 1+{ x }_{ 1 } \right)$
Area so formed is

$A=\cfrac { 1 }{ 2 } { e }^{ -{ x }_{ 1 } }{ \left( 1+{ x }_{ 1 } \right) }^{ 2 }$
And this is maximum at

${ x }_{ 1 }=1\quad \Rightarrow { y }_{ 1 }=\cfrac { 1 }{ e }$
Similarly for

$y={ e }^{ x }$ we get

${ x }_{ 1 }=-1\Rightarrow { y }_{ 1 }=\cfrac { 1 }{ e }$
Hence, option 'B' correct

A point

$\displaystyle P(a, a^n)$ on the graph of

$\displaystyle y = x^n (n \: \epsilon \: N)$ in the first quadrant a normal is drawn. The normal intersects the y-axis at the point

$(0, b)$ . If

$\displaystyle _{a \rightarrow 0}^{Lim \: b} \textrm{=} \frac {1}{2}$ , then n equals

0%
$1$
0%
$3$
0%
$2$
0%
$4$

Explanation

$y={ x }^{ n }$
Slope of normal is

$-\cfrac { dx }{ dy } =-\cfrac { 1 }{ n{ \alpha }^{ n-1 } }$
So the equation of normal is

$\left( y-{ \alpha }^{ n } \right) =-\cfrac { 1 }{ n{ \alpha }^{ n-1 } } \left( x-\alpha \right) \\ y+\cfrac { x }{ n{ \alpha }^{ n-1 } } =n{ \alpha }^{ 2-n }+{ \alpha }^{ n }$
Therefore

$b=n{ \alpha }^{ 2-n }+{ \alpha }^{ n }$
Hence to exists

$\displaystyle\lim _{ \alpha \rightarrow 0 }{ b }$ must be equal to

$2$

The normal curve

$xy = 4$ at the point

$(1, 4)$ meets the curve again at

0%
$(-4, -1)$
0%
$\displaystyle \left(-8, \frac {1}{2}\right)$
0%
$\displaystyle \left(-16, -\frac {1}{4}\right)$
0%
$(-1, -4)$

Explanation

Differentiating the equation of the curve with respect to x, we get

$y+xy'=0$
Or

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

$=\dfrac{-4}{x^{2}}$

$y'_{x=1}=-4$
Hence slope of normal is

$=\dfrac{1}{4}$ .
Hence equation of normal will be

$y-4=\dfrac{1}{4}(x-1)$
Or

$4y-16=x-1$
Or

$4y-x=15$ ...(i)

$xy=4$ ...(ii)
Solving the above two equations

$4y-\dfrac{4}{y}=15$
Or

$4y^{2}-4=15y$

$4y^{2}-15y-4=0$

$y=4$ and

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

$y=\dfrac{-1}{4}$ .
Thus

$x=\dfrac{4}{y}=-16$ .
Hence the required point is

$(-16,\dfrac{-1}{4})$ .

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

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

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

Explanation

Eliminating parameter t we get

$x = K(K/y) \Rightarrow y = K^2/x$

$\dfrac{dy}{dx} = -\dfrac{K^2}{x^2} < 0 \forall x\epsilon R$
Now slope of given line is

$=-\dfrac{b}{a}$

$\Rightarrow \dfrac{b}{a} > 0$ Hence either

$a>0, b>0$ or

$a<0, b<0$

For the curve represented parametrically by the equations,

$\displaystyle x = \displaystyle\frac{2}{cot\:t} + 1$ &

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

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tangent at $\displaystyle t = \displaystyle\frac{\pi}{4}$ is parallel to x - axis
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normal at $\displaystyle t =\displaystyle\frac{\pi}{4}$ is parallel to y - axis
0%
tangent at $\displaystyle t =\displaystyle\frac{\pi}{4}$ is parallel to the line $y = x$
0%
tangent and normal intersect at the point $(2, 1)$

Explanation

$dy=sec^{2}t-cosec^{2}t.dt$ ...(i)

$dx=2sec^{2}t.dt$ ...(ii)

$\dfrac{dy}{dx}$

$=\dfrac{sec^{2}t-cosec^{2}t.dt}{2sec^{2}t.dt}$

$=\dfrac{sec^{2}t-cosec^{2}t}{2sec^{2}t}$ .

Hence

$\dfrac{dy}{dx}_{t=\frac{\pi}{4}}$

$=0$ slope of tangent

Slope of normal

$=\dfrac{-dx}{dy}$

$=\dfrac{-2sec^{2}t}{sec^{2}t-cosec^{2}t}$

Hence

$=\dfrac{-dx}{dy}_{t=\frac{\pi}{4}}=\infty$

Hence the normal is perpendicular to x axis and thus parallel to y axis.

and tangent is parallel to x - axis.

The straight line which is both a tangent and normal to the curve

$\displaystyle x = 3t^2, y = 2t^3$ is

0%
$\displaystyle y + \sqrt 2 (x - 2) = 0$
0%
$\displaystyle y - \sqrt 2 (x - 2) = 0$
0%
$\displaystyle y + \sqrt 3 (x - 1) = 0$
0%
$\displaystyle y - \sqrt 3 (x - 1) = 0$

Explanation

The slope of the tangent at

$P(t)=\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{\mathrm{dy} }{\mathrm{d} t}\dfrac{\mathrm{d} t}{\mathrm{d} x}=t$
Let the tangent cuts the curve and normal at

$Q(t_{1})$ , then the coordinate of the point is

$(3t_{1}^{2},2t_{1}^{3})$
The slope of the line passing through

$P$ and

$Q$ =

$\dfrac{(2t_{1}^{3}-2t^{3})}{3t_{1}^{2}-3t^{2}}=t$

$\Rightarrow t_{1}=-\dfrac{t}{2}$ and

$Q=(3\dfrac{t^{2}}{4},-\dfrac{t^{3}}{4})$
Slope of normal at

$Q=-\dfrac{\mathrm{d}x }{\mathrm{d} y}=\dfrac{2}{t}$
The equation of straight line having slope

$t$ and passing through

$Q$ is

$y-(-\dfrac{t^{3}}{4})=t(x-3\dfrac{t^{2}}{4})$

$\Rightarrow tx-y=t^{3}\cdots \cdots (1)$
The equation of line having slope

$\dfrac{2}{t}$ and passing through

$Q$ is

$y-(-\dfrac{t^{3}}{4})=\dfrac{2}{t}(x-3\dfrac{t^{2}}{4})$

$\Rightarrow \dfrac{2x}{t}-y=\dfrac{3t}{2}+\dfrac{t^{3}}{4}\cdots \cdots (2)$
On solving

$(1)$ and

$(2)$ we have

$t=\sqrt{2}$

$\Rightarrow$ Option

$(A)$ and

$(B)$ are correct

The tangent to the curve

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

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

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

$\left(c+1,{e}^{c+1}\right)$

0%
on the left of $x=c$
0%
on the right of $x=c$
0%
at no point
0%
at all points.

Explanation

The equation of tangent to the curve

$y={e}^{x}$ at

$\left(c.{e}^{c}\right)$ is

$y-{e}^{c}={e}^{c}\left(x-c\right)$ ...(1)

and the equation of line joining

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

$\left(c+1,{e}^{c+1}\right)$ is

$\displaystyle y-{ e }^{ c-1 }=\dfrac { { e }^{ c+1 }-{ e }^{ c-1 } }{ \left( c+1 \right) -\left( c-1 \right) } \left( x-\left( x-1 \right) \right)$

$\displaystyle \Rightarrow y-{ e }^{ c-1 }=\dfrac { { e }^{ c }\left( e-{ e }^{ -1 } \right) }{ 2 } \left( x-\left( c+1 \right) \right)$ ...(2)

Subtracting equation (1) and (2), we get

$\displaystyle { e }^{ c }-{ e }^{ c-1 }={ e }^{ c }\left( x-c \right) \left[ \dfrac { e-{ e }^{ -1 }-2 }{ 2 } \right] +{ e }^{ c }\left( \dfrac { e-e^{ -1 } }{ 2 } \right)$

$\displaystyle \Rightarrow x-c=\dfrac { \left[ 1-{ e }^{ -1 }-\left( \dfrac { e-{ e }^{ -1 } }{ 2 } \right) \right] }{ \dfrac { e-{ e }^{ -1 }-2 }{ 2 } } =\dfrac { 2-e-{ e }^{ -1 } }{ e-{ e }^{ -1 }-2 }$

$\displaystyle =\dfrac { e+{ e }^{ -1 }-2 }{ 2\left( e-{ e }^{ -1 } \right) } =\dfrac { \dfrac { e+{ e }^{ -1 } }{ 2 } -1 }{ 1-\dfrac { e-{ e }^{ -1 } }{ 2 } } =\dfrac { +ve }{ -ve } =-ve$

$\Rightarrow x-c<0\Rightarrow x<c$

$\therefore$ the two lines meet on the left of line

$x=c$

Find the equations of tangents to the curve y=x

$^{4}$ which are drawn from the point

$(2,0)$

0%
Only $\displaystyle y-\left ( \frac{8}{3} \right )^{4}=4\left ( \frac{8}{3} \right )^{3}\left ( x-\frac{8}{3} \right )$
0%
Only $y=0$
0%
Both A and B
0%
Only $\displaystyle y-\left ( \frac{8}{3} \right )=4\left ( \frac{8}{3} \right )\left ( x-(\frac{8}{3})^{4} \right )$

Explanation

Given,

$y=x^4$

$\therefore \dfrac{dy}{dx}=4x^{3}$ .
Now equation of the tangent through

$(2,0)$ .

$\dfrac{y}{x-2}=4x^{3}$

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

$y=4x^{4}-8x^{3}$

$x^{4}=4x^{4}-8x^{3}$

$3x^{4}-8x^{3}=0$

$x^{3}[3x-8]=0$

$x=0$ and

$x=\dfrac{8}{3}$ .
If

$x=0$ then

$y=0$ .

$\dfrac{y-0}{x-2}=\dfrac{y-0}{x-0}$

$xy=xy-2y$

$-2y=0$

$\therefore y=0$ ... (x-axis).
Now considering

$x=\dfrac{8}{3}$ we get

$y=\dfrac{4096}{81}$
Hence

$\dfrac{y-0}{x-2}=\dfrac{y-\dfrac{4096}{81}}{x-\dfrac{8}{3}}$

$xy-\dfrac{8y}{3}=xy-\dfrac{4096x}{81}-2y+\dfrac{8192}{81}$

$-\dfrac{8y}{3}+2y=-\dfrac{4096x}{81}+\dfrac{8192}{81}$

$\dfrac{2y}{3}=-\dfrac{4096x}{81}+\dfrac{8192}{81}$

If the line

$ax+by+c=0$ is a normal to the rectangular hyperbola

$xy=1$ then

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

Explanation

$xy=1$

$xy'+y=0$

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

Hence

Slope of normal will be

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

Since

$xy=1$ hence any

$(x,y)$ lying on the hyperbola has the sign of the abscissa and ordinate as same.

That sign of x is same as sing of y.

Hence

$\dfrac{x}{y}$ will always be positive.

Or in other words

$\dfrac{-1}{y'}>0$

Hence

$\dfrac{-a}{b}>0$

Hence if

$a<0,b>0$ or if

$a>0,b<0$ .

A figure is bounded by the curve

$\displaystyle y=x^{2}+1,$ the axes of co-ordinates and the line x=Determine the co-ordinates of a point P at which a tangent be drawn to the curve so as to cut off a trapezium of greatest area from the figure.

0%
$\displaystyle \left ( \frac{1}{2}, \frac{5}{2} \right )$
0%
$\displaystyle \left ( \frac{1}{2}, \frac{5}{4} \right ).$
0%
$\displaystyle \left ( -\frac{1}{2}, \frac{5}{2} \right ).$
0%
$\displaystyle \left ( \frac{1}{2}, -\frac{5}{4} \right ).$

Explanation

The given equation

$\displaystyle x^{2}=y-1$ represents a parabola with vertex at

$(0,1)$
Let the co-ordinates of any point p on the curve

$\displaystyle y=1+x^{2}be\left ( h,1+h^{2} \right ).$
Now

$\displaystyle \dfrac{dy}{dx}=2x=2h$ Tangent at P is

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

$\displaystyle y-2xh=1-h^{2}$
Substituting

$x=0$ , we get

$\displaystyle y=\left ( 1-h^{2} \right )OL$
Substituting

$x=1$ , we get

$\displaystyle y=1+2h-h^{2}$

$\displaystyle \therefore M is \left ( 1,1+2h-h^{2} \right ),N\left ( 1,0 \right )$
A=area of trapezium

$\displaystyle =\dfrac{1}{2}\left ( OL+MN \right ).ON$

$\displaystyle =\dfrac{1}{2}\left [ 1-h^{2}+12h-h^{2} \right ].1$

$\displaystyle A=(1+h-h^{2})$

$\displaystyle \therefore \dfrac{dA}{dh}=1-2h=0$

$\therefore h=\dfrac{1}{2}$ and

$\displaystyle \dfrac{d^{2}A}{dh^{2}}=-2=-ive$ Hence max.

$\displaystyle \therefore$ Point P is

$\displaystyle \left ( \dfrac{1}{2}.\dfrac{5}{4} \right ).$

Ans: B

At which point the tangent to

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

$\displaystyle \left ( 4am^{2},8am^{3} \right )$ cuts the curve again.

0%
$\displaystyle \left ( am^{4},-am^{2} \right ).$
0%
$\displaystyle \left ( am^{3},-am^{2} \right ).$
0%
$\displaystyle \left ( am^{2},-am^{3} \right ).$
0%
$\displaystyle \left ( am^{1},-am^{3} \right ).$

Which of tangents to the curve

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

$x+2y=0$ .

0%
$\displaystyle 2x+4y+3\pi = 0$
0%
$\displaystyle x+4y-\pi = 0.$
0%
$\displaystyle 2x+4y-\pi = 0.$
0%
$\displaystyle x-4y-3\pi = 0.$

Explanation

Given

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

$\displaystyle \dfrac{dy}{dx}= -\sin \left ( x+y \right ).\left [ 1+\dfrac{dy}{dx} \right ]$ ...(1)

Since the tangent is parallel to

$x+2y=0$

So, slope of tangent

$=-\dfrac{1}{2}$

$\therefore \displaystyle \dfrac{dy}{dx}= slope= -\dfrac{1}{2}.$

$\displaystyle \sin \left ( x+y \right ) = \sin \left ( \dfrac{\pi}{2} \right ).$

$\Rightarrow x+y=\dfrac{\pi}{2}$

$\displaystyle \therefore \cos \left ( x+y \right )= 0$

$\displaystyle \therefore y= \cos \left ( x+y \right )= 0$

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

$\Rightarrow \sin x= 1$ (

$\because y= 0$ )

$\displaystyle \therefore x= \dfrac{\pi }{2}, -\dfrac{3\pi }{2}$ as

$\displaystyle -2\pi < x< 2\pi$

Hence the points are

$\displaystyle \left [\dfrac{-3\pi }{2} ,0 \right ]$ and

$\displaystyle \left [\dfrac{\pi}{2},0 \right ]$ where the tangents are parallel to the line x+2y=0.

$\displaystyle \therefore$ The equations of tangents are:

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

$\Rightarrow \displaystyle 2x+4y+3\pi = 0$

and

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

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

The point (s) on the curve

$\displaystyle y^{3}+3x^{2}= 12y,$ where the tangent is vertical (i.e., parallel to the y-axis), is / true

0%
$\displaystyle \left ( \pm \frac{4}{\sqrt{3}},-2 \right )$
0%
$\displaystyle \left ( \pm \frac{\sqrt{11}}{3},1 \right )$
0%
$\displaystyle \left ( 0, 0 \right )$
0%
$\displaystyle \left ( \pm \frac{4}{\sqrt{3}},2 \right )$

Explanation

We require the point where

$\dfrac{dx}{dy}=0$
Or

$3y^{2}.dy+6x.dx=12dy$
Or

$dy(3y^{2}-12)=-6xdx$
Or

$\dfrac{-dx}{dy}=\dfrac{3y^{2}-12}{6}$

$=0$
Hence

$3y^{2}-12=0$

$y^{2}-4=0$

$y=\pm 2$
Now

$y=-2$ does not satisfy the equation the curve.
Substituting in the equation of the curve, y=2,

$8+3x^{2}=24$

$3x^{2}=16$

$x=\pm\dfrac{4}{\sqrt{3}}$ .

Let the equation of a curve be

$x=a\left ( \theta +\sin \theta \right )$ ,

$y=a\left ( 1-\cos \theta \right )$ . If

$\theta$ changes at a constant rate

$k$ then the rate of change of slope of the tangent to the curve at

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

0%
$\displaystyle \frac{2k}{\sqrt{3}}$
0%
$\displaystyle \frac{k}{\sqrt{3}}$
0%
$k$
0%
none of these

Explanation

$\theta$ change with constant rate

$\Rightarrow \dfrac{d\theta}{dt}=k$

$x=a\left( \theta +sin\theta \right)$

$\cfrac { dx }{ d\theta } =a\left( 1+cos\theta \right)$

$y=a\left( 1-cos\theta \right)$

$\cfrac { dy }{ d\theta } =asin\theta$

Then,

$\cfrac { dy }{ dx } =\cfrac { asin\theta }{ a\left( 1+cos\theta \right) }$

Rate of change of slope

$=\dfrac{d}{dt}. \dfrac{dy}{dx} = \dfrac{1}{1+\cos\theta}.\dfrac{d\theta}{dt} = k$ , at

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

The positive value of

$k$ for which

$ke^x-x=0$ has only one real solution is

0%
$\dfrac{1}{e}$
0%
1
0%
e
0%
$\log_{e}2$

Explanation

$ke^{ x }-x=0$ will have only one solution when

$g(x)=x$ is tangent to

$f(x)=ke^{x}$
let

$P(a,b)$ be a point on

$f(x)$ and it also satisfy

$g(x)$
Therefore,

$ke^{ a }=a$ ...(1)
slope of

$g(x)=$ slope of

$f(x)$ at point P

$\Rightarrow 1=\dfrac { dy }{ dx }$ at point P

$=k{ e }^{ a }$
From (1), we get

$a=1$
Therefore,

$k=1/e$

Ans: A

Find the co-ordinates of the point (s) on the curve

$\displaystyle y= \frac{x^{2}-1}{x^{2}+1}, x> 0$ such that tangent at these point (s)have the greatest slope.

0%
$\displaystyle \left ( \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{2}} \right ).$
0%
$\displaystyle \left ( \frac{1}{\sqrt{3}},-\frac{1}{{2}} \right ).$
0%
$\displaystyle \left ( \frac{1}{{3}},-\frac{4}{{5}} \right ).$
0%
$\displaystyle \left ( {\sqrt{3}},\frac{1}{{2}} \right ).$

Explanation

$\displaystyle y= \dfrac{x^{2}-1}{x^{2}+1}= \dfrac{x^{2}+1-2}{x^{2}+1}= 1-\dfrac{2}{x^{2}+1}$

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

S=slope

$\displaystyle = \dfrac{dy}{dx}= \dfrac{4x}{\left ( x^{2}+1 \right )^{2}}$

For maximum and minimum of S,

$\displaystyle \dfrac{dS}{dx}= 0$

$\displaystyle \dfrac{dS}{dx}= 4\dfrac{\left ( x^{2}+1 \right )^{2}.1-x.2\left ( x^{2}+1 \right ).2x}{\left ( x^{2}+1 \right )^{4}}$

$\displaystyle \dfrac{dS}{dx}= 4\dfrac{x^{2}+1-4x^{2}}{\left ( x^{2}+1 \right )^{3}}$

$\displaystyle \dfrac{dS}{dx}= 4\dfrac{1-3x^{2}}{\left ( 1+x^{2} \right )^{3}}=0$

$\displaystyle \therefore x= \dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}$

Now consider change of sign at

$\displaystyle x= \dfrac{1}{\sqrt{3}}, \dfrac{dS}{dx}$ changes from +ve to -ve.

Hence S has maximum at

$\displaystyle x= \dfrac{1}{\sqrt{3}}$ .

When

$\displaystyle x= \dfrac{1}{\sqrt{3}}$ the value of

$\displaystyle y= -\dfrac{1}{2}$

Hence the point is

$\displaystyle \left ( \dfrac{1}{\sqrt{3}},-\dfrac{1}{\sqrt{2}} \right ).$

The normal to the curve given by

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

$\displaystyle \theta$ is such that it

0%
makes a constant angle with x-axis
0%
is at a constant distance from the origin
0%
touches a fixed circle
0%
passes through the origin

Explanation

$x=a\left( \cos { \theta } +\theta \sin { \theta } \right)$ and

$y=a\left( \sin { \theta } -\theta \cos { \theta } \right)$

$\displaystyle \dfrac { dx }{ d\theta } =a\left( -\sin { \theta } +\sin { \theta } +\theta \cos { \theta } \right) =a\theta \cos { \theta }$

$\displaystyle \dfrac { dy }{ d\theta } =a\left( \cos { \theta } -\cos { \theta } +\theta \sin { \theta } \right) =a\theta \sin { \theta }$

$\displaystyle \therefore \dfrac { dy }{ dx } =\tan { \theta }$

So, the equation of the normal is

$\displaystyle y-a\sin { \theta } +a\theta \cos { \theta } =-\dfrac { \cos { \theta } }{ \sin { \theta } } \left( x-a\cos { \theta } -a\theta \sin { \theta } \right)$

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

$\Rightarrow x\cos { \theta } +y\sin { \theta } =a$

It is always at a constant distance

$a$ from the origin.

The point of intersection of the tangents drawn to the curve

$\displaystyle x^{2}y= 1-y$ at the point where it is intersected by the curve xy

$=1-y,$ is given by

0%
$(0, 1)$
0%
$(1, 1)$
0%
$(0, -1)$
0%
none of these

Explanation

$point\quad of\quad intersection\quad of\quad tangents$

$to\quad the\quad curve\quad will\quad be\quad the\quad intersection\quad point\quad of\quad curves.\\ xy=1-y\\ y=\dfrac { 1 }{ 1+x } \\ another\quad curve\quad { x }^{ 2 }y=1-y\\ { x }^{ 2 }.\dfrac { 1 }{ 1+x } =1-\dfrac { 1 }{ 1+x } \\ \dfrac { { x }^{ 2 } }{ 1+x } =\dfrac { x }{ 1+x } \\ x=0\quad so\quad y=1.$

If the sum of the squares of the intercepts on the axes cut off by the tangent to the cuve

$\displaystyle x^{1/3}+y^{1/3}= a^{1/3}\left ( a> 0 \right )$ at

$\displaystyle \left ( \dfrac{a}{8}, \dfrac{a}{8} \right )$ is

$2$ , then

$a$ has the value

0%
$1$
0%
$2$
0%
$4$
0%
$8$

Explanation

$\displaystyle x^{1/3}+y^{1/3}= a^{1/3}\left ( a> 0 \right )$

$y'(x)= -\cfrac{y^{2/3}}{x^{2/3}}$

$\left.y'\right|_{(x,y)=\left(\dfrac{a}{8}, \dfrac{a}{8}\right)} = -1$

Equation of tangent

$x+y =c$

$x +y = \dfrac{a}{4};\>\>\>\>\>\> \because \left(\dfrac{a}{8}, \dfrac{a}{8}\right)$ satisfies it

$x-$ intercept

$= a\dfrac{a}{4} = y-$ intercept
Given,

$\cfrac{a^2}{16}+\cfrac{a^2}{16} = 2$

$a = \pm 4$

$a =4;\>\>\>\>\>\because a>0$

The point(s) on the curve

$y^{3}+3x^{2}=12y$ the tangent is vertical is (are)

0%
$\left ( \pm 4/\sqrt{3}\: -2 \right )$
0%
$\left ( \pm \sqrt{11/3},1 \right )$
0%
$\left ( 0,0 \right )$
0%
$\left ( \pm 4/\sqrt{3}\: ,2 \right )$

Explanation

$3y^{2}.y'+6x=12y'$
Or

$y'(3y^{2}-12)=-6x$
Or

$y'(y^{2}-4)=-2x$
Or

$y'=\dfrac{-2x}{y^{2}-4}$
Hence for the tangent to be vertical

$y^{2}-4=0$

$y=\pm2$
Substituting

$y=2$ , we get

$3x^{2}=12y-y^{3}$

$=24-8$

$=16$
Hence

$x=\dfrac{\pm4}{\sqrt{3}}$

$y=-2$ gives

$3x^{2}=-16$
This is not possible.
Hence the required point is

$(\dfrac{\pm4}{\sqrt{3}},2)$ .

The families of curves defined by the equations

$\displaystyle y= ax, y^{2}+x^{2}= c^{2}$ are perpendicular for

0%
a $= 2,$ c $= 4$
0%
a $= -2,$ c $= 3$
0%
a $= 3,$ c $= 2$
0%
a $= 3,$ c $= -2$

Explanation

Since, the straight line

$y=ax$ passes through the center of the circle

$x^{2}+y^{2}=c^{2}$
Therefore,

$y=ax$ is normal to circle

$x^{2}+y^{2}=c^{2}$ always for all values of

$a$ &

$c$

Ans: A,B,C,D

The coordinates of the point

$P$ on the curve

$y^{2}= 2x^{3}$ the tangent at which is perpendicular to the line

$4x-3y + 2 = 0$ , are given by

0%
$\left ( 2,4 \right )$
0%
$\left ( 1,\sqrt{2} \right )$
0%
$\left ( 1/2,-1/2 \right )$
0%
$\left ( 1/8,-1/16 \right )$

Explanation

$4x-3y+2=0$
Or

$3y=4x+2$
Or

$y=\dfrac{4}{3}x+\dfrac{2}{3}$ .
Hence slope of the tangent will be

$=\dfrac{-3}{4}$
Now

$\dfrac{dy}{dx}=\dfrac{-3}{4}$

$y=\pm\sqrt{2}.x^{\frac{3}{2}}$
Hence

$\dfrac{dy}{dx}$

$=\pm\sqrt{2}.\dfrac{3}{2}.x^{\frac{1}{2}}$

$=\dfrac{-3}{4}$

Or

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

$2\sqrt{2}.x^{\dfrac{1}{2}}=1$
Or

$x=\dfrac{1}{8}$ .
Hence

$y=-\sqrt{2}.x^{\dfrac{3}{2}}$

$=-\sqrt{2}.\dfrac{1}{16\sqrt{2}}$

$=-\dfrac{1}{16}$
Hence the co-ordinates are

$(\dfrac{1}{8},\dfrac{-1}{16})$

The tangent to the curve

$x= a\sqrt{\cos 2\theta }\cos \theta$ ,

$y= a\sqrt{\cos 2\theta }\sin \theta$ at the point corresponding to

$\theta = \pi /6$ is

0%
parallel to the $x$ -axis
0%
parallel to the $y$ -axis
0%
parallel to line $y = x$
0%
none of these

Explanation

$\displaystyle \dfrac{dx}{d\theta }= -a\sqrt{\cos 2\theta }\sin \theta +\dfrac{-a\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$

$\displaystyle = -a\dfrac{\cos 2\theta \sin \theta +\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$

$\displaystyle = \dfrac{-a\sin 3\theta }{\sqrt{\cos 2\theta }}$

$\displaystyle \dfrac{dy}{d\theta }= a\sqrt{\cos 2\theta }\, \cos \theta -a\dfrac{\sin \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$

\displaystyle = \dfrac{a\cos 3\theta }{\sqrt{\cos 2\theta }}$$

Hence

$\displaystyle \dfrac{dy}{dx}= -\cot 3\theta \Rightarrow \dfrac{dy}{dx}|_{\theta = \pi /6 }= 0$

So the tangent to the curve at

$\theta = \pi /6$ is parallel to the

$x$ -axis.

The equation of the tangent line at an inflection point of

$\displaystyle f\left ( x \right )=x^{4}-6x^{3}+12x^{2}-8x+3$ is

0%
y $= 3x+4$
0%
y $= 4$
0%
y $= 3x+2$
0%
none of these

Explanation

Given,

$\displaystyle f\left ( x \right )=x^{4}-6x^{3}+12x^{2}-8x+3$
At inflection point

$f''(x) = 0$

$12x^2-36x+24 = 0$

$(x-2)(x-1) = 0$

$x =2,1$

$f'(x) = 4x^3-18x^2+24x-8$

$f'(1) = 2$ and

$f'(2) = 0$

$f(1) = 2$ and

$f(2) = 3$
So the equations are

$y=3$ and

$y=2x$

The equation of the common tangent to the curves

$\displaystyle y^{2}= 8x$ and

$\displaystyle xy= -1$ is

0%
$3y =9x\:+\:2$
0%
$y =2x\:+\:1$
0%
$2y = x\:+ \:8$
0%
$y = x\:+\:2$

Explanation

A point on the curve

$xy = -1$ is of the form

$\left ( t,-1/t \right )$ .Now

$\displaystyle y= -\dfrac{1}{x}\Rightarrow \dfrac{dy}{dx}= \dfrac{1}{x^{2}}\Rightarrow \dfrac{dy}{dx}|_{\left ( t,-1/t \right )}= \dfrac{1}{t^{2}}$

$\therefore$ Equation of tangent to the curve

$xy = -1$ at

$\left ( t,-1/t \right )$ is

$\displaystyle y+\dfrac{1}{t}= \dfrac{1}{t^{2}}\left ( x-t \right )\Rightarrow y= \dfrac{1}{t^{2}}\, x-\dfrac{1}{t}$

For this line to be tangent to the parabola

$y^{2}= 8x$ it should

be of the form

$y=mx+\displaystyle \dfrac{2}{m}$ so

$\displaystyle \dfrac{1}{t^{2}}$ and

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

$m=-1$

$\displaystyle \therefore -t= \dfrac{1}{t^{2}}\Rightarrow -t^{3}= 1\Rightarrow t= -1$

Thus the required tangent is

$y = x + 2$ .

The equation of the tangents to

$\displaystyle 4x^{2}-9y^{2}=36$ which are perpendicular to the straight line

$\displaystyle 2y+5x= 10$ are

0%
$\displaystyle 5\left ( y-3 \right )=2 \left ( x-\sqrt{\frac{117}{4}} \right )$
0%
$\displaystyle 5\left ( y-2 \right )=2 \left ( x-\sqrt{-18} \right )$
0%
$\displaystyle 5\left ( y+2 \right )=2 \left ( x-\sqrt{-18} \right )$
0%
none of these

Explanation

Given,

$2y+5x=10$

$2y'+5=0$

$y'=\dfrac{-5}{2}$ .
Hence the required slope is

$\dfrac{2}{5}$ .
Now

$4x^{2}-9y^{2}=36$ .

$4x^{2}-36=9y^{2}$

$y=\dfrac{1}{3}.\sqrt{4x^{2}-36}$ .

$y'=\dfrac{8x}{6\sqrt{4x^{2}-36}}$

$=\dfrac{2}{5}$

$40x=12\sqrt{4x^{2}-36}$

$10x=3\sqrt{4x^{2}-36}$

$100x^{2}=36x^{2}-324$

$64x^{2}=-324$

$x^{2}=\dfrac{-81}{16}$ .
Hence no real values of x.
Hence answer is none of these.

Of all the line tangent to the graph of the curve

$\displaystyle y=\frac{6}{x^{2}+3},$ find the equations of the tangent lines of minimum and maximum slope respectively.

0%
$3x + 4y - 9 = 0; 3x - 4y + 9 =0$
0%
$3x + 4y + 9 = 0; 3x - 4y + 9 =0$
0%
$4x + 3y - 9 = 0; 4x - 3y - 9 =0$
0%
$4x - 3y - 9 = 0; 4x + 3y - 9 =0$

Explanation

Slope of the tangents will be

$y'=\dfrac{-12x}{(x^{2}+3)^{2}}$
Now for maximum and minimum slopes

$y"=0$
Or

$-12(x^{2}+3)^{2}+24x(x^{2}+3).2x=0$
Or

$(x^{2}+3)[48x^{2}-12x^{2}-36]=0$
Or

$36x^{2}-36=0$
Or

$x=\pm1$ .
Hence

$y'(1)=\dfrac{-12}{16}=\dfrac{-3}{4}$
And

$y'(-1)=\dfrac{3}{4}$ .
Hence

$y(1)=y(-1)=\dfrac{6}{4}=\dfrac{3}{2}$ .
Hence the equations are

$y-\dfrac{3}{2}=\dfrac{3}{4}(x+1)$
Or

$4y-6=3x+3$
Or

$3x-4y+9=0$
And

$y-\dfrac{3}{2}=\dfrac{-3}{4}(x-1)$
Or

$4y-6=-3x+3$
Or

$3x+4y-6=0$ .

The equations of the tangents to the curve

$y=x^{4}$ from the point

$(2, 0)$ not on the curve, are given by

0%
$y = 0$
0%
$y - 1 = 5(x -1)$
0%
$\displaystyle y-\frac{4098}{81}=\frac{2048}{27}\left ( x-\frac{8}{3} \right )$
0%
$\displaystyle y-\frac{32}{243}=\frac{80}{81}\left ( x-\frac{2}{3} \right )$

Explanation

Let

$\left ( x_{0},x_{0}^{4} \right )$ be the point of tangency.

Then the equation of the tangent will be

$y-x_{0}^{4}=y(x_{0})\left ( x-x_{0} \right )$ . Since this tangent passes through the point (2, 0), we have

$-x_{0}^{4}=4x_{0}^{3}\left ( 2-x_{0} \right )$ , or

$3x_{0}^{4}-8x_{0}^{3}=0$

That is,

$x_{0}=0$ or

$x_{0}=8/3$ , so that the points of tangency are (0, 0) and (8/3, 4096/81). Therefore, the equations of the tangents are

$y = 0$ and

$\displaystyle y-\frac{4098}{81}=\frac{2048}{27}\left ( x-\frac{8}{3} \right )$

For the curve

${x}^{2}+4xy+8{y}^{2}=64$ the tangents are parallel to the

$x$ -axis only at the points

0%
$(0,2\sqrt { 2 } )$ and $(0,-2\sqrt { 2 } )$
0%
$(8,-4)$ and $(-8,4)$
0%
$(8\sqrt { 2 } ,-2\sqrt { 2 } )$ and $(-8\sqrt { 2 } ,2\sqrt { 2 } )$
0%
$(9,0)$ and $(-8,0)$

Explanation

Given curve is

${x}^{2}+4xy+8{y}^{2}=64.....(i)$
On differentiating w.r.t

$x$ ,we get

$2x+4(y+x\cfrac{dy}{dx})+16y\cfrac{dy}{dx}=0$

$\Rightarrow$

$2x+4y+(4x+16y)\cfrac{dy}{dx}=0$

$\Rightarrow$

$\cfrac{dy}{dx}=-\cfrac{(x+2y)}{2(x+4y)}$
Since, tangent are parallel to

$x$ -axis only
ie.,

$\cfrac{dy}{dx}=0$

$\Rightarrow$

$-\cfrac{(x+2y)}{2(x+4y)}=0$

$\Rightarrow$

$x+2y=0.....(ii)$
Now, on putting the values of

$x$ from eqs.

$(i)$ in

$(ii)$ we get

$4{y}^{2}-8{y}^{2}+8{y}^{2}=64$

$\Rightarrow$

${y}^{2}=16$

$\Rightarrow$

$y=\pm 4$
from eq.

$(ii)$
When

$y=4,x=-8$
and when

$y=-4, x=8$
Hence, required points are

$(-8,4)$ and

$(8,-4)$

$y = 4x - 5$ is a tangent to the curve

$y^{2} = px^{3} + q$ at

$(2, 3)$ , then

0%
$p = 2, q = -7$
0%
$p = -2, q = 7$
0%
$p = -2, q = -7$
0%
$p = 2, q = 7$

Explanation

Since,

$(2, 3)$ lies on the curve

$y^{2} = px^{3} + q$
Therefore,

$9 = 8p + q ..... (i)$

$y^{2} = px^{3} + q$

$\Rightarrow 2y \dfrac {dy}{dx} = 3px^{2}$

$\dfrac {dy}{dx} = \dfrac {3px^{2}}{2y}$

$\Rightarrow \left (\dfrac {dy}{dx}\right )_{(2, 3)} = \dfrac {12p}{6} = 2p$
Since,

$y = 4x - 5$ is tangent to

$y^{2} = px^{3} + q$ at

$(2, 3)$ . Therefore,

$\therefore \left (\dfrac {dy}{dx}\right )_{(2, 3)} =$ Slope of the line

$y = 4x - 5$

$\Rightarrow 2p = 4 \Rightarrow p = 2$
Putting

$p = 2$ in Eq. (i), we get

$q = -7$ .

Find the point of intersections of the tangets drawn to the curve

$\displaystyle x^{2}y=1-y$ at the points where it is intersected by the curve

$xy = 1 - y$

0%
$(1, 0)$
0%
$(0, 1)$
0%
$(\dfrac{2}{3},\dfrac{1}{3})$
0%
$(\dfrac{1}{3},\dfrac{2}{3})$

Explanation

Given equation of curves

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

$xy=1-y$ ....(2)

Let

$P(x_1,y_1)$ be the point of intersection of the curves.

$\Rightarrow x_1^2y_1=x_1y_1$

$\Rightarrow x_1y_1(x_1-1)=0$

$\Rightarrow x_1=0, x_1=1$

$\Rightarrow y_1=1, \dfrac{1}{2}$
So, the point of intersection of the curves are

$(0,1)$ and

$(1,\dfrac{1}{2})$

Slope of tangent to curve (1),

$2xy+x^2y'=-y'$

$\Rightarrow y'=\dfrac{-2xy}{x^2+1}$
Slope of tangent to curve (1) at

$(0,1)$ is

$0.$
Slope of tangent to curve (1) at

$(1,\dfrac{1}{2})$ is

$-\dfrac{1}{2}$

Equation of tangent to curve (1) at

$(0,1)$ is

$y-1=0$ .....(3)

Equation of tangent to curve (1) at

$(1,\dfrac{1}{2})$ is

$y-\dfrac{1}{2}=-\dfrac{1}{2}(x-1)$

$\Rightarrow 2y+x-2=0$ ....(4)

Slope of tangent to curve (2),

$xy'+y=-y'$

$y'=-\dfrac{y}{1+x}$

Slope of tangent to curve (2) at

$(0,1)$ is

$-1$ .
Slope of tangent to curve (2) at

$(1,\dfrac{1}{2})$ is

$-\dfrac{1}{4}$

Equation of tangent to curve (2) at

$(0,1)$ is

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

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

Equation of tangent to curve (2) at

$(1,\dfrac{1}{2})$ is

$y-\dfrac{1}{2}=-\dfrac{1}{4}(x-1)$

$\Rightarrow x+4y-3=0$ ....(6)

Solving (3) and (5),we get

$x=0,y=1$

Solving (4) and (6), we get

$x=\dfrac{1}{3}, y=\dfrac{2}{3}$

So, the required points are

$(0,1)$ and

$(\dfrac{1}{3},\dfrac{2}{3})$

If the tangent is drawn to the curve y = f(x) at a point where it crosses the y - axis then its equation is

0%
$x - 4y = 2$
0%
$x + 4y = 2$
0%
$x + 4y +2 = 0$
0%
none of these

Explanation

It is given that

$\dfrac{df(x)}{dx}=f(x)^{2}$
Let

$f(x)=y$
Then

$y'=y^{2}$
Or

$\int y^{-2}.dy=\int dx$
Or

$\dfrac{-1}{y}=x+c$
Now

$y=\dfrac{-1}{2}$ at x=0.
Hence

$c=2$
Or

$-\dfrac{1}{y}=x+2$
Or

$xy+2y+1=0$ ...(i)

$\dfrac{dy}{dx}_{x=0}$

$=y^{2}_{y=\frac{-1}{2}}$
Or

$y'=\dfrac{1}{4}$ .
Now equation of the tangent is

$y-(\dfrac{-1}{2})=\dfrac{1}{4}(x-0)$
Or

$4y+2=x$
Or

$x-4y=2$ is the required equation of tangent.

If the line

$ax + by + c = 0$ is a normal to the curve

$xy = 1$ . Then

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

Explanation

$ax+by+c=0$
Hence

$by=-ax-c$
Or

$y=\dfrac{-a}{b}x-c$
Hence

$m=\dfrac{-a}{b}$
= Slope of normal.
Now

$xy'+y=0$
Or

$y'=\dfrac{-y}{x}$
Or

$y'=\dfrac{-1}{x^{2}}$
Thus

$\dfrac{-1}{y'}$

$=x^{2}$
Hence

$x^{2}=\dfrac{-a}{b}$
Or
Now

$x^{2}>0$ thus

$\dfrac{-a}{b}$ has to be positive.
Hence we are left with 2 options.
Either

$a<0$ and

$b>0$
Or

$a>0$ and

$b<0$ .

The coordinates of the points(s) at which the tangents to the curve

$\displaystyle y=x^{3}-3x^{2}-7x+6$ cut the positive semi axis OX a line segment half that on the negative semi axis OY is/are given by

0%
$(-1, 9)$
0%
$(3, -15)$
0%
$(1, -3)$
0%
none

Explanation

Let the point of tangency be

$(x_{1}+y_{1})$

Then the equation of tangent is

$\dfrac{y-y_{1}}{x-x_{1}}=3x_{1}^{2}-6x_{1}-7$

when

$y=0,x=x_{1}+\dfrac{-y}{3x_{1}-6x_{1}-7}$

when

$x=0,y=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$

$-2(x_{1}+\dfrac{-y}{3x_{1}-6x_{1}-7})=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$

$-2(\dfrac{x_{1}(3x_{1}-6x_{1}-7)-y_{1}}{3x_{1}-6x_{1}-7})=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$

$2=3x_{1}-6x_{1}-7$

$3x_{1}-6x_{1}-9=0$

$x_{1}=-1 or 3$

when

$x_{1}=-1,y_{1}=(-1)^{3}-3(-1)^{2}-7(-1)+6=9$

and the y-intercept is

$9-(-1)[3(-1)^{2}-6(-1)-7]=11> 0$

when

$x_{1}=3,y_{1}=(3)^{3}-3(3)^{2}-7(3)+6=-15$

and the y-intercept is

$-15-(3)[3(3)^{2}-6(3)-7]=-21< 0$

So,the point of tangency is

$(-1,9)$ , this is the correct answer.

What normal to the curve

$\displaystyle y=x^{2}$ form the shortest chord?

0%
$\displaystyle x+\sqrt{2}y=\sqrt{2}$ or $\displaystyle x-\sqrt{2}y=-\sqrt{2}$
0%
$\displaystyle x+\sqrt{2}y=\sqrt{2}$ or $\displaystyle x-\sqrt{2}y=\sqrt{2}$
0%
$\displaystyle x+\sqrt{2}y=-\sqrt{2}$ or $\displaystyle x-\sqrt{2}y=-\sqrt{2}$
0%
$\displaystyle x+\sqrt{2}y=-\sqrt{2}$ or $\displaystyle x-\sqrt{2}y=\sqrt{2}$

Explanation

Any point on curve

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

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

$\displaystyle \frac{dy}{dx}=2x$ equation of normal at

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

$\displaystyle y-t^{2}=-\frac{1}{2t}\left ( x-t \right )$
solving with

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

$\displaystyle x^{2}-t^{2}=\frac{-1}{2t}\left ( x-t \right )\Rightarrow \left ( x-t \right )\left ( x+t+\frac{1}{2t} \right )=0\Rightarrow x=-t-\frac{1}{2t}$
So normal cuts the curve again at

$\displaystyle Q\left ( -t-\frac{1}{2t},\left ( -t-\frac{1}{2t} \right )^{2} \right )$

$\displaystyle z=PQ^{2}=4t^{2}\left ( 1+\frac{1}{4t^{2}} \right )^{3}$

$\displaystyle \frac{dz}{dt}=0\Rightarrow t=\pm \frac{1}{\sqrt{2}}=0,\frac{dz}{dt}$

changes sign from negative to positive about

$\displaystyle t=\frac{1}{\sqrt{2}}$ as well as

$\displaystyle t=-\frac{1}{\sqrt{2}}$ (No chord is format for t=0)
z is minimum

at

$\displaystyle t=\pm \frac{1}{\sqrt{2}}$ & minimum value

of

$\displaystyle z=PQ^{2}=3$ shortest normal chord has

length

$\displaystyle \sqrt{3}$ & its equation is

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

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

The tangent to the curve

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

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

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

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

0%
on the left of $x = c$
0%
on the right of $x = c$
0%
at no paint
0%
at all points

Explanation

Given equation of curve

$y=e^x$

$\dfrac{dy}{dx}=e^x$

Slope of tangent at point

$(c,e^c)$ is

$e^c$

Equation of tangent at

$(c,e^{c})$ is

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

Equation of straight line joining

$A\left ( c+1,e^{c+1} \right )$ and

$B\left ( c-1,e^{c-1} \right )$ is

$\displaystyle y-e^{c+1}=\dfrac{e^{c+1}-e^{c-1}}{2}\left ( x-c-1 \right )$ .....(2)

Subtracting (2) from (1), we get

$\displaystyle e^{c}\left ( e-1 \right )=e^{c}\left [ \left ( x-c \right )-\dfrac{1}{2} \left ( e-e^{-1} \right )\left ( x-c \right )+\dfrac{1}{2}\left ( e-e^{-1} \right )\right ]$

$\displaystyle \Rightarrow \dfrac{1}{2}\left ( e+e^{-1} \right )-1=\left ( x-c \right )\left [ 1-\dfrac{1}{2}\left ( e-e^{-1} \right ) \right ]$

$\displaystyle \Rightarrow x-c=\dfrac{e+e^{-1}-2}{2-e+e^{-1}}< 0$

$[ \because e+e^{-1}>2$ and

$2+e^{-1}-e<0 ]$

$\Rightarrow x<c$

Thus the two lines meet to the left of

$x = c$ .

The curve

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

0%
$(1,1)$
0%
At no point
0%
$(0,1)$
0%
$(1,0)$

The equation of the normal to the curve

$y(1+{x}^{2})=2-x$ where the tangent crosses

$x$ -axis is

0%
$5x-y-10=0$
0%
$x-5y-10=0$
0%
$5x+y+10=0$
0%
$x+5y+10=0$

Explanation

Given curve can be written as,

$y=\dfrac{2-x}{1+x^2}$

To get the point where the above curve will cross the x-axis, put

$y=0$

$\Rightarrow 0=\dfrac{2-x}{1+x^2}\Rightarrow x=2$

Thus the required point is

$(2,0)$

Now differentiate the given curve using quotient rule,

$\dfrac{dy}{dx}=\dfrac{(1+x^2)(-1)-(2-x)(2x)}{(1+x^2)^2}$

$\quad =\dfrac{x^2-4x-1}{(1+x^2)^2}$

Thus the slope of a tangent to this curve at point

$(2,0)$ is

$m=\dfrac{dy}{dx}\bigg|_{(2,0)}=\dfrac{2^2-4(2)-1}{(1+2^2)^2}=-\dfrac{5}{25}=-\dfrac{1}{5}$

Therefore the slope of normal at this point will be

$5$

Hence required equation of normal is given by,

$(y-0)=5(x-2)$

$\Rightarrow 5x-y-10=0$

The curve given by

$x+y={ e }^{ xy }$ has a tangent parallel to the y-axis at the point

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

Explanation

Given curve is

$x+y={ e }^{ xy }\quad$
On differentiating w.r.t

$x$ we get

$1+\cfrac { dy }{ dx } ={ e }^{ xy }\left\{ y+x\cfrac { dy }{ dx } \right\}$

$\Rightarrow \cfrac { dy }{ dx } =\cfrac { y{ e }^{ xy }-1 }{ 1-x{ e }^{ xy } }$

Since, tangent is parallel to y-axis
Here,

$\cfrac { dy }{ dx } =\infty \Rightarrow 1-x{ e }^{ xy }=0$

$\Rightarrow 1-x(x+y)=0$
This holds for

$x=1,y=0$

Abscissa of

$p_{1}, p_{2}, p_{3} .... p_{n}$ are in

0%
A.P.
0%
G.P.
0%
H.P
0%
None

If the slope of the curve

$y=\cfrac { ax }{ b-x }$ at the point

$(1,1)$ is

$2$ , then the values of

$a$ and

$b$ are respectively

0%
$1,-2$
0%
$-1,2$
0%
$1,2$
0%
None of these

Explanation

$y=\cfrac { ax }{ b-x }$

Slope=

$\cfrac { dy }{ dx }$

So,

$\cfrac { dy }{ dx }$ =

$\cfrac { a(b-x)-(-1)(ax) }{ { (b-1) }^{ 2 } } \quad \longrightarrow (1)$

for

$(1,1)$

put in equation of curve-

$1=\cfrac { a }{ (b-1) } \quad \Rightarrow (b-1)=a\quad \longrightarrow (2)$

From slop after putting (1,1)

We get,

$\\ 2=\cfrac { a(b-1)+a }{ { (b-1) }^{ 2 } }$

put

$(b-1)=a$ from equation(2)

$2=\cfrac { { a }^{ 2 }+a }{ { a }^{ 2 } } \quad \quad \\ \Rightarrow 2{ a }^{ 2 }={ a }^{ 2 }+a\\ \quad \Rightarrow a(a-1)=0\\ \quad \Rightarrow a=0,1$

put in (2),

We get

$b=1,2$

In option only one value is given

$\therefore a=1,b=2$

The curve that passes through the point

$(2,3)$ and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by

0%
${ \left( \cfrac { x }{ 2 } \right) }^{ 2 }+{ \left( \cfrac { y }{ 3 } \right) }^{ 2 }=2\quad$
0%
$2y-3x=0$
0%
$y=\cfrac { 6 }{ x }$
0%
${ x }^{ 2 }+{ y }^{ 2 }=13$

Explanation

Let

$B$ be the pt. of contact

$y=f(x)$ be the curve

then

$AB=BC$ (given condition)

$C\equiv (2x,0)$

$A\equiv (0,2y)$

Slope

$m$ of line

$AC$ is

$m=\dfrac{dy}{dx}=\dfrac{0.2y}{2x-0}$

$\dfrac{dy}{dx}=\dfrac{-y}{x}$

$\dfrac{dy}{y}=\dfrac{-dx}{x}$

$\ln y=-\ln x+c$

$yx=c$

$(2, 3)$ satisfies the curve

$\therefore 2\times 3=c$

$\therefore y=\dfrac{6}{x}$ is the curve

$\therefore$ Option

$C$ is correct

1453826_765333_ans_aed4145d8126434bb1c77800e58e1cb6.png

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 12 - 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 12 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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