MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 5

The angle at which the curve

$\displaystyle y=ke^{kx}$ intersects the y - axis is

0%
$\displaystyle \tan ^{-1}k^{2}$
0%
$\displaystyle \cot ^{-1}\left ( k^{2} \right )$
0%
$\displaystyle \sin ^{-1\left ( \dfrac{1}{\sqrt{1+k^{4}}} \right )}$
0%
$\displaystyle \sec {-1\left ( \sqrt{1+k^{4}} \right )}$

Explanation

$\displaystyle \dfrac{dy}{dx}=K^{2}e^{kx}$

$\displaystyle \dfrac{dy}{dx}|_{x=0}=K^{2}=\tan \theta$

(where

$\displaystyle \phi$ is angle made by x-axis)

Let

$\displaystyle \phi$ be the gngle made by y-axis

$\displaystyle \tan \theta =\tan \left ( \dfrac{3\pi }{2}-\phi \right )=\cot \phi$

$\displaystyle \cot \phi =K^{2}$

$\displaystyle \phi =\cot ^{-1}\left ( K^{2} \right )$

$\displaystyle \Rightarrow \phi =\sin ^{-1}\left ( \dfrac{1}{\sqrt{1+K^{4}}} \right )$

The abscissa of the point on the curve

$\displaystyle \sqrt{xy}=a+x$ the tangent at which cuts off equal intercepts from the co-ordinate axes is (a > 0)

0%
$\displaystyle \dfrac{a}{\sqrt{2}}$
0%
$\displaystyle- \dfrac{a}{\sqrt{2}}$
0%
$\displaystyle a\sqrt{2}$
0%
$\displaystyle -a\sqrt{2}$

Explanation

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

$\displaystyle \dfrac{dy}{dx}=\dfrac{-a^{2}}{x^{2}}+1=-1$ ( for equal intercepts)

$\displaystyle x^{2}=\dfrac{a^{2}}{2}\Rightarrow x=\pm \dfrac{a}{\sqrt{2}}$

The curve

$\displaystyle y=ax^{3}+bx^{2}+cx+5$ touches the

$x$ - axis at

$P(-2, 0)$ and cuts the

$y$ -axis at a point

$Q$ , where its gradient is

$3$ . Find

$a, b, c$ .

0%
$\displaystyle a=-\frac{1}{5}, b=1,c=3$
0%
$\displaystyle a=-\frac{1}{4}, b=-1,c=4$
0%
$\displaystyle a=-\frac{1}{4}, b=0,c=3$
0%
$\displaystyle a=-\frac{1}{3}, b=1,c=-3$

Explanation

Given,

$y= ax^3+bx^2+cx+5$

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

Since the curve touches

$x$ -axis at

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

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

$\left ( i \right )$

The curve cut the

$y$ -axis at

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

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

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

$\left ( ii \right )$

$a = -\dfrac {1}{4}$ ,

$b =0$

Let h be a twice continuously differentiable positive function on an open interval

$J$ . Let

$\displaystyle g\left ( x \right )=ln\left ( h(x) \right )$ for each

$\displaystyle x\epsilon J$
Suppose

$\displaystyle \left ( h'\left ( x \right )\right )^{2}> h''\left ( x \right )h\left ( x \right )$ for each

$\displaystyle x\epsilon J$ Then

0%
$g$ is increasing on $J$
0%
$g$ is decreasing on $J$
0%
$g$ is concave up on $J$
0%
$g$ is concave down on $J$

Explanation

Given,

$\displaystyle g(x)=\ln(h(x))$

$\Rightarrow g'(x)=\cfrac{h'(x)}{h(x)}$

$\Rightarrow g''(x)=\cfrac{h(x).h''(x)-(h'(x))^2}{(h(x))^2}$
Using given condition, Clearly

$g''(x)< 0, x \epsilon J$
Hence

$g(x)$ is concave down on

$J.$

If the curve

$\displaystyle { \left( \frac { x }{ a } \right) }^{ n }+{ \left( \frac { y }{ b } \right) }^{ n }=2$ touches the straight line

$\displaystyle \frac { x }{ a } +\frac { y }{ b } =2$ , then find the value of

$n$ .

0%
$2$
0%
$3$
0%
$4$
0%
any real number

Explanation

Given

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

Differentiating both sides w.r.t

$x,$ we get

$\displaystyle \frac { n }{ a } { \left( \frac { x }{ a } \right) }^{ n-1 }+\frac { b }{ b } { \left( \frac { y }{ b } \right) }^{ n-1 }\times \frac { dy }{ dx } =0$

$\displaystyle \Rightarrow \frac { dy }{ dx } =-\frac { n }{ a } { \left( \frac { x }{ a } \right) }^{ n-1 }\times \frac { b }{ a } { \left( \frac { b }{ y } \right) }^{ n-1 }$

$\displaystyle \therefore \frac { dy }{ dx }$ at

$\displaystyle \left( a,b \right) =\frac { b }{ a }$

$\therefore$ Tangent is

$\displaystyle y-b=-\frac { b }{ a } \left( x-a \right) \Rightarrow bx+ay=2ab\Rightarrow \frac { x }{ a } +\frac { y }{ b } =2$

for all values of

$n$

$(\because\displaystyle\frac{dy}{dx}$ is independent of

$n)$

For the curve represented parametrically by the equations

$\displaystyle x=2\ln\cot t+1$ and

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

0%
tangent at $t = \displaystyle \dfrac {\pi }4$ is parallel to x-axis
0%
normal at $t = \displaystyle \dfrac {\pi }4$ is parallel to y=axis
0%
tangent at is parallel to the line $y = x$
0%
normal at is parallel to the line $y = x$

Explanation

Given,

$x= 2 \ln \cot t+1$ and

$y=\tan t+\cot t$

$\therefore \displaystyle \frac{dx}{dt}=\frac{2\left ( -\mathrm{cosec} ^{2t} \right )}{\cot t}$
at

$\displaystyle t = \frac{\pi }{4},\frac{dx}{dt}=-4$
and

$\displaystyle \frac{dy}{dt}=\sec ^{2}t-\mathrm{cosec} ^{2}t$

$\displaystyle at t =\frac{\pi }{4}\:\:\;\frac{dy}{dt}=0$

$\displaystyle \frac{dy}{dx}=0$ for tangent & hence it is parallel to x-axis & its normal is parallel to y axis

The coordinates of the point(s) on the graph of the function

$\displaystyle f(x)=\frac{x^{3}}3{-\frac{5x^{2}}{2}}+7x-4$ where the tangent drawn cut off intercepts from the coordinate axes which are equal in magnitude but opposite in sign is

0%
$(2,\dfrac 83)$
0%
$(3, \dfrac 72)$
0%
$(1,\dfrac 56)$
0%
none

Explanation

Hence

$\dfrac{dy}{dx}=1$ at that point.
Hence

$x^{2}-5x+7=1$
Or

$x^{2}-5x+6=0$
Or

$(x-2)(x-3)=0$
Or

$x=2$ and

$x=3$

$f(2)=\dfrac{8}{3}$
And

$f(3)=\dfrac{7}{2}$
Hence the points are

$(2,\dfrac{8}{3})$ and

$(3,\dfrac{7}{2})$ .

If the radius of a sphere is measured as

$9 \ cm$ with an error of

$0.03 \ cm$ then, find the approximate error in calculating its volume.

0%
$\displaystyle 9.72\pi\:\: cm^{3}$
0%
$\displaystyle 7.92\pi\:\: cm^{3}$
0%
$\displaystyle 8.72\pi\:\: cm^{3}$
0%
None of these

Explanation

Given,

$r=9 cm, \Delta r=0.03cm$

We know Volume of sphere with radius 'r' is

$V=\cfrac{4}{3}\pi r^3$

$\therefore \Delta V=4\pi r^2\Delta r$

$\Rightarrow \Delta V=4\pi\times 81\times .03=9.72\pi cm^3$ (using given values)

At what point of the curve

$\displaystyle y=2x^{2}-x+1$ tangent is parallel to

$y = 3x + 4$

0%
$(0, 1)$
0%
$(1, 2)$
0%
$(-1, 4)$
0%
$(2, 7)$

Explanation

Let the point is

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

$y=2x^2-x+1$
Differentiating w.r.t

$x$

$\cfrac{dy}{dx}=4x-1$
Thus slope of tangent at P is

$=\left(\cfrac{dy}{dx}\right)_{(a,b)}=4a-1$
But given the tangent is parallel to line

$y=3x+4$

$\Rightarrow 4a-1=3\Rightarrow a=1$
Also the point P lies on the given curve,

$b=2a^2-a+1=2$
Therefore, the point P is

$(1,2)$
Hence, option 'B' is correct.

The slope of the normal to the curve

$\displaystyle x=a\left ( \theta -\sin \theta \right ),\: \: y=a\left ( 1-\cos \theta \right )$ at point

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

0%
$0$
0%
$1$
0%
$-1$
0%
$\dfrac 1{\displaystyle \sqrt{2}}$

Explanation

The slope of normal

$\displaystyle x=a\left ( \theta -\sin \theta \right );\: \: \: \: y=a\left ( 1-\cos \theta \right )\: \: \: at\: \: \: \theta =\dfrac{\pi }{2}$

$\displaystyle \left ( \dfrac{dx}{d\theta } \right )_{\theta =\dfrac{\pi }{2}}=a\left ( 1-\cos \theta \right )=a$

$\displaystyle \left ( \dfrac{dy}{d\theta } \right )_{\theta =\dfrac{\pi }{2}}=a\sin \theta =a$

$\therefore \left ( \dfrac{dy}{dx} \right )=1$
Therefore, slope of normal at the given point is,

$m= \left ( -\dfrac{dx}{dy} \right )=-1$
Hence, option 'C' is correct.

asymptotes of the graph

0%
$\displaystyle x=\frac{3\pi }{2}$
0%
$\displaystyle x=-\frac{\pi }{2}$
0%
$\displaystyle x=\frac{\pi }{2}$
0%
$\displaystyle x=-\frac{3\pi }{2}$

Explanation

$f'(x)=\dfrac{\cos(x)}{1+\sin(x)}$
Or

$\dfrac{dy}{dx}=\dfrac{\cos x}{1+\sin x}$
For asymptotes,

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

$dx=0$
Or

$\sin(x)=-1$
Or

$x=\dfrac{3\pi}{2},\dfrac{-\pi}{2}$ .

The slope of the tangents to the curve

$y = (x + 1) (x - 3)$ at the points where it crosses x - axis are

0%
$\displaystyle \pm 2$
0%
$\displaystyle \pm 3$
0%
$\displaystyle \pm 4$
0%
None of these

Explanation

$y = (x + 1)(x - 3) ..(1)$

Substitute

$y = 0$ to get the point of intersection of this curve with

$x-axis$

$0 = (x + 1) (x - 3)$

$\Rightarrow x = -1, 3$

So the points are

$(-1, 0)$ and

$(3, 0)$

Now differentiating eq. (1), we get

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

$\Rightarrow \displaystyle \left ( \dfrac{dy}{dx} \right )=2\left ( x-1 \right )$

Thus slope at

$(-1,0)$ is

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

and at

$(3,0)$ is

$=\displaystyle \left ( \dfrac{dy}{dx} \right )_{\left ( 3,0 \right )}=4$

Hence, option 'C' is correct.

Let

$f$ be a continuous, differentiable and bijective function. If the tangent to

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

$x=b$ , then there exists at least one

$c\in \left( a,b \right)$ such that

0%
$f'\left( c \right) =0$
0%
$f'\left( c \right) >0$
0%
$f'\left( c \right) <0$
0%
none of these

Explanation

Since the same line is tangent at one point

$x=a$ and normal at other point

$x=b$

$\Rightarrow$ Tangent at

$x=b$ will be perpendicular to tangent at

$x=a$

$\Rightarrow$ Slope of tangent changes from positive to negative or negative to positive. Therefore, it takes the value zero somewhere. Thus, there exists a point

$c\in \left( a,b \right)$ where

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

The normal to the curve

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

$x$ axis at the point

0%
$(0, a)$
0%
$(a, 0)$
0%
$(\dfrac a 4, \dfrac a 4)$
0%
No where

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 normal at P is

$=\displaystyle -( \frac{dx}{dy} )_{( x_{1},y_{1} )}=-\sqrt{\frac{x_{1}}{y_{1}}}$
If normal

$\displaystyle \perp$ to x axis then

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

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

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

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 slope of normal to the curve

$\displaystyle y^{2}=4ax$ at a point

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

0%
$\dfrac 1 t$
0%
$t$
0%
$-t$
0%
$-\dfrac 1 t$

Explanation

$y^2=4ax$

$\Rightarrow 2y\cfrac{dy}{dx}=4a\Rightarrow \cfrac{dy}{dx}=\cfrac{2a}{y}$
Therefore, slope of normal to the given curve is,

$=\left(-\cfrac{dx}{dy}\right)_{(at^2,2at)}=-\cfrac{2at}{2a}=-t$
Hence, option 'C' is correct.

On the ellipse,

$4x^2\, +\, 9y^2\, =\, 1$ , the points at which the tangents are parallel to the line

$8x = 9y$ are

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

Explanation

We have

${ 4x }^{ 2 }+9{ y }^{ 2 }=1$ ...(1)
Differentiating w.r.t x, we get

$\displaystyle 8x+18y\frac { dy }{ dx } =0\Rightarrow \frac { dy }{ dx } =-\frac { 4x }{ 9y }$
The tangent at point

$\left( x,y \right)$ will be parallel to the line

$8x=9y$ if

$\displaystyle \frac { -4x }{ 9y } =\frac { 8 }{ 9 } \Rightarrow x=-2y$
Substituting this in (1), we get

$\displaystyle 4{ \left( -2y \right) }^{ 2 }+{ 9y }^{ 2 }=1\Rightarrow { 25y }^{ 2 }=1\Rightarrow y=\pm \frac { 1 }{ 5 }$
Thus the point where the tangent are parallel to

$8x=9y$ are

$\displaystyle \left( \frac { -2 }{ 5 } ,\frac { 1 }{ 5 } \right)$ and

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

The slope of the tangent to the curve

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

$(1, 1)$ is

$2$ then values of

$a$ and

$b$ are respectively -

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

Explanation

Given curve is

$xy + ax - by = 0$
Slope of tangent at

$(1, 1)$ is

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

$\displaystyle \left ( \frac{dy}{dx} \right )_{\left ( 1,1 \right )}=-\left [ \frac{\left ( a+y \right )}{x-b} \right ]_{\left ( 1,1 \right )}=-\frac{\left ( a+1 \right )}{1-b}$

$\displaystyle \therefore -\frac{\left ( a+1 \right )}{1-b}=2$
So,

$2b - a = 3$ ..........(1)

$\displaystyle \because (1, 1)$ lies on curve

$xy + ax - by = 0$

$\displaystyle \therefore a-b=-1$ .........(2)
From (1) and (2), we get

$a = 1, b = 2$
Hence, option 'A' is correct.

At what point the tangent line to the curve

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

$x + 2y = 0$

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

Explanation

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

$\displaystyle \left ( -2\pi \leq x\leq 2\pi \right )$

$\Rightarrow \displaystyle \frac{dy}{dx}=-\sin \left ( x+y \right )\left ( 1+\frac{dy}{dx} \right )$

$\Rightarrow \displaystyle \frac{dy}{dx}\left ( 1+\sin \left ( x+y \right ) \right )=-\sin \left ( x+y \right )$

$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{\sin \left ( x+y \right )}{1+\sin \left ( x+y \right )}$
Given tangent is parallel to the line,

$x + 2y = 0$

$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{\sin \left ( x+y \right )}{1+\sin \left ( x+y \right )}=-\frac{1}{2}$

$\displaystyle 2\sin \left ( x+y \right )=1+\sin \left ( x+y \right )$

$\displaystyle \sin \left ( x+y \right )=1\Rightarrow \cos(x+y)=0\Rightarrow x+y=\cfrac{\pi}{2}$
Also the point lies on the given curve,

$y=0\Rightarrow x= \dfrac{\pi }{2}$
Thus the point is,

$\displaystyle \left ( \frac{\pi }{2},0 \right )$
Hence, option 'A' 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 point at which the tangent to the curve

$\displaystyle y=x^{3}+5$ is perpendicular to the line

$x + 3y = 2$ are

0%
$(6, 1), (-1, 4)$
0%
$(6, 1) (4, -1)$
0%
$(1, 6), (1, 4)$
0%
$(1, 6), (-1, 4)$

Explanation

Let point be

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

$y = \displaystyle x^{3}+5$

$\Rightarrow \displaystyle \dfrac{dy}{dx}=3x^{2}$
Now, the slope of tangent at this point is

$=\displaystyle \left ( \frac{dy}{dx} \right )_{x_{1}y_{1}}=3x_{1}^{2}$
It is

$\displaystyle \perp$ to

$x + 3y - 2 = 0$
So

$\displaystyle 3x_{1}^{2}\times -\frac{1}{3}=-1\Rightarrow x_1=\pm 1$
Thus,
at

$\displaystyle x_{1}=1$ ,

$\Rightarrow$

$\displaystyle y_{1}=6$
and at

$\displaystyle x_{1}=-1$

$\Rightarrow$

$\displaystyle y_{1}=4$
Thus, the points are

$(1, 6)$ and

$(-1, 4)$
Hence, option 'D' is correct.

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.

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 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.

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,

$2< e<3$

$\Rightarrow 4<b<6$

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.

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 function

$\displaystyle f\left( x \right)=\left( { a }^{ 2 }-3a+2 \right) \cos { \frac { x }{ 2 } } +\left( a-1 \right) x$ possesses critical points, then

$a$ belongs to the interval

0%
$\left( -\infty ,0 \right) \cup \left( 4,\infty \right)$
0%
$(-\infty ,0]\cup [4,\infty )$
0%
$(-\infty ,0]\cup \left\{ 1 \right\} \cup [4,\infty )$
0%
None of these

Explanation

Given

$\displaystyle f\left( x \right) =\left( { a }^{ 2 }-3a+3 \right) \cos { \frac { x }{ 2 } } +\left( a-1 \right) x$

$\displaystyle \Rightarrow f'\left( x \right) =\frac { -1 }{ 2 } \left( a-1 \right) \left( a-2 \right) \sin { \left( \frac { x }{ 2 } \right) } +\left( a-1 \right)$

$\displaystyle \Rightarrow f'\left( x \right) =\left( a-1 \right) \left[ 1-\frac { 1 }{ 2 } \left( a-2 \right) \sin { \left( \frac { x }{ 2 } \right) } \right]$

$f(x)$ possesses critical points, then

$f'(x)=0$

$\displaystyle \Rightarrow \left( a-1 \right) \left[ 1-\left( \frac { a-2 }{ 2 } \right) \sin { \frac { x }{ 2 } } \right] =0$

$\Rightarrow a=1$ and

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

$\Rightarrow a=1$ and

$\displaystyle \sin { \left( \frac { x }{ 2 } \right) } =\frac { 2 }{ a-2 }$

$\Rightarrow a=1$ and

$\displaystyle \left| \frac { 2 }{ a-2 } \right| \le 1$

$\Rightarrow a=1$ and

$\left| a-2 \right| \ge 2$

$\Rightarrow a=1$ and

$a-2\ge 2$ or

$a-2\le -2$

$\Rightarrow a=1$ and

$a\ge 4$ or

$a\le 0$

$\Rightarrow a=1$ and

$a\in (-\infty ,0]\cup [4,\infty )$

Therefore,

$a\in (-\infty ,0]\cup \left\{ 1 \right\} \cup [4,\infty )$

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.

If f(x) =

$\dfrac{x}{ sin x}$ and g(x) =

$\dfrac{x}{tanx}$ where 0<x

$\leq$ 1, then in this interval

$f(x)$ is

0%
both f(x) and g(x) are increasing functions
0%
both f(x) and g(x) are decreasing functions
0%
f(x) is an increasing function
0%
g(x) is an increasing function

Explanation

We have f(x)

$= \displaystyle \dfrac{x}{ sin x }$ and g(x)

$=\displaystyle \dfrac{x}{tan x}$

$\therefore f'(x) =\displaystyle \dfrac{sin x - x cos x }{sin^2 x }$ and

$g'(x)=\displaystyle \dfrac{tan x - x sec^2 x}{tan^2 x}$
Let

$\Phi (x0) = sin x - x cos x$ and

$\Psi (x) = tan x - x sec^2 x$
Then, f'(x) =

$\Phi (x) / sin^2 x$ and

$g'(x) = \Psi (x)/ tan^2$
Now,

$\Phi(x) = cos x - cos x + x sin x = x sin x$
and

$\Psi (x) = sec^2 x - sec^2 x - 2x sec^2 x tan x = -2x sec^2 x tan x$
For

$0 < x \leq 1$ , we have x > 0 , sin x >0, tan x >0, sec x >0

$\therefore \Phi (x) = x sin x >x$ and

$\Psi (x( <0$ for

$0<x \leq 1$

$\Rightarrow \Phi (x)$ is increasing on (0,1) and

$\Psi (x)$ is decreasing on (0,1)

$\Rightarrow \Phi (x) > Phi(0)$ and

$\Psi (x) < \Psi (0) \Rightarrow \Phi (x) >0$ and

$\Psi (x) <0$

$\therefore f(x) =\Phi (x) /sin^2 x >0$ & g(x) =

$\Phi (x)/ tan^2 x <0$

$\Rightarrow$ f(x) is increasing on (0,1) and g(x) is decreasing on (0,1)

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

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

$\Rightarrow \quad y=4$

Determine the intervals over which the function is decreasing, increasing, and constant.

0%
Increasing $[3, \infty )$ ; Decreasing $(-\infty , 3]$
0%
Increasing $(-\infty , 3]$ ; Decreasing $[3, \infty )$
0%
Increasing $(-\infty , 3]$ ; Decreasing $(-\infty , 3]$
0%
Increasing $[3, \infty)$ ; Decreasing $[3, \infty )$

Explanation

As we can see from the graph, when

$x$ is increasing till

$3$ ,

$y$ is also increasing

Further increment in

$x$ reduces the

$y$ 's values.

It means that upto

$x=3$ , function is increasing and after

$x=3$ function is decreasing.

Hence, increasing in the interval

$(-\infty,3]$ ; decreasing in the interval

$[3,\infty)$

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.

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

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

$\Rightarrow m=1$

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 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

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

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

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

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

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

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

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

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

$a < 0,\,b < 0$

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 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

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

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

$\Rightarrow c=2$

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

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

$y+2x$ is

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

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 function

$f(x)=ax+b$ is strictly increasing for all real

$x$ , if

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$a> 0$
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$a< 0$
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$a=0$
0%
$a\le 0$

Explanation

Given,

$f(x)=ax+b$
On differentiating w.r.t

$x$ we get

$f'(x)=a$
For strictly increasing,

$f'(x)> 0$

$\Rightarrow$

$a> 0$

The equation of normal of

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

$(2,\,1)$ is

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

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

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

$y-1=3x-6$

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

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

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

Explanation

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

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

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

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

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

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

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

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

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

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

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 5 - MCQExams.com

0:0:3

Practice Class 12 Commerce Maths Quiz Questions and Answers

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