MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 10

If tangent at any point on the curve

${ y }^{ 2 }=1+{ x }^{ 2 }\ makes\ an\ angle\ \theta$ with positive direction of the x-axis then

0%
$\left| \tan { \theta } \right| >1$
0%
$\left| \tan { \theta } \right| <1$
0%
$\left| \tan { \theta } \right| \ge1$
0%
$\left| \tan { \theta } \right| \le 1$

The equation of the normal to the curve

$y=(1+x)^{ y }+\sin { ^{ -1 }(\sin ^{ 2 }{ x)\ at\ x=0\ is } }$ .

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

If the line

$ax+y=c$ , touches both the curves

${x}^{2}+{y}^{2}=1$ and

${y}^{2}=4\sqrt{2}x$ , then

$\left| c \right|$ is equal to:

0%
$1/2$
0%
$2$
0%
$\sqrt{2}$
0%
$\cfrac { 1 }{ \sqrt { 2 } }$

Explanation

Tangent to

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

$y=mx+\cfrac { \sqrt { 2 } }{ m }$ is also tangent to

${x}^{2}+{y}^{2}=1$

$\Rightarrow \left| \cfrac { \sqrt { 2 } /m }{ \sqrt { 1+{ m }^{ 2 } } } \right| =1\Rightarrow m=\pm 1$

Tangent will be

$y=x+\sqrt{2}$ or

$y=-x-\sqrt{2}$

compare with

$y=-ax+c$

$\Rightarrow a=\pm 1;c=\pm \sqrt { 2 }$

$\therefore |c|=\sqrt 2$

The equation of the normal to the curve

$y=\left( 1+x \right) ^{ y }+{ sin }^{ -1 }\left( { sin }^{ 2 }x \right) at\quad x=0$ is

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

Explanation

Given,

$y=(1+x)^y+\sin ^{-1}(\sin ^2x)$

when

$x=0$

$y=1^y+0=1$

$\therefore (x_1,y_1)=(0,1)$

$\dfrac{dy}{dx}=(1+x)^y\left [ \ln (1+x)\dfrac{dy}{dx}+\dfrac{y}{1+x} \right ]+\dfrac{2\sin x\cos x}{\sqrt{1-\sin ^4x}}$

$\dfrac{dy}{dx}_{(0,1)}=1[1]+0$

$\therefore \dfrac{dy}{dx}_{(0,1)}=1$

Therefore, slope of normal is

$-1$

Equation of normal,

$y-y_1=\dfrac{-1}{\frac{dy}{dx}}(x-x_1)$

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

$y-1=-x$

$x+y=1$ is the equation of normal

The sum of the length of sub tangent of sub tangent and tangent to the curve

$x=c\left[ 2cos\theta -log\left( cos\quad ec\theta +cot\theta \right) \right] ,y=csin2\theta \quad at\quad \theta =\frac { \pi }{ 3 } is$

0%
$\dfrac { c }{ 2 }$
0%
$2c$
0%
$\dfrac { 3c }{ 2 }$
0%
$\dfrac { 5c }{ 2 }$

Explanation

Given,

$x=c[2\cos \theta −\log (\csc \theta +\cot \theta )]$

$\Rightarrow \dfrac {dx}{d\theta }=c(-2\sin \theta +\csc \theta )$

$y=c\sin 2\theta$

$\dfrac{dy}{d\theta }=2c\cos 2\theta$

Also,

$y_1=c\sin \dfrac{2\pi }{3}=\dfrac{c\sqrt 3}{2}$

Slope of tangent at

$\theta =\dfrac{\pi }{3}$ is

$m=\dfrac{dy}{dx}_{\theta =\frac{\pi }{3}}=\sqrt 3$

Length of tangent

$=\left | \dfrac{y_1\sqrt{1+m^2}}{m} \right |=c$

Length of subtangent

$=\left | \dfrac{y_1}{m} \right |=\dfrac{c}{2}$

Required sum

$=c+\dfrac{c}{2}=\dfrac{3c}{2}$

If the tangent to the curve

$x=at^2, y=2at$ is perpendicular to

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

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

Explanation

Let the required point be

$(x_1,y_1).$

It is given that points lies on the curve.

$\therefore$

$x_1=at^2$ and

$y_1=2at$

Now,

$x=at^2$ and

$y=2at$

Differentiating w.r.t.

$t,$ we get

$\Rightarrow$

$\dfrac{dx}{dt}=2at$ and

$\dfrac{dy}{dt}=2a$

$\Rightarrow$

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

$\Rightarrow$

$\dfrac{dy}{dx}=\dfrac{2a}{2at}$

$\Rightarrow$

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

$\Rightarrow$

$\dfrac{dy}{dx}=\dfrac{2a}{y}$ [ Since,

$y=2at$ ]

Slope of the tangent

$=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=\dfrac{2a}{y_1}$

It is given that the tangent is perpendicular to the

$y-axis.$

It means theta it is parallel to the

$x-axis.$

$\therefore$ Slope of the tangent

$=$ Slope of the

$x-axis.$

$\Rightarrow$

$\dfrac{2a}{y_1}=0$

$\Rightarrow$

$a=0$

Now,

$x_1=at^2=(0)t^2=0$

$y_1=2at=2(0)t=0$

$\therefore$

$(x_1,y_1)=(0,0)$

The slope of the tangent to the curve

$x=t^2+3t-8, y=2t^2-2t-5$ at point

$(2, -1)$ is

0%
$\dfrac {22}{7}$
0%
$\dfrac {6}{7}$
0%
$-6$
0%
$\dfrac {7}{6}$

Explanation

$x=t^2+3t-8$ ........

$(1)$

$y=2t^2-2t-5$ ..........

$(2)$

Differentiate

$(1)$ , we get

$\dfrac{dx}{dt}=2t+3$

Differentiate

$(2)$ , we get

$\dfrac{dy}{dx}=4t-2$

$m=\dfrac{dy}{dx}=\dfrac{4t-2}{2t+3}$

Given point is

$(2, -1)$

Put the point in original x and y, we get

$\Rightarrow x=t^2+3t-8$

$2=t^2+3t-8$

$t^2+3t-10=0$

$(t-2)(t+5)=0$

$t=2, t=-5$

$\Rightarrow y=2t^2-2t-5$

$(-1)=2t^2-2t-5$

$2t^2-2t-4=0$

$(t+1)(t-2)=0$

$t=-1, t=2$

Since,

$t=2$ common in both parts, so we take

$\dfrac{dy}{dx}=\dfrac{4t-2}{2t-3}$ at

$t=2$

$\dfrac{dy}{dx}=\dfrac{4(2)-2}{2(2)-3}=\dfrac{8-2}{4+3}=\dfrac{6}{7}$

$m=\dfrac{dy}{dx}=\dfrac{6}{7}$ .

The equation of tangent at those points where the curve

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

$x$ -axis are

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

Explanation

solution:

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

Let the tangent meet the x-axis at point

$(x, 0)$

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

The tangent passes through point

$(x, 0)$

$\therefore 0=x^2-3x+2=0$

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

$\Rightarrow x=2, x=1$

Case I: when

$x=2$

Slope of tangent,

$\left.\dfrac{dy}{dx}\right|_{(2, 0)}=2(2)-3=4-3=1$

$\therefore (x_1, y_1)=(2, 0)$

Equation of tangent

$y-y_1=m(x-x_1)$

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

$\Rightarrow x-y-2=0$

Case II: When

$x=1$

Slope of tangent

$\left.\dfrac{dy}{dx}\right|_{(2, 0)}=2-3=-1$

$\therefore (x_1, y_1)=(1, 0)$

Equation of tangent

$y-y_1=m(x-x_1)$

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

$\Rightarrow x+y-1=0$ .

The equation of the normal to the curve

$y=x+\sin x\cos x$ at

$x=\dfrac {\pi}{2}$ is

0%
$x=2$
0%
$x=\pi$
0%
$x+\pi =0$
0%
$2x=\pi$

Explanation

$y=x+\sin x cos x$

Differentiating both sides w.r.t.

$x,$

$\Rightarrow$

$\dfrac{dy}{dx}=1+\cos^2x-\sin^2x$

Slope of the tangent

$(m)=\left(\dfrac{dy}{dx}\right)_{x=\frac{\pi}{2}}=1+\cos^2\dfrac{\pi}{2}-\sin^2\dfrac{\pi}{2}$

$=1-1$

$=0$

Slope of the normal

$=\dfrac{-1}{m}=\dfrac{-1}{0}$

When,

$x=\dfrac{\pi}{2},$

$y=\dfrac{\pi}{2}+\cos\dfrac{\pi}{2}\sin\dfrac{\pi}{2}$

$\therefore$

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

Now,

$(x_1,y_1)=\left(\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

$\therefore$ Equation of the normal.

$\Rightarrow$

$y-y_1=\dfrac{-1}{m}(x-x_1)$

$\Rightarrow$

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

$\Rightarrow$

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

$\Rightarrow$

$2x=\pi$

The point on the curve

$y=12x-x^2$ , where the slope of the tangent is zero will be

0%
$(0, 0)$
0%
$(2, 16)$
0%
$(3, 9)$
0%
$(6, 36)$

Explanation

Let the point be

$P(x, y)$

$y=12x-x^2$

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

$\left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=12-2x_1$

since slope of tangent is zero

$\left(\dfrac{dy}{dx}\right)_{x_1, y_1}=0$

$12-2x_1=0$

$2x_1=12$

$x_1=6$

Also curve passing through tangent

$y_1=12x_1-x^2_1$

$y_1=12\times 6-36$

$y_1=72-36$

$y_1=36$

The points are

$(6, 36)$ .

The slope of the tangent to the curve

$x=3t^2+1, y=t^3-1$ at

$x=1$ is

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

Explanation

Given curves are,

$x=3t^2+1$ ---- ( 1 )

$y=t^3-1$ ---- ( 2 )

Substituting

$x=1$ in ( 1 ) we get,

$\Rightarrow$

$3t^2+1=1$

$\Rightarrow$

$3t^2=0$

$\Rightarrow$

$t=0$

Differentiate ( 1 ) w.r.t.

$t,$ we get

$\Rightarrow$

$\dfrac{dx}{dt}=6t$

Differentiate ( 2 ) w.r.t. $$t,$ we get

$\Rightarrow$

$\dfrac{dy}{dt}=3t^2$

$\Rightarrow$

$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{3t^2}{6t}$

$\therefore$

$\dfrac{dy}{dx}=\dfrac{t}{2}$

Slope of the tangent

$=\left(\dfrac{dy}{dx}\right)_{t=0}=\dfrac{0}{2}=0$

The point on the curve

$y=x^2-3x+2$ where tangent is perpendicular to

$y=x$ is

0%
$(0, 2)$
0%
$(1, 0)$
0%
$(-1, 6)$
0%
$(2, -2)$

Explanation

$y=x^2-3x+2$ [ Given ]

$y=x$ [ Given ]

Differentiating both sides w.r.t.

$x,$ we get

$\Rightarrow$

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

Let

$(x_1,y_1)$ be the required point.

It is given that point lies on the curve.

$\therefore$

$y_1=x_1^2-3x_1+2$

$\Rightarrow$

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

Differentiating both sides w.r.t.

$x,$ we get

$\therefore$

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

Slope of the tangent

$=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=2x_1-3$

The tangent is perpendicular to the line.

$\therefore$ Slope of the tangent

$=\dfrac{-1}{slope\,of\,the\,line}=\dfrac{-1}{1}=-1$

Now,

$2x_1-3=-1$

$\Rightarrow$

$2x_1=2$

$\Rightarrow$

$x_1=1$

$\Rightarrow$

$y_1=x_1^2-3x_1+2$

$\Rightarrow$

$y_1=(1)^2-3(1)+2$

$\Rightarrow$

$y_1=1-3+2$

$\therefore$

$y_1=0$

$\therefore$

$(x_1,y_1)=(1,0)$

The equation of the normal to the curve

$y=x(2-x)$ at the point

$(2, 0)$ is

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

Explanation

The given curve is

$y=2x-x^2$

i.e,

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

Now the equation of the tangent at the point (

$x_{1},y_{1}$ )=

$(2,0)$

or,

$xx_1-(x+x_1)+\dfrac{1}{2}(y+y_1)=0$

i.e

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

i.e,

$2x-4+y=0$

i.e,

$y=-2x+4$

So, the slope of the tangent is

$(-2)$

Therefore the slope of the normal=

$\dfrac{1}{2}$

So,the equation of the normal at the point

$(2,0)$ is

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

i.e,

$2y=x-2$

i.e,

$x-2y=2$

At what points the slope of the tangent to the curve

$x^2+y^2-2x-3=0$ is zero?

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

Explanation

$x^2+y^2-2x-3=0$ is zero.

Differentiate w.r.t. x

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

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

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

$\dfrac{dy}{dx}=\dfrac{1-x}{y}$ ........

$(1)$

If line is parallel to x-axis

Angle with x-axis

$=\theta =0$

Slope of x-axis

$=\tan\theta =\tan 0^o=0$

Slope of tangent

$=$ Slope of x-axis

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

$\dfrac{1-x}{y}=0$

$x=1$

Finding y when

$x=1$

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

$(1)^2+y^2-2(1)-3=0\\$

$1+y^2-2-3=0\\$

$y=\pm 2$

Hence, the points are

$(1, 2)$ and

$(1, -2)$ .

Any tangent to the curve

$y=2x^7+3x+5$

0%
Is parallel to $x$ -axis
0%
Is parallel to $y$ -axis
0%
Makes an acute angle with $x$ -axis
0%
Makes an obtuse angle with $x$ -axis

Explanation

We have,

$y=2x^7+3x+5$

$\dfrac{dy}{dx}=14x^6+3$

$\Rightarrow \dfrac{dy}{dx} > 3$ (

$\because x^6$ is always positive for any real value)

$\Rightarrow \dfrac{dy}{dx} > 0$ so,

$\tan \theta > 0$

Hence,

$\theta$ lies in first quadrant.

Thus, the tangent to curve makes an acute angle.

The normal to the curve

$x^2=4y$ passing through

$(1, 2)$ is

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

Explanation

Curve is

$x^2=4y$

Diff wrt to x

$2x=\dfrac{4dy}{dx}\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{2}$

Slope of normal

$=\dfrac{-1}{dy/dx}=\dfrac{-2}{x}$

Let (h, k) be the point where normal and curve intersects

$\therefore$ Slope of normal at (h, k)

$=-2/h$

Equation of normal passes through (h, k)

$y-y_1=m(x-x_1)$

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

Since normal passes through

$(1, 2)$

$2-k=\dfrac{-2}{h}(1-h)$

$k=2+\dfrac{2}{h}(1-h)$ ..........

$(1)$

since (h, k) lies on the curve

$x^2=4y$

$h^2=4k$

$k=\dfrac{h^2}{4}$ .......

$(2)$

using

$(1)$ and

$(2)$

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

$\dfrac{2}{h}=\dfrac{h^2}{4}$

$h=2$

Putting

$h=2$ in

$(2)$

$k=\dfrac{h^2}{4}, k=1$

$h=2$ and

$k=1$ putting in equation of normal

$\Rightarrow y-k=\dfrac{-2(k-h)}{h}$

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

$\Rightarrow y-1=-x+2\\$

$\Rightarrow x+y=2+1\\$

$\Rightarrow x+y=3$ .

The equation of the normal to the curve

$x=a\cos^3\theta, y=a\sin^3\theta$ at the point

$\theta =\dfrac {\pi}{4}$ is

0%
$x=0$
0%
$y=0$
0%
$x=y$
0%
$x+y=a$

Explanation

Slope of normal to the curve =

$- \dfrac{1}{\left(\dfrac{dy}{dx} \right)_{\theta = \pi/4}}$

$x = a \cos^3 \theta, \, y = a \sin^3 \theta$

$\dfrac{dx}{d \theta} = 3 a \cos^2 \theta (- \sin \theta) , \dfrac{dy}{d \theta} = 3 a \sin^2 \theta \cos \theta$

$\therefore \dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx / d \theta} = \dfrac{+3a \sin^2 \theta \cos \theta}{-3 a \cos^2 \theta \sin \theta} = -\tan \theta$

$\therefore \left(\dfrac{dy}{dx}\right)_{\theta = \pi/4} = -1$

$\therefore$ Equation of normal

$= (y - y_0) = \dfrac{-1}{\left(\dfrac{dy}{dx}\right)_{\theta = \pi/4}} [ x - x_0]$

$y - \dfrac{a}{2 \sqrt{2}} = \dfrac{-1}{-1} \left(x - \dfrac{a}{2 \sqrt{2}} \right)$

$[x_0 = a \cos^3 \dfrac{\pi}{4} = \dfrac{a}{2 \sqrt{2}}$

$y - \dfrac{a}{2 \sqrt{2}} = x- \dfrac{a}{2 \sqrt{2}}$

$y_0 = a \sin^3 \dfrac{\pi}{4} = \dfrac{a}{2 \sqrt{2}}]$

$y = x$

The number of tangents to the cure

$x^{3/2}+y^{3/2}=2a^{3/2}, a> 0$ , which are equally inclined to the axes, is

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

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%
None of these

Explanation

Differentiating w.r.t.x, we get

$1 + \dfrac{dy}{dx} = e^{xy}\left ( y + x\dfrac{dy}{dx} \right )$ or

$\dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$

As the tangent is parallel along

$y-axis$

$\dfrac{dy}{dx} = \infty=\dfrac{ye^{xy}-1}{1-xe^{xy}}$

$\Rightarrow 1-xe^{xy}=0$ This holds for x = 1, y = 0

Mark the correct alternative of the following.
The point on the curve

$9y^2=x^3$ , where the normal to the curve makes equal intercepts with the axes is?

0%
$(4, \pm 8/3)$
0%
$(-4, 8/3)$
0%
$(-4, -8/3)$
0%
$(8/3, 4)$

Explanation

Let the required point be

$(x_1, y_1)$

equation of the curve is

$9y^2=x^2$

since

$(x_1, y_1)$ lies on the curve, therefore

$9y^2_1=x^3_1$ .......

$(1)$

Now

$9y^2=x^3$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{x^2}{6y}$

$\Rightarrow \left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=\dfrac{x^2}{6y^1}$

since normal to the curve at

$(x_1, y_1)$ makes equal intersepts with the coordinate axis, therefore slope of the normal

$=\pm 1$

$\Rightarrow \dfrac{1}{-(dy/dx)_{(x_1, y_1)}}=\pm 1$

$\Rightarrow (dy/dx)_{(x_1, y_1)}=\pm 1$

$\rightarrow \dfrac{x^2_1}{6y^1}=\pm 1$

$\Rightarrow x^4_1=36y^2_1=36\left(\dfrac{x^3_1}{9}\right)$ (using

$1$ )

$\Rightarrow x^4_1=4x^3_1\Rightarrow x^3_1(x_1-4)=0$

$\Rightarrow x_1=0, 4$

Putting

$x_1=0$ in

$(1)$ we get

$9y^2_1=0\Rightarrow y_1=0$

Putting

$x_1=4$ in

$(1)$ we get

$9y^2_1=(4)^3\Rightarrow y_1=\pm \dfrac{8}{3}$

But the line making equal intersepts with the coordinate axes cannot pass through origin.

Hence, the required points are

$\left(4, \dfrac{8}{3}\right)$ and

$\left(4, \dfrac{-8}{3}\right)$ .

Mark the correct alternative of the following.
The line

$y=mx+1$ is a tangent to the curve

$y^2=4x$ , if the value of m is?

0%
$1$
0%
$2$
0%
$3$
0%
$1/2$

Explanation

Given equation of the tangent to the given curve

$y=mx+1$

Now substituting the value of y in

$y^2=4x$ , we get

$\Rightarrow (mx+1)^2=4x$

$\Rightarrow m^2x^2+1+2mx-4x=0$

$\Rightarrow m^2x^2+x(2m-4)+1=0$ ..........

$(1)$

Since, a tangent touches the curve at one point, the root of equation

$(1)$ must be equal.

Thus, we get

Discriminant,

$D=b^2-4ac=0$

$(2m-4)^2-4(m^2)(1)=0$

$\Rightarrow 4m^2-16m+16-4m^2=0$

$\Rightarrow -16m+16=0$

$\Rightarrow m=1$ .

The slope of the tangent to the curve

$x=t^2+3t-8, y=2t^2-2t-5$ at the point

$(2, -1)$ is

0%
$\dfrac {22}{7}$
0%
$\dfrac {6}{7}$
0%
$\dfrac {7}{6}$
0%
$-\dfrac {6}{7}$

Explanation

Given curve

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

$y = 2t^2 - 2t - 5$

slope of tangent to the curve

$= \dfrac{dy}{dx} = \dfrac{dy/dt}{dx / dt}$

$\therefore \dfrac{dx}{dt} = 2t + 3 , \, \dfrac{dy}{dt} = 4t - 2$

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{4t - 2}{2t + 3}$

$\left(\dfrac{dy}{dx} \right)_{t = 2}$

$\dfrac{4(2) - 2}{2(2) + 3} = \dfrac{6}{7}$

The normal at the point

$(1, 1)$ on the curve

$2y+x^2=3$ is

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

Explanation

$f(x)=x(x-4)^2, x\in 1$

The equation of the givne curve is

$2y+x^2=3$

Differentiating wrt to x, we have

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

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

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

The slope of the normal to the given curve at point

$(1, 1)$ is

$\dfrac{-1}{(dy/dx)_{(1, 1)}}=1$

Hence the equation of the normal to the given curve at

$(1, 1)$ is given as

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

$\Rightarrow y-1=x-1$

$\Rightarrow x-y=0$ .

Consider the equation

$x^y=e^{x-y}$
What is

$\dfrac{d^2y}{dx^2}$ at

$x=1$ equal to ?

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

Explanation

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

Taking

$\log$ both sides, we have

$y \log{x} = \left( x - y \right) \log{e}$

$y \log{x} = x - y ..... \left( 1 \right)$

$\dfrac{y}{x}=1+\log x$

$x = 1 \Rightarrow y = 1$

Differentiating equation

$\left( 1 \right)$ w.r.t.

$x$ , we have

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

$\Rightarrow \left( 1 + \log{x} \right) \cfrac{dy}{dx} = 1 - \cfrac{y}{x}$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{x - y}{x \left( 1 + \log{x} \right)}$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{y \log{x}}{x \left( 1 + \log{x} \right)}$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{\log{x}}{{\left( 1 + \log{x} \right)}^{2}}$

Again differentiating above equation w.r.t.

$x$ , we have

$\cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{{\left( 1 + \log{x} \right)}^{2} \cdot \frac{1}{x} - \log{x} \cdot 2 \left( 1 + \log{x} \right) \frac{1}{x}}{{\left( 1 + \log{x} \right)}^{4}}$

$\Rightarrow \cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{\frac{1}{x} \left( 1 - \log{x} \right)}{{\left( 1 + \log{x} \right)}^{3}}$

$x = 1$ ,

$\cfrac{{d}^{2}y}{d{x}^{2}} = 1$

Consider the equation

$x^y=e^{x-y}$
What is

$\dfrac{dy}{dx}$ at

$x=1$ equal to ?

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

Explanation

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

Taking

$\log$ both sides, we have

$y \log{x} = \left( x - y \right) \log{e}$

$y \log{x} = x - y ..... \left( 1 \right)$

$x = 1 \Rightarrow y = 1$

Differentiating equation

$\left( 1 \right)$ w.r.t.

$x$ , we have

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

$\Rightarrow \left( 1 + \log{x} \right) \cfrac{dy}{dx} = 1 - \cfrac{y}{x}$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{x - y}{x \left( 1 + \log{x} \right)}$

$x = 1, \; y = 1$

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

A curve

$y=me^{mx}$ where

$m > 0$ intersects y-axis at a point

$P$ .
How much angle does the tangent at

$P$ make with y-axis ?

0%
$\tan^{-1}m^2$
0%
$\cot^{-1}(a+m^2)$
0%
$\sin^{-1}(\dfrac{1}{\sqrt{1+m^4}})$
0%
$\sec^{-1}\sqrt{1+m^4}$

A curve

$y=me^{mx}$ where

$m > 0$ intersects y-axis at a point

$P$ .
What is the slope of the curve at the point of intersection

$P$ ?

0%
$m$
0%
$m^2$
0%
$2m$
0%
$2m^2$

Explanation

$y = m {e}^{mx}, \; m > 0$

Therefore,

Slope

$= \cfrac{dy}{dx} = {m}^{2} {e}^{mx}$

Substituting

$x = 0$ , we have

$Slope={m}^{2} {e}^{m \cdot 0} = {m}^{2}$

For

$x > 1, y=\log_e x$ satisfies the inequality

0%
$x-1 > y$
0%
$x^2 -1 >y$
0%
$y > x-1$
0%
$\dfrac {x-1}{x} < y$

The slope of the tangent to the curve

$y = \sqrt{4-x^{2}}$ at the point, where the ordinate and the abscissa are equal , is

0%
-1
0%
1
0%
0
0%
None of these

Explanation

Putting

$y=x$ in

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

$x = \sqrt{2}, -\sqrt{2}$ .

So, the point is

$(\sqrt{2}, \sqrt{2})$ .

Differentiating

$y^{2}+x^{2} = 4$ w.r.t. x,

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

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

$\Rightarrow at(\sqrt{2}, \sqrt{2}), \dfrac{dy}{dx} = -1$

If m is the slope of a tangent to the curve

$e^{y}=1+x^{2},$ then

0%
$\left | m \right |> 1$
0%
$m> 1$
0%
$m> -1$
0%
$\left | m \right |\leq 1$

Explanation

Differentiating w.r.t.x, we get

$e^{y} \dfrac{dy}{dx} = 2x$

$\Rightarrow \dfrac{dy}{dx} = \dfrac{2x}{1+x^{2}} (\because e^{y} = 1 +x^{2})$

$\Rightarrow m = \dfrac{2x}{1 + x^{2}} or \left | m \right |= \dfrac{2\left | x \right |}{1 + \left | x \right |^{2}}$

But

$1 + \left | x \right |^{2} - 2\left | x \right |=(1-\left | x \right |)^{2}\geq 0$

$\Rightarrow 1 + \left | x \right |^{2} \geq 2\left | x \right |$

$\therefore \left | m \right |\leq 1$

The abscissa of points P and Q in the curve

$y = e^{x}+e^{-x}$ such that tangents at P and Q make

$60^{o}$ with the x-axis

0%
ln $\left ( \dfrac{\sqrt{3}+\sqrt{7}}{7} \right )$ and ln $\left ( \dfrac{\sqrt{3}+\sqrt{5}}{2} \right )$
0%
ln $\left ( \dfrac{\sqrt{3}+\sqrt{7}}{2} \right )$
0%
ln $\left ( \dfrac{\sqrt{7}+\sqrt{3}}{2} \right )$
0%
$\pm$ ln $\left ( \dfrac{\sqrt{3}+\sqrt{7}}{2} \right )$

If x=4 y = 14 is a normal to the curve

$y^{2}=ax^{3}-\beta$ at (2,3) then the value of

$\alpha +\beta$ is

0%
9
0%
-5
0%
7
0%
-7

At what points of curve

$y = \dfrac{2}{3}x^{3}+\dfrac{1}{2}x^{2}$ , the tangent makes the equal with the axis?

0%
$(\dfrac{1}{2},\dfrac{5}{24})$ and $\left ( -1,\dfrac{-1}{6} \right )$
0%
$(\dfrac{1}{2},\dfrac{4}{9})$ and $( -1,0)$
0%
$\left ( \dfrac{1}{3},\dfrac{1}{7} \right )$ and $\left ( -3, \dfrac{1}{2} \right )$
0%
$\left ( \dfrac{1}{3},\dfrac{4}{47} \right )$ and $\left ( -1, \dfrac{1}{2} \right )$

The curve represented parametrically by the equations x = 2 in

$\cot t+1$ and

$y=\tan t+\cot t$

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

If a variable tangent to the curve

$x^{2}y=c^{3}$ makes intercepts a, b on x-and y-axes, respectively, then the value of

$a^{2}b$ is

0%
$27c^{3}$
0%
$\dfrac{4}{27}c^{3}$
0%
$\dfrac{27}{4}c^{3}$
0%
$\dfrac{4}{9}c^{3}$

The angle between the tangent to the curves

$y = x^{2}$ and

$x = y^{2}$ at (1, 1) is

0%
$\cos ^{-1}\dfrac{4}{5}$
0%
$\sin ^{-1}\dfrac{3}{5}$
0%
$\tan ^{-1}\dfrac{3}{4}$
0%
$\tan ^{-1}\dfrac{1}{3}$

At the point

$P(a, a^{n})$ on the graph of

$y = x^{n}(n \epsilon n)$ in the first quadrant, a normal is drawn. the normal intersects the y-axis at the point (0, b) . if

$\underset{a\rightarrow b}{lim}b=\dfrac{1}{2}$ , then n equals

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

Point on the curve

$f(x)=\dfrac{x}{1-x^{2}}$ where the tangent is inclined at an angle of

$\dfrac{\pi }{4}$ ot the x-axis are

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

The x-intercept of the tangent at any arbitrary point of the curve

$\dfrac{a}{x^{2}}+\dfrac{b}{y^{2}}=1$ is proportion to

0%
square of the abscissa of the point of tangency
0%
square root of the abscissa of the point of tangency
0%
cube of the abscissa pf the point of tangency
0%
cube root of the abscissa of the point of tangency

If the tangent at any point

$P(4m^{2}, 8m^{3})$ of

$x^{3}-y^{3}=0$ is also a normal to the curve

$x^{3}-y^{3}=0$ , then value of m is

0%
$m = \dfrac{\sqrt{2}}{3}$
0%
$m = -\dfrac{\sqrt{2}}{3}$
0%
$m = \dfrac{3}{\sqrt{2}}$
0%
$m = -\dfrac{3}{\sqrt{2}}$

The slope of the tangent to the curve

$y = f(x)$ at

$\left [ x, f(x) \right ]$ is 2x +If the curve passes through the point (1, 2)then the area bounded by the curve, the x-axis and the line x = 1 is

0%
$\dfrac{5}{6}$
0%
$\dfrac{6}{5}$
0%
$\dfrac{1}{6}$
0%
6

The normal to the curve

$x = a (\cos 0 + 0\sin 0), y= a (\sin 0- 0\cos 0)$ at any point 0 is such that

0%
it makes a constant angle with x-axis
0%
it passes through the origin
0%
it is at a constant distance from the origin
0%
none of these

A curve passes through

$(2,1)$ and is such that the square of the ordinate is twice the contained by the abscissa and the intercept of the normal. Then the equation of curve is

0%
$x^2 +y^2=9x$
0%
$4x^2 +y^2=9x$
0%
$4x^2 +2y^2=9x$
0%
None of these

The curve for which the ratio of the length of the segment by any tangent on the

$Y-$ axis to the length of the radius vector is constant

$(K)$ , is

0%
$(y+\sqrt {x^2 -y^2})x^{k-1}=c$
0%
$(y+\sqrt {x^2 +y^2})x^{k-1}=c$
0%
$(y-\sqrt {x^2 -y^2})x^{k-1}=c$
0%
$(y+\sqrt {x^2 +y^2})x^{k-1}=c$

$f(x) = \displaystyle \int_{1}^{x} e^{t^2/2}(1-t^2)dt,$ then

$\dfrac{d}{dx} f(x)$ at x=1 is

0%
$0$
0%
$1$
0%
$2$
0%
$-1$

Explanation

We have,

$f(x)=\displaystyle \int_{1}^{x}e^{t^2/2}(1-t^2)dt$

$f'(x)=[e^{x^2/2}(1-x^2)]^2$

$f'(1)=e^{1/2}.0=0$

$y=\displaystyle \int_{x}^{x^2}t^2dt ,$ then the equation of tangent at x=1 is

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

Explanation

$x=1,y=0,$

$\dfrac{dy}{dx}=2x.(x^4)^2-(x^2)^2=1$

$\,\therefore$ Equation of tangent is

$y=x-1.$

The tangent to the curve

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

$(c, e^{c})$ 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 point
0%
at all point

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%
$a < 0, b < 0$

The equation of the curves through the point

$(1,0)$ and whose slope is

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

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

Explanation

Slope

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

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

$\implies \int\dfrac{1}{y-1}dy = \int(\dfrac{1}{x} - \dfrac{1}{x+1})dx + C$

$\implies \log(y-1)=\log(\dfrac{x}{x+1})+\log{c}$

$\implies \dfrac{(y-1)(x+1)}{x} = k$
Putting

$x=1, y=0$ , we get

$k=-2$
The equation is

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

The point(s) on the curve

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

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

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.