MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 2

The equation of the tangent to the curve

$\displaystyle 6y=7-x^{3}$ at point

$(1, 1)$ is

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

Explanation

Given curve is,

$6y=7-x^3$

$\Rightarrow 6\cfrac{dy}{dx}=-3x^2\Rightarrow \cfrac{dy}{dx}=-\cfrac{1}{2}x^2$
Thus slope of tangent to this curve at

$(1,1,)$ is

$m=-\cfrac{1}{2}$
The equation of required tangent is,

$(y-1)=-\cfrac{1}{2}(x-1)\Rightarrow x+2y=3$
Hence, option 'B' is correct.

The slope of the curve

$\displaystyle y=\sin x+\cos ^{2}x$ is zero at the point where -

0%
$\displaystyle x=\frac{\pi }{4}$
0%
$\displaystyle x=\frac{\pi }{2}$
0%
$\displaystyle x=\pi$
0%
No where

The equation of the tangent to the curve

$\displaystyle y=\cos x$ at

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

0%
$\displaystyle 3x-2\sqrt{3}y=\pi +\sqrt{3}$
0%
$\displaystyle 3x+2\sqrt{3}y=\pi +\sqrt{3}$
0%
$\displaystyle 3x+2\sqrt{3}y=\pi -\sqrt{3}$
0%
None of these

Explanation

$\displaystyle y=\cos x$

$\displaystyle \frac{dy}{dx}=-\sin x$
The slope of tangent at

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

$m=\displaystyle \left ( \dfrac{dy}{dx} \right )_{x=\dfrac{\pi }{3}}=-\cfrac{\sqrt{3}}{2}$
Now at

$\displaystyle x=\dfrac{\pi }{3};\: \: \: y=\cfrac{1}{2}$
So the point is

$\displaystyle \left ( \cfrac{\pi }{3},\cfrac{1}{2} \right )$
Therefore, required tangent is,

$\displaystyle \left ( y-\cfrac{1}{2} \right )=-\cfrac{\sqrt{3}}{2}(x-\cfrac{\pi}{3})$

$\Rightarrow 3x+2\sqrt{3}y=\pi +\sqrt{3}$
Hence, option 'B' is correct.

The slope of the tangent to the curve

$\displaystyle y=\sin x$ at point

$(0, 0)$ is

0%
$1$
0%
$0$
0%
$\displaystyle \infty$
0%
None of these

Explanation

Given

$y=\sin x$

$\Rightarrow \cfrac{dy}{dx}=\cos x$
Thus, slope of tangent at

$(0,0)$ is,

$m=\left (\cfrac{dy}{dx}\right )_{(0.0)}=\cos (0)=1$
Hence, option 'A' is correct.

If tangent at a point of the curve

$y = f(x)$ is perpendicular to

$2x - 3y = 5$ then at that point

$\displaystyle \dfrac{dy}{dx}$ equals

0%
$\dfrac 2 3$
0%
$-\dfrac 2 3$
0%
$\dfrac 3 2$
0%
$-\dfrac 3 2$

Explanation

We know slope of tangent to the curve is

$\displaystyle \frac{dy}{dx}$
Now slope of line

$2x - 3y = 5$ is

$\displaystyle \frac{2}{3}$
and tangent to

$y=f(x)$ is perpendicular to this line.

$\displaystyle \therefore$

$\displaystyle \frac{dy}{dx}=-\frac{3}{2}$
Hence, option 'D' is correct.

The inclination of the tangent w.r.t.

$x$ - axis to the curve

$\displaystyle x^{2}+2y=8x-7$ at the point

$x = 5$ is

0%
$\displaystyle\dfrac{ \pi }4$
0%
$\displaystyle\dfrac{ \pi }3$
0%
$\displaystyle\dfrac{3 \pi }4$
0%
$\displaystyle\dfrac{ \pi }2$

Explanation

Given,

$\displaystyle x^{2}+2y=8x-7$

$\Rightarrow 2y=8x-x^2-7$

$\therefore 2\cfrac{dy}{dx}=8-2x\Rightarrow \cfrac{dy}{dx}=4-x$
Thus slope of tangent to the given curve at

$x=5$ is,

$m=-1$
If

$'\theta '$ is the inclination of the tangent w.r.t

$x-axis$ then,

$\tan\theta=-1\Rightarrow \theta=\cfrac{3\pi}{4}$
Hence, option 'C' is correct.

The equation of the tangent to the curve

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

$(1,2)$ is

0%
$y = 2x$
0%
$x + 2y = 5$
0%
$2x + y = 4$
0%
None of these

Explanation

$y=x^2+1$

$\cfrac{dy}{dx}=2x$
Thus slope of tangent at

$(1,2)$ is,

$m=2\times 1=2$
Therefore, the required tangent is given by,

$(y-2)=2(x-1)\Rightarrow y=2x$
Hence, option 'A' is correct.

The point on the curve

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

$y = x$ is -

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

Explanation

Let the point be

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

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

$x$

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

$=\left(\cfrac{dy}{dx}\right)_{(a,b)}=2a-3$
But given the tangent is perpendicular to line

$y=x$

$\Rightarrow 2a-3=-1\Rightarrow a=1$
Also the point P lies on the given curve,

$b=a^2-3a+2=0$
Therefore, the point P is

$(1,0)$
Hence, option 'B' is correct.

The equation of normal to the curve

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

$x = 2$ is-

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

The sum of the intercepts made by a tangent to the curve

$\displaystyle \sqrt{x}+\sqrt{y}=4$ at point

$(4, 4)$ on coordinate axes is -

0%
$\displaystyle 4\sqrt{2}$
0%
$\displaystyle 6\sqrt{3}$
0%
$\displaystyle 8\sqrt{2}$
0%
$\displaystyle \sqrt{256}$

The equation of normal to the curve

$\displaystyle y=\tan x$ at the point

$(0, 0)$ is -

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

Explanation

$\displaystyle y=\tan x$

$\Rightarrow \displaystyle \frac{dy}{dx}=\sec x$

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

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

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

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

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

The equation of normal to the curve

$\displaystyle x^{\tfrac 23}$ +

$\displaystyle y^{\tfrac 23}$ =

$\displaystyle a^{\tfrac 23}$ at the point

$(a, 0)$ is -

0%
$x = a$
0%
$x = -a$
0%
$y = a$
0%
$y = -a$

Explanation

Given curve is,

$\displaystyle x^{\dfrac{2}{3}}+y^{\dfrac{2}{3}}=a^{\dfrac{2}{3}}$
Differentiating w.r.t

$x$

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

$\Rightarrow \cfrac{dy}{dx}=\left(\cfrac{y}{x}\right)^{\dfrac{1}{3}}$
Thus slope of normal at

$(a,0)$ is

$m=-\cfrac{dx}{dy}=\left(\cfrac{a}{0}\right)^{\dfrac{1}{3}}\to \infty$
Therefore, equation of normal is,

$x=a$
Hence, option 'A' is correct.

At what point the tangent to the curve

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

$x$ - axis

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

Explanation

Let the point be

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

$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$
Differentiating w.r.t

$x$

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

$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{y}{x}}$
Thus slope of tangent at P is

$=\displaystyle \left ( \frac{dy}{dx} \right )_{\left ( x_{1},y_{1} \right )}=-\sqrt{\frac{y_{1}}{x_{1}}}$
if tangent

$\displaystyle \perp$ to x axis then

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

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

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

The equation of the normal to the curve

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

$(1, 1)$ is -

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

Explanation

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

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

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

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

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

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

$\Rightarrow x - y =0$
Hence, option 'D' is correct.

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%
$x-y=1$
0%
answer required

Explanation

Equation of the curve is

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

Here slope of tangent is given by

$m=\dfrac{dy}{dx}=-x=-1$
So slope of normal will be

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

Hence, equation is given by

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

$\Rightarrow y=x$

The line

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

$y^2 = 4x$ at the point.

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

Explanation

We have

$y=x+1, y^2=4x$
Solving these,

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

$\Rightarrow (x-1)^2=0\Rightarrow x=1\therefore y = (x+1)_{x=1}=2$
Thus required point of contact is

$(1,2)$

If a tangent to the curve

$\displaystyle y=6x-{ x }^{ 2 }$ is parallel to the line

$\displaystyle 4x-2y-1=0$ , then the point of tangency on the curve is:

0%
(2, 8)
0%
(8, 2)
0%
(6, 1)
0%
(4, 2)

Explanation

$y=6x-x^2$

$\Rightarrow y' =6-2x\ as\ tangent \ is\ parallel \ to \ line 4x-2y-1=0 \$ slope

$=2$

$\Rightarrow 6-2x=2$

$\Rightarrow x=2$

$y=6\cdot 2-2^2=8$

Hence the point of tangency is

$(2,8)$

The slope of the tangent to the curve

$y = \int_{0}^{x} \dfrac {dt}{1 + t^{3}}$ at the point where

$x = 1$ is

0%
$\dfrac {1}{4}$
0%
$\dfrac {1}{3}$
0%
$\dfrac {1}{2}$
0%
$1$

Explanation

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

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

The equation of the tangent to the curve

$y=4xe^x$ at

$\left(-1, \displaystyle\frac{-4}{e}\right)$ is

0%
$y=-1$
0%
$y=-\displaystyle\frac{4}{e}$
0%
$x=-1$
0%
$x=\displaystyle\frac{-4}{e}$

Explanation

Given curve is

$y=4xe^x$

$\displaystyle\frac{dy}{dx}=4e^x+4xe^x$
At

$\left(-1, -\displaystyle\frac{4}{e}\right)\left(\displaystyle\frac{dy}{dx}\right)_{(\displaystyle -1, -4/e)}=4e^{-1}+4(-1)e^{-1}$

$\therefore$ Equation of tangent is

$\left(y+\displaystyle\frac{4}{e}\right)=0(x+1)$

$\Rightarrow y=-\displaystyle\frac{4}{e}$

The equation of the normal to the curve

${ y }^{ 4 }=a{ x }^{ 3 }$ at

$\left( a,a \right)$ is

0%
$x+2y=3a$
0%
$3x-4y+a=0$
0%
$4x+3y=7a$
0%
$4x-3y=0$

Explanation

Given curve is

${ y }^{ 4 }=a{ x }^{ 3 }$
On differentiating with respect to

$x$ , we get

$4{ y }^{ 3 }\dfrac { dy }{ dx } =3a{ x }^{ 2 }$

$\Rightarrow { \left( \dfrac { dy }{ dx } \right) }_{ \left( a,a \right) }=\dfrac { 3{ a }^{ 3 } }{ 4{ a }^{ 3 } } =\dfrac { 3 }{ 4 }$

$\therefore$ Equation of normal at point

$\left( a,a \right)$ is

$y-a=-\dfrac { 4 }{ 3 } \left( x-a \right)$

$\Rightarrow 4x+3y=7a$

The slope of the tangent to the curve

$y=\displaystyle\int_{0}^{x}\dfrac{dt}{1+t^3}$ at the point where x=1 is

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
1

Explanation

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

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

Consider the curve

$y = e^{2x}$ .What is the slope of the tangent to the curve at (0, 1) ?

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

Explanation

We know that slope of tangent to a curve

$y=f(x)$ at a point

$(x_1,y_1)$ equals

$f^{'}(x_1)$

$\Rightarrow$ Slope of tangent to the curve

$y=e^{2x}$ equals

$\cfrac{d(e^{2x})}{dx} =2e^{2x}$

$\Rightarrow$ Slope of tangent at

$(0,1)=f^{'}(0)=2$

The gradient of the tangent line at the point

$(a cos \alpha, a sin \alpha)$ to the circle

$x^2 + y^2 = a^2$ , is

0%
$tan (\pi - \alpha)$
0%
$tan \alpha$
0%
$cot \alpha$
0%
- $cot \alpha$

Explanation

Equation of the circle,

$x^2+y^2=a^2$

At a point

$(a \cos \alpha, a \ sin \alpha)$

Gradient of radius

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

$= \dfrac{a \ sin \alpha-0}{a \ cos \alpha-0}$

$=\ tan \alpha$

gradient of radius

$\times$ gradient of tangent = -1

$\ tan \alpha \times$ gradient of tangent = -1

$\Rightarrow$ gradient of tangent

$= -\ tan \alpha$

$\Rightarrow$ gradient of tangent

$= \ tan (\pi - \alpha)$

If normal is drawn to

${ y }^{ 2 }=12x$ making an angle

${45}^{o}$ with the axis then the foot of the normal is

0%
$\left( 3,8 \right)$
0%
$\left( 3,-6 \right)$
0%
$\left( 12,-12 \right)$
0%
$\left( 8,-8 \right)$

Explanation

Given that,

${y^2} = 12x - - \left( 1 \right)$

Diff. w.r.t. to

$x$

$2y\dfrac{{dy}}{{dx}} = 12$

$y\dfrac{{dy}}{{dx}} = 6 - - - - \left( 2 \right)$

Slope tangent

$\dfrac{{dy}}{{dx}} = \tan 45$

$m = 1$

Slope at Normal from

$(2)$

$\Rightarrow y\left( { - 1} \right) = 6$

$\Rightarrow y =- 6$

$\therefore x = 3$

Point

$=(3,-6)$

Option

$(A)$ is correct answer.

Which one of the following be the gradient of the hyperbola

$xy=1$ at the point

$\left(t,\dfrac{1}{t}\right)$

0%
$-\dfrac{1}{t}$
0%
$-\dfrac{1}{t^2}$
0%
$\dfrac{1}{t}$
0%
$-\dfrac{2}{t^2}$

Explanation

Given the hyperbola

$xy=1$ .

Now differentiate both sides with respect to

$x$ we get,

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

or,

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

Now the gradient of the given hyperbola at

$\left(t,\dfrac{1}{t}\right)$ is

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

If the product of the slope of tangent to curve at

$(x,y)$ and its y-co-ordinate is equal to the x-co-ordinate of the point, then it represent.

0%
circle
0%
parabola
0%
ellipse
0%
rectangular hyperbola

Explanation

Let slope of a tangent at point

$(x,y)$ to the curve is

$=\cfrac { dy }{ dx }$

$\therefore y\cfrac { dy }{ dx } =x$ (given)

$\therefore ydy=xdx\quad$
Integrating both side

$\int { y } dy=\int { x } dx$

$\therefore \cfrac { { y }^{ 2 } }{ 2 } =\cfrac { { x }^{ 2 } }{ 2 } +c'$

$\therefore { y }^{ 2 }-{ x }^{ 2 }+2c'$

$\therefore { y }^{ 2 }-{ x }^{ 2 }=c$
where

$2c'=c$
Which is a rectangular hyperbola

The slope of the tangent to the curve

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

$(1,1)$ is

$2$ , then value of

$a$ and

$b$ are respectively:

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

Explanation

$xy+ax-by=0$

$(1,1)$ lies on curve

$1+a-b=0$

$a-b=-1$ .......

$(i)$

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

$1+2+a-2b=0$

$a-2b=-3$ ........

$(ii)$

Solving

$(i)$ and

$(ii)$ , we get

$a=1,b=2$

The Equation of the tangent to the curves

${y^2} = 8x$ and

$xy = -1$ is

0%
$3y=9x+2$
0%
$y=2x+1$
0%
$2y=x+1$
0%
$y=x+2$

Explanation

Any tangent to

${y^2} = 8x$ is

$y = mx + \dfrac{2}{m}$ {equation of tangent of parabola}
it should also be tangent to

$xy=-1$
so

$x\left( {mx + \dfrac{2}{m}} \right) = - 1$

$m^2x^2+2x+m=0$

$\therefore D=0$

$4-4m^3=0$

$\therefore m=1$
so tangent

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

$\Rightarrow y = x + 2$

The values of

$x$ for which the tangents to the curves

$y=x\cos{x},y=\cfrac{\sin{x}}{x}$ are parallel to the axis of

$x$ are roots of (respectively)

0%
$\sin{x}=x,\tan{x}=x$
0%
$\cot{x}=x,\sec{x}=x$
0%
$\cot{x}=x,\tan{x}=x$
0%
$\tan{x}=x,\cot{x}=x$

Explanation

$y=x\cos x$

$\dfrac{dy}{dx}=\cos x-x\sin x=0$

$\implies x=\cot x$

$y=\dfrac{\sin x}{x}$

$\dfrac{dy}{dx}=\dfrac{x\cos x-\sin x}{x^{2}}=0$

$\implies x=\tan x$

A curve with equation of the form

$y=a{x}^{4}+b{x}^{3}+cx+d$ has zero gradient at the point

$(0,1)$ and also touches the x-axis at the point

$(-1,0)$ then

0%
$a=3$
0%
$b=4$
0%
$c+d=1$
0%
for $x< -1$ the curve has a negative gradient

Explanation

given

$y=ax^4+bx^3+cx+d$

$y_{}{'}=4ax^3+3bx^2+c=0$ at (0,1)

$c=0$

curve passes through point (-1,0)

$0=a-b-c+d=>a+d=b$

here

$y_{}{'}<0$ for

$x<-1$

The equation of the tangent to the curve

$y=2\sin{x}+\sin{2x}$ at

$x=\cfrac{\pi}{3}$ on it is

0%
$y-3=0$
0%
$y+\sqrt{3}=0$
0%
$2y-3=0$
0%
$2y-3\sqrt{3}=0$

Explanation

$x=\dfrac{\pi }{3}\Rightarrow y=2\left(\dfrac{\sqrt{3}}{2}\right)+\dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}$

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

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

$=0$

$\therefore$ equation of tangent is

$\dfrac{y-3\sqrt{3}/2}{x=\pi /3}=0$

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

The coordinates of the feet of the normals drawn from the point (14, 7) to the curve

$y^2 - 16 x - 8y = 0$ are

0%
(0, 0)
0%
(3, 4)
0%
(3, -4)
0%
(8, 16)

The slope of the tangent to the curve

$y=sinx$ where it crosses the

$x-axis$ is

0%
$1$
0%
$-1$
0%
$\pm 1$
0%
$\pm 2$

Explanation

Consider the given equation of the curve.

$y=\sin x$ ……(1)

On differentiating equation (1) with respect to $x$ , we get

$\dfrac{dy}{dx}=\cos x$

When given curve crosses the x-axis then.

$y=0$

$\sin x=0$

$x=0$

Therefore,

${{\left( \dfrac{dy}{dx} \right)}_{y=0}}=\cos 0$

${{\left( \dfrac{dy}{dx} \right)}_{y=0}}=1$

Hence, this is the answer.

The equation of normal to the curve

$y=\left| { x }^{ 2 }-\left| x \right| \right|$ at

$x=-2$ is

0%
$3y=2x+10$
0%
$3y=x+8$
0%
$2y=x+6$
0%
$2y=3x+10$

Explanation

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

$x=-2,y=2$

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

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

The equation of normal is

$y=\dfrac{x}{3}+\bigg(2+\dfrac{2}{3}\bigg)$

$3{y}=x+8$

The equation of the normal at

$t=\dfrac{\pi}{2}$ to the curve

$x=2\sin t, y=2\cos t$ is?

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

Explanation

$dx = 2\cos t , dy = -2\sin t$

$-\cfrac{dx}{dy} = -\cot t$

$\cfrac{dx}{dy} = \cot t$

Slope =

$0$ as it is x-axis

Hence,

$y = 0$

The intercept on x-axis made by tangent to the curve,

$\displaystyle y=\int _{ 0 }^{ x }{ \left| t \right| } dt,x\in R$ , which are parallel to the line

$y=2x$ , are equal to

0%
$\pm 1$
0%
$\pm 2$
0%
$\pm 3$
0%
$\pm 4$

Explanation

$y = \int_{0}^{x}{\left| t \right| dt}$

$\Rightarrow \cfrac{dy}{dx} = \left| x \right|$

Since tangent to the curve is parallel to the line

$y = 2x$ .

$\therefore \cfrac{dy}{dx} = 2$

$\Rightarrow x = \pm 2$

Therefore,

$y = \int_{0}^{\pm 2}{\left| t \right| dt}$

$\Rightarrow y = \pm 2$

Therefore,

Equation of tangents are-

$y - 2 = 2 \left( x - 2 \right)$

$y + 2 = 2 \left( x + 2 \right)$

For

$x$ -intercept, substitute

$y = 0$ , we have

$0 - 2 = 2 \left( x - 2 \right) \Rightarrow x = +1$

$0 + 2 = 2 \left( x + 2 \right) \Rightarrow x = -1$

Hence the intercept on

$x$ -axis are

$\pm 1$ .

The Point (s) on the cure

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

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

Explanation

curve given

$y^3+3x^2=12y$

Thus differentiating w.r.t

$x$

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

$\dfrac {dy}{dx}(y^2-4)+2x=0$

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

For tangent to be

$\parallel$ to

$y-$ axis

$\dfrac {dy}{dx}\rightarrow \infty$

Thus

$y^2-4=0$

$y=\pm 2$

Putting

$y=+2$ in curve we get

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

Putting

$y=-2$ in curve

$x$ does not exit

Thus

$y=-2$ not in range as

$x^2 < 0$

Thus points where tangent is

$\parallel$ to

$y-$ axis is

$\left [\pm \dfrac {4}{\sqrt {3}},\ 2 \right]$ Answer.

Tangent drawn to

$y=ax^2+bx+c$ at

$(5,4)$ is parallel to

$x-axis.$ If

$a$

$\epsilon$

$[2,4]$ . Then maximum value of c.

0%
$54$
0%
$56$
0%
$104$
0%
$106$

Explanation

$y=a{x^2}+b{x}+c\implies \dfrac{d y}{d x}=2{a}{x}+b$

$x=5,y=4\implies 25{a}+5{b}+c=4$ and also tangent is parallel to

$x-axis$

$\implies 10{a}+b=0\implies b=-10{a}\implies c=4-5{b}-25{a}=4+25{a}$

$a\in [2,4]\implies c\in [54,104]$

The angle made by the tangent line at (1, 3) on the curve

$y=4x-{ x }^{ 2 }$ with

$\overset { - }{ OX }$ is

0%
${ tan }^{ -1 }(2)$
0%
${ tan }^{ -1 }(1/3)$
0%
${ tan }^{ -1 }(3)$
0%
$\pi /4$

Explanation

Given,

$y=4x-x^2$

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

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

Therefore, angle made by tangent,

$\tan \theta =2$

$\therefore \theta =\tan ^{-1}2$

The equation of normal to the curve

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

$\left(1,1\right)$

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

Explanation

The curve is

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

Slope

$=2{y}^{\prime}=-2x$ or

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

Slope at

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

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

Equation of normal is

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

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

$\Rightarrow x-y=0$ is the equation of the normal.

equation of tangent at

$(0,0)$ for the equation

$y^2=16x$

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

Explanation

The equation is

$y^2=16x$

The point is

$(0,0)$

The slope of tangent

$2y \dfrac{dy}{dx}=16\\\dfrac{dy}{dx}=\dfrac 8y\\\left.\dfrac{dy}{dx}\right|_{(0,0)}=\infty$

Equation of tangent is

$y-0=\dfrac{8}{0}(x-0)\\x=0$

The area of triangle formed by tangent and normal at point

$(\sqrt{3}, 1)$ of the curve

$x^2+y^2=4$ and x-axis is?

0%
$\dfrac{4}{\sqrt{3}}$
0%
$\dfrac{2}{\sqrt{3}}$
0%
$\dfrac{8}{\sqrt{3}}$
0%
$\dfrac{5}{\sqrt{3}}$

Explanation

Slope of OP

$=\dfrac{1}{\sqrt{3}}$ , slope of PQ

$=-\sqrt{3}$

$y-1=-\sqrt{3}(x-\sqrt{3})=-\sqrt{3}x+3$

$\Rightarrow \sqrt{3}x+y=4$ and

$Q\left(\dfrac{4}{\sqrt{3}}, 0\right)$

$\Delta OPQ=\dfrac{2}{\sqrt{3}}$ .
1245724_1611460_ans_76b3a467db73433698dc5776e9a56489.PNG

The tangent to the curve

$y=e^{kx}$ at a point (0, 1) meets the x-axis at (q, 0) where

$a \epsilon \left [ -2,-1 \right ]$ then

$k \epsilon$

0%
[-1/2,0]
0%
[-1,-1/2]
0%
[0,1]
0%
[1/2, 1]

A curve

$y=me^{mx}$ where

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

$P$ .

What is the equation of tangent to the curve at

$P$ ?

0%
$y=mx+m$
0%
$y=-mx+2m$
0%
$y=m^2x+2m$
0%
$y=m^2x+m$

Explanation

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

Substituting

$x = 0$ , we have

$y = m {e}^{m \cdot 0} = m$

Therefore,

Slope

$= \cfrac{dy}{dx} = {m}^{2} {e}^{mx} = {m}^{2} {e}^{m \cdot 0} = {m}^{2}$

Therefore, equation of tangent at

$P$ is given by-

$y - m = \cfrac{dy}{dx} \left( x - 0 \right)$

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

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

If the slope of the tangent to the curve

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

$(1, 1)$ on it is

$2$ , then values of

$a$ and

$b$ are

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

Explanation

$a+b +1=0$

${xy}'+y+a+{b y}'=0$

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

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

$2+2b +a+1=0$

$2+2 b -b =0$

$b =-2$

$a=1$

If the slope of the tangent to the curve

$y = x^{3}$ at a point on it is equal to the ordinate of the point then the point is

0%
$(27, 3)$
0%
$(3, 27)$
0%
$(3, 3)$
0%
$(1, 1)$

Explanation

Differntiating the given equation

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

$3a^{2}=a^{3}$

$a=3$

$(3,3^{3})$

$(3,27)$

The area of the triangle formed by the tangent to the curve

$\displaystyle y=\frac{8}{4+x^{2}}$ at

$x=2$ on it and the

$x$ -axis is

0%
$2$ sq.units
0%
$4$ sq.units
0%
$8$ sq.units
0%
$16$ sq.units

Explanation

${y}'=\dfrac{8\times 24}{(4+x^{2})^{2}}$

$y=1=\dfrac{-8\times 2\times 2 }{(4+4)^{2}}$

$=\dfrac{-5\times 4}{64}=\dfrac{1}{2}$

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

$2y-2=-x+2$

$2y+x=4$

$A=\dfrac{1}{2}\times 2\times 4=4$

The two curves

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

$y=8x-x^{2}-9$ at the point

$(2, 3)$ have common

0%
tangent as $4x-y-5=0$
0%
tangent as $x+4y-14=0$
0%
normal as $4x+y=11$
0%
normal as $x-4y=10$

Explanation

$y_1 = x^2-1$

$\therefore y'_1=2x$

$y_2 = 8x-x^2-9$

$\therefore y'_2=8-2x$

At the common tangent point, slope will be same.

$2x=8-2x$

$4x=8$

$x=2, \therefore y=3$
This point,

$(2,3)$ , satisfies both curves.
At

$(2,3)$

$y'_1=4$
Equation of tangent at

$(2,3)$

$(y-3)=4(x-2)$

$4x-y-5=0$

For normal, slope will be

$-\dfrac{1}{4}$

$\therefore$ Equation of normal at

$(2,3)$ is

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

$4y-12+x-2=0$

$4y+x-14=0$

Hence, option A.

Equation of the tangent to the curve

$y(x-2)(x-3)-x+7=0$ at the point where
the curve cuts x-axis is

0%
$x-20y=7$
0%
$x-20y+7=0$
0%
$x+20y-7=0$
0%
$x+20y+7=0$

Explanation

$y=0$ (at x-axis)

$0-x+7=0$

$x=7$

$y=\dfrac{x-7}{(x-2)(x-3)}$

${y}'=\dfrac{(x-2)(x-3)-(x-7)(2x-5)}{(x-2)^{2}(x-3)^{2}}$
put

$x=7$

$\dfrac{5\times 4-(0)}{(5\times 4)^{2}}$

${y}'=\dfrac{1}{20}$

$(y-0)=\dfrac{1}{20}(x-7)$

$x-20y-7=0$

P(1, 1) is a point on the parabola

$y=x^{2}$ whose vertex is A. The point on the curve at which the tangent drawn is parallel to the chord

$\overline{AP}$ is

0%
$(\displaystyle \frac{1}{2}, \frac{1}{4})$
0%
$(\displaystyle \frac{-1}{2}, \frac{1}{4})$
0%
$(2, 4)$
0%
$(4, 2)$

Explanation

$tan \theta =\dfrac{(1-0)}{(1-0)}=1$

${y}'=2x$

$2x=1$

$x=\dfrac{1}{2}, y=\dfrac{1}{4}$

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 2 - 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 2 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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