MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 7

If equation of normal at a point

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

$\displaystyle x^{3}-y^{2}=0\: \: is\: \: y=3mx-4m^{3}$ then

$\displaystyle m^{2}$ equals

0%
$0$
0%
$1$
0%
$-\dfrac 2 9$
0%
$\dfrac 2 9$

Explanation

$\displaystyle x^{3}-y^{2}=0$

$\Rightarrow \displaystyle 3x^{2}-2y\left ( \frac{dy}{dx} \right )=0$

$\Rightarrow \displaystyle 3x^{2}=2y\left ( \frac{dy}{dx} \right )$

$\Rightarrow \displaystyle \frac{3x^{2}}{2y}=\frac{dy}{dx}$

$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( m^{2},-m^{3} \right )}=\frac{3\times m^{4}}{-2\times m^{3}}$

$\therefore \displaystyle -\left ( \frac{dx}{dy} \right )=\frac{2}{3m}$
Therefore, equation of normal at given point is,

$\Rightarrow \displaystyle \left ( y+m^{3} \right )=\frac{2}{3m}\left ( x-m^{2} \right )$

$\Rightarrow \displaystyle 3my+3m^{4}=2x-2m^{2}$

$\Rightarrow \displaystyle 3my=2x-3m^{4}-2m^{2}$

$\Rightarrow \displaystyle y=\frac{2x}{3m}-m^{3}-\frac{2}{3}m ..(1)$
but given equation of normal is,

$\displaystyle y=3mx-4m^{3} ..(2)$
Thus comparing (1) and (2), we get

$\displaystyle 3m=\frac{2}{3m}$

$\Rightarrow \displaystyle m^{2}=\frac{2}{9}$
Hence, option 'D' is correct.

The coordinates of the points on the curve

$\displaystyle x=a\left ( \theta +\sin \theta \right ),y=a\left ( 1-\cos \theta \right )$ where tangent is inclined an angle

$\displaystyle \dfrac{\pi }4$ to the

$x-$ axis are -

0%
$(a, a)$
0%
$\displaystyle \left ( a\left ( \frac{\pi }{2}-1 \right ),a \right )$
0%
$\displaystyle \left ( a\left ( \frac{\pi }{2}+1 \right ),a \right )$
0%
$\displaystyle \left ( a,a\left ( \frac{\pi }{2}+1 \right ) \right )$

Explanation

The co-ordinates are

$x = \displaystyle a\left ( \theta +\sin \theta \right );\: \: y=a\left ( 1-\cos \theta \right )$
Thus,

$\displaystyle \frac{dx}{d\theta }=a\left ( 1+\cos \theta \right )\displaystyle =a\left ( 2\cos^{2}\frac{\theta }{2}\right )$ and

$\displaystyle \frac{dy}{d\theta }=a\sin \theta =2a\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$

$\therefore \displaystyle \dfrac{dy}{dx}=\tan \dfrac{\theta }{2}$
but,

$\displaystyle \frac{dy}{dx}=\tan \frac{\pi }{4}$ (given)

$\therefore 1 = \tan \dfrac{\theta }{2}$

$\Rightarrow \displaystyle \theta =\frac{\pi }{2}$
So the point is,

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

$x = \displaystyle a\left ( \frac{\pi }{2}+1 \right )$

$y = a(1 - 0)=a$
Therefore, the required point is

$\displaystyle \left ( a\left ( \frac{\pi }{2}+1 \right ),a \right )$
Hence, option 'C' is correct.

The line

$\dfrac x a +\dfrac y b = 1$ touches the curve

$\displaystyle y=be^{-\tfrac xa}$ at the point -

0%
$(0, a)$
0%
$(0. 0)$
0%
$(0, b)$
0%
$(b, 0)$

Explanation

$\displaystyle y=b\times e^{-\frac{x}{a}}$

$\Rightarrow \displaystyle \frac{dy}{dx}=b\times e^{-\tfrac{x}{a}}\times -\frac{1}{a}=-\frac{b}{a}e^{-\frac{x}{a}}$

Now the slope of given line is,

$\displaystyle m=-\frac{1}{a}\times \frac{b}{1}=-\cfrac{b}{a}$

Thus for the given line to be tangent to the given curve,

$\displaystyle -\frac{b}{a}=-\frac{b}{a}\times e^{-\dfrac{x}{a}}$

$\Rightarrow \displaystyle e^{\dfrac{x}{a}}=1$

$\Rightarrow x = 0\Rightarrow y = b$

Therefore, the point is

$(0, b)$

Hence, option 'C' is correct.

At what values of

$a$ , the curve

$x^4+3ax^3+6x^2+5$ is not situated below any of its tangent lines

0%
$|a|\,>\,\displaystyle\frac{4}{3}$
0%
$|a|\,<\,\displaystyle\frac{4}{3}$
0%
$|a|\,>\,1$
0%
$|a|\,<\,\displaystyle\frac{1}{3}$

Explanation

$y=4x^3+3ax^3+6x^2+5$
For given situation curve must be concave so,

$\displaystyle\frac{d^2y}{dx^2}\,>\,0$

$y'=4x^3+9ax^2+12x$

$y''=12x^2+18ax+12\,>\,0=6(2x^2+3ax+2)\,>\,0$

$D\,<\,0\;\;\Rightarrow\;\;9a^2-16\,<\,0$

$\Rightarrow |a|\,<\,\displaystyle\frac{4}{3}$

The points on the curve

$\displaystyle y^{2}=4a\left ( x+a\sin \frac{x}{a} \right )$ at which the tangent is parallel to x axis lie on -

0%
a straight line
0%
a parabola
0%
a circle
0%
an ellipse

Explanation

$\displaystyle y^{2}=4a\left ( x+a\sin \frac{x}{a} \right )$

$\Rightarrow \displaystyle 2y\frac{dy}{dx}=4a\left ( 1+a\cos \frac{x}{a}\times \frac{1}{a} \right )$

$\Rightarrow \displaystyle \frac{dy}{dx}=\frac{2a}{y}\left ( 1+\cos \frac{x}{a} \right )$
Now for tangent to be parallel to x-axis,

$\displaystyle \frac{dy}{dx}=0=\frac{2a}{y}\left ( 1+\cos \frac{x}{a} \right )$

$\Rightarrow \displaystyle \frac{x}{a}=\pi$

$\Rightarrow \displaystyle x=a\pi$

$\therefore \displaystyle y^{2}=4a\left ( a\pi +a\times 0 \right )$

$\Rightarrow \displaystyle y^{2}=4a^{2}\pi =4ax$
Clearly, this point lie on a parabola.
Hence, option 'B' is correct.

If the tangent at

$(1,\,1)$ on

$y^2=x(2-x)^2$ meets the curve again at

$P(a,\,b)$ then

$a/b$ is equal to

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

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

$\Rightarrow 2yy'=(2-x)^2+2x(x-2)$

$\Rightarrow y'=\displaystyle\frac{3x^2-8x+4}{2y}\;\Rightarrow y'_{(1,\,1)}=-\displaystyle\frac{1}{2}$

$\Rightarrow$ Equation of tangent is

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

$\Rightarrow 2y-2=1-x\;\Rightarrow x+2y=3$
Solve with

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

$\Rightarrow (x-3)^2=4x(x-2)^2$

$\Rightarrow x^2-6x+9=4x^3-16x^2+16x$

$\Rightarrow 4x^3-17x^2+22x-9=0$

$\Rightarrow (x-1)^2(4x-9)=0$

$\Rightarrow x=\displaystyle\frac{9}{4}\;\;\Rightarrow y=\dfrac 38$

$\Rightarrow \displaystyle\frac{a}{b}=\displaystyle\frac{9\times8}{4\times3}=6$

The abcissa of the point on the curve

$\displaystyle ay^{2}=x^{3}$ the normal at which cuts off equal intercepts from the axes is -

0%
$1$
0%
$\dfrac {4a } 3$
0%
$3$
0%
$\dfrac {4a } 9$

Explanation

Let point

$\displaystyle \left ( x_{1},y_{1} \right )$ on the curve

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

$\Rightarrow \displaystyle 2ay\times \frac{dy}{dx}=3x^{2}$

$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )_{x_{1},y_{1}}=\frac{3x_{1}^{2}}{2ay_{1}}$
Thus slope of Normal is

$=\displaystyle -\left ( \frac{dy}{dx} \right )_{x_{1},y_{1}}=-\frac{2ay_{1}}{3x_{1}^{2}}$
For equal intercept slope of normal should be

$-1$ ,

$\displaystyle -\frac{2ay_{1}}{3x_{1}^{2}}=-1$

$\Rightarrow \displaystyle 3x_{1}^{2}=2ay_{1}$ ..........(1)
Also point lies on curve

$\displaystyle ay_{1}^{2}=x_{1}^{3}$ .......(2)
Solve (1) and (2), we get

$\displaystyle x_{1}=\frac{4a}{9}$
Hence, option 'D' is correct.

The area of triangle formed by tangent to the hyperbola

$\displaystyle 2xy=a^{2}$ and coordinates axes is -

0%
$\displaystyle a^{2}$
0%
$\displaystyle 2a^{2}$
0%
$\displaystyle\dfrac{ a^{2} } 2$
0%
$\displaystyle\dfrac{ 3a^{2} } 2$

Explanation

Given curve is,

$2xy=a^2$
Differentiating w.r.t

$x$

$2y+2x\cfrac{dy}{dx}=0\Rightarrow \cfrac{dy}{dx}=-\cfrac{y}{x}$
Thus equation of tangent to this curve at any point

$P(x_1,y_1)$ is given by,

$(y-y_1)=-\cfrac{y_1}{x_1}(x-x_1)$
The intercepts on the axis are,

$2x_1$ and

$2y_1$
Hence are of triangle formed by tangent at P on the coordinate axes is,

$\Delta =\cfrac{1}{2}\times 2x_1\times 2y_1=2x_1y_1=a^2$ since point P also lie on the given curve

$(2x_1 y_1=a^2)$
Hence, option 'A' is correct.

If the tangent at any point on the curve

$\displaystyle x^{4}+y^{4}=a^{4}$ cuts off intercept

$p$ and

$q$ on the axes, the value of

$\displaystyle p^{-\frac43}+q^{-\frac43}$ is

0%
$\displaystyle a^{-\frac43}$
0%
$\displaystyle a^{-\frac13}$
0%
$\displaystyle a^{-\frac12}$
0%
None of these

Explanation

$\displaystyle x^{4}+y^{4}=a^{4}$
Let point be

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

$\displaystyle 4x^{3}+4y^{3}\dfrac{dy}{dx}=0$

$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )=-\dfrac{4x^{3}}{4y^{3}}$

$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )_{x_{1}y_{1}}=\dfrac{-x_{1}^{3}}{y_{1}^{3}}$

Intercepts are

$p, q$ , then

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

$\Rightarrow \displaystyle y\times y_{1}^{3}-y_{1}^{4}=-xx_{1}^{3}+x_{1}^{4}$

$\Rightarrow \displaystyle xx_{1}^{3}+yy_{1}^{3}=x_{1}^{4}+y_{1}^{4}$

$\Rightarrow \displaystyle p=\dfrac{a^{4}}{x_{1}^{3}},\: \: \: q=\dfrac{a^{4}}{y_{1}^{3}}$

$\Rightarrow \displaystyle p^{-\frac{4}{3}}+q^{-\frac{4}{3}}=\dfrac{a^{-\frac{16}{3}}}{\left ( x_{1}^{3} \right )^{-\frac{4}{3}}}+\dfrac{a^{-\frac{16}{3}}}{\left ( y_{1}^{3} \right )^{-\frac{4}{3}}}$

$= \displaystyle a^{-\frac{16}{3}}\left ( x_{1}^{4}+y_{1}^{4} \right )=a^{-\frac{16}{3}}\times a^{4}$

$\Rightarrow \displaystyle a^{4-\frac{16}{3}}=a^{-\frac{4}{3}}$

$\Rightarrow \displaystyle p^{-\frac{4}{3}}+q^{-\frac{4}{3}}=a^{-\frac{4}{3}}$
Hence, option 'A' is correct.

The equation of normal to the curve

$x+y=x^{y}$ , where it cuts x-axis is

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

Explanation

$ln(x+y)=yln(x)$

$\dfrac{1}{x+y}.(1+\dfrac{dy}{dx})=\dfrac{y}{x}+\dfrac{dy}{dx}.ln(x)$

Now

$y=0$ and

$\dfrac{dy}{dx}=m$ we get

$\dfrac{1+m}{x}=m.ln(x)$

Now

$x=1$ when

$y=0$

Hence

$1+m=0$

$m=-1$

Slope of normal

$=1.$

Hence

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

$y=x-1$ .

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,

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

Slope of the tangent to the given curve is

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

$= 3t^2 \times \dfrac{1}{6t}$

$=\dfrac{t}{2}$

Since the slope has to be calculated at

$x = 1,$ i.e. at

$3t^2 + 1 = 1$ , we get

$t = 0$

Thus, the required slope is

$0.$

If the tangent to the curve

$\displaystyle 2y^{3}=ax^{2}+x^{3}$ at a point

$(a, a)$ cuts off intercepts p and q on the coordinates axes where

$\displaystyle p^{2}+q^{2}=61$ then

$a$ equals to

0%
$30$
0%
$-30$
0%
$0$
0%
$\displaystyle \pm 30$

Explanation

Given Curve is,

$\displaystyle 2y^{3}=ax^{2}+x^{3}$

Differentiating w.r.t

$x$ , we get

$\displaystyle 6y^{2}\times \frac{dy}{dx}=2ax+3x^{2}$

Thus slope of tangent at

$(a,a)$ is

$=\displaystyle \left ( \frac{dy}{dx} \right )_{\left ( a,a \right )}=\frac{2ax+3x^{2}}{6y^{2}}=\frac{2a^{2}+3a^{2}}{6a^{2}}=\frac{5}{6}$

Since equation of tangent from point

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

is $$y-y_{1}=m(x-x_{1})

Thus equation of tangent is,

$(y - a) = \displaystyle \frac{5}{6}\left ( x-a \right )$

$\Rightarrow 6y - 6a = 5x - 5a$

$\Rightarrow 5x - 6y + a = 0$

$\Rightarrow \displaystyle \frac{5}{-\dfrac{a}{5}}+\frac{y}{\dfrac{a}{6}}=1$

$\therefore \displaystyle p=\frac{-a}{5},\: \: \: \: q=\frac{a}{6}$

Given,

$\displaystyle p^{2}+q^{2}=\frac{a^{2}}{25}+\frac{a^{2}}{36}=61$

$\Rightarrow\displaystyle a^{2}\left ( \frac{61}{25\times 36} \right )=61$

$\Rightarrow a = \displaystyle \pm 5\times 6 \displaystyle =\pm 30$

Hence, option 'D' is correct.

The points on the curve

$\displaystyle 9y^2=x^{3}$ where the normal to the curve makes equal intercepts with coordinates axes is :

0%
$\displaystyle \left ( 4,\frac{8}{3} \right )\: \: or\: \: \left ( 4,-\frac{8}{3} \right )$
0%
$\displaystyle \left ( -4,\frac{8}{3} \right )$
0%
$\displaystyle \left ( -4,-\frac{8}{3} \right )$
0%
None of these

Explanation

Let the point on the curve

$\displaystyle 9y^{2}=x^{3}$ be

$P\left ( x_{1},y_{1} \right )$
differentiating the curve w.r.t

$x$

$\displaystyle 2\times 9\times y\times \frac{dy}{dx}=3x^{2}$

$\Rightarrow \displaystyle \frac{dy}{dx}=\frac{x^{2}}{2\times 3y}$

$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( x_{1},y_{1} \right )}=\frac{x_{1}^{2}}{2\times 3y_{1}}$
Thus slope of Normal at P

$=\displaystyle -\frac{dy}{dx}=\frac{-3y_{1}\times 2}{x_{1}^{2}}$
Given normal makes equal intercept

$\Rightarrow \displaystyle \frac{-3y_{1}\times 2}{x_{1}^{2}}=\pm 1$

$\Rightarrow \displaystyle \frac{-3y_{1}\times 2}{x_{1}^{2}}=1$

$\Rightarrow \displaystyle -3y_{1}\times 2=x_{1}^{2}$ so

$9y_1^2\times 4=x_{1}^{4}\Rightarrow x_1^3 \times 4 = x_1^4$
So

$\displaystyle x_{1}=4,\: \: \: y=-\frac{8}{3}$

$\Rightarrow \displaystyle \frac{-3y_{1}\times 2}{x_{1}^{2}}=-1$

$\Rightarrow \displaystyle 2\times 3y_{1}=x_{1}^{2}$ ..........(1)
Also point lies point on the curve

$\displaystyle 9y^{2}=x^{3}$

$\Rightarrow \displaystyle 9y_{1}^{2}=x_{1}^{3}$

$\Rightarrow \displaystyle 9\times \frac{x_{1}^{4}}{36}=x_{1}^{3}$

$\Rightarrow \displaystyle x_{1}=4$

$\Rightarrow \displaystyle x_{1}=4,\: \: \: y_{1}=\frac{16}{6}=\frac{8}{3}$

$\Rightarrow \displaystyle \left ( 4,\frac{8}{3} \right )$
Finally the points are

$\displaystyle \left ( 4,\frac{8}{3} \right )\: \: \: or\: \: \: \left ( 4,\frac{-8}{3} \right )$
Hence, option 'A' is correct.

If the line

$x -y = 0$ is tangent to

$f(x) = b \ln x - x$ , then

$b$ lies in the interval

0%
$(1, 3)$
0%
$(0, 1)$
0%
$(4, 6)$
0%
$(6, 8)$

Explanation

Given equation of curve is

$y=b\ln x-x$
Also, given

$x-y=0$ is tangent to the curve.
So, let

$P(x_1,x_1)$ be the point of tangency.

$x_1=b\ln {x_1}-{x_1}$

$2x_1= b\ln {x_1}$ ....(1)

Slope of tangent to the curve at P

$=\dfrac{b}{x_1}-1$
Slope of given tangent

$=1$

$\Rightarrow \dfrac{b}{x_1}-1=1$

$\Rightarrow x_1=\dfrac{b}{2}$
Substitute this value in (1), we get

$b=b\ln {\dfrac{b}{2}}$

$\Rightarrow \ln {\dfrac{b}{2}}=1$

$\Rightarrow \dfrac{b}{2}=e$

$\Rightarrow b=2e$
Since,

$2< e<3$

$\Rightarrow 4<b<6$

If the curve

$y^2=ax^3-6x^2+b$ passes through

$(0,\,1)$ and has its tangent parallel to y-axis at

$x=2$ , then

0%
$a=2,\,b=1$
0%
$a=\displaystyle\frac{23}{8},\,b=1$
0%
$a=-\displaystyle\frac{8}{23},\,b=1$
0%
$a=-\displaystyle\frac{23}{8},\,b=1$

Explanation

Given equation of curve

$y^2=ax^3-6x^2+b$
Since, the curve passes through

$(0,1)$

$\Rightarrow b=1$

Since, the tangent is parallel to y-axis at

$x=2$

$\Rightarrow y=8a-23$
So, point of tangency is

$(2,8a-23)$

Slope of tangent to curve

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

$\Rightarrow \dfrac{dy}{dx}=\dfrac{3ax^2-12x}{2y}$
Slope of tangent to curve at

$(2,8a-23)$ is

$\dfrac{12a-24}{16a-46}$

Since, the tangent is parallel to y-axis

$\dfrac{12a-24}{16a-46}=\dfrac{1}{0}$

$\Rightarrow 16a-46=0$

$\Rightarrow a=\dfrac{23}{8}$

Let tangent at a point P on the curve

$\displaystyle { x }^{ 2m }={ y }^{ \tfrac { n }{ 2 } }={ a }^{ \tfrac { 4m+n }{ 2 } }$ meets the x-axis and y-axis at A and B respectively, If AP:PB is

$\displaystyle \frac { n }{ \lambda m }$ , where P lies between A and B, then find the value of

$\displaystyle \lambda$

0%
$4$
0%
$3$
0%
$-4$
0%
$-3$

Explanation

$\displaystyle { x }^{ 2m }{ y }^{ \tfrac { n }{ 2 } }={ a }^{ \tfrac { 4m+n }{ 2 } }$

$\displaystyle 2m\ln x +\frac { n }{ 2 } \ln y=\frac { 4m+n }{ 2 }\ln a$

$\displaystyle \frac { \left( 2m \right) }{ x } +\left( \frac { n }{ 2 } \right) \frac { 1 }{ y } \frac { dy }{ dx } =0$

$\displaystyle \frac { dy }{ dx } =\left( \frac { -2m }{ x } \right) \frac { 2y }{ n }$
Let

$P$ be the point

$\displaystyle \left( { x }_{ 1 },{ y }_{ 1 } \right)$
So equation of tangent is

$\displaystyle y-{ y }_{ 1 }=\left( \frac { -4m }{ n } .\frac { { y }_{ 1 } }{ { x }_{ 1 } } \right) \left( x-{ x }_{ 1 } \right)$ .......(i)
Let Tangents meets the axes at

$A$ and

$B$ respectively

$\displaystyle A=\left( \frac { 4m+n }{ 4m } { x }_{ 1 },0 \right) \quad B=\left( 0,\frac { 4m+n }{ n } .{ y }_{ 1 } \right)$
Let

$P$ divides

$AB$ in ratio

$\displaystyle \mu :1$

$\displaystyle \therefore \quad { x }_{ 1 }=\frac { \mu .0+1.\left( \dfrac { 4m+n }{ 4m } \right) { x }_{ 1 } }{ \mu +1 }$

$\displaystyle\therefore 4m+n=4m\left( \mu +1 \right)$
Also

$\displaystyle \mu \left( 4m+n \right) =n\left( \mu +1 \right)$

$\displaystyle\therefore \quad \mu =\frac { n }{ 4m }$

$\displaystyle \therefore \quad \lambda =4$

The minimum value of the polynomial.

$p(x)=3{ x }^{ 2 }-5x+2$

0%
$-\frac { 1 }{ 6 }$
0%
$\frac { 1 }{ 6 }$
0%
$\frac { 1 }{ 12 }$
0%
$-\frac { 1 }{ 12 }$

Explanation

$p(x)=3{ x }^{ 2 }-5x+2$

$=3\left( x2-\frac { 5 }{ 3 } x+\frac { 2 }{ 3 } \right)$

$=3\left[ { x }^{ 2 }-\frac { 5 }{ 3 } x+\left( { \frac { 5 }{ 6 } } \right) ^{ 2 }-\left( { \frac { 5 }{ 6 } } \right) ^{ 2 }+\frac { 2 }{ 3 } \right]$

$=3\left[ \left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }-\frac { 25 }{ 36 } +\frac { 2 }{ 3 } \right]$

$=3\left[ \left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }+\frac { -25+24 }{ 36 } \right]$

$=3\left[ \left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }-\frac { 1 }{ 36 } \right] =3\left( { x-\frac { 5 }{ 6 } } \right) ^{ 2 }-\frac { 1 }{ 12 }$

If the tangent to the curve

$x = a(8 + sin \theta), y = a(1 + cos \theta )$ at

$\theta = \displaystyle \frac{\pi}{3}$ makes an angle

$\alpha$ with x-axis, then

$\alpha$ is equal to

0%
$\displaystyle \frac{\pi}{3}$
0%
$\displaystyle \frac{2\pi}{3}$
0%
$\displaystyle \frac{\pi}{6}$
0%
$\displaystyle \frac{5\pi}{6}$

Explanation

$x=a(8+\sin \theta)$

$\frac{d x}{d \theta}=a \cos \theta$

$y=a(1+\cos \theta)$

$\frac{d y}{d \theta}=-a \sin \theta$

Now,

$\frac{d y}{d \theta}=-a \sin \theta$

$\frac{d x}{d \theta}=a \cos \theta$

$\frac{d y}{d x}=-\tan \theta$

$\frac{d y}{d x}\left(\operatorname{at} \theta=\frac{\pi}{3}\right)=-\sqrt{3}=m$

Now,

$m=\tan \alpha$

$\Rightarrow \tan \alpha=-\sqrt{3}$

Hence,

$\alpha=\frac{2 \pi}{3}$

The curve which passes through

$(1, 2)$ and whose tangent at any point has a slope that is half of slope of the line joining origin to the point of contact, is -

0%
A rectangle hyperbola
0%
A circle
0%
A parabola
0%
A straight line through origin
0%
Answer required

Explanation

Let the arbitrary point be $x,y.$
Now the slope of the line joining the origin and the point is
$=\dfrac{y-0}{x-0}$
$=\dfrac{y}{x}$ .
The equation of the curve
$y=f(x)$ .
Now slope of the tangent
$=\dfrac{dy}{dx}$
$=\dfrac{y}{2x}$ ... as per the given condition.
Hence
$2\dfrac{dy}{y}=\dfrac{dx}{x}$
Integrating both sides we get
$2\ln y=\ln x+\ln c$
Now $y_{x=1}=2$
Hence
$2\ln (2)=\ln (1)+\ln (c)$
Or
$c=2^{2}=4$
Hence
$2\ln y=\ln x+\ln 4$
Or
$\ln (y^{2})=\ln 4x$
Or
$y^{2}=4x$ is the required equation.
This is an equation of parabola.

The lines tangent to the curve

$x^3-y^3+x^2y-yx^2+3x-2y=0$ and

$x^5-y^4+2x+3y=0$ at the origin intersect at an angle

$\theta$ equal to

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

Explanation

Given equation of curve

$x^3-y^3+x^2y-yx^2+3x-2y=0$
Differentiating we get

$\displaystyle 3{ x }^{ 2 }-3{ y }^{ 2 }\frac { dy }{ dx } +x^{ 2 }\frac { dy }{ dx } +2xy-2xy-x^{ 2 }\frac { dy }{ dx } +3-2\frac { dy }{ dx } =0$

$(3{ y }^{ 2 }+2)\dfrac { dy }{ dx } =3{ x }^{ 2 }+3$

$\Rightarrow \dfrac { dy }{ dx } =\dfrac { 3{ x }^{ 2 }+3 }{ (3{ y }^{ 2 }+2) }$

Slope of tangent to curve at (0,0) is

$m_1=\dfrac{3}{2}$

Given equation of other curve

$x^5-y^4+2x+3y=0$
On differentiation ,

$\displaystyle 5{ x }^{ 4 }-4{ y }^{ 3 }\frac { dy }{ dx } +2+3\frac { dy }{ dx } =0$

$\Rightarrow \dfrac { dy }{ dx } =\dfrac { 5{ x }^{ 4 }+2 }{ (4{ y }^{ 3 }-3) }$
Slope of tangent to the curve at (0,0) is

$m_2=-\dfrac{2}{3}$

Here,

$m_1m_2=-1$
Hence, the lines are perpendicular.

A curve

$\displaystyle y=f\left( x \right) ;\left( y>0 \right)$ passes thorugh

$(1,1)$ and at point

$\displaystyle P(x,y)$ tangents cuts x-axis and y-axis at A and B respectively. If P divides AB internally in the ratio

$3 : 2$ , then the value of

$\displaystyle f\left( \frac { 1 }{ 8 } \right)$ is

0%
$4$
0%
$\displaystyle \frac { 1 }{ 4 }$
0%
$\displaystyle 16\sqrt { 2 }$
0%
$\displaystyle \frac { 1 }{ 16\sqrt { 2 } }$

Explanation

$\displaystyle \frac { dy }{ dx } =-\frac { k }{ h } =-\frac { 2y }{ 3x }$

$\displaystyle \Rightarrow \quad 3\frac { dy }{ y } +2\frac { dx }{ x } =0$

$\displaystyle \Rightarrow \quad 3\ln { y }^{ 3 }{ x }^{ 3 }=C$

$\displaystyle \because \quad$ passing through

$(1,1)$

$\displaystyle \Rightarrow \quad c=0$

$\displaystyle \Rightarrow \quad { y }^{ 3 }{ x }^{ 2 }=1$
at

$\displaystyle x=\frac { 1 }{ 8 }$

$\Rightarrow \quad y=4$

The slope of the tangent to the curve

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

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

$(2,-1)$ is

0%
$\cfrac{22}{7}$
0%
$\cfrac{6}{7}$
0%
$\cfrac{7}{6}$
0%
$\cfrac{-6}{7}$
0%
answer required

Explanation

Slope to any curve is given by

$\dfrac{dy}{dx}$

Here,

$\cfrac{dx}{dt}=2t+3$ ...(1)

And

$\cfrac{dy}{dt}=4t-2$ ...(2)

Dividing (2) by (1), we get

$\cfrac {dy}{dx} = \cfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

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

At point

$(2,-1)$ ,

$t=2$

So slope is given by,

$\Rightarrow \dfrac{dy}{dx}=\dfrac{4\times 2-2}{2\times 2+3}=\dfrac{6}{7}$

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%
$\cfrac{1}{2}$
0%
answer required

Explanation

Lets put

$y=mx+1$ in

${y}^{2}=4x$ we get

${(mx+1)}^{2}=4x$

$\Rightarrow {m}^{2}{x}^{2}+1+2mx-4x=0$
Tangent touches a curve at one point so descriminant of the equation should be zero.

$\Rightarrow {(2m-4)}^{2}-4\times 1\times {m}^{2}=0$

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

$\Rightarrow m=1$

Let

$y=e^{x^2}$ and

$y=e^{x^2}\sin\, x$ be two given curves. Then, angle between the tangents to the curves at any point their intersection is

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

Explanation

For intersecting points,

$e^{x^2}=e^{x^2}\sin\,x$

$\Rightarrow e^{x^2}(\sin x-1)=0$

$\Rightarrow e^{x^2}=0$ or

$\sin\,x=1$

But

$e^{x^2}\neq 0\Rightarrow \sin x=1$

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

Now,

$y=e^{x^2}$

$\therefore \dfrac{dy}{dx}=e^{x^2}.2x=2xe^{x^2}$

$[\because \sin\,x=1, \cos \, x = 0]$

Since, both the curves has equal slope.
Hence, angle between the tangents at intersecting point is

$0$ .

The abscissa of the points, where the tangent to curve

$y={x}^{3} - 3{x}^{2} - 9x+5$ is parallel to x-axis, are

0%
$x=0$ and $0$
0%
$x=1$ and $-1$
0%
$x=1$ and $-3$
0%
$x=-1$ and $3$

Explanation

Given,

$y={ x }^{ 3 }-3{ x }^{ 2 }-9x+5$

$\Rightarrow \dfrac { dy }{ dx } =3{ x }^{ 2 }-6x-9$
We know that, this equation gives the slope of the tangent to the curve. The tangent is parallel to x-axis,

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

$\Rightarrow 3{ x }^{ 2 }-6x-9=0$

$\Rightarrow x=-1,3$

The equation of normal of

$x^2+y^2-2x+4y-5=0$ at

$(2,\,1)$ is

0%
$y=3x-5$
0%
$2y=3x-4$
0%
$y=3x+4$
0%
$y=x+1$

Explanation

Given equation is

$x^2+y^2-2x+4y-5=0$
On differentiating, we get

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

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

$\Rightarrow\,\begin{pmatrix}\dfrac{dy}{dx}\end{pmatrix}_{(2,\,1)}=\dfrac{1-2}{1+2}=\dfrac{-1}{3}$

$\Rightarrow\,-\begin{pmatrix}\dfrac{dx}{dy}\end{pmatrix}_{(2,\,1)}=3$
Now, equation of normal is

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

$y-1=3x-6$

$\Rightarrow\;y=3x-5$

If the tangent at a point P, with parameter t, on the curve

$x = 4t^{2} + 3, y = 8t^{3} - 1, t\epsilon R$ meets the curve again at a point Q, then the coordinates of Q are:

0%
$(t^{2} + 3, -t^{3} - 1)$
0%
$(4t^{2} + 3, -8t^{3} - 1)$
0%
$(16t^{3} + 3, -64 t^{3} - 1)$
0%
$(t^{2} + 3, t^{3} - 1)$

Suppose that the equation

$f\left( x \right) ={ x }^{ 2 }+bx+c=0$ has two distinct real roots

$\alpha$ and

$\beta$ . The angle between the tangent to the curve

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

$\left( \dfrac { \alpha +\beta }{ 2 } ,f\left( \dfrac { \alpha +\beta }{ 2 } \right) \right)$ and the positive direction of the

$x$ -axis is

0%
${ 0 }^{ }$
0%
${ 30 }^{ }$
0%
${ 60 }^{ }$
0%
${ 90 }^{ }$

Explanation

Since,

$\alpha$ and

$\beta$ are the roots of

$f\left( x \right) ={ x }^{ 2 }+bx+c$

$\therefore \alpha +\beta =-b$ and

$\alpha \beta =c$

$\therefore f\left( \dfrac { \alpha +\beta }{ 2 } \right) ={ \left( \dfrac { \alpha +\beta }{ 2 } \right) }^{ 2 }+b\left( \dfrac { \alpha +\beta }{ 2 } \right) +c$

$={ \left( -\dfrac { b }{ 2 } \right) }^{ 2 }+b\left( -\dfrac { b }{ 2 } \right) +c$

$=\dfrac { { b }^{ 2 } }{ 4 } -\dfrac { { b }^{ 2 } }{ 2 } +c=-\dfrac { { b }^{ 2 } }{ 4 } +c$

Now,

$\dfrac { dy }{ dx } =f^{ \prime }\left( x \right) =2x+b$

At point,

$\left( \dfrac { \alpha +\beta }{ 2 } ,f\left( \dfrac { \alpha +\beta }{ 2 } \right) \right)$ , i.e.,

$\left( -\dfrac { b }{ 2 } ,-\dfrac { { b }^{ 2 } }{ 4 } +c \right)$ ,

$\dfrac { dy }{ dx } =2\left( -\dfrac { b }{ 2 } \right) +b=0$

Hence, the slope of the tangent to the curve and the positive direction of

$x$ -axis is

${ 0 }^{ }$ .

The points on the curve

$9{y}^{2}={x}^{3}$ , where the normal to the curve makes equal intercepts with the axes are

0%
$\left( 4,\pm \cfrac { 8 }{ 3 } \right)$
0%
$\left( 4,\cfrac { -8 }{ 3 } \right)$
0%
$\left( 4,\pm \cfrac { 3 }{ 8 } \right)$
0%
$\left( \pm 4,\cfrac { 3 }{ 8 } \right)$
0%
answer required

Explanation

Let the point on curve be

$(h,k)$

$y^2= \dfrac{x^3}{9}\\ 2y.\dfrac{dy}{dx}= \dfrac{1}{9} .3 x^2\\ \therefore \dfrac{dy}{dx}= \dfrac{x^2}{6y}$

$\dfrac{dy}{dx}$ gives slope of curve at given point

$\dfrac{dy}{dx}\;at\;(h,k) = \dfrac{h^2}{6k}$
but this is the slope of tangent and we need slope of normal
As we know that slope of normal

$\times$ slope of tangent is

$-1$

$\therefore$ slope of normal is

$\dfrac{-6k}{h^2}$
Given in question that normal makes equal intercepts on axes which means slope of normal is either

$1$ or

$-1$

$-6k=h^2$ and

$6k=h^2$
Now go through the option you will that option A satisfy our equation

The normal to the curve

${x}^{2}=4y$ passing

$(1,2)$ is

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

Explanation

Given,

$x^2=4y$

Slope of tangent to the curve will be

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

$\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{2}$
So slope of normal will be

$-\dfrac{dx}{dy}=\dfrac{-2}{x}$
Let the point on the curve be

$(h,k)$
So slope of normal will be

$\dfrac{-2}{h}$
Equation of normal is given by

$(y-k)=\dfrac{-2}{h}(x-h)$
It is given that the normal passes through

$(1,2)$

$\Rightarrow (2-k)=\dfrac{-2}{h}(1-h)$ .............(1)
Also,

$(h,k)$ lies on the curve so

${h}^{2}=4k$
From (1) we have

$\dfrac{{h}^{3}}{4}=2h+2-2h=2$

$\Rightarrow {h}^{3}=8$

$\Rightarrow h=2$

$k=\dfrac{{h}^{2}}{4} \Rightarrow k=1$
So the equation will be

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

$x+y=3$

The coordinates of the point P on the curve

$x = a(\theta + \sin \theta), y = a(1 - \cos \theta)$ where the tangent is inclined at an angle

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

0%
$\left (a\left (\dfrac {\pi}{2} - 1\right ), a\right )$
0%
$\left (a\left (\dfrac {\pi}{2} + 1\right ), a\right )$
0%
$\left (a \dfrac {\pi}{2}, a\right )$
0%
$(a, a)$

Explanation

$x = a(\theta + \sin \theta), y = a(1 - \cos \theta)$

$\dfrac {dy}{d\theta} = a\sin \theta, \dfrac {dx}{d\theta} = a(1 + \cos \theta)$

$\dfrac {dy}{dx} = \dfrac {\sin \theta}{1 + \cos \theta} = \tan \dfrac {\theta}{2}$

$\Rightarrow \tan \dfrac {\theta}{2} = 1$

$\Rightarrow \dfrac {\theta}{2} = \dfrac {\pi}{4} \Rightarrow \theta = \dfrac {\pi}{2}$

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

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

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

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

$(x, y) \equiv \left (a \left (\dfrac {\pi}{2} + 1\right ), a\right )$ .

Angle between

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

${ x }^{ 2 }=y$ at the origin is

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

Explanation

It is clear from the graph that both the curves have a tangent at the coordinate axes, so the angle between the curves is

$\dfrac { \pi }{ 2 }$ .

Alternative
Given curves are

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

${ x }^{ 2 }=y$

On differentiating with respect to

$x$ , we get

$2y\dfrac { dy }{ dx } =1$ and

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

$\Rightarrow \dfrac { dy }{ dx } =\dfrac { 1 }{ 2y }$ and

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

$\left( 0,0 \right)$

${ m }_{ 1 }=\dfrac { dy }{ dx } =\infty$ and

${ m }_{ 2 }=\dfrac { dy }{ dx } =0$

$\therefore \tan { \theta } =\dfrac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } =\infty$

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

Equation of normal to the circle

$\displaystyle x=6\cos { \theta } ,y=6\sin { \theta }$ at

$\displaystyle p\left( \frac { 2\pi }{ 3 } \right)$ is

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

Explanation

The equation of the circle is

$x^2 + y^2 = 36$

Differentiating both sides w.r.t. x, we get

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

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

Since the normal is perpendicular to the tangent, the slope of the normal will be

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

Here, the slope will be equal to

$\tan{\theta}$ and since

$\theta$ is

$\dfrac{2\pi}{3}$ , the slope of the normal becomes

$\tan{\dfrac{2\pi}{3}} = -\sqrt{3}$

Also, the normal passes through the center of the circle, which is origin here.

The equation of the normal becomes

$y = -\sqrt{3} x$ , i.e.

$\sqrt{3} x + y = 0$

If the line

$\alpha\,x+by+c=0$ is a tangent to the curve

$xy=4$ , then

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

Explanation

Slope of line

$ax+by+c=0$ is

$\Rightarrow\;-\dfrac{a}{b}$

$y=\dfrac{4}{x}=1,\,\dfrac{dy}{dx}=-\dfrac{4}{x^2}$ ,

$-\dfrac{a}{b}=-\dfrac{4}{x^2}\Rightarrow\,\dfrac{a}{b}=\dfrac{4}{x^2} > 0$

$a < 0,\,b < 0$

The slope at any point of a curve

$y=f\left( x \right)$ is given by

$\dfrac { dy }{ dx } =3{ x }^{ 2 }$ and it passes through

$\left( -1,1 \right)$ . The equation of the curve is

0%
$y={ x }^{ 3 }+2$
0%
$y=-{ x }^{ 3 }-2$
0%
$y=3{ x }^{ 3 }+4$
0%
$y=-{ x }^{ 3 }+2$

Explanation

Given,

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

$\Rightarrow dy=3{ x }^{ 2 }dx$
On integrating, we get

$y=\dfrac { 3{ x }^{ 3 } }{ 3 } +c$

$\Rightarrow y={ x }^{ 3 }+c$
It passes through

$\left( -1,1 \right)$ .

$\therefore 1={ \left( -1 \right) }^{ 3 }+c$

$\Rightarrow c=2$

$\therefore y={ x }^{ 3 }+2$

The equation of one of the curves whose slope at any point is equal to

$y+2x$ is

0%
$y=2(e^x+x-1)$
0%
$y=2(e^x-x-1)$
0%
$y=2(e^x-x+1)$
0%
$y=2(e^x+x+1)$

Explanation

Given,

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

$y+2x=z$

$\Rightarrow\;\dfrac{dy}{dx}+2=\dfrac{dz}{dx}$

$\Rightarrow\;\dfrac{dy}{dx}=\dfrac{dz}{dx}-2\;\dots(ii)$
From Eqs. (i) and (ii)

$\dfrac{dz}{dx}-2=z$

$\Rightarrow\;\int\dfrac{dz}{.z+2}=\int\,dx$

$\Rightarrow\;log(z+2)=x+c$

$\Rightarrow\;log(y+2x+2)=x+c$

$\Rightarrow\;y+2x+2=e^{x+c}$

$\Rightarrow\;y+2x+2=e^x.e^c$

$\Rightarrow\;y=2[e^x-x-1]$

The slope of the normal to the curve

$y = 3x^2$ at the point whose

$x$ -coordinate

$2$ is

0%
$\dfrac{1}{13}$
0%
$\dfrac{1}{14}$
0%
$\dfrac{-1}{12}$
0%
$\dfrac{1}{12}$

Explanation

$y=3x^{2}$ . Differentiating the equation of the curve with respect to

$x$ .

$\dfrac{dy}{dx}=6x$
Now
Slope

$_{x=2}=\dfrac{dy}{dx}|_{x=2}$

$=6(2)$

$=12$ .
Hence the slope of the tangent at

$x=2$ is

$12$ . Since the normal is perpendicular to the tangent, therefore, the

Slope

$_\text{normal}\times$ Slope

$_\text{tangent}=-1$ or Slope

$_\text{normal}=\dfrac{-1}{\text{Slope}_\text{tangent}}=\dfrac{-1}{12}$

$\Delta$ is the area of the triangle formed by the positive x-axis and the normal and tangent to the circle

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

$(1, \sqrt {3})$ , then

$\Delta =$

0%
$\dfrac {\sqrt {3}}{2}$
0%
$\sqrt {3}$
0%
$2\sqrt {3}$
0%
$6$

Explanation

$x^{2} + y^{2} = 4$

Differentiating wrt x, we get

$\dfrac {dy}{dx} = -\dfrac {x}{y}, m = \dfrac {-1}{\sqrt {3}}$

Equation of tangent:

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

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

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

$Area = \left |\dfrac {y_{1}^{2}(1 + m^{2})}{2m}\right | = 2\sqrt {3}$

If the tangent to the curve

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

$(a, a)$ cuts off intercepts

$\alpha$ and

$\beta$ on the coordinate axes where

$\alpha^{2} + \beta^{2} = 61$ then the value of '

$a$ ' is equal to

0%
$25$
0%
$36$
0%
$\pm 30$
0%
$\pm 40$

Explanation

$2y^{3} = ax^{2} + x^{3}$

Diff. w.r.t

$x$

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

$\dfrac {dy}{dx}|_{(a, a)} = \dfrac {2a^{2} + 3a^{2}}{6a^{2}} = \dfrac {5}{6}$

Hence tangent at

$(a,a)$ is,

$y-a =\dfrac{5}{6}(x-a)\Rightarrow 6y-6a=5x-5a\Rightarrow 5x-6y+a=0$

$\therefore x - intercept = \dfrac {-a}{5} = \alpha$

and

$y - intercept = \dfrac {a}{6} = \beta$

$\therefore \alpha^{2} + \beta^{2} = 61\Rightarrow \dfrac {a^{2}}{25} + \dfrac {a^{2}}{36} = 61$

$\dfrac{61a^2}{900}=61\Rightarrow a^2=900$

$\therefore a = \pm 30$

The tangent to the curve

$y=x^3+1$ at (1, 2) makes an agnle

$\theta$ with y axis, then the value of tan

$\theta$ is.

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

Explanation

Given curve is

$y=x^3+1$

Differentiate it

$\Rightarrow \dfrac{dy}{dx}=3x^2+0=3x^2$

Thus slope of tangent to the curve at the point

$(1,2)$ is

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

If line makes an angle

$\theta$ with y-axis then it will make angle

$\pi-\theta$ with positive x-axis

Thus slope

$=\tan(\pi-\theta)=3$

$\Rightarrow \tan\theta=-3$

The slope of the tangent to the curve

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

$x=0$ is

0%
$3$
0%
$2$
0%
$1$
0%
$-1$

Explanation

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

$\Rightarrow \dfrac{dy}{dx}=6x+3\cos x$

Thus slope of tangent to the given curve at

$x=0$ is

$=\dfrac{dy}{dx}\bigg|_{x=0}=(6x+3\cos x)\bigg|_{x=0}=6(0)+3\cos 0=0+3\cdot 1=3$

What is the slope of the tangent to the curve

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

$x=0$ ?

0%
0
0%
1
0%
2
0%
None of the above

Explanation

Slope of tangent to a curve f(x), at a point

$({x}_{o},{y}_{o})$ is given by

$f'(x)$ ,

where

$f'(x)$ is derivative of f(x) at

$x={x}_{o}$

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

${x}_{o}=0$

$f'\left( x \right) =\cfrac { df\left( x \right) }{ dx } =\cfrac { df(x) }{ d\sin ^{ 2 }{ x } } \times \cfrac { d\sin ^{ 2 }{ x } }{ d\sin x } \times \cfrac { d\sin x }{ dx }$

$f'\left( x \right) =\cfrac { 1 }{ \sqrt { 1-\sin ^{ 2 }{ x } } } \times 2\sin x\times \cos x$

$f'\left( 0 \right) =0$

Slope of tangent to the curve is 0

Find the slope of the normal to the curve

$4x^3+6x^2-5xy-8y^2+9x+14=0$ T the point

$-2, 3$ .

0%
$\infty$
0%
$11$
0%
$\displaystyle\frac{9}{19}$
0%
$\displaystyle-\frac{19}{9}$

Explanation

The given curve is

$4x^3+6x^2-5xy-8y^2+9x+14=0$

Differentiating above equation w.r.t

$x$ , we get

$12{ x }^{ 2 }+12x-5x\cfrac { dy }{ dx } -5y-16y\cfrac { dy }{ dx } +9=0$

$\Rightarrow \cfrac { dy }{ dx } =\cfrac { 12x^{ 2 }+12x-5y+9 }{ \left( 5x+16y \right) }$

$\left( x,y \right) \equiv \left( -2,3 \right)$

$m=\cfrac { dy }{ dx } =\dfrac{12(4)+12(-2)-5(3)+9}{5(-2)+16(3)}=\dfrac{18}{38}=\dfrac{9}{19}$ is the slope of tangent to the curve

Slope of normal is

$-\dfrac{1}{m}=-\dfrac{19}{9}$

A mirror in the first quadrant is in the shape of a hyperbola whose equation is xy =A light source in the second quadrant emits a beam of light that hits the mirror at the point (2,1/2). If the reflected ray is parallel to the y-axis the slope of the incident beam is

0%
$\dfrac{13}{8}$
0%
$\dfrac{7}{4}$
0%
$\dfrac{15}{8}$
0%
$2$

Explanation

Slope of tangent at

$\left ( 2, \frac{1}{2} \right )$

$m=-\dfrac{1}{4}$

$tan \theta =-\dfrac{1}{y}$

$\theta =\phi -90^o$

$tan\theta =4$
Slope of incident ray

$= m$

$\left | \dfrac{m-\left ( -\dfrac{1}{4} \right )}{1+m\left ( -\dfrac{1}{4} \right )} \right |=tan \theta =4$

$\left | \dfrac{4m+1}{4-m} \right |=4$

$4m + 1 = 16 - 4m$

$m=\dfrac{15}{8}$

What is the slope of the tangent to the curve

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

0%
$\dfrac{7}{6}$
0%
$\dfrac{6}{7}$
0%
1
0%
$\dfrac{5}{6}$

Explanation

Differentiating

$x,y$ w.r.t

$t,$ we get

$\dfrac { dy }{ dt } =\dfrac { d }{ dt } (2t^2-2t-5)=4t-2\\ \dfrac { dx }{ dt } =\dfrac { d }{ dt } (t^2+3t-8)=2t+3\\$

$\therefore \dfrac { dy }{ dx } =\dfrac { \left( \dfrac { dy }{ dt } \right) }{ \left( \dfrac { dx }{ dt } \right) } =\dfrac { 4t-2 }{ 2t+3 }$

$t=2$ derivative will be

$\dfrac {dy}{dx}=\dfrac {4\times 2-2}{2\times 2+3}=\dfrac {6}{7}$

Thus, the slope of tangent is

$\dfrac{6}{7}$ .

If the tangent to the function

$y = f(x)$ at

$(3, 4)$ makes an angle of

$\dfrac {3\pi}{4}$ with the positive direction of x-axis in anticlockwise direction then

$f'(3)$ is

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

Explanation

Given

$y=f(x)$

Differentiating w.r.t x, we get

$y' = f'(x)$ which is the slope of the tangent

$f'(3) = \tan \dfrac {3\pi}{4} = -\tan \dfrac {\pi}{4} = -1$

Hence,

$f'(3)=-1$ .

How many tangents are parallel to x-axis for the curve

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

0%
1
0%
2
0%
3
0%
No tangent is parallel to x-axis.

Explanation

Slope of the tangent is

$\dfrac {dy} {dx}$

So,

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

Tangent parallel to x-axis. So, slope of tangent should be

$0.$

$\Rightarrow 2x-4=0$

$\Rightarrow x=2$ is the only point where slope is parallel to x-axis.

So, only 1 tangent exists.

Consider the curve

$y = e^{2x}$ .Where does the tangent to the curve at (0, 1) meet the x-axis ?

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

Explanation

Slope of tangent at any point to the curve is given by

$\left|{ \cfrac { df\left( x \right) }{ dx } }\right|_{ ({ x }_{ 0 }, { y }_{ 0 }) }$

Here

$f(x)=y={e}^{2x}$

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

At point

$(0,1)$

$\cfrac{dy}{dx}=2{e}^{2(0)}$

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

Equation of tangent is at point

$({x}_{0},{y}_{0})$

$y-{y}_{0}=m(x-{x}_{0})$

So, tangent at (0,1) is

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

$\therefore y=2x+1$

When this tangent meets x-axis,

$y$ co-ordinate becomes zero

Putting

$y$ in above equation as zero, we get

$0=2x+1$

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

Thus, the tangent to the curve meets at

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

The slope of the tangent to the curve given by

$x = 1 - \cos { \theta }$ ,

$y = \theta -\sin { \theta }$ at

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

0%
$0$
0%
$-1$
0%
$1$
0%
Not defined

Explanation

$y=\theta -\sin\theta$

$\Longrightarrow \dfrac { dy }{ d\theta } =1-\cos\theta$

$x=1-\cos\theta$

$\Longrightarrow \dfrac { dx }{ d\theta } =sin\theta$

Slope of the tangent is

$\dfrac{dy}{dx}$

$\Longrightarrow \dfrac{dy}{dx} =\dfrac { 1-\cos\theta}{\sin\theta}$

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

$\Longrightarrow \dfrac { dy }{ dx } =\dfrac { 1-\cos\frac { \pi }{ 2 } }{ \sin\frac { \pi }{ 2 } } =1$

Let

$f(x)=2{ x }^{ 3 }-5{ x }^{ 2 }-4x+3,\cfrac { 1 }{ 2 } \le x\le 3$ . The point at which the tangent to the curve is parallel to the X-axis is

0%
$(1,-4)$
0%
$(2,-9)$
0%
$(2,-4)$
0%
$(2,-1)$
0%
$(2,-5)$

Explanation

Given,

$f(x)=2{ x }^{ 3 }-5{ x }^{ 2 }-4x+3$

$f'(x)=6{x}^{2}-10x-4$

If tangent is parallel to X-axis, then

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

$\Rightarrow 6{ x }^{ 2 }-10x-4=0\quad$

$\Rightarrow x=\cfrac { 5\pm \sqrt { 25+24 } }{ 6 } =\cfrac { 5\pm 7 }{ 6 }$

$\Rightarrow x=2,\cfrac { -1 }{ 3 }$

$\quad \therefore y=f(x)=2{ (2) }^{ 3 }-5{ (2) }^{ 2 }-4(2)+3=-9$

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 7 - 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 7 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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