If $$f(x)=e^x(x-2)^2$$ then

f is increasing in $$(-\infty, 0)$$ and $$(2, \infty)$$ deceasing in $$(0, 2)$$

f is increasing in $$(-\infty, 0)$$ and deceasing in $$(0, \infty)$$

f is increasing in $$(2, \infty)$$ and deceasing in $$(-\infty, 0)$$

f is increasing in $$(0, 2)$$ and deceasing in $$(-\infty, 0)$$ and $$(2, \infty)$$

MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 11

The number of points on the curve

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

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

Explanation

If the tangents are equally inclined to the axes we get

$y'_{x,y}=1$ .

Therefore

$\dfrac{3}{2}\sqrt{x}+\dfrac{3}{2}\sqrt{y}.y'=0$

$y'=1$ implies

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

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

$x=y$ .

$x_{1}=y_{1}$ .

Substituting in the equation we get

$2x^{\frac{3}{2}}=a^{\frac{3}{2}}$

$x=\dfrac{a}{2^{\frac{2}{3}}}$

Therefore

$(x_{1},y_{1})=\left(\dfrac{a}{2^{\frac{2}{3}}},\dfrac{a}{2^{\frac{2}{3}}}\right)$

Therefore there exists only one point.

Let

$\displaystyle f(x) = ln \: mx (m > 0)$ and

$g(x) = px$ . Then the equation

$\displaystyle |f(x)| = g(x)$ has only one solution for

0%
$\displaystyle 0 < p < \frac {m}{e}$
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$\displaystyle p < \frac {e}{m}$
0%
$\displaystyle 0 < p < \frac {e}{m}$
0%
$\displaystyle p > \frac {m}{e}$

Explanation

Let

$P(h,k)$ be a point on

$f(x)=\ln mx$
then,

$k=\ln mh$

$\dfrac { dy }{ dx }$ at point P

$=\dfrac { 1 }{ h }$
Equation of tangent at point P

$L_{1}: y-k=\dfrac { \left( x-h \right) }{ h }$
Since, it passes through origin
Therefore,

$k=1$ &

$h=e/m$
and

$L_{1}: y=x/h$

$\Rightarrow L_{1}: y=mx/e$

From the figure:

$|f(x)|=g(x)$ have one solution when slope of

$g(x)$ is greater than slope of

$L_{1}$
Therefore,

$p>m/e$

Ans: D

Let the equation of a curve be

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

$y=a\left ( 1-\cos \theta \right )$ . If

$\theta$ changes at a constant rate

$k$ then the rate of change of slope of the tangent to the curve at

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

0%
$\displaystyle \frac{2k}{\sqrt{3}}$
0%
$\displaystyle \frac{k}{\sqrt{3}}$
0%
$k$
0%
none of these

Explanation

$\theta$ change with constant rate

$\Rightarrow \dfrac{d\theta}{dt}=k$

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

$\cfrac { dx }{ d\theta } =a\left( 1+cos\theta \right)$

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

$\cfrac { dy }{ d\theta } =asin\theta$

Then,

$\cfrac { dy }{ dx } =\cfrac { asin\theta }{ a\left( 1+cos\theta \right) }$

Rate of change of slope

$=\dfrac{d}{dt}. \dfrac{dy}{dx} = \dfrac{1}{1+\cos\theta}.\dfrac{d\theta}{dt} = k$ , at

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

The real number

$\displaystyle '\alpha'$ such that the curve

$\displaystyle f(x) = e^x$ is tangent to the curve

$\displaystyle g(x) = \alpha x^2$ .

0%
$\displaystyle \frac{e^2}{4}$
0%
$\displaystyle \frac{e^2}{2}$
0%
$\displaystyle \frac{e}{4}$
0%
$\displaystyle \frac{e}{2}$

Explanation

At the point of contact both the curves must have the same slope.
Hence

$f'(x)_{x=x_{1}}=g'(x)_{x=x_{1}}$
Or

$e^{x_{1}}=2\alpha.x_{1}$
Or

$\alpha=\dfrac{e^{x_{1}}}{2x_{1}}$
Now, considering

$x_{1}=1$ and

$x_{1}=2$ we get

$\alpha=\dfrac{e}{2}$ and

$\alpha=\dfrac{e^{2}}{4}$ .

The area of a triangle is computed using the formula

$S=\dfrac {1}{2}$ bc sin A. If the relative errors made in measuring b, c and calculating S are respectively

$0.02$ ,

$0.01$ and

$0.13$ the approximate error in A when

$A=\pi /6$ is

0%
$0.05$ radians
0%
$0.01$ radians
0%
$0.05$ degree
0%
$0.01$ degree

Explanation

Error formula for given equation is

$\dfrac{\Delta s}{s}=\dfrac{\Delta b}{b}+\dfrac{\Delta c}{c}+\dfrac{\Delta sinx}{sinx}$

$0.13=0.02+0.01+\dfrac{\Delta sinx}{1/2}$

$\Delta sinx=0.05$

If the circle

$x^2+y^2+2gx+2fy+c=0$ is touched by

$y=x$ at P such that OP =

$6\sqrt{2}$
then the value of c is

0%
36
0%
144
0%
72
0%
None of these

Explanation

Let the point of contact be

$x_{1},y_{1}$
However it lies on the line

$y=x$
Hence

$x_{1}=y_{1}$
Applying distance formula, we get

$\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}$

$2x_{1}^2=72$

$x_{1}=6$ ...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.

$2x+2yy'+2g+2fy'=0$
Now

$y'=1$ since

$y'$ is the slope of the line

$y=x$ .
By substituting, we get

$x+y=-(g+f)$
Now

$x_{1}=y_{1}=6$
Hence

$12=-(g+f)$ ...(i)
Substituting

$x=y=6$ in the equation of the circle, we get

$36+36+2(6)(g+f)+c=0$

$72+12(-12)+c=0$

$c=144-72$

$c=72$

The angle made by the tangent of the curve

$\displaystyle x = a(t + \sin t \cos t); y = a (1 + \sin t)^2$ with the x-axis at any point on it is

0%
$\displaystyle \frac {1}{4} (\pi + 2t)$
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$\displaystyle \frac {1 - \sin t}{ \cos t}$
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$\displaystyle \frac {1}{4} (2t - \pi)$
0%
$\displaystyle \frac {1 + \sin t}{\cos 2 t}$

Explanation

$x=a\left(t+\sin{t}\cos{t}\right)$

$\dfrac{dx}{dt}=a\left(1+{\cos}^{2}{t}-{\sin}^{2}{t}\right)$

$\quad \ =a\left(1+\left({\cos}^{2}{t}-{\sin}^{2}{t}\right)\right)$

$\dfrac{dx}{dt}=a\left(1+\cos{2t}\right)$

$y=a{\left(1+\sin{t}\right)}^{2}$

$\dfrac{dy}{dt}=2a\left(1+\sin{t}\right)\cos{t}$

$\dfrac{dy}{dt}=2a\cos{t}\left(1+\sin{t}\right)$

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

$=\dfrac{2a\cos{t}\left(1+\sin{t}\right)}{a\left(1+\cos{2t}\right)}\\$

$=\dfrac{2\cos{t}\left(1+\sin{t}\right)}{2{\cos}^{2}{t}}\\$

$=\dfrac{\left(1+\sin{t}\right)}{\cos{t}}\\$

$=\dfrac{\left({\cos}^{2}{\dfrac{t}{2}}+{\sin}^{2}{\dfrac{t}{2}}+2\sin{\dfrac{t}{2}}\cos{\dfrac{t}{2}}\right)}{{\cos}^{2}{\dfrac{t}{2}}-{\sin}^{2}{\dfrac{t}{2}}}\\$

$=\dfrac{{\left(\cos{\dfrac{t}{2}}+\sin{\dfrac{t}{2}}\right)}^{2}}{{\cos}^{2}{\dfrac{t}{2}}-{\sin}^{2}{\dfrac{t}{2}}}\\$

$=\dfrac{\left(\cos{\dfrac{t}{2}}+\sin{\dfrac{t}{2}}\right)}{\left(\cos{\dfrac{t}{2}}-\sin{\dfrac{t}{2}}\right)}\\$

$=\dfrac{1+\tan{\dfrac{t}{2}}}{1-\tan{\dfrac{t}{2}}}\\$

$=\dfrac{\tan{\dfrac{\pi}{4}}+\tan{\dfrac{t}{2}}}{1-\tan{\dfrac{\pi}{4}}\tan{\dfrac{t}{2}}}\\$

$=\tan{\left(\dfrac{\pi}{4}+\dfrac{t}{2}\right)}\\$

$\theta$ is the angle which the tangent at any point

$t$ makes with the

$X$ axis

then

$\tan{\theta}=\tan{\left(\dfrac{\pi}{4}+\dfrac{t}{2}\right)}$

$\therefore\,\theta=\dfrac{\pi}{4}+\dfrac{t}{2}=\dfrac{1}{4}\left(\pi+2t\right)$

Let

$\displaystyle \:f : R\rightarrow R$ be a function such that

$\displaystyle \:f \left ( x \right )= ax+3\sin x+4\cos x.$ Then

$\displaystyle \:f \left ( x \right )$ is invertible if

0%
$\displaystyle \:a \epsilon \left ( -5, 5 \right )$
0%
$\displaystyle \:a \epsilon \left ( -\infty , 5 \right )$
0%
$\displaystyle \:a \epsilon \left ( -5, +\infty \right )$
0%
none of these

Explanation

$f(x)=ax+3sinx+4cosx$

$f'(x)=a+3cosx-4sinx$

$-5\leq3cosx-4sinx \geq5$
hence

$a\varepsilon (-5,5)$

For the curve represented parametrically by the equations,

$\displaystyle x = \displaystyle\frac{2}{cot\:t} + 1$ &

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

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

Explanation

$dy=sec^{2}t-cosec^{2}t.dt$ ...(i)

$dx=2sec^{2}t.dt$ ...(ii)

$\dfrac{dy}{dx}$

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

$=\dfrac{sec^{2}t-cosec^{2}t}{2sec^{2}t}$ .

Hence

$\dfrac{dy}{dx}_{t=\frac{\pi}{4}}$

$=0$ slope of tangent

Slope of normal

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

$=\dfrac{-2sec^{2}t}{sec^{2}t-cosec^{2}t}$

Hence

$=\dfrac{-dx}{dy}_{t=\frac{\pi}{4}}=\infty$

Hence the normal is perpendicular to x axis and thus parallel to y axis.

and tangent is parallel to x - axis.

Consider the curve represented parametrically by the equation

$\displaystyle x = t^3 - 4t^2 - 3t$ and

$\displaystyle y = 2t^2 + 3t - 5$ where

$\displaystyle t \: \epsilon \: R$ .
If

$H$ denotes the number of point on the curve where the tangent is horizontal and

$V$ the number of point where the tangent is vertical then

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$H = 2$ and $V = 1$
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$H = 1$ and $V = 2$
0%
$H = 2$ and $V = 2$
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$H = 1$ and $V = 1$

If the line

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

$xy=1$ then

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

Explanation

$xy=1$

$xy'+y=0$

$y'=\dfrac{-y}{x}$

Hence

Slope of normal will be

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

Since

$xy=1$ hence any

$(x,y)$ lying on the hyperbola has the sign of the abscissa and ordinate as same.

That sign of x is same as sing of y.

Hence

$\dfrac{x}{y}$ will always be positive.

Or in other words

$\dfrac{-1}{y'}>0$

Hence

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

Hence if

$a<0,b>0$ or if

$a>0,b<0$ .

The point (s) on the curve

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

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

Explanation

We require the point where

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

$3y^{2}.dy+6x.dx=12dy$
Or

$dy(3y^{2}-12)=-6xdx$
Or

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

$=0$
Hence

$3y^{2}-12=0$

$y^{2}-4=0$

$y=\pm 2$
Now

$y=-2$ does not satisfy the equation the curve.
Substituting in the equation of the curve, y=2,

$8+3x^{2}=24$

$3x^{2}=16$

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

For a

$\displaystyle a\epsilon \left [ \pi , 2\pi \right ],$ the function

$\displaystyle f\left ( x \right )= \frac{1}{3}\sin a \tan ^{3}x+\left ( \sin a-1 \right )\tan x+ \frac{\sqrt{a-2}}{\sqrt{8-a}}$

0%
$\displaystyle x= n\pi \left ( n\epsilon I \right )$ as critical points
0%
no critical points
0%
$\displaystyle x= 2n\pi \left ( n\epsilon I \right )$ as critical points
0%
$\displaystyle x= \left ( 2n+1 \right )\pi \left ( n\epsilon I \right )$ as critical points.

Explanation

$f\left( x \right) =\dfrac { 1 }{ 3 } \sin a\tan ^{ 3 } x+\left( \sin a-1 \right) \tan x+\dfrac { \sqrt { a-2 } }{ \sqrt { 8-a } }$

$f'\left( x \right) =\sin { a } \tan ^{ 2 }{ x } \sec ^{ 2 }{ x } +\left( \sin a-1 \right) \sec ^{ 2 }{ x } =\sec ^{ 2 }{ x } \left( \sin { a } \left( \sec ^{ 2 }{ x } -1 \right) +\left( \sin a-1 \right) \right)$

$\Rightarrow f'\left( x \right) =\sec ^{ 2 }{ x } \left( \sin { a } \sec ^{ 2 }{ x } -1 \right)$
for critical points

$f'(x)=0$

$\Rightarrow \sec ^{ 2 }{ x } =0$

$\Rightarrow x\in \emptyset$
or

$\sec ^{ 2 }{ x } =\dfrac { 1 }{ \sin { a } }$
since, for

$a\epsilon \left[ \pi ,2\pi \right], \quad \sin { a } <0$
Therefore,

$x\in \emptyset$
Thus,

$f(x)$ have no critical points.

Ans: B

The point(s) on the curve

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

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

Explanation

$3y^{2}.y'+6x=12y'$
Or

$y'(3y^{2}-12)=-6x$
Or

$y'(y^{2}-4)=-2x$
Or

$y'=\dfrac{-2x}{y^{2}-4}$
Hence for the tangent to be vertical

$y^{2}-4=0$

$y=\pm2$
Substituting

$y=2$ , we get

$3x^{2}=12y-y^{3}$

$=24-8$

$=16$
Hence

$x=\dfrac{\pm4}{\sqrt{3}}$

$y=-2$ gives

$3x^{2}=-16$
This is not possible.
Hence the required point is

$(\dfrac{\pm4}{\sqrt{3}},2)$ .

The set of values of

$\displaystyle \lambda$ for which the function

$\displaystyle f\left ( x \right )= \left ( 4\lambda -3 \right )\left ( x+\log 5 \right )+2\left ( \lambda -7 \right ). \cot \dfrac{x}{2}\sin ^{2}\dfrac{x}{2}.$ does not posses critical point is:

0%
$\displaystyle \left ( 1,\infty \right )$
0%
$\displaystyle \left ( 2,\infty \right )$
0%
$\displaystyle \left ( -\infty ,-4/3 \right )$
0%
$\displaystyle \left ( -\infty ,-1 \right )$

Explanation

$\displaystyle 2\cot \frac{x}{2}\sin ^{2}\frac{x}{2}= 2\cos \frac{x}{2}\sin \frac{x}{2}= \sin x$

$\displaystyle \therefore f\left ( x \right )= \left ( 4\lambda -3 \right )\left ( x+\log 5 \right )+\left ( \lambda -7 \right )\sin x$

$\displaystyle {f}'x= \left ( 4\lambda -3 \right )+\left ( \lambda -7 \right )\cos x= 0$ ...(1)

$\displaystyle \therefore \cos x= \frac{4\lambda -3}{\lambda -7}$ ...(2)Now

$\displaystyle -1\leq \cos x\leq 1 \therefore -1\leq \frac{4\lambda -3}{\lambda -7}\leq 1$ Above gives us two

inequalities

$\displaystyle -1-\frac{4\lambda -3}{\lambda -7}\leq 0$

or

$\displaystyle \frac{4\lambda -3}{\lambda -7}\leq 0$

$\displaystyle \frac{-5\lambda +10}{\lambda -7}\leq 0$ or

$\displaystyle \frac{3\lambda +4}{\lambda -7}\leq 0$ or

$\displaystyle \frac{5\left ( \lambda -2 \right )}{\lambda -7}\geq 0$ or

$\displaystyle \frac{3\left ( \lambda +4/3 \right )}{\lambda -7}\leq 0$ Above are the

conditions for

$f(x)$ to have critical points. But the function does not possess critical points. Therefore we must have

$\displaystyle \frac{5\left ( \lambda -2 \right )}{\lambda -7}< 0$ or

$\displaystyle \frac{3\left ( \lambda +4/3 \right )}{\lambda -7}> 0$ or

$\displaystyle \frac{5\left ( \lambda -2 \right )\left ( \lambda -7 \right )}{\left ( \lambda -7 \right )^{2}}< 0$ or

$\displaystyle \frac{3\left ( \lambda \lambda +4/3 \right )\left ( \lambda -7 \right )}{\left ( \lambda -7 \right )^{2}}> 0$

$\displaystyle \therefore \lambda \epsilon \left ( 2,7 \right )$ or

$\displaystyle \lambda < -4/3$ or

$\displaystyle > 7$

$\displaystyle \therefore \lambda \epsilon \left ( 2,7 \right )$ and

$\displaystyle \lambda \epsilon \left ( -\infty ,-4/3 \right ) or \left ( 7,\infty \right )$

$\displaystyle \therefore \lambda \epsilon \left ( 2,\infty \right )$ or

$\displaystyle \lambda \epsilon \left ( -\infty ,-4/3 \right )$

The positive value of

$k$ for which

$ke^x-x=0$ has only one real solution is

0%
$\dfrac{1}{e}$
0%
1
0%
e
0%
$\log_{e}2$

Explanation

$ke^{ x }-x=0$ will have only one solution when

$g(x)=x$ is tangent to

$f(x)=ke^{x}$
let

$P(a,b)$ be a point on

$f(x)$ and it also satisfy

$g(x)$
Therefore,

$ke^{ a }=a$ ...(1)
slope of

$g(x)=$ slope of

$f(x)$ at point P

$\Rightarrow 1=\dfrac { dy }{ dx }$ at point P

$=k{ e }^{ a }$
From (1), we get

$a=1$
Therefore,

$k=1/e$

Ans: A

The coordinates of the point

$P$ on the curve

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

$4x-3y + 2 = 0$ , are given by

0%
$\left ( 2,4 \right )$
0%
$\left ( 1,\sqrt{2} \right )$
0%
$\left ( 1/2,-1/2 \right )$
0%
$\left ( 1/8,-1/16 \right )$

Explanation

$4x-3y+2=0$
Or

$3y=4x+2$
Or

$y=\dfrac{4}{3}x+\dfrac{2}{3}$ .
Hence slope of the tangent will be

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

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

$y=\pm\sqrt{2}.x^{\frac{3}{2}}$
Hence

$\dfrac{dy}{dx}$

$=\pm\sqrt{2}.\dfrac{3}{2}.x^{\frac{1}{2}}$

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

Or

$-\sqrt{2}.\dfrac{3}{2}.x^{\dfrac{1}{2}}=\dfrac{-3}{4}$
Or

$2\sqrt{2}.x^{\dfrac{1}{2}}=1$
Or

$x=\dfrac{1}{8}$ .
Hence

$y=-\sqrt{2}.x^{\dfrac{3}{2}}$

$=-\sqrt{2}.\dfrac{1}{16\sqrt{2}}$

$=-\dfrac{1}{16}$
Hence the co-ordinates are

$(\dfrac{1}{8},\dfrac{-1}{16})$

Find the co-ordinates of the point (s) on the curve

$\displaystyle y= \frac{x^{2}-1}{x^{2}+1}, x> 0$ such that tangent at these point (s)have the greatest slope.

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

Explanation

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

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

S=slope

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

For maximum and minimum of S,

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

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

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

$\displaystyle \dfrac{dS}{dx}= 4\dfrac{1-3x^{2}}{\left ( 1+x^{2} \right )^{3}}=0$

$\displaystyle \therefore x= \dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}$

Now consider change of sign at

$\displaystyle x= \dfrac{1}{\sqrt{3}}, \dfrac{dS}{dx}$ changes from +ve to -ve.

Hence S has maximum at

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

When

$\displaystyle x= \dfrac{1}{\sqrt{3}}$ the value of

$\displaystyle y= -\dfrac{1}{2}$

Hence the point is

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

The equation of the tangents to

$\displaystyle 4x^{2}-9y^{2}=36$ which are perpendicular to the straight line

$\displaystyle 2y+5x= 10$ are

0%
$\displaystyle 5\left ( y-3 \right )=2 \left ( x-\sqrt{\frac{117}{4}} \right )$
0%
$\displaystyle 5\left ( y-2 \right )=2 \left ( x-\sqrt{-18} \right )$
0%
$\displaystyle 5\left ( y+2 \right )=2 \left ( x-\sqrt{-18} \right )$
0%
none of these

Explanation

Given,

$2y+5x=10$

$2y'+5=0$

$y'=\dfrac{-5}{2}$ .
Hence the required slope is

$\dfrac{2}{5}$ .
Now

$4x^{2}-9y^{2}=36$ .

$4x^{2}-36=9y^{2}$

$y=\dfrac{1}{3}.\sqrt{4x^{2}-36}$ .

$y'=\dfrac{8x}{6\sqrt{4x^{2}-36}}$

$=\dfrac{2}{5}$

$40x=12\sqrt{4x^{2}-36}$

$10x=3\sqrt{4x^{2}-36}$

$100x^{2}=36x^{2}-324$

$64x^{2}=-324$

$x^{2}=\dfrac{-81}{16}$ .
Hence no real values of x.
Hence answer is none of these.

For the curve

${x}^{2}+4xy+8{y}^{2}=64$ the tangents are parallel to the

$x$ -axis only at the points

0%
$(0,2\sqrt { 2 } )$ and $(0,-2\sqrt { 2 } )$
0%
$(8,-4)$ and $(-8,4)$
0%
$(8\sqrt { 2 } ,-2\sqrt { 2 } )$ and $(-8\sqrt { 2 } ,2\sqrt { 2 } )$
0%
$(9,0)$ and $(-8,0)$

Explanation

Given curve is

${x}^{2}+4xy+8{y}^{2}=64.....(i)$
On differentiating w.r.t

$x$ ,we get

$2x+4(y+x\cfrac{dy}{dx})+16y\cfrac{dy}{dx}=0$

$\Rightarrow$

$2x+4y+(4x+16y)\cfrac{dy}{dx}=0$

$\Rightarrow$

$\cfrac{dy}{dx}=-\cfrac{(x+2y)}{2(x+4y)}$
Since, tangent are parallel to

$x$ -axis only
ie.,

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

$\Rightarrow$

$-\cfrac{(x+2y)}{2(x+4y)}=0$

$\Rightarrow$

$x+2y=0.....(ii)$
Now, on putting the values of

$x$ from eqs.

$(i)$ in

$(ii)$ we get

$4{y}^{2}-8{y}^{2}+8{y}^{2}=64$

$\Rightarrow$

${y}^{2}=16$

$\Rightarrow$

$y=\pm 4$
from eq.

$(ii)$
When

$y=4,x=-8$
and when

$y=-4, x=8$
Hence, required points are

$(-8,4)$ and

$(8,-4)$

Given function

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

0%
Increasing
0%
Decreasing
0%
Even
0%
None of these

Explanation

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

$f\left( x \right) =\cfrac { \left[ { e }^{ 2x }+1 \right] \left[ 2{ e }^{ 2x } \right] -\left[ { e }^{ 2x }-1 \right] \left[ 2{ e }^{ 2x } \right] }{ { \left[ { e }^{ 2x }+1 \right] }^{ 2 } }$

$f^{ 1 }\left( x \right) =\cfrac { 2{ e }^{ 2x }\left[ { e }^{ 2x }+1-{ e }^{ 2x }+1 \right] }{ { \left[ { e }^{ 2x }+1 \right] }^{ 2 } }$

$f^{ 1 }\left( x \right) =\cfrac { 4{ e }^{ 2x } }{ { \left[ { e }^{ 2x }+1 \right] }^{ 2 } }$

Now,

${ \left[ { e }^{ 2x }+1 \right] }^{ 2 }>0$ and

${ e }^{ 2x }>0$

$\therefore f^{ 1 }\left( x \right) >0$

$\therefore f\left( x \right)$ is increasing function.

If the line

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

$xy = 1$ . Then

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

Explanation

$ax+by+c=0$
Hence

$by=-ax-c$
Or

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

$m=\dfrac{-a}{b}$
= Slope of normal.
Now

$xy'+y=0$
Or

$y'=\dfrac{-y}{x}$
Or

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

$\dfrac{-1}{y'}$

$=x^{2}$
Hence

$x^{2}=\dfrac{-a}{b}$
Or
Now

$x^{2}>0$ thus

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

$a<0$ and

$b>0$
Or

$a>0$ and

$b<0$ .

The coordinates of the points(s) at which the tangents to the curve

$\displaystyle y=x^{3}-3x^{2}-7x+6$ cut the positive semi axis OX a line segment half that on the negative semi axis OY is/are given by

0%
$(-1, 9)$
0%
$(3, -15)$
0%
$(1, -3)$
0%
none

Explanation

Let the point of tangency be

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

Then the equation of tangent is

$\dfrac{y-y_{1}}{x-x_{1}}=3x_{1}^{2}-6x_{1}-7$

when

$y=0,x=x_{1}+\dfrac{-y}{3x_{1}-6x_{1}-7}$

when

$x=0,y=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$

$-2(x_{1}+\dfrac{-y}{3x_{1}-6x_{1}-7})=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$

$-2(\dfrac{x_{1}(3x_{1}-6x_{1}-7)-y_{1}}{3x_{1}-6x_{1}-7})=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$

$2=3x_{1}-6x_{1}-7$

$3x_{1}-6x_{1}-9=0$

$x_{1}=-1 or 3$

when

$x_{1}=-1,y_{1}=(-1)^{3}-3(-1)^{2}-7(-1)+6=9$

and the y-intercept is

$9-(-1)[3(-1)^{2}-6(-1)-7]=11> 0$

when

$x_{1}=3,y_{1}=(3)^{3}-3(3)^{2}-7(3)+6=-15$

and the y-intercept is

$-15-(3)[3(3)^{2}-6(3)-7]=-21< 0$

So,the point of tangency is

$(-1,9)$ , this is the correct answer.

The tangent to the curve

$x= a\sqrt{\cos 2\theta }\cos \theta$ ,

$y= a\sqrt{\cos 2\theta }\sin \theta$ at the point corresponding to

$\theta = \pi /6$ is

0%
parallel to the $x$ -axis
0%
parallel to the $y$ -axis
0%
parallel to line $y = x$
0%
none of these

Explanation

$\displaystyle \dfrac{dx}{d\theta }= -a\sqrt{\cos 2\theta }\sin \theta +\dfrac{-a\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$

$\displaystyle = -a\dfrac{\cos 2\theta \sin \theta +\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$

$\displaystyle = \dfrac{-a\sin 3\theta }{\sqrt{\cos 2\theta }}$

$\displaystyle \dfrac{dy}{d\theta }= a\sqrt{\cos 2\theta }\, \cos \theta -a\dfrac{\sin \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$

\displaystyle = \dfrac{a\cos 3\theta }{\sqrt{\cos 2\theta }}$$

Hence

$\displaystyle \dfrac{dy}{dx}= -\cot 3\theta \Rightarrow \dfrac{dy}{dx}|_{\theta = \pi /6 }= 0$

So the tangent to the curve at

$\theta = \pi /6$ is parallel to the

$x$ -axis.

The tangent to the curve

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

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

$(c -1,e^{c-1})$ and

$(c +1,e^{c+1})$

0%
on the left of $x = c$
0%
on the right of $x = c$
0%
at no paint
0%
at all points

Explanation

Given equation of curve

$y=e^x$

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

Slope of tangent at point

$(c,e^c)$ is

$e^c$

Equation of tangent at

$(c,e^{c})$ is

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

Equation of straight line joining

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

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

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

Subtracting (2) from (1), we get

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

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

$\displaystyle \Rightarrow x-c=\dfrac{e+e^{-1}-2}{2-e+e^{-1}}< 0$

$[ \because e+e^{-1}>2$ and

$2+e^{-1}-e<0 ]$

$\Rightarrow x<c$

Thus the two lines meet to the left of

$x = c$ .

$f(x)=e^x(x-2)^2$ then

0%
f is increasing in $(-\infty, 0)$ and $(2, \infty)$ deceasing in $(0, 2)$
0%
f is increasing in $(-\infty, 0)$ and deceasing in $(0, \infty)$
0%
f is increasing in $(2, \infty)$ and deceasing in $(-\infty, 0)$
0%
f is increasing in $(0, 2)$ and deceasing in $(-\infty, 0)$ and $(2, \infty)$

Explanation

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

For increasing or decreasing, we need to find

$f'\left( x \right)$

$f'\left( x \right) ={ e }^{ x }\times 2(x-2)+{ e }^{ x }{ (x-2) }^{ x }$

$={ e }^{ x }(x-2)\left[ 2+x-2 \right]$

$f'\left( x \right) ={ e }^{ x }x(x-2)$

It will have two solutions,

$x=0,x=2$
plotting it on number line

(Image)

Therefore

$f'\left( x \right) >0$ for

$\left( -\infty ,0 \right) \bigcup { (2,\infty ) }$ and

$f'\left( x \right) <0$ for

$(0,2)$

Thus function increases for

$\left( -\infty ,0 \right) \bigcup { (2,\infty ) }$ and function decreases for

$(0,2)$

The value of

$a$ for which the function

$f(x)=(4a-3)(x+\log 5)+2(a-7)\cot \dfrac x2\sin ^2\dfrac x2$ does not possess critical points is

0%
$(-\infty,-\dfrac 43)$
0%
$(-\infty,-1)$
0%
$[1,\infty)$
0%
$(2,\infty)$

Explanation

$f(x)=(4a−3)(x+\log5)+2(a−7)\cot(x/2){ \sin }^{ 2 }x/2\\ f(x)=(4a−3)(x+\log5)+(a−7)(2\times \cot(x/2){ \sin }^{ 2 }x/2)\\ f(x)=(4a−3)(x+\log5)+(a−7)(\sin\ x)\\ f'(x)=(4a−3)+(a−7)(\cos\ x)$

$f'(x)$ will not be zero

$(4a-3)>(a-7)$

$2a>-4$

$a>-2$

$4a-3<-(a-7)$

$5a<10$

$a<2$

All the critical points of

$f(x)=\dfrac {|2-x|}{x^2}$ is/are:

0%
$x=0,2$
0%
$x=2,4$
0%
$x=2,-4$
0%
None of the above.

Explanation

The critical point for

$f(x)$ is a value

$x_0$ where its derivative is zero. i.e.,

$f^{'}(x_0)=0$

Given that

$f(x)=\dfrac{|2-x|}{x^2}$

Therefore,

$f^{'}(x)=\dfrac{x^2(\dfrac{|2-x|}{2-x})-|2-x|(2x)}{x^4}=0$

$\implies \dfrac{|2-x|(x^2-2x(2-x))}{x^4(2-x)}=0$

$\implies |2-x|(x^2-2x(2-x))=0$ and

$x^4\neq 0$ and

$2-x\neq 0$

$\implies |2-x|=0$ and

$(3x-4)x=0$ and

$x\neq 0$ and

$x\neq 2$

$\implies x=2$ and

$x=0$ and

$x=\dfrac 43$ and

$x\neq 0$ and

$x\neq 2$

Therefore,

$x=\dfrac 43$ is the critical point.

If the slope of the curve

$y=\cfrac { ax }{ b-x }$ at the point

$(1,1)$ is

$2$ , then the values of

$a$ and

$b$ are respectively

0%
$1,-2$
0%
$-1,2$
0%
$1,2$
0%
None of these

Explanation

$y=\cfrac { ax }{ b-x }$

Slope=

$\cfrac { dy }{ dx }$

So,

$\cfrac { dy }{ dx }$ =

$\cfrac { a(b-x)-(-1)(ax) }{ { (b-1) }^{ 2 } } \quad \longrightarrow (1)$

for

$(1,1)$

put in equation of curve-

$1=\cfrac { a }{ (b-1) } \quad \Rightarrow (b-1)=a\quad \longrightarrow (2)$

From slop after putting (1,1)

We get,

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

put

$(b-1)=a$ from equation(2)

$2=\cfrac { { a }^{ 2 }+a }{ { a }^{ 2 } } \quad \quad \\ \Rightarrow 2{ a }^{ 2 }={ a }^{ 2 }+a\\ \quad \Rightarrow a(a-1)=0\\ \quad \Rightarrow a=0,1$

put in (2),

We get

$b=1,2$

In option only one value is given

$\therefore a=1,b=2$

$f:[1, 10]\rightarrow[1,10]$ is a non-decreasing function and

$g:[1,10] \rightarrow [1,10]$ is a non-increasing function, Let

$h(x) = f(g(x))$ with

$h(1)=1$ . then,

$h(2)$

0%
less than $1$
0%
is more than two
0%
is equal to $2$
0%
is not defined

$y = 4x - 5$ is a tangent to the curve

$y^{2} = px^{3} + q$ at

$(2, 3)$ , then

0%
$p = 2, q = -7$
0%
$p = -2, q = 7$
0%
$p = -2, q = -7$
0%
$p = 2, q = 7$

Explanation

Since,

$(2, 3)$ lies on the curve

$y^{2} = px^{3} + q$
Therefore,

$9 = 8p + q ..... (i)$

$y^{2} = px^{3} + q$

$\Rightarrow 2y \dfrac {dy}{dx} = 3px^{2}$

$\dfrac {dy}{dx} = \dfrac {3px^{2}}{2y}$

$\Rightarrow \left (\dfrac {dy}{dx}\right )_{(2, 3)} = \dfrac {12p}{6} = 2p$
Since,

$y = 4x - 5$ is tangent to

$y^{2} = px^{3} + q$ at

$(2, 3)$ . Therefore,

$\therefore \left (\dfrac {dy}{dx}\right )_{(2, 3)} =$ Slope of the line

$y = 4x - 5$

$\Rightarrow 2p = 4 \Rightarrow p = 2$
Putting

$p = 2$ in Eq. (i), we get

$q = -7$ .

The curve given by

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

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

Explanation

Given curve is

$x+y={ e }^{ xy }\quad$
On differentiating w.r.t

$x$ we get

$1+\cfrac { dy }{ dx } ={ e }^{ xy }\left\{ y+x\cfrac { dy }{ dx } \right\}$

$\Rightarrow \cfrac { dy }{ dx } =\cfrac { y{ e }^{ xy }-1 }{ 1-x{ e }^{ xy } }$

Since, tangent is parallel to y-axis
Here,

$\cfrac { dy }{ dx } =\infty \Rightarrow 1-x{ e }^{ xy }=0$

$\Rightarrow 1-x(x+y)=0$
This holds for

$x=1,y=0$

Abscissa of

$p_{1}, p_{2}, p_{3} .... p_{n}$ are in

0%
A.P.
0%
G.P.
0%
H.P
0%
None

$g(x)$ is continuous function at

$x = a$ , such that

$g(a) > 0$ and

$f'(x) (g(x))(x^{2} - ax + a^{2}) \forall x\epsilon R$ , then

$f(x)$ is

0%
Increasing in the neighbourhood of $x = a$
0%
Decreasing in the neighbourhood of $x = a$
0%
Constant in the neighbourhood of $x = a$
0%
Maximum at $x = a$

Let

$f'(\sin { x } )<0$ and

$f''(\sin { x } )>0>\forall x\in \left( 0,\cfrac { \pi }{ 2 } \right)$ and

$g(x)=f'(\sin { x } )+f'(\cos { x } )$ , then

$g(x)$ is decreasing in

0%
$\left( \cfrac { \pi }{ 4 } ,\cfrac { \pi }{ 2 } \right)$
0%
$\left( 0,\cfrac { \pi }{ 4 } \right)$
0%
$\left( 0,\cfrac { \pi }{ 2 } \right)$
0%
$\left( \cfrac { \pi }{ 6 } ,\cfrac { \pi }{ 2 } \right)$

$f(x)=e^{3x}-sinx+x^{2}$ , Find

$f '(x)$

0%
$3e^{3x}-cosx+2x$
0%
$e^{3x}+cosx+2x$
0%
$3e^{3x}+cosx+x$
0%
$3e^{3x}+sinx+2x$

Given that f(x) is a differentiable function of x and that

$f(x).f(y)=f(x)-4-f(y)+f(xy)-2$ and that

$f(2)=5$ . Then

$f'(3)$ is equal to

0%
2
0%
24
0%
15
0%
19

The curve that passes through the point

$(2,3)$ and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by

0%
${ \left( \cfrac { x }{ 2 } \right) }^{ 2 }+{ \left( \cfrac { y }{ 3 } \right) }^{ 2 }=2\quad$
0%
$2y-3x=0$
0%
$y=\cfrac { 6 }{ x }$
0%
${ x }^{ 2 }+{ y }^{ 2 }=13$

Explanation

Let

$B$ be the pt. of contact

$y=f(x)$ be the curve

then

$AB=BC$ (given condition)

$C\equiv (2x,0)$

$A\equiv (0,2y)$

Slope

$m$ of line

$AC$ is

$m=\dfrac{dy}{dx}=\dfrac{0.2y}{2x-0}$

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

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

$\ln y=-\ln x+c$

$yx=c$

$(2, 3)$ satisfies the curve

$\therefore 2\times 3=c$

$\therefore y=\dfrac{6}{x}$ is the curve

$\therefore$ Option

$C$ is correct

1453826_765333_ans_aed4145d8126434bb1c77800e58e1cb6.png

If the line

$y=4x-5$ touches to the curve

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

$(2,3)$ then

$7a+2b=$

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

Explanation

The slope of given line

$y=4x-5$ is

$m=4$ .

Now, since the given line touches the curve at

$(2,3)$ , the slope of curve

$\dfrac{dy}{dx}$ at the given point is equal to the slope

$m$ of the line.

Hence, for slope of curve, differentiating both sides w.r.t x:-

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

Putting the value

$x=2$ and

$y=3$

$6\dfrac{dy}{dx}=3a\times 4 \implies 6\times4=12a$

$\implies a=2$

Now, the given point lies on the curve, so it must satisfy the equation of the curve:-

$\implies (3)^2=2\times (2)^3+b$

$\implies b=-7$

$7a+2b=(7\times 2-2\times 7)=0$

Hence, answer is option-(A).

$f(x)$ is an even function, where

$f(x)\ne 0$ , then which one of the following is correct?

0%
$f'(x)$ is an even function
0%
$f'(x)$ is an odd function
0%
$f'(x)$ may be an even or odd function depending on the type of function
0%
$f'(x)$ is a constant function

Explanation

Let

$f(x)$ be an even function

$\Rightarrow f(x) = f(-x)$

Differentiate both sides, we get

$\dfrac{df}{dx} = \dfrac{d(f(-x))}{dx} \quad ...(1)$

To differentiate

$f(-x)$ , we use the chain rule formula as follows:

Let

$u = - x$ , hence

$\dfrac{d(f(-x))}{dx}=\dfrac{df(u)}{du}\cdot \dfrac{du}{dx}=f'(u)\cdot (-1)=-f'(-x)$

Substituting in

$(1)$ we get

$f '(x) = - f '(- x)$

That is

$f'(x)$ is an odd function.

For the curve

$x=t^2-1$ ,

$y=t^2-t$ , the tangent is perpendicular to

$x$ -axis then

0%
$t=0$
0%
$t=\dfrac{1}{2}$
0%
$t=1$
0%
$t=\dfrac{1}{\sqrt{3}}$

Explanation

Given curve is

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

Derivating w.r.to

$t$ we get

$\dfrac{dx}{dt}=2t$ -------(1)

and

$\dfrac{dy}{dt}=2t-1$ -------(2)

dividing (2) by (1) we get

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

Therefore, the slope of the tangent is

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

Given that the tangent is perpendicular to x-axis. Therefore, tangent is parallel to y-axis.

We know that slope of y-axis is infinity and the slopes of the two parallel lines are equal.

Therefore, slope of the tangent is infinity.

Hence,

$\dfrac{2t-1}{2t}=\dfrac{1}{0}$

$\implies 2t=0 \implies t=0$

Let

$f\left( x \right) = {\tan ^{ - 1}}x - \frac{{In\left| x \right|}}{2},x \ne 0.$ . Then

$f\left( x \right)$ is increasing in

0%
$\left( {0,\infty } \right)$
0%
$\left( { - \infty ,0} \right)$
0%
$\left( {1,\infty } \right)$
0%
none of these

Explanation

$f(x)=\tan ^{-1} x-\frac{\ln |x|}{2}, x \neq 0$

$f^{\prime}(x)=\frac{1}{1+x^{2}}-\frac{1}{2 x}$

$\Rightarrow f^{\prime}(x)=0$

$\Rightarrow \frac{1}{1+x^{2}}-\frac{1}{2 x}=0$

$\Rightarrow \frac{1}{1+x^{2}}=\frac{1}{2 x}$

$\Rightarrow x^{2}+1=2 x \\$

$\Rightarrow x^{2}-2 x+1=0 \\$

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

$\therefore x=1$

Hence

$f(x)$ is increasing in

$(-\infty, 0)$

If the tangent at

$({x_1},{y_1})$ to the curve

${x^3} + {y^3} = {a^3}$ meets the curve again at

$({x_2},{y_2})$ , then

0%
${{{x_2}} \over {{x_1}}} + {{{y_2}} \over {{y_1}}} = - 1$
0%
${{{x_2}} \over {{y_1}}} + {{{x_1}} \over {{y_2}}} = - 1$
0%
${{{x_1}} \over {{x_2}}} + {{{y_1}} \over {{y_2}}} = - 1$
0%
${{{x_2}} \over {{x_1}}} + {{{y_2}} \over {{y_1}}} = 1$

Explanation

Given curve equation is

$x^3+y^3=a^3$ .

The curve passes through the points

$(x_1,y_1)$ and

$(x_2,y_2)$ .

Therefore,

$x_1^2+y_1^2=a^3$ -------(1)

and

$x_2^3+y_2^3=a^3$ -------(2)

subtracting (1) from (2) we get

$y_2^3-y_1^3=-(x_2^3-x_1^3)$ -------(3)

Therefore,

$3x^2+3y^2(\dfrac{dy}{dx})=0$

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

Therefore, the slope of the tangent at

$(x_1,y_1)$ is

$-\dfrac{x_1^2}{y_1^2}$

The tangent passes through the points

$(x_1,y_1)$ and

$(x_2,y_2)$ .

Therefore, the slope of the tangent is

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

On comparing two slopes we get,

$-\dfrac{x_1^2}{y_1^2}=\dfrac{y_2-y_1}{x_2-x_1}$

we know that

$a^3-b^3=(a-b)(a^2+ab+b^2)$

$\implies -\dfrac{x_1^2}{y_1^2}=\dfrac{y_2^3-y_1^3}{x_2^3-x_1^3}\times \dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}$

$\implies -\dfrac{x_1^2}{y_1^2}=-\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}$

$\implies x_1^2y_1^2+x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1^2+x_1^2y_1y_2+x_1^2y_1^2$

$\implies x_1^2y_2^2-x_2^2y_1^2=x_1x_2y_1^2-x_1^2y_1y_2$

$\implies (x_1y_2-x_2y_1)(x_1y_2+x_2y_1)=x_1y_1(x_2y_1-x_1y_2)$

$\implies x-1y_2+x_2y_1=-x_1y_1$

$\implies \dfrac{x_2}{x_1}+\dfrac{y_2}{y_1}=-1$

If the error committed in measuring the radius of the circle is

$0.05\%$ , then the corresponding error in calculating the area is:

0%
$0.05\%$
0%
$0.025\%$
0%
$0.25\%$
0%
$0.1\%$

Explanation

$\dfrac { dr }{ r } =0.05\Rightarrow dr=(0.05)r$

Area of circle

$=\pi r^2$

$\ A=\pi r^{ 2 }\Rightarrow \dfrac { dA }{ dr } =2\pi r\\ dA=2\pi rdr\Rightarrow \dfrac { dA }{ A } =\dfrac { 2\pi rdr }{ \pi r^{ 2 } } =\dfrac { 2dr }{ r } \\ \therefore \dfrac { dA }{ A } =2(0.05)^{ 2 }=0.1$

$\therefore$ Corresponding error in area

$= 0.1\%$

A point on the curve

$y = 2{x^3} + 13{x^2} + 5x + 9$ , the tangent at which passes through the origin is

0%
$(1, 15)$
0%
$(1, -15)$
0%
$(15, 1)$
0%
$(-1, 15)$

Explanation

Let

$P(a,\,b)$ be a point on the given curve

$y=2x^3+13x^2+5x+9$

$b=2a^3+13a^2+5a+9$ --------- ( 1 )

Now, taking derivative of

$y$ .

$\therefore$

$y'=6x^2+26x+5\Rightarrow y'\,at\,P=6a^2+26a+5$ .

Equation of tangent at

$P$ is

$y-b=(6a^2+26a+5)(x-a)$

This tangent passes through origin means

$b=a(6a^2+26a+5)$

$2a^3+13a^2+5a+9=6a^3+26a^2+5a$

$4a^3+13a^2-9=0$

$(a+1)(a+3)(4a-3)=0$

Taking,

$a+1=0$

We get,

$a=-1$

Substituting

$a=-1$ in equation (1) we get,

$b=2(-1)^3+13(-1)^2+5(-1)+9$

$\therefore$

$b=15$

$\therefore$

$P(a,\,b)=(-1,\,15)$

$\therefore$ A point on the curve

$y=2x^3+13x^2+5x+9$ , the tangent at which passes through the origin is

$(-1,\,15)$ .

The value of n for which the length of the sub-normal at any point of the curve

$y^3= a^{1-n}x^{2n}$ must be constant, is

0%
$-1$
0%
$-\frac{1}{2}$
0%
$\frac{3}{4}$
0%
$1$

The value of 'a' for which the function

$f\left( x \right) = \left( {a + 2} \right){x^3} - 3a{x^2} + 9ax - 1$ decreases for all real values of x is

0%
$( - \infty , - 3]$
0%
$\left( { - \infty , - 3} \right)$
0%
$\left( { - \infty , - 2} \right)$
0%
$( - \infty , - 3] \cup [0,\infty )$

Explanation

decreasing or strictly decreasing =>

$f'(x)<0$

monotonically decreasing =>

$f'(x)<=0$ .

$f'(x)=3(a+2)x^{2}-6ax+9a$

Given

$f'(x)<0$ =>

$a+2<0 , (6a)^{2}-4(3(a+2))(9a)<0$ (from quadratic equations graph concept)

after simplifying

$a\epsilon(-\infty,-3)$

If the tangent at any point on the curve

$x^{4} +y^{4}=a^{4}$ cuts off intercepts

$p$ and

$q$ on the coordinate axes the value of

$p^{-4/3}+q^{-4/3}$ is

0%
$a^{-4/3}$
0%
$a^{-1/3}$
0%
$a^{1/2}$
0%
$None\ of\ these$

Explanation

Consider the given expression.

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

Let the point on the curve be $M\left( {{x}_{0}},{{y}_{0}} \right)$ .

Therefore,

$x_{0}^{4}+x_{0}^{4}={{a}^{4}}$

Differentiate the expression given in the question with respect to $x$ .

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

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

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

So, at the point $M$ , the slope is,

$\Rightarrow -\dfrac{x_{0}^{3}}{y_{0}^{3}}$

Therefore, equation of the tangent is,

$$\begin{align}

$y-{{y}_{0}}=-\dfrac{x_{0}^{3}}{y_{0}^{3}}\left( x-{{x}_{0}} \right)$

$yy_{0}^{3}-y_{0}^{4}=-x_{0}^{3}x+x_{0}^{4}$

\end{align}$$

Let $p$ and $q$ be the $x$ and $y$ intercept, respectively. So, at $x$ intercept,

$-y_{0}^{4}=-x_{0}^{3}p+x_{0}^{4}$

$x_{0}^{3}p=x_{0}^{4}+y_{0}^{4}$

$p=\dfrac{x_{0}^{4}+y_{0}^{4}}{x_{0}^{3}}$

$p=\dfrac{{{a}^{4}}}{x_{0}^{3}}$

Similarly,

$q=\dfrac{{{a}^{4}}}{y_{0}^{3}}$

Now, calculate the value of ${{p}^{-4/3}}+{{q}^{-4/3}}$ .

$={{\left( \dfrac{{{a}^{4}}}{x_{0}^{3}} \right)}^{-4/3}}+{{\left( \dfrac{{{a}^{4}}}{y_{0}^{3}} \right)}^{-4/3}}$

$={{\left( \dfrac{x_{0}^{3}}{{{a}^{4}}} \right)}^{4/3}}+{{\left( \dfrac{y_{0}^{3}}{{{a}^{4}}} \right)}^{4/3}}$

$=\left( \dfrac{x_{0}^{4}}{{{a}^{16/3}}} \right)+\left( \dfrac{y_{0}^{4}}{{{a}^{16/3}}} \right)$

$=\dfrac{x_{0}^{4}+y_{0}^{4}}{{{a}^{16/3}}}$

$=\dfrac{{{a}^{4}}}{{{a}^{16/3}}}$

$={{a}^{-4/3}}$

Hence, this is the required result.

The slope of the tangent to the curve at a point

$(x,y)$ on it is proportional to

$(x-2).$ If the slope of the tangent to the curve at

$(10,-9)$ on it is

$-3$ . The equation of the curves is .

0%
$y=k(x-2)^2$
0%
$y=\dfrac{-3}{16}(x-2)^2+1$
0%
$y=\dfrac{-3}{16}(x-2)^2+3$
0%
$y=K(x+2)^2$

At any two points of the curve represented parametrically by

$x = a\left( {2\cos t - \cos 2t} \right);y = a\left( {2\sin t - \sin 2t} \right)$ the tangent are parallel to the axis of

$x$ corresponding to the values of the parameter

$1$ differing from each other by

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

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 11 - 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 11 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

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