MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 5

The sum of the intercepts made on the axes of coordinates by any tangent to the curve

$\sqrt{x}+\sqrt{y}=2$ is equal to

0%
$4$
0%
$2$
0%
$8$
0%
none of these

Explanation

Let point

$\left( p,q \right)$ lies on the curve

$\sqrt { x } +\sqrt { y } =2$ , such that

$\sqrt { p } +\sqrt { q } =2$
Slope is

$\cfrac { dy }{ dx } =-\sqrt { \cfrac { p }{ q } }$
And equation of tangent is

$y-q=-\sqrt { \cfrac { p }{ q } } (x-p)$
Therefore X intercept is

$\sqrt { p } \left( \sqrt { p } +\sqrt { q } \right)$
abd Y intercept is

$\sqrt { q } \left( \sqrt { p } +\sqrt { q } \right)$
Hence sum of intercept is

$\sqrt { p } \left( \sqrt { p } +\sqrt { q } \right) +\sqrt { q } \left( \sqrt { p } +\sqrt { q } \right) \\ =2\left( \sqrt { p } +\sqrt { q } \right) =4$

A tangent to the curve

$y=\displaystyle \int_{0}^{x}\left | t \right |dt$ , which is parallel to the line y=x, cuts off an intercept from the y-axis equal to

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

Explanation

Since the integral is in the first quadrant

$|t|=t$
Thus

$\int_{x=0} ^{x} |t|.dt$

$=\int_{x=0} ^{x} t.dt$

$=\dfrac{x^{2}}{2}$
Or

$2y=x^{2}$ is the equation of the curve.
Now

$2y'=2x$
Or

$y'=x$ ...(i)
Now the tangent is parallel to

$y=x$
Hence slope

$=y'=1$
Thus the x-coordinate of point of contact is 1.
Hence

$y=\dfrac{1}{2}$ .
Hence equation of the tangent will be

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

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

$y=x-\dfrac{1}{2}$ .
Hence intercept cut on the y axis is

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

Slope of Normal to the curve

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

$(-1, 0)$ is

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

Explanation

We have:

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

Now differentiate,

$\dfrac{dy}{dx}=2x+2x^{-3}$

So slope of tangent at

$(-1,0)$ is

$\dfrac{dy}{dx}\bigg|_{x=-1}=2(-1)+2(-1)^3=-2-2=-4$

Hence slope of normal at the same point is

$=\dfrac{1}{4}$

Note: Normal and tangent are perpendicular to each other

If the normal to the curve

$\displaystyle y= f\left ( x \right )$ at the point

$\displaystyle \left ( 3, 4 \right )$ makes an angle

$\displaystyle \frac{3\pi}{4}$ with the positive x-axis then

$\displaystyle f'\left ( 3 \right )$ is equal to

0%
$-1$
0%
$\displaystyle -\frac{3}{4}$
0%
$\displaystyle \frac{4}{3}$
0%
$1$

Explanation

Slope of normal

$=\dfrac{-1}{f'(x)}$

$=tan(\dfrac{3\pi}{4})$

$=-1$
Hence

$f'(x)=1$
Or

$\dfrac{dy}{dx}=1$
Or

$y=x+c$
Hence

$y=f(x)$ is an equation of a straight line parallel to

$y=x$ .
Hence

$f'(x)$ is independent of x and its value is 1.

Angle between the tangents to the curve

$y= x^{2}-5x+6$ at the points

$(2,0)$ and

$\left ( 3,0 \right )$ is

0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{\pi}{3}$
0%
$\displaystyle \frac{\pi}{6}$
0%
$\displaystyle \frac{\pi}{4}$

Explanation

Given equation of curve

$\displaystyle y=x^{2}-5x+6$

$\displaystyle \Rightarrow \frac{dy}{dx}=2x-5$

Slope of tangent to the curve at

$(2,0)$ is

$\displaystyle \left(\frac{dy}{dx}\right)_{(2,0)}=2(2)-5=-1=m_{1}$

Slope of tangent to the curve at

$(3,0)$ is

$\displaystyle \left(\frac{dy}{dx}\right)_{(3,0)}=2(3)-5=1=m_{2}$

Since

$\displaystyle m_{1}m_{2}=-1$

$\therefore$ Angle between the tangents to the curve at

$(2,0)$ and

$(3, 0)$ is

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

The number of tangents to the curve

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

$(0,1)$ is

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

Explanation

For

$x>0$ ,

$y=e^{x}$ .

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

$\dfrac{dy}{dx}_{x=0}=1$ .
Hence

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

$x-y+1=0$ is the required equation of tangent.
Similarly for

$x<0$

$y=e^{-x}$ .

$\dfrac{dy}{dx}=-e^{-x}$ .
Now

$\dfrac{dy}{dx}_{x=0}=-1$ .
Hence

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

$x+y-1=0$ is the required equation of tangent.
Hence at

$x=0$ we will have 2 tangents to the curve

$y=e^{|x|}$ which are mutually perpendicular.

The curve

$y+e^{xy}+x= 0$ has a tangent parellel to y-axis at a point

0%
$\left ( -1,\:0 \right )$
0%
$\left ( 1,\:0 \right )$
0%
$\left ( 1,\:1 \right )$
0%
$\left ( 0,\:0 \right )$

Explanation

$\dfrac{dy}{dx}+e^{xy}[x\dfrac{dy}{dx}+y]+1=0$
Or

$\dfrac{dy}{dx}[1+x.e^{xy}]+1+y.e^{xy}=0$
Or

$\dfrac{dy}{dx}=\dfrac{-(1+y.e^{xy})}{1+x.e^{xy}}$
Now

$1+xe^{xy}=0$ since the slope of the tangent is

$90^{0}$ .
Hence

$xe^{xy}=-1$
Or

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

$x^{2}+xy=1$
Or

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

$y=\dfrac{1}{x}-x$ ...(i)
Considering

$y=0$ ,

$x^{2}=1$

$x=\pm1$ .
Hence we get 2 points

$(1,0)$ and

$(-1,0)$ .
Out of these 2, only

$(-1,0)$ lies on the curve.
Hence the required point is

$(-1,0)$ .

The tangent to the curve

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

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

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

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

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

Explanation

Equation of straight line joining

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

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

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

Equation of tangent at

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

$y-{ e }^{ c }={ e }^{ c }\left( x-c \right)$ (2)

Subtracting (1) from (2),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

$x=c$

0%
Assertion is true and Reason is true; Reason is a correct explanation for Assertion.
0%
Assertion is True, Reason is true; Reason is not a correct explanation for Assertion.
0%
Assertion is true, Reason is false
0%
Assertion is false, Reason is true

Explanation

We have

$\displaystyle y={ x }^{ 2 }+bx+c\Rightarrow \dfrac { dy }{ dx } =2x+b$

Since the curve touches the line

$y=x$ at the point

$\left( 1,1 \right)$

${ \left[ 2x+b \right] }_{ \left( 1,1 \right) }^{ }=1\Rightarrow 2+b=1\Rightarrow b=-1$

Also, the curve passes through the point

$\left( 1,1 \right)$

$\therefore 1=1+b+c\Rightarrow c=-b=1$

$\displaystyle \therefore y={ x }^{ 2 }-x+1\Rightarrow \dfrac { dy }{ dx } =2x-1$

Now,

$\displaystyle \dfrac { dy }{ dx } <0\Rightarrow 2x-1<0\Rightarrow x<\dfrac { 1 }{ 2 }$

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

$\displaystyle y= x^{2}.$
The two curves

0%
intersect each other
0%
touch each other
0%
do not meet
0%
represent parabola

Explanation

$y=4x^{2}$ and

$y=x^{2}$ represent two parabola concave upward and symmetric about positive y-axis. Both the curves do not intersect, however they touch each other at

$x=0$ .

$y'=8x$ and

$y'=2x$
Now

$8x=2x$
Or

$x=0$
Hence both curves have equal slope at

$x=0$ and are horizontal (parallel to x axis)at

$x=0$

The normal to the curve

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

$y=a\sin \theta$ at

$\theta$ always passes through the fixed point

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

Explanation

$dx=a\sin \theta.d\theta$
$dy=a\cos \theta.d\theta$
Or
$\dfrac{-dx}{dy}=-\tan\theta$ .
Thus
Equation of normal will be
$y-a\sin \theta=-\tan\theta(x-a+a\cos \theta)$
Or
$\cos \theta.y-a\sin \theta.\cos \theta=-x\sin \theta+a\sin \theta-a\sin \theta \cos \theta$
$\cos \theta.y+\sin \theta.x=a\sin \theta$
Hence it passes through the fixed point $(a,0)$ .

The normal to the curve

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

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

$\theta$ is such that

0%
it makes angle $\displaystyle \frac{\pi }{2}+\theta$ with x-axis
0%
it passes through the origin
0%
it is at a constant distance from the origin
0%
it passes through $\displaystyle \left ( a\frac{\pi }{2}, -a \right )$

Explanation

$dx=a(-\sin \theta+\theta.\cos \theta+\sin \theta).d\theta$
$dx=a(\theta.\cos \theta)$
$dy=a(\theta.\sin \theta)d\theta$ .
Hence
$\dfrac{-dx}{dy}=-\cot\theta.$
$=\tan(\dfrac{\pi}{2}+\theta)$
Hence it makes an angle of $\dfrac{\pi}{2}+\theta$ with the x axis.
Now equation of normal
$y-a\sin \theta+a\theta.\cos \theta=-\cot\theta(x-a\cos \theta-\theta \sin \theta).$
Hence
$\sin \theta y-a\sin ^{2}\theta+a\theta.\cos \theta.\sin \theta=-x\cos \theta+a\cos ^{2}\theta+a\theta.\sin \theta.\cos \theta$
$y.\sin \theta+x\cos \theta=a$
Hence Distance from origin is ' $a$ '.

The curve possessing the property text the intercept made by the tangent at any point of the curve on the

$y-$ axis is equal to square of the abscissa of the point of tangency, is given by

0%
$y^{2}=x+C$
0%
$y=2x^{2}+C$
0%
$y=-x^{2}+cx$
0%
$None\ of\ these$

Which of this/these is/are tangent(s) to

$\displaystyle 3x^{2}+y^{2}+x+2y=0$ and also is/are perpendicular to the line 4x-2y=1 ?

0%
$2y+x=0$
0%
$2y-x = 2$
0%
$2y +x +4 = 0$
0%
$\displaystyle 2y+x+\frac{13}{3}=0$

Explanation

Tangent being perpendicular to given line of slope 2, will have its slope as -

$\displaystyle \dfrac{1}{2}$ .

Slope of tangent=

$\displaystyle -\dfrac{dy}{dx}=-\dfrac{6x+1}{2(y+1)}=-\dfrac{1}{2}\therefore y=6x$ .

Sloping with the given curve, we have

$\displaystyle 3x^{2}+36x^{2}+x+12x=0$

$13x(3x+1)=0\therefore x=0, -1/3 \therefore y=0, -2$

Hence the two points are

$\displaystyle (0,0),\left ( -\dfrac{1}{3},-2 \right )$

$\displaystyle \therefore y=-\dfrac{1}{2}x$ and

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

$2y+x=0$ and

$\displaystyle 2y+x+\dfrac{13}{3}=0$

Ans: D

If the tangent at P on the curve

$\displaystyle x^{m}y^{n}=a^{m+n}$ meets the co-ordinates axes at A and B, then

$AP: PB=$

0%
$m^2:n^2$
0%
$m^3:n^3$
0%
$m:n$
0%
$2m:n$

Explanation

$\displaystyle x^{m}y^{n}=a^{m+n}.$

Take

$log$ both sides

$\displaystyle m\log x+n\log y=\left ( m+n \right )\log a$

Differentiating w.r.t

$x$ , we get

$\displaystyle m\cdot \dfrac{1}{x}+n\cdot \dfrac{1}{y}\dfrac{dy}{dx}=0$

$\displaystyle \therefore \dfrac{dy}{dx}=-\dfrac{m}{n}\dfrac{y}{x}$

Hence the equation of the tangent is

$\displaystyle Y-y= -\dfrac{m}{n}\dfrac{y}{x}\left ( X-x \right )$

$\Rightarrow\displaystyle myX+nxY= \left ( m+n \right )\cdot xy$ ...(1)

Let the tangent (1) meet the axes in points

$A$ and

$B$ .

Putting

$Y= 0,$ we get

$\displaystyle A= \left [ \dfrac{m+n}{m}x, 0 \right ]$

Putting

$X=0$ we get

$\displaystyle B=\left ( 0, \dfrac{m+n}{m}y \right )$

Point of contact

$P$ is

$(x, y).$

Let

$P$ divide

$AB$ in the ratio

$\displaystyle \lambda :1$

$\displaystyle \therefore x= \dfrac{\lambda .0+1\cdot \dfrac{m+n}{n}x}{\lambda +1}$

$\Rightarrow\displaystyle \left ( \lambda +1 \right )x= \left ( \dfrac{m}{n}+1 \right )x$

$\Rightarrow\displaystyle \lambda +1= \dfrac{m}{n}+1$

$\displaystyle\therefore \lambda =\dfrac{m}{n}$

Find the equation of the tangent to the curve at any point

$(X, Y)$ .

$\displaystyle \frac{x^{m}}{a^{m}}+\frac{y^{m}}{b^{m}}=1.$

0%
$\displaystyle \frac{X}{a}\left ( \frac{x}{a} \right )^{m-1}+\frac{Y}{b}\left ( \frac{y}{b} \right )^{m-1}=1$
0%
$\displaystyle \frac{X}{b}\left ( \frac{x}{a} \right )^{m-1}+\frac{Y}{a}\left ( \frac{y}{b} \right )^{m-1}=1$
0%
$\displaystyle \frac{X}{a}\left ( \frac{x}{b} \right )^{m-1}+\frac{Y}{b}\left ( \frac{y}{a} \right )^{m-1}=1$
0%
$\displaystyle \frac{X}{b}\left ( \frac{x}{b} \right )^{m-1}+\frac{Y}{b}\left ( \frac{y}{b} \right )^{m-1}=1$

Explanation

Given,

$\displaystyle \frac{x^{m}}{a^{m}}+\frac{y^{m}}{b^{m}}=1.$ ..(1)
Differentiating w.r.t. x, we get

$\displaystyle m.\frac{x^{m-1}}{a^{m}}+m.\frac{y^{m-1}}{b^{m}}\cdot \frac{dy}{dx}=0.$

$\displaystyle \therefore\frac{dy}{dx} =\frac{1}{a}\left ( \frac{x}{a} \right )^{m-1}.b\left ( \frac{b}{y} \right )^{m-1}$

$\displaystyle \therefore$ Equation of tangent is

$\displaystyle Y-y=\frac{dy}{dx}\left ( X-x \right )$
or

$\displaystyle Y-y=-\frac{1}{a}\left ( \frac{x}{a} \right )^{m-1}.b\left ( \frac{b}{y} \right )^{m-1}\left ( X-x \right )$
or

$\displaystyle \frac{X}{a}\left ( \frac{x}{a} \right )^{m-1}-\left ( \frac{x}{a} \right )^{m}=-\frac{Y}{b}\left ( \frac{y}{b} \right )^{m-1}+\left ( \frac{y}{b} \right )^{m}$
or

$\displaystyle \frac{X}{a}\left ( \frac{x}{a} \right )^{m-1}+\frac{Y}{b}\left ( \frac{y}{b} \right )^{m-1}=\left ( \frac{x}{a} \right )^{m}+\left ( \frac{y}{b} \right )^{m}$
or

$\displaystyle \frac{X}{a}\left ( \frac{x}{a} \right )^{m-1}+\frac{Y}{b}\left ( \frac{y}{b} \right )^{m-1}=1,$ by (1)

For the equation

$\displaystyle x^{2/3}+y^{2/3}=a^{2/3}$ , find the equation of tangent at the point

$\displaystyle x=a\sin ^{3}\theta, y=a\cos ^{3} \theta$ .

0%
$\displaystyle y-a\cos ^{3}\theta =\frac{\cos \theta }{\sin \theta }(x-a \sin ^{3}\theta )$
0%
$\displaystyle y-a\cos ^{3}\theta =-\frac{\cos \theta }{\sin \theta }(x-a \sin ^{3}\theta )$
0%
$\displaystyle y-a\cos ^{3}\theta =-\frac{\cos \theta }{\sin \theta }(x+a \sin ^{3}\theta )$
0%
none of these

Explanation

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

Differentiating with respect to x, gives us

$\dfrac{2}{3}[x^{-1/3}+y^{-1/3}y']=0$

Therefore

$y'=-\dfrac{x^{-1/3}}{y^{-1/3}}$

$=-\left(\dfrac{y}{x}\right)^{1/3}$

$x=asin^{3}\theta$ and

$y=acos^{3}\theta$

$y'=-\dfrac{cos\theta}{sin\theta}$

This is the slope of the tangent at the given point.

Hence the equation of the tangent is

$(y-acos^{3}\theta)=\dfrac{-cos\theta}{sin\theta}(x-asin^{3}\theta)$ [slope-point form]

Find the condition that the line

$\displaystyle Ax+By= 1$ may be a normal to the curve

$\displaystyle a^{n-1}y=x^{n}.$

0%
$\displaystyle a^{n}B\left ( B^{2}+nA^{2} \right )^{n}=A^{n}n^{n}.$
0%
$\displaystyle a^{n-1}B\left ( B^{2}+nA^{2} \right )^{n-1}=A^{n}n^{n}.$
0%
$\displaystyle a^{n}B\left ( B^{2}+nA^{2} \right )^{n-1}=A^{n}n^{n}.$
0%
$\displaystyle a^{n-1}B\left ( B^{2}-nA^{2} \right )^{n-1}=A^{n}n^{n}.$

Explanation

Given,

$\displaystyle a^{n-1}y=x^{n}$

$\displaystyle \therefore \dfrac{dy}{dx}=n\dfrac{x^{n-1}}{a^{n-1}}=n\dfrac{x^{n-1}}{a^{n-1}}\cdot \dfrac{1}{x}=n\dfrac{y}{x}.$

$\displaystyle \therefore$ Normal is:

$\displaystyle Y-y=-\dfrac{1}{dy/dx}\left ( X-x \right) =-\dfrac{x}{ny}\left ( X-x \right )$

$\displaystyle \therefore Xx+Yny=ny^{2}+x^{2}.$

Compare with

$AX+BY=1$

$\displaystyle \therefore \dfrac{X}{A}=\dfrac{ny}{B}= \dfrac{ny^{2}+x^{2}}{1}=k,$ say.

$\displaystyle \therefore x= Ak, y=(Bk/n)$ and

$\displaystyle ny^{2}+x^{2}=k$ or

$\displaystyle k^{2}\left [ \left ( B^{2}/n+A^{2} \right ) \right ]=k$

$\displaystyle \therefore k=\dfrac{n}{B^{2}+nA^{2}}$ ..(1)

Now

$\displaystyle a^{n-1}y=x^{n}.$

Put for

$x$ and

$y$ .

$\displaystyle a^{n-1}y\cdot \dfrac{Bk}{n}= A^{n}k^{n}$

$\Rightarrow\displaystyle a^{n-1}B= nA^{n}k^{n-1}$

$\Rightarrow\displaystyle a^{n-1}B=nA^{n}\cdot \left ( \dfrac{n}{B^{2}+nA^{2}} \right )^{n-1},$ by (1)

$\Rightarrow\displaystyle a^{n-1}B\left ( B^{2}+nA^{2} \right )^{n-1}=A^{n}n^{n}.$

Above is the required condition.

If the normal to the curve

$y=f(x)$ at the point

$(3,4)$ makes an angle

$3\pi /4$ with the positive x-axis, then

$f'(3)=$

0%
$-1$
0%
$0$
0%
$1$
0%
$\sqrt 3$

Explanation

Tangent being perpendicular to given line of slope 2, will have its slope as -

$\displaystyle \dfrac{1}{2}$ .

Slope of tangent=

$\displaystyle -\dfrac{fx}{fy}=-\dfrac{6x+1}{2(y+1)}=-\dfrac{1}{2}\therefore y=6x$ .

Sloping with the given curve, we have

$\displaystyle 3x^{2}+36x^{2}+x+12x=0$

$13x(3x+1)=0\therefore x=0, -1/3 \therefore y=0, -2$

Hence the two points are

$\displaystyle (0,0),\left ( -\dfrac{1}{3},-2 \right )$

$\displaystyle \therefore y=-\dfrac{1}{2}x$ and

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

$2y+x=0$ and

$\displaystyle 2y+x+\dfrac{13}{3}=0$

Ans: D

Find the slopes of the tangents of the curve

$y=(x+1)(x-3)$ at the points where it cuts the X-axis.

0%
$4$
0%
$-4$
0%
$2$
0%
$-2$

Explanation

$f(x)=(x+1)(x-3)$
Now

$f(x)=0$

$\Rightarrow x=-1,x=3$
Differentiating

$f(x)$ with respect to x

$\dfrac{dy}{dx}=x+1+x-3$

$=2x-2$

$=2(x-1)$
Now slope of the tangent at

$(h,k)$ will be

$\dfrac{dy}{dx}_{h,k}$
Hence slopes of the tangent at

$x=-1$ and

$x=3$ , will be

$\dfrac{dy}{dx}_{x=-1}=2(x-1)_{x=-1}=-4$
And

$\dfrac{dy}{dx}_{x=3}=2(x-1)_{x=3}=4$

If the normal to the curve y = f(x) at the point (3, 4) makes an angle

$\dfrac{3\pi }{4}$ with the positive x-axis, then f'(3) is equal to

0%
-1
0%
$-\dfrac{3 }{4}$
0%
$\dfrac{4}{5}$
0%
1

If the tangent at the point

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

$\displaystyle ay^{2}= x^{3}$ meets the curve again at Q.then the co-ordinates of Q is/are

0%
$\displaystyle \left ( -\frac{1}{4}at^{2},-\frac{1}{8}at^{3} \right ).$
0%
$\displaystyle \left ( \frac{1}{4}at^{2},-\frac{1}{8}at^{3} \right ).$
0%
$\displaystyle \left ( \frac{1}{4}at^{3},-\frac{1}{8}at^{2} \right ).$
0%
$\displaystyle \left ( -\frac{1}{4}at^{3},-\frac{1}{8}at^{3} \right ).$

Explanation

Given,

$ay^2 = x^3$

$\Rightarrow 2ay\cfrac{dy}{dx} = 3x^2$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{3x^2}{2ay}$
Thus slope of tangent at point

$(at^2, at^3)$ is

$m = \cfrac{3.a^2t^4}{2a.at^3} = \cfrac{3}{2}t$
Therefore equation of tangent at the same point is

$(y-at^3) = \cfrac{3}{2}t(x-at^2)$
Now let this line again intersect the given curve at

$(at_1^2, at_1^3)$

$\Rightarrow (at_1^3-at^3) = \cfrac{3}{2}t(at_1^2-at^2)$

$\Rightarrow (t_1^3-t^3) = \cfrac{3}{2}t(t_1^2-t^2)$

$\Rightarrow 2(t_1^2+t^2-tt_1) =3t(t_1-t)\Rightarrow 2t_1^2-t^2-tt_1=0$

$\Rightarrow (t-t_1)(2t_1+t)=0\Rightarrow t_1 = -\cfrac{1}{2}t$
Hence required point is,

$\left(\cfrac{1}{4}at^2, -\cfrac{1}{8}at^3\right)$

What are the tangent and normal to the curve

$x=\displaystyle \frac{2at^{2}}{1+t^{2}}$ ,

$y= \displaystyle \frac{2at^{3}}{a+t^{2}}$ at the point for which

$\displaystyle t=\frac{1}{2}$

0%
$\displaystyle 16x+13y=9a$ , $\displaystyle 13x-16y=2a$
0%
$\displaystyle 13x-16y=2a$ , $\displaystyle 16x+13y=9a$
0%
$\displaystyle 16x-13y=9a$ , $\displaystyle 13x+16y=2a$
0%
$\displaystyle 13x+16y=2a$ , $\displaystyle 16x-13y=9a$

Explanation

Given

$x=\displaystyle \frac{2at^{2}}{1+t^{2}}$ ,

$y= \displaystyle \frac{2at^{3}}{1+t^{2}}$
At

$t=\dfrac{1}{2}$ we get

$x=\dfrac{2a}{5}$ ,

$y=\dfrac{a}{5}$
Hence, the point is

$(\dfrac{2a}{5},\dfrac{a}{5})$

Now,

$\dfrac { dx }{ dt } =2a\dfrac { (1+t^{ 2 })2t-{ 2t }^{ 3 } }{ { (1+t^{ 2 }) }^{ 2 } } =\dfrac { 4at }{ { (1+t^{ 2 }) }^{ 2 } }$

$\dfrac { dy }{ dt } =2a\dfrac { (1+t^{ 2 })3t^{ 2 }-{ t }^{ 3 }(2t) }{ { (1+t^{ 2 }) }^{ 2 } } =\dfrac { 2a(3t^{ 2 }+{ t }^{ 4 }) }{ { (1+t^{ 2 }) }^{ 2 } }$

Now,

$\dfrac{dy}{dx}=\dfrac { (3t^{ 2 }+{ t }^{ 4 }) }{ 2t }$

Slope of tangent at

$t=\dfrac{1}{2}$ is

$\dfrac { 13 }{ 16a }$

Equation of tangent at

$(\dfrac{2a}{5},\dfrac{a}{5})$ is

$y-\dfrac{a}{5}=\dfrac { 13 }{ 16a } (x-\dfrac{2a}{5})$

$\Rightarrow 13x-16y=2a$

Equation of normal at

$(\dfrac{2a}{5},\dfrac{a}{5})$ is

$y-\dfrac{a}{5}=-\dfrac { 16a }{ 13 } (x-\dfrac{2a}{5})$

$\Rightarrow 16x+13y=9a$

A and B are points

$(-2,0)$ and

$(1,3)$ on the curve

$\displaystyle y=4-x^{2}$ . If the tangent at P on the curve be parallel to chord AB, then co-ordinates of point P are

0%
$\displaystyle \left ( -\frac{1}{3}, \frac{5}{3} \right )$
0%
$\displaystyle \left ( \frac{1}{2}, -\frac{15}{4} \right )$
0%
$\displaystyle \left ( -\frac{1}{2}, \frac{15}{4} \right )$
0%
$\displaystyle \left ( -\frac{1}{3}, \frac{1}{5} \right )$

Explanation

Given equation of curve

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

$\dfrac{dy}{dx}=-2x$

Given points

$A(-2,0)$ and

$(1,3)$
Slope of AB

$=\dfrac{3}{3}=1$

$\Rightarrow -2x=1$

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

So, by equation (i), we get

$y=\dfrac{15}{4}$
Hence, the point is

$(-\dfrac{1}{2},\dfrac{15}{4})$

Normal to the curve

$y=\displaystyle x^{3}-2x^{2}+4$ at the point where

$x=2$

0%
$3x+4y=18$
0%
$x+4y=18$
0%
$4x+3y=18$
0%
$4x+y=18$

Explanation

Substituting

$x=2$ in the equation, we get

$y=8-8+4=4$
Thus the point is

$(2,4)$ .
Now

$dy=3x^{2}-4x.dx$
Or

$\dfrac{-dx}{dy}=\dfrac{-1}{3x^{2}-4x}$
Now

$\dfrac{-dx}{dy}_{x=2}=\dfrac{-1}{12-8}$

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

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

$4y-16=2-x$
Or

$x+4y=18$

Find the points on the curve

$y=x^{3}$ , the tangents at which are inclined at an angle of

$60^{\circ}$ to x-axis.

0%
$x=\pm\dfrac{1}{\sqrt{\sqrt{3}}}, y =\dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$ .
0%
$x=\dfrac{1}{\sqrt{\sqrt{3}}}, y =\dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$ .
0%
$x=\pm\dfrac{1}{\sqrt{\sqrt{3}}}, y =\pm \dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$ .
0%
$x=-\dfrac{1}{\sqrt{\sqrt{3}}}, y =\pm \dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$ .

Explanation

Hence slope of tangents

$=tan60^{0}$

$=\sqrt{3}$

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

$\dfrac{dy}{dx}$

$=3x^{2}$

$=\sqrt{3}$
Or

$x^{2}=\dfrac{1}{\sqrt{3}}$
Hence

$x=\pm\dfrac{1}{\sqrt{\sqrt{3}}}$ .
Thus y

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

Find the points on the curve

$y=x/(1-x^{2})$ where the tangents makes an angle of

$\pi /4$ with x-axis

0%
$(\sqrt { 3 } ,-\sqrt { \dfrac { 2 }{ 3 } } ),(-\sqrt { 2 } ,\sqrt { \dfrac { 2 }{ 3 } } )$
0%
$(\sqrt { 3 } ,-\sqrt { \dfrac { 3 }{ 4 } } ),(-\sqrt { 3 } ,\sqrt { \dfrac { 3 }{ 4 } } )$
0%
$(\sqrt { 3 } ,-\sqrt { \dfrac { 3 }{ 2 } } ),(-\sqrt { 3 } ,\sqrt { \dfrac { 3 }{ 2 } } )$
0%
none of these

Explanation

$\dfrac{dy}{dx}=tan45^{0}$
Or

$\dfrac{dy}{dx}=1$
Or

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

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

$1+x^{2}=1-2x^{2}+x^{4}$
Or

$x^{4}-3x^{2}=0$
Or

$x^{2}[x^{2}-3]=0$

$x=0$ or

$x=\pm\sqrt{3}$
Hence

$y=\dfrac{\sqrt{3}}{1-3}=-\dfrac{\sqrt{3}}{2}$

$y=\dfrac{-\sqrt{3}}{1-3}=\dfrac{\sqrt{3}}{2}$

Hence

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

Find the equations of the tangents drawn to the curve

$\displaystyle y= x^{4}$ which are drawn from the point (2,0).

0%
$\displaystyle y= 0$ and $\displaystyle y-\left ( \frac{4}{3} \right )^{4}= 4\left ( \frac{4}{3} \right )^{3}\left ( x-\frac{4}{3} \right )$
0%
$\displaystyle y= 0$ and $\displaystyle y-\left ( \frac{8}{3} \right )^{4}= 4\left ( \frac{8}{3} \right )^{3}\left ( x-\frac{8}{3} \right )$
0%
$\displaystyle y= 0$ and $\displaystyle y-\left ( \frac{4}{3} \right )^{4}= 4\left ( \frac{8}{3} \right )^{2}\left ( x-\frac{8}{3} \right )$
0%
$\displaystyle y= 0$ and $\displaystyle y-\left ( \frac{8}{3} \right )^{4}= 4\left ( \frac{8}{3} \right )^{2}\left ( x-\frac{8}{3} \right )$

Explanation

Let the equation of curve be

$y=x^4$
Let the point of contact be

$(h,k)$

$\Rightarrow \displaystyle k= h^{4}$ ...(1)

Now,

$\dfrac{dy}{dx}=4x^3$
Slope of tangent to the curve at

$(h,k)$ is

$4h^3$
Tangent is

$\displaystyle y-k= 4h^{3}\left ( x-h \right )$ ...(2)
It passes through (2,0)

$\displaystyle \therefore -k= 4h^{3}\left ( 2-h \right )$

$\displaystyle -h^{4}= 8h^{3}-4h^{4}$ by (1)

$\displaystyle 3h^{4}-8h^{3}= 0$

$\displaystyle \therefore h= 0$ or

$\dfrac{8}{3}$

$\displaystyle \therefore k= 0$ or

$\displaystyle \left ( 8/3 \right )^{4}$

$\displaystyle \therefore$ Points are (0,0) and

$\displaystyle \left [ 8/3, \left ( 8/3 \right )^{4} \right ]$
Putting in (2), tangents are

$\displaystyle y= 0$
and

$\displaystyle y-\left ( \frac{8}{3} \right )^{4}= 4\left ( \frac{8}{3} \right )^{3}\left ( x-\frac{8}{3} \right )$

The points of contact of the tangents drawn from the origin to the curve

$y=\sin{x}$ lie on the curve

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

Explanation

Let the tangent be drawn at the point

$\left(x,y\right).$

Its equation is

$\displaystyle Y-y=\frac { dy }{ dx } \left( X-x \right)$

But

$y=\sin{x}$

$\therefore\displaystyle\frac{dy}{dx}=\cos{x}$

$\Rightarrow Y-y=\cos{x}\left(X-x\right)$

Since it passes through

$\left(0,0\right)$ , therefore substituting

$\left(x,y\right)$ by

$\left(0,0\right)$ we get

$-y=-x\cos{x}\Rightarrow\displaystyle\frac{y}{x}=\cos{x}$ and

$y=\sin{x}$

$\displaystyle \therefore \frac { { y }^{ 2 } }{ { x }^{ 2 } } +{ y }^{ 2 }=\cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } =1$

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

hence the points of contact lie on

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

The curve

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

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

Explanation

$\dfrac{dy}{dx}-e^{xy}[x\dfrac{dy}{dx}+y]+1=0$
Or

$\dfrac{dy}{dx}[1-x.e^{xy}]+1-y.e^{xy}=0$
Or

$\dfrac{dy}{dx}=\dfrac{-(1-y.e^{xy})}{1-x.e^{xy}}$
Now

$1-xe^{xy}=0$ since the slope of the tangent is

$90^{0}$ .
Hence

$xe^{xy}=1$
Or

$x(x+y)=1$
Or

$x^{2}+xy=1$
Or

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

$y=\dfrac{1}{x}-x$ ...(i)
Considering

$y=0$ ,

$x^{2}=1$

$x=\pm1$ .
Hence we get 2 points

$(1,0)$ and

$(-1,0)$ .
Out of these 2, only

$(1,0)$ lies on the curve.
Hence the required point is

$(1,0)$ .

Let

$f(x, y)$ be a curve in the

$x-y$ plane having the property that distance from the origin of any tangent to the curve is equal to distance of point of contact from the

$y-$ axis. Of

$f(1, 2)=0$ , then all such possible curves are

0%
$x^2+y^2=5x$
0%
$x^2-y^2=5x$
0%
$x^2y^2=5x$
0%
$All\ of\ these$

At what point p(x,y)of the curve

$\displaystyle y=e^{-\left | x \right |}$ should a tangent be drawn so that area of the triangle bounded by the tangent and the co-ordinate axes be greatest ?

0%
$\displaystyle \left ( \pm e, 1 \right )$
0%
$\displaystyle \left ( \pm 1,\dfrac 1e \right )$
0%
$\displaystyle \left ( 1,\pm \dfrac 1e \right )$
0%
$\displaystyle \left ( \pm 1, e \right )$

Explanation

The equation of the curve does not change if

$x$ be + ive or-ive.
Hence we consider only +ive values of

$x$ i.e

$\displaystyle x> 0,$

$\displaystyle \therefore \left | x \right |=x \therefore y={-x}$
Consider any point (x,y) tangent at which is

$\displaystyle Y-y=-e^{-x}\left ( X-x \right )$
It meets the axis

$\displaystyle Y=0 at \left ( x=ye^{x},0 \right )$
It meets the axis

$\displaystyle X=0 at (0,y+\dfrac x{e^{x}})$
Area of the triangle

$\displaystyle =\frac{1}{2}AB$

$\displaystyle \Delta =\frac{1}{2}\left ( x=ye^{x} \right )\left ( y+\frac{x}{e^{x}} \right )$

$\displaystyle =\frac{1}{2}\left ( xy+xy+y^{2}e^{x}+\frac{x^{2}}{e^{x}} \right )$ Substitute

$\displaystyle y=\frac{1}{e^{x}}$

$\displaystyle \Delta =\frac{1}{2}\left [ \frac{2x}{e^{x}}+\frac{e^{x}}{e^{2x}}+\frac{x^{2}}{e^{x}} \right ]=\frac{1}{2}e^{-x}\left ( x+1 \right )^{2}$
For max. value of

$\displaystyle \Delta$
we have

$\displaystyle \frac{d\Delta }{dx}=0 \therefore \frac{1}{2}e^{-x\left [ -\left ( x+1 \right )^{2}+2\left ( x+1 \right ) \right ]}=0$
or

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

$\displaystyle \therefore \frac{d\Delta }{dx}=0\Rightarrow x=1-1.$
We shall consider only x=1as x is +ive.

$\displaystyle \frac{d^{2}\Delta }{dx^{2}}=\frac{1}{2}e^{x}\left [ -1\left ( 1-x^{2} \right )-2x \right ]$

$\displaystyle =-\frac{x}{e^{x}}=-\frac{1}{e}=-ive$ at

$x=1$
Hence

$\displaystyle \Delta$ is max.
When x=1

$\displaystyle \therefore y=e^{-1}=\dfrac 1e.$

$\displaystyle \therefore$
Required point is

$(1,\dfrac 1e)$ . we have already stated that if

$x$ be changed to

$-x$ the equation of the curve does not change.
Hence the required points are

$\displaystyle \left ( \pm 1,\dfrac 1e \right )$

$\displaystyle x\cos \alpha +y\sin \alpha =p$ touches

$\displaystyle x^{2}+a^{2}y^{2}=a^{2},$ then

0%
$\displaystyle p^{2}=a^{2}\sin^{2}\alpha +\cos^{2}\alpha$
0%
$\displaystyle p^{2}=a^{2}\cos^{2}\alpha +\sin^{2}\alpha$
0%
$\displaystyle 1/p^{2}=\sin^{2}\alpha +\alpha^{2}\cos^{2}\alpha$
0%
$\displaystyle 1/p^{2}=\cos^{2}\alpha +\alpha^{2}\sin^{2}\alpha$

Explanation

Solving for y, we have

$\displaystyle y=\dfrac { p }{ \sin { \alpha } } -x\cot { \alpha }$

Putting this value in

${ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }={ a }^{ 2 }$ , we have

$\displaystyle { x }^{ 2 }+{ a }^{ 2 }{ \left( \dfrac { p }{ \sin { \alpha } } -x\cot { \alpha } \right) }^{ 2 }={ a }^{ 2 }$

$\displaystyle \Rightarrow { x }^{ 2 }\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) -\dfrac { 2a^{ 2 }xp\cot { \alpha } }{ \sin { \alpha } } +\dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha } } -{ a }^{ 2 }=0$

The discriminant of this equation must be zero. So

$\displaystyle { a }^{ 4 }\dfrac { { p }^{ 2 }\cot ^{ 2 }{ \alpha } }{ \sin ^{ 2 }{ \alpha } } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) \left( \dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha } } -{ a }^{ 2 } \right) \\ \Rightarrow { a }^{ 2 }{ p }^{ 2 }\cot ^{ 2 }{ \alpha } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) \left( { p }^{ 2 }-\sin ^{ 2 }{ \alpha } \right) \\ \Rightarrow { p }^{ 2 }\left( { a }^{ 2 }\cot ^{ 2 }{ \alpha } -1-{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) =\sin ^{ 2 }{ \alpha } -{ a }^{ 2 }\cos ^{ 2 }{ \alpha } \\ \Rightarrow { p }^{ 2 }={ a }^{ 2 }\cos ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \alpha }$

The point of intersection of the tangents drawn to the curve

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

$\displaystyle xy=1-y$ is given by

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

Explanation

Intersection of

${ x }^{ 2 }y=1-y$ are

$xy=1-y$

are given by

${ x }^{ 2 }y=xy$ .

Since

$y\neq 0$ on any of two curves

So we must have

$x=0,1$ .

Thus the curve intersect at

$\displaystyle \left( 0,1 \right) ;\left( 0,\dfrac { 1 }{ 2 } \right)$

Now differentiating

${ x }^{ 2 }y=1-y$ . we have

$\displaystyle \dfrac { dy }{ dx } =\dfrac { 2xy }{ 1+{ x }^{ 2 } }$

Thus for

$\left( x,y \right) =\left( 0,1 \right)$

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

And for

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

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

Thus the equation of tangent are

$Y=1$ and

$\displaystyle Y-\dfrac { 1 }{ 2 } =\left( -\dfrac { 1 }{ 2 } \right) \left( X-1 \right)$

Their intersection is given by

$\left( 0,1 \right)$

If the line

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

$xy=1$ , then

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

Explanation

Given equation of curve

$xy=1$

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

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{y}{x}$
Slope of tangent at

$P(x_1,y_1)=-\dfrac{1}{x_1^2}$
Slope of normal at P is

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

Given equation of normal

$ax+by+c=0$
Slope of the normal at P is

$-\dfrac{a}{b}$ .....(2)

From (1) and (2),

$x^2=-\dfrac{a}{b}$
Since,

$x^2 >0$

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

$\Rightarrow \dfrac{a}{b}<0$
i.e.

$\dfrac{a}{b}$ should be negative.

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

$a>0 , b<0$

Find the co-ordinates of the points on the curve

$\displaystyle y= x/\left ( 1+x^{2} \right )$ where the tangent to the curve has greatest slope.

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

Explanation

Given curve

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

Here slope

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

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

Now,

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

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

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

For maximum or minimum of S,

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

$\displaystyle \Rightarrow x= -\sqrt{3}, 0, \sqrt{3}$ .

Now, at

$x=0$ ,

$\dfrac{ds}{dx}$ changes from +ive to -ive

$\displaystyle x= \pm \sqrt{3}$ . it changes from -ive to +ive .

Hence slope S is maximum when x=0 and min. when

$\displaystyle x= \pm \sqrt{3}$ , thus for greatest slope, we have x=0 and y=0.

Hence the required point is (0, 0), that is, the origin.

The sum of the intercepts of a tangent to

$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}, a> 0$ upon the coordinate axes is

0%
$2a$
0%
$a$
0%
$a/2$
0%
$\displaystyle \sqrt{a}$

Explanation

$\sqrt { x } +\sqrt { y } =\sqrt { a } ,a>0$

Differentiating w.r.t

$x$ , we get

$\displaystyle \dfrac { 1 }{ 2\sqrt { x } } +\dfrac { 1 }{ 2\sqrt { y } } \dfrac { dy }{ dx } =0\Rightarrow \dfrac { dy }{ dx } =-\dfrac { \sqrt { y } }{ \sqrt { x } }$

Equation of tangent at any point

$\left(x,y\right)$ of the curve is

$\displaystyle Y-y=-\dfrac { \sqrt { y } }{ \sqrt { x } } \left( X-x \right)$

So intercepts of x-axis and y-axis are

$x+\sqrt{xy}$ and

$y+\sqrt{xy}$ .

Therefore, the sum of intercepts

$=x+y+2\sqrt { xy } ={ \left( \sqrt { x } +\sqrt { y } \right) }^{ 2 }=a$

The equation of tangent to the curve

$\displaystyle y=1-e^{\tfrac x2}$ at the point where it meets y - axis is -

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

Explanation

Substituting

$y=0$

$\displaystyle 0=1-e^{\dfrac{x}{2}}\Rightarrow x=0$
Thus the point is

$(0, 0)$
Now,

$\displaystyle \dfrac{dy}{dx}=-e^{\dfrac{x}{2}}\times \dfrac{1}{2}$

$\displaystyle \left ( \dfrac{dy}{dx} \right )_{\left ( 0,0 \right )}=-\dfrac{1}{2}\times e^{\dfrac{0}{2}}=-\dfrac{1}{2}$
Therefore, the required equation is given by,

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

$2y = -x\Rightarrow x + 2y = 0$
Hence, option 'A' is correct.

The line

$y=x$ is a tangent to the parabola

$\displaystyle y= ax^{2}+bx+c$ at the point

$x=1$ .If the parabola passes through the point

$(-1,0)$ , then determine

$a, b, c.$

0%
$\displaystyle a= \frac{1}{2}, b= \frac{1}{4}, c= \frac{1}{3}.$
0%
$\displaystyle a= \frac{1}{4}, b= \frac{1}{2}, c= \frac{1}{4}.$
0%
$\displaystyle a= 2, b= 1, c= 4.$
0%
$\displaystyle a= 4, b= 2, c= 4.$

Explanation

Given equation of parabola is

$\displaystyle y= ax^{2}+bx+c$

$\displaystyle \frac{dy}{dx}= 2ax+b$
Slope of tangent to the curve at

$x=1$ is

$2a+b$
Given tangent is

$y=x$ . Slope of this tangent is

$1.$
So,

$2a+b=1$ ...(1)

Since, the parabola passes through

$(-1,0)$

$\displaystyle \therefore a-b+c= = 0$ ...(2)

Given

$y=x$ is a tangent at

$x=1$

$\displaystyle \therefore y= 1.$
Hence

$(1,1)$ lies both on tangent and parabola

$\displaystyle \therefore a+b+c= 1$ ...(3)

Solving (1), (2) and (3), we get

$\displaystyle a= \frac{1}{4}, b= \frac{1}{2}, c= \frac{1}{4}.$

The parametric equations of a curve are given by

$\displaystyle x= \sec ^{2}t, y= \cot t.$ Tangent at

$\displaystyle P \,t=\dfrac { \pi }4$ meets the curve again at Q; then

$\displaystyle PQ=?$

0%
$\left ( 2\sqrt{5} \right ).$
0%
$\dfrac {\left ( 5\sqrt{5} \right )}2.$
0%
$\dfrac {\left ( 3\sqrt{5} \right )}2.$
0%
$\left ( 3\sqrt{5} \right ).$

Explanation

$\displaystyle x= \sec ^{2}t, y= \cot t.$
Given

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

$\Rightarrow x=\sec^2 (\dfrac{\pi}{4}), y=\cot \dfrac{\pi}{4}$

$\Rightarrow x=2, y=1$
So, the point P is

$(2,1)$

Now,

$\dfrac{dx}{dt}=2\sec t \sec t\tan t$

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

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{2}\cot^3 t$
Slope of tangent at P is

$-\dfrac{1}{2}$

Equation of tangent at P is

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

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

Since,

$\sec^2 t -\tan^2 t=1$

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

Solving (1) and (2), we get

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

$\Rightarrow \displaystyle y= 1, 1,-\frac{1}{2}$

$\Rightarrow x=2, 5$ (by (1))

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

$\therefore PQ= \sqrt{\dfrac{45}{4}}= \dfrac{3\sqrt{5}}{2}.$
The values of

$y=1,1$ correspond to point P.

The line

$\dfrac xa+\dfrac yb=1$ touches the curve

$\displaystyle y=be^{-x/a}$ at the point

0%
$(a,b/a)$
0%
$(-a,b/a)$
0%
$(a,a/b)$
0%
None of these

Explanation

Simplifying the equation of the line we get

$bx+ay=ab$

$ay=-bx+ab$
Or

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

$\dfrac{dy}{dx}=\dfrac{-b}{a}$
Or

$-\dfrac{b}{a}.e^{-x/a}.=\dfrac{-b}{a}$
Or

$e^{-x/a}=1$
Or

$x=0$
Hence

$y=b$ .
Therefore the point is

$(0,b)$

If the tangent at

$(1, 1)$ on

$\displaystyle y^{2}= x\left ( 2-x^{2} \right )$ meets the curve again at

$P$ , then

$P$ is

0%
$\displaystyle \left (4, 4 \right )$
0%
$\displaystyle \left (-1, 2 \right )$
0%
$\displaystyle \left (9/4, 3/8 \right )$
0%
$None\ of\ these$

Explanation

$2y.\dfrac{dy}{dx}=(2-x)^{2}-2x(2-x)$
At

$(1,1)$ we get

$2.\dfrac{dy}{dx}=1-2$
Or

$\dfrac{dy}{dx}=\dfrac{-1}{2}$
Now Equation of the tangent is

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

$2y-2=-x+1$
Or

$x+2y=3$
Or

$x=3-2y$ .
Substituting in the equation we get

$y^{2}=(3-2y)(2-3+2y)^{2}$
Or

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

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

$y^{2}=12y^{2}-12y+3-8y^{3}+8y^{2}-2y$

$y^{2}=-8y^{3}+20y^{2}-14y+3$
Or

$8y^{3}-19y^{2}+14y-3=0$
Solving the above cubic, we get

$y=\dfrac{3}{8},\dfrac{3}{8}$ and

$y=1$
Considering

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

$x=3-2y$

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

$=\dfrac{9}{4}$
Hence the point is

$(\dfrac{9}{4},\dfrac{3}{8})$

If the line,

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

$xy=2,$ then

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

Explanation

$xy'+y=0$
Or

$y'=\dfrac{-y}{x}$
Hence slope of normal

$=\dfrac{x}{y}$

$=\dfrac{2}{y^{2}}$
Hence the slope is always positive.
Now
slope of the line is

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

$a<0$ and

$b>0$
Or

$a>0$ and

$b<0$ .

The coordinates of the point

$M(x, y)$ on

$\displaystyle y= e^{-\left | x \right |}$ so that the area formed by the coordinates axes and the tangent at

$M$ is greatest, are

0%
$(e, 1)$
0%
$\displaystyle \left (1, e^{-1} \right )$
0%
$\displaystyle \left (-1, e^{-1} \right )$
0%
$(0, 1)$

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

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

The slope of the tangent to the curve represented by

$x= t^{2}+3t-8$ and

$y= 2t^{2}-2t-5$ at the point

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

0%
7/6
0%
2/3
0%
3/2
0%
6/7

Explanation

We first determine the value of

$t$ corresponding to the given values ofx and

$y$ . From

$t^{2}+3t-8= 2$ , we get

$t = 2, -5$ , and from

$2t^{2}-2t-5= 2$ we get

$t = 2, -1$ . Hence to the given point there corresponds the value

$t = 2$ . Therefore, the slope of the tangent at

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

$\displaystyle \left | y' \right |_{t=2}=\left | \dfrac{dy/dt}{dx/dt} \right |_{t=2}\:=\left | \dfrac{4t-2}{2t+3} \right |_{t=2}=\:\dfrac{6}{7}$

The curve

$y= ax^{3}+bx^{2}+cx+8$ touches

$x$ -axis at

$P\left ( -2,0 \right )$ and cuts the

$y$ -axis at a point

$Q(0,8)$ where its gradient isThe values of

$a$ ,

$b$ ,

$c$ are respectively

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$-\displaystyle \frac{1}{2},-\frac{3}{4},3$
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$\displaystyle 3, -\frac{1}{2},-4$
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$\displaystyle -\frac{1}{2},-\frac{7}{4},2$
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none of these

Explanation

Given,

$y= a x^3+b x^2+cx+8$

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

Since the curve touches

$x$ -axis at

$\left ( -2,0 \right )$ so

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

$\left ( i \right )$

The curve cut the

$y$ -axis at

$\left ( 0,8 \right )$ so

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

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

$\left ( ii \right )$

$a = -1/4$ ,

$b =0$

The value of m for which the area of the triangle included between the axes and any tangent to the

$\displaystyle x^{m}y= b^{m}$ curve is constant, is

0%
$\dfrac 12$
0%
$1$
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$\dfrac 32$
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$2$

Explanation

Given,

$y =b^mx^{-m}$

$y' = -b^mx^{-(m+1)}$
At any point

$(x1,y1)$
Equation of tangent is given by

$\cfrac{y-y1}{x-x1} = -b^mx1^{-(m+1)}$
Y intercept

$= -m$
X intercept

$= (m-1)\times \dfrac 1m$
Area bounded

$= \cfrac{1}{2}\cfrac{(m-1)x1}{m} \times m$
For area to be constant,

$m =1$

If the tangent at any point on the curve

$\displaystyle x^{4}+y^{4}= a^{4}$ cuts off intercepts p and q on the coordinate axes, the value of

$\displaystyle p^{-\tfrac 43}+q^{-\tfrac 43}$ is

0%
$\displaystyle a^{-\tfrac 43}$
0%
$\displaystyle a^{-\tfrac 12}$
0%
$\displaystyle a^{\tfrac 12}$
0%
none of these

Explanation

Let
$x=a\sqrt{\cos \theta}$
$y=a\sqrt{\sin \theta}$
Hence
$\dfrac{dy}{dx}$

$=-\dfrac{\cos \theta.\sqrt{\cos \theta}}{\sin \theta.\sqrt{\sin \theta}}$
$=\cot\theta^{\dfrac{3}{2}}$ .
Hence equation of tangent will be
$y-a\sqrt{\sin \theta}=-\dfrac{\cos \theta.\sqrt{\cos \theta}}{\sin \theta.\sqrt{\sin \theta}}(x-a\sqrt{\cos \theta})$
$\sin \theta.\sqrt{\sin \theta}y-a\sin ^{2}\theta=-\cos \theta\sqrt{\cos \theta}x+a\cos ^{2}\theta$
$(\sin \theta)^{\dfrac{3}{2}}.y+(\cos \theta)^{\dfrac{3}{2}}x=a$
Hence the intercepts are
$Q=\dfrac{a}{(\sin \theta)^{\dfrac{3}{2}}}$ and $P=\dfrac{a}{(\cos \theta)^{\dfrac{3}{2}}}$
Hence
$Q^{\dfrac{-4}{3}}+P^{\dfrac{-4}{3}}$
$=a^{-\dfrac{4}{3}}[\cos ^{2}\theta+\sin ^{2}\theta]$
$=a^{-\dfrac{4}{3}}$ .

The equation of the tangent to the curve

$\displaystyle y= \left ( 2x-1 \right )e^{2\left ( 1-x \right )}$ at the point of its maximum is

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

Explanation

Given,

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

$\therefore y'\left( x \right) =-2\left( 2x-1 \right) { e }^{ 2\left( 1-x \right) }+2{ e }^{ 2\left( 1-x \right) }\\ =2{ e }^{ 2\left( 1-x \right) }\left( -2x+2 \right)$
Thus

$y'\left( x \right) =0\Rightarrow x=1$
Since

$y''\left( 1 \right) <0$ so

$\left( 1,1 \right)$ is the point of maximum and the equation of tangent is

$y-1=0\left( x-1 \right)$ ,i.e.

$y=1$

The coordinates of the point on the curve

$\displaystyle \left ( x^{2}+1 \right )\left ( y-3 \right )=x$ where a tangent to the curve has the greatest slope are given by

0%
$\displaystyle \left ( \sqrt{3}, 3+\sqrt{3}/4 \right )$
0%
$\displaystyle \left ( -\sqrt{3}, 3-\sqrt{3}/4 \right )$
0%
$\displaystyle \left ( 0, 3 \right )$
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none of these

Explanation

Solving for y the given equation we have

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

Now

$\displaystyle f'\left( x \right) =\dfrac { -2x\left( 3-{ x }^{ 2 } \right) }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } }$

For extremum of

$f\left( x \right)$ , we have

$f'\left( x \right) =0\Rightarrow x=0,x=\pm \sqrt { 3 }$

$x=0,f'\left( x \right)$ changes sign from positive to negative

$\Rightarrow f\left( x \right)$ has maxima at

$x=0$ .

Thus the required point is

$\left( 0,3 \right)$

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 5 - 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 5 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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