MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 4

Let the parabolas

$y=x^{2}+ax+b$ and

$y=x(c-x)$ touch each other at the point

$(1, 0)$ . Then

0%
$a=-3$
0%
$b=1$
0%
$c=2$
0%
$b+c=3$

Explanation

As point

$\left( 1,0 \right)$ lies on

$y={ x }^{ 2 }+ax+b\Rightarrow a+b=-1$ ...(1)
And also lies on

$y=x\left( c-x \right) \Rightarrow c=1$ ...(2)
They touch each other, so slope of tangents drawn from

$(1, 0)$ will be same, then

$\cfrac { dy }{ dx } =2x+a=c-2x\Rightarrow a=c-4\Rightarrow a=-3$ ...(3)
Using (1) and (3) we get

$-3+b=1\Rightarrow b=2$ ...(4)
Now using (2) and (4) we get

$b+c=2+1=3$
Hence, options 'A' and 'D' are correct.

The curve

$\displaystyle \frac{x^{n}}{a^{n}}+\frac{y^{n}}{b^{n}}=2$ touches the line

$\displaystyle \frac{x}{a}+\frac{y}{b}=2$ at the point

0%
$(b, a)$
0%
$(a, b)$
0%
$(1, 1)$
0%
$\displaystyle \left ( \frac{1}{a}, \frac{1}{b} \right )$

Explanation

Slope of curve

$\cfrac { { x }^{ n } }{ { a }^{ n } } +\cfrac { { y }^{ n } }{ { b }^{ n } } =2$ is

$\cfrac { dy }{ dx } =-\cfrac { { b }^{ n } }{ { a }^{ n } } \cfrac { { x }^{ n-1 } }{ { y }^{ n-1 } }$
Slope of tangent

$\cfrac { { x } }{ { a } } +\cfrac { { y } }{ { b } } =2$ is

$\cfrac { dy }{ dx } =-\cfrac { { b } }{ { a } }$
Equating both slopes, we get

$-\cfrac { { b }^{ n } }{ { a }^{ n } } \cfrac { { x }^{ n-1 } }{ { y }^{ n-1 } } =-\cfrac { { b } }{ { a } }$
Only

$x=a\quad y=b$ satisfy the above equation.
Hence, option 'B' is correct.

If the line joining the points

$(0, 3)$ and

$(5, -2)$ is the tangent to the curve

$\displaystyle y=\frac{c}{x+1}$ then the value of

$c$ is

0%
$1$
0%
$-2$
0%
$4$
0%
none of these

Explanation

Slop of line joining the points

$(0,3)\quad and\quad (5,2)$ is given by

$\dfrac { 3+2 }{ 0-5 } =-1$

This is equal to the slop of tangent on the curve and that is given by

$\dfrac { dy }{ dx } =\dfrac { -c }{ { \left( x+1 \right) }^{ 2 } } \\ \dfrac { dy }{ dx } =-1\\ \Rightarrow c={ \left( x+1 \right) }^{ 2 }...................(1)$ \

Equation of line joining the points

$(0,3)\quad and\quad (5,2)$ is given by

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

Solving equation of tangent and the curve for point of intersection

$\dfrac { c }{ x+1 } +x=3$ .............(2)

Solving (1) and (2)

$x=1$

Putting this in (2), we get c=4

A point on the ellipse

$4x^{2}+9y^{2}=36$ where the tangent is equally inclided to the axes is

0%
$\displaystyle \left ( \frac{9}{\sqrt{13}}, \frac{4}{\sqrt{13}} \right )$
0%
$\displaystyle \left ( -\frac{9}{\sqrt{13}}, \frac{4}{\sqrt{13}} \right )$
0%
$\displaystyle \left ( \frac{9}{\sqrt{13}}, -\frac{4}{\sqrt{13}} \right )$
0%
$\displaystyle \left ( \frac{4}{\sqrt{13}}, -\frac{9}{\sqrt{13}} \right )$

Explanation

Slope of tangent to the curve

$4{ x }^{ 2 }+9{ y }^{ 2 }=36$ is

$1$ . since it is equally inclined to the axes.

$\cfrac { dy }{ dx } =\cfrac { 4x }{ 9y } =1\\ \Rightarrow x=\cfrac { 9y }{ 4 }$
On solving obtained equation with the equation of curve, we get

$y=\pm \cfrac { 4 }{ \sqrt { 13 } }$
Substituting value of y we get

$(x,y)=\left( \cfrac { 9 }{ \sqrt { 13 } } ,\cfrac { 4 }{ \sqrt { 13 } } \right) ,\left( -\cfrac { 9 }{ \sqrt { 13 } } ,\cfrac { 4 }{ \sqrt { 13 } } \right) ,\left( \cfrac { 9 }{ \sqrt { 13 } } ,-\cfrac { 4 }{ \sqrt { 13 } } \right)$
Hence, options 'A', 'B' and 'C' are correct.

If error in measuring the edge of a cube is

$k$ % then the percentage error in estimating its volume is

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

Explanation

Let the actual length of the cube be a.

Therefore the measured length of the cube will be

$=a(1\pm0.0k)$

$=a(1\pm\dfrac{k}{100})$

Considering positive error,

$a'=a(1+\dfrac{k}{100})$

$V'=a^{3}(1+\dfrac{k}{100})^{3}$

$=a^{3}(1+3(\dfrac{k}{100})+3(\dfrac{k}{100})^{2}+(\dfrac{k}{100})^{3})$

Since

$\dfrac{k}{100}<<1$ , hence we neglect the higher order terms.

Thus

$V'=a^{3}(1+3(\dfrac{k}{100}))$

Actual volume V

$V=a^{3}$

Therefore

$V'-V=a^{3}(1+\dfrac{3k}{100})-a^{3}$

$=a^{3}(\dfrac{3k}{100})$

$\dfrac{V'-V}{V}=\dfrac{a^{3}\dfrac{3k}{100}}{a^{3}}$

$=\dfrac{3k}{100}$

$\dfrac{V'-V}{V}\times 100=3k$

Therefore percentage error in volume is

$3k$ .

The angle between two tangents to the ellipse

$\displaystyle \frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at the points where the line

$y=1$ cuts the curve is

0%
$\displaystyle \frac{\pi }{4}$
0%
$\tan^{-1}\displaystyle \frac{6\sqrt{2}}{7}$
0%
$\displaystyle \frac{\pi }{2}$
0%
none of these

Explanation

Substituting

$y=1$ in

$\cfrac { { x }^{ 2 } }{ 16 } +\cfrac { { y }^{ 2 } }{ 9 } =1$
We get

$x=\pm \cfrac { 8\sqrt { 2 } }{ 3 }$
And slope of tangents

$\cfrac { dy }{ dx } =\cfrac { 9x }{ 16y } =m_{1}, m_{2} =\pm \cfrac { 3\sqrt { 2 } }{ 2 }$
Therefore

$tan\left( \theta \right) =\dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}}=\left| -\cfrac { 6\sqrt { 2 } }{ 7 } \right| \Rightarrow \theta ={ tan }^{ -1 }\left( \cfrac { 6\sqrt { 2 } }{ 7 } \right)$

A tangent to the curve

$y=\displaystyle \int_{0}^{x}\left | t \right |dt$ , which is parallel to the line y=x, cuts off an intercept from the y-axis equal to

0%
$1$
0%
$\displaystyle -\frac{1}{2}$
0%
$\displaystyle \frac{1}{2}$
0%
$-1$

Explanation

Since the integral is in the first quadrant

$|t|=t$
Thus

$\int_{x=0} ^{x} |t|.dt$

$=\int_{x=0} ^{x} t.dt$

$=\dfrac{x^{2}}{2}$
Or

$2y=x^{2}$ is the equation of the curve.
Now

$2y'=2x$
Or

$y'=x$ ...(i)
Now the tangent is parallel to

$y=x$
Hence slope

$=y'=1$
Thus the x-coordinate of point of contact is 1.
Hence

$y=\dfrac{1}{2}$ .
Hence equation of the tangent will be

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

$x-y=\dfrac{1}{2}$
Or

$y=x-\dfrac{1}{2}$ .
Hence intercept cut on the y axis is

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

Angle between the tangents to the curve

$y= x^{2}-5x+6$ at the points

$(2,0)$ and

$\left ( 3,0 \right )$ is

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

Explanation

Given equation of curve

$\displaystyle y=x^{2}-5x+6$

$\displaystyle \Rightarrow \frac{dy}{dx}=2x-5$

Slope of tangent to the curve at

$(2,0)$ is

$\displaystyle \left(\frac{dy}{dx}\right)_{(2,0)}=2(2)-5=-1=m_{1}$

Slope of tangent to the curve at

$(3,0)$ is

$\displaystyle \left(\frac{dy}{dx}\right)_{(3,0)}=2(3)-5=1=m_{2}$

Since

$\displaystyle m_{1}m_{2}=-1$

$\therefore$ Angle between the tangents to the curve at

$(2,0)$ and

$(3, 0)$ is

$\displaystyle \dfrac{\pi}{2}$

The number of tangents to the curve

$\displaystyle y= e^{\left | x \right |}$ at the point

$(0,1)$ is

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

Explanation

For

$x>0$ ,

$y=e^{x}$ .

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

$\dfrac{dy}{dx}_{x=0}=1$ .
Hence

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

$x-y+1=0$ is the required equation of tangent.
Similarly for

$x<0$

$y=e^{-x}$ .

$\dfrac{dy}{dx}=-e^{-x}$ .
Now

$\dfrac{dy}{dx}_{x=0}=-1$ .
Hence

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

$x+y-1=0$ is the required equation of tangent.
Hence at

$x=0$ we will have 2 tangents to the curve

$y=e^{|x|}$ which are mutually perpendicular.

The curve

$y+e^{xy}+x= 0$ has a tangent parellel to y-axis at a point

0%
$\left ( -1,\:0 \right )$
0%
$\left ( 1,\:0 \right )$
0%
$\left ( 1,\:1 \right )$
0%
$\left ( 0,\:0 \right )$

Explanation

$\dfrac{dy}{dx}+e^{xy}[x\dfrac{dy}{dx}+y]+1=0$
Or

$\dfrac{dy}{dx}[1+x.e^{xy}]+1+y.e^{xy}=0$
Or

$\dfrac{dy}{dx}=\dfrac{-(1+y.e^{xy})}{1+x.e^{xy}}$
Now

$1+xe^{xy}=0$ since the slope of the tangent is

$90^{0}$ .
Hence

$xe^{xy}=-1$
Or

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

$x^{2}+xy=1$
Or

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

$y=\dfrac{1}{x}-x$ ...(i)
Considering

$y=0$ ,

$x^{2}=1$

$x=\pm1$ .
Hence we get 2 points

$(1,0)$ and

$(-1,0)$ .
Out of these 2, only

$(-1,0)$ lies on the curve.
Hence the required point is

$(-1,0)$ .

The tangent to the curve

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

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

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

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

0%
on the left of $\displaystyle x=c$
0%
on the right of $\displaystyle x=c$
0%
at no point
0%
at all points

Explanation

Equation of straight line joining

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

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

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

Equation of tangent at

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

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

Subtracting (1) from (2),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

$x=c$

For

$a\in \left[ \pi ,2\pi \right]$ and

$n\in Z$ , the critical points of

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

0%
$x=n\pi$
0%
$x=2n\pi$
0%
$x=(2n+1)\pi$
0%
None of these

Explanation

Given,

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

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

At critical points, we must have

$f'(x)=0$

$\Rightarrow \sin { a } \tan ^{ 2 }{ x } +\sin { a } -1=0\left( \because \sec ^{ 2 }{ x } \neq 0\:for\:any\:x\in R \right)$

$\displaystyle \Rightarrow \tan ^{ 2 }{ x } =\dfrac { 1-\sin { a } }{ \sin { a } }$

Since

$\displaystyle a\in \left[ \pi ,2\pi \right] ,\dfrac { 1-\sin { a } }{ \sin { a } } <0$

$\displaystyle \therefore \tan ^{ 2 }{ x } =\dfrac { 1-\sin { a } }{ \sin { a } }$ has no solution in

$R$

$\Rightarrow f(x)$ has no critical points

0%
Assertion is true and Reason is true; Reason is a correct explanation for Assertion.
0%
Assertion is True, Reason is true; Reason is not a correct explanation for Assertion.
0%
Assertion is true, Reason is false
0%
Assertion is false, Reason is true

Explanation

We have

$\displaystyle y={ x }^{ 2 }+bx+c\Rightarrow \dfrac { dy }{ dx } =2x+b$

Since the curve touches the line

$y=x$ at the point

$\left( 1,1 \right)$

${ \left[ 2x+b \right] }_{ \left( 1,1 \right) }^{ }=1\Rightarrow 2+b=1\Rightarrow b=-1$

Also, the curve passes through the point

$\left( 1,1 \right)$

$\therefore 1=1+b+c\Rightarrow c=-b=1$

$\displaystyle \therefore y={ x }^{ 2 }-x+1\Rightarrow \dfrac { dy }{ dx } =2x-1$

Now,

$\displaystyle \dfrac { dy }{ dx } <0\Rightarrow 2x-1<0\Rightarrow x<\dfrac { 1 }{ 2 }$

$\displaystyle y=4x^{2}$ and

$\displaystyle y= x^{2}.$
The two curves

0%
intersect each other
0%
touch each other
0%
do not meet
0%
represent parabola

Explanation

$y=4x^{2}$ and

$y=x^{2}$ represent two parabola concave upward and symmetric about positive y-axis. Both the curves do not intersect, however they touch each other at

$x=0$ .

$y'=8x$ and

$y'=2x$
Now

$8x=2x$
Or

$x=0$
Hence both curves have equal slope at

$x=0$ and are horizontal (parallel to x axis)at

$x=0$

If the normal to the curve

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

$\displaystyle \left ( 3, 4 \right )$ makes an angle

$\displaystyle \frac{3\pi}{4}$ with the positive x-axis then

$\displaystyle f'\left ( 3 \right )$ is equal to

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

Explanation

Slope of normal

$=\dfrac{-1}{f'(x)}$

$=tan(\dfrac{3\pi}{4})$

$=-1$
Hence

$f'(x)=1$
Or

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

$y=x+c$
Hence

$y=f(x)$ is an equation of a straight line parallel to

$y=x$ .
Hence

$f'(x)$ is independent of x and its value is 1.

Find the slopes of the tangents of the curve

$y=(x+1)(x-3)$ at the points where it cuts the X-axis.

0%
$4$
0%
$-4$
0%
$2$
0%
$-2$

Explanation

$f(x)=(x+1)(x-3)$
Now

$f(x)=0$

$\Rightarrow x=-1,x=3$
Differentiating

$f(x)$ with respect to x

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

$=2x-2$

$=2(x-1)$
Now slope of the tangent at

$(h,k)$ will be

$\dfrac{dy}{dx}_{h,k}$
Hence slopes of the tangent at

$x=-1$ and

$x=3$ , will be

$\dfrac{dy}{dx}_{x=-1}=2(x-1)_{x=-1}=-4$
And

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

Find the points on the curve

$y=x^{3}$ , the tangents at which are inclined at an angle of

$60^{\circ}$ to x-axis.

0%
$x=\pm\dfrac{1}{\sqrt{\sqrt{3}}}, y =\dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$ .
0%
$x=\dfrac{1}{\sqrt{\sqrt{3}}}, y =\dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$ .
0%
$x=\pm\dfrac{1}{\sqrt{\sqrt{3}}}, y =\pm \dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$ .
0%
$x=-\dfrac{1}{\sqrt{\sqrt{3}}}, y =\pm \dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$ .

Explanation

Hence slope of tangents

$=tan60^{0}$

$=\sqrt{3}$

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

$\dfrac{dy}{dx}$

$=3x^{2}$

$=\sqrt{3}$
Or

$x^{2}=\dfrac{1}{\sqrt{3}}$
Hence

$x=\pm\dfrac{1}{\sqrt{\sqrt{3}}}$ .
Thus y

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

Find the points on the curve

$y=x/(1-x^{2})$ where the tangents makes an angle of

$\pi /4$ with x-axis

0%
$(\sqrt { 3 } ,-\sqrt { \dfrac { 2 }{ 3 } } ),(-\sqrt { 2 } ,\sqrt { \dfrac { 2 }{ 3 } } )$
0%
$(\sqrt { 3 } ,-\sqrt { \dfrac { 3 }{ 4 } } ),(-\sqrt { 3 } ,\sqrt { \dfrac { 3 }{ 4 } } )$
0%
$(\sqrt { 3 } ,-\sqrt { \dfrac { 3 }{ 2 } } ),(-\sqrt { 3 } ,\sqrt { \dfrac { 3 }{ 2 } } )$
0%
none of these

Explanation

$\dfrac{dy}{dx}=tan45^{0}$
Or

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

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

$1-x^{2}+2x^{2}=(1-x^{2})^{2}$
Or

$1+x^{2}=1-2x^{2}+x^{4}$
Or

$x^{4}-3x^{2}=0$
Or

$x^{2}[x^{2}-3]=0$

$x=0$ or

$x=\pm\sqrt{3}$
Hence

$y=\dfrac{\sqrt{3}}{1-3}=-\dfrac{\sqrt{3}}{2}$

$y=\dfrac{-\sqrt{3}}{1-3}=\dfrac{\sqrt{3}}{2}$

Hence

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

Find the condition that the line

$\displaystyle Ax+By= 1$ may be a normal to the curve

$\displaystyle a^{n-1}y=x^{n}.$

0%
$\displaystyle a^{n}B\left ( B^{2}+nA^{2} \right )^{n}=A^{n}n^{n}.$
0%
$\displaystyle a^{n-1}B\left ( B^{2}+nA^{2} \right )^{n-1}=A^{n}n^{n}.$
0%
$\displaystyle a^{n}B\left ( B^{2}+nA^{2} \right )^{n-1}=A^{n}n^{n}.$
0%
$\displaystyle a^{n-1}B\left ( B^{2}-nA^{2} \right )^{n-1}=A^{n}n^{n}.$

Explanation

Given,

$\displaystyle a^{n-1}y=x^{n}$

$\displaystyle \therefore \dfrac{dy}{dx}=n\dfrac{x^{n-1}}{a^{n-1}}=n\dfrac{x^{n-1}}{a^{n-1}}\cdot \dfrac{1}{x}=n\dfrac{y}{x}.$

$\displaystyle \therefore$ Normal is:

$\displaystyle Y-y=-\dfrac{1}{dy/dx}\left ( X-x \right) =-\dfrac{x}{ny}\left ( X-x \right )$

$\displaystyle \therefore Xx+Yny=ny^{2}+x^{2}.$

Compare with

$AX+BY=1$

$\displaystyle \therefore \dfrac{X}{A}=\dfrac{ny}{B}= \dfrac{ny^{2}+x^{2}}{1}=k,$ say.

$\displaystyle \therefore x= Ak, y=(Bk/n)$ and

$\displaystyle ny^{2}+x^{2}=k$ or

$\displaystyle k^{2}\left [ \left ( B^{2}/n+A^{2} \right ) \right ]=k$

$\displaystyle \therefore k=\dfrac{n}{B^{2}+nA^{2}}$ ..(1)

Now

$\displaystyle a^{n-1}y=x^{n}.$

Put for

$x$ and

$y$ .

$\displaystyle a^{n-1}y\cdot \dfrac{Bk}{n}= A^{n}k^{n}$

$\Rightarrow\displaystyle a^{n-1}B= nA^{n}k^{n-1}$

$\Rightarrow\displaystyle a^{n-1}B=nA^{n}\cdot \left ( \dfrac{n}{B^{2}+nA^{2}} \right )^{n-1},$ by (1)

$\Rightarrow\displaystyle a^{n-1}B\left ( B^{2}+nA^{2} \right )^{n-1}=A^{n}n^{n}.$

Above is the required condition.

If the normal to the curve

$y=f(x)$ at the point

$(3,4)$ makes an angle

$3\pi /4$ with the positive x-axis, then

$f'(3)=$

0%
$-1$
0%
$0$
0%
$1$
0%
$\sqrt 3$

Explanation

Tangent being perpendicular to given line of slope 2, will have its slope as -

$\displaystyle \dfrac{1}{2}$ .

Slope of tangent=

$\displaystyle -\dfrac{fx}{fy}=-\dfrac{6x+1}{2(y+1)}=-\dfrac{1}{2}\therefore y=6x$ .

Sloping with the given curve, we have

$\displaystyle 3x^{2}+36x^{2}+x+12x=0$

$13x(3x+1)=0\therefore x=0, -1/3 \therefore y=0, -2$

Hence the two points are

$\displaystyle (0,0),\left ( -\dfrac{1}{3},-2 \right )$

$\displaystyle \therefore y=-\dfrac{1}{2}x$ and

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

$2y+x=0$ and

$\displaystyle 2y+x+\dfrac{13}{3}=0$

Ans: D

The curve

$\displaystyle y-e^{xy}+x=0$ has a vertical tangent at the point

0%
$(1,\ 1)$
0%
$no\ point$
0%
$(0,\ 1)$
0%
$(1,\ 0)$

Explanation

$\dfrac{dy}{dx}-e^{xy}[x\dfrac{dy}{dx}+y]+1=0$
Or

$\dfrac{dy}{dx}[1-x.e^{xy}]+1-y.e^{xy}=0$
Or

$\dfrac{dy}{dx}=\dfrac{-(1-y.e^{xy})}{1-x.e^{xy}}$
Now

$1-xe^{xy}=0$ since the slope of the tangent is

$90^{0}$ .
Hence

$xe^{xy}=1$
Or

$x(x+y)=1$
Or

$x^{2}+xy=1$
Or

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

$y=\dfrac{1}{x}-x$ ...(i)
Considering

$y=0$ ,

$x^{2}=1$

$x=\pm1$ .
Hence we get 2 points

$(1,0)$ and

$(-1,0)$ .
Out of these 2, only

$(1,0)$ lies on the curve.
Hence the required point is

$(1,0)$ .

The set of all values of x for which the function

$\displaystyle f\left ( x \right )= \left ( k^{2}-3k+2 \right )\left ( \cos ^{2}\frac{x}{4}-\sin ^{2}\frac{x}{4} \right )+\left ( k-1 \right )x+\sin 1$ does not posses critical points is

0%
$\displaystyle \left ( -4,4 \right )$
0%
$\displaystyle \left ( 0,4 \right)$
0%
$\displaystyle \left ( 0,1 \right )\cup \left ( 1,4 \right )$
0%
$\displaystyle \left ( 0,2 \right )\cup \left ( 2,4 \right )$

Explanation

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

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

$=\displaystyle \left ( k-1 \right )\left [ 1-\frac{k-2}{2}\sin \frac{x}{2} \right ]$
Since f(x) does not possess critical points therefore f'(x) is not equal to zero.
i.e.,

$\displaystyle k\neq 1$ or

$\displaystyle 1-\frac{k-2}{2}\sin \frac{x}{2}= 0$ does not posses a solution or

$\displaystyle \sin \frac{x}{2}= \frac{2}{k-2}$ does not have a solution.
Hence we must

have

$\displaystyle \left | \frac{2}{k-2} \right |> 1$ as

$\displaystyle \left | \sin \frac{x}{2} \right |< 1.$
Above implies that

$\displaystyle \left | k-2 \right |^{2}\leq 4$
or

$\displaystyle -2< \left ( k-2 \right )< 2$

$\displaystyle \because x^{2}< a^{2}\Rightarrow \left ( x^{2}-a^{2} \right )= -ive$ or

$\displaystyle -a< x< a$

$\displaystyle \therefore 0 < k< 4$ . Also

$\displaystyle k\neq 1.$

$\displaystyle \therefore k \epsilon \left ( 0,1 \right )\cup \left ( 1,4 \right )$

Determine the intervals of monotonicity of

$\displaystyle f \left ( x \right )= \log \left | x \right |.$

0%
increasing for $x>0$
0%
increasing for $x<0$
0%
decreasing for $x>0$
0%
decreasing for $x<0$

Explanation

$\displaystyle f'\left ( x \right )=\frac{1}{x}, x \neq 0$ , at

$x=0,$

$f(x)$ is not defined.
Hence

$f (x)$ is increasing for

$\displaystyle x> 0$ since

$f'(x) > 0 \forall x \in (0, \infty)$
and

$f(x)$ is decreasing for

$x< 0$ since

$f'(x) < 0 \forall x \in (-\infty, 0)$

$\displaystyle x\cos \alpha +y\sin \alpha =p$ touches

$\displaystyle x^{2}+a^{2}y^{2}=a^{2},$ then

0%
$\displaystyle p^{2}=a^{2}\sin^{2}\alpha +\cos^{2}\alpha$
0%
$\displaystyle p^{2}=a^{2}\cos^{2}\alpha +\sin^{2}\alpha$
0%
$\displaystyle 1/p^{2}=\sin^{2}\alpha +\alpha^{2}\cos^{2}\alpha$
0%
$\displaystyle 1/p^{2}=\cos^{2}\alpha +\alpha^{2}\sin^{2}\alpha$

Explanation

Solving for y, we have

$\displaystyle y=\dfrac { p }{ \sin { \alpha } } -x\cot { \alpha }$

Putting this value in

${ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }={ a }^{ 2 }$ , we have

$\displaystyle { x }^{ 2 }+{ a }^{ 2 }{ \left( \dfrac { p }{ \sin { \alpha } } -x\cot { \alpha } \right) }^{ 2 }={ a }^{ 2 }$

$\displaystyle \Rightarrow { x }^{ 2 }\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) -\dfrac { 2a^{ 2 }xp\cot { \alpha } }{ \sin { \alpha } } +\dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha } } -{ a }^{ 2 }=0$

The discriminant of this equation must be zero. So

$\displaystyle { a }^{ 4 }\dfrac { { p }^{ 2 }\cot ^{ 2 }{ \alpha } }{ \sin ^{ 2 }{ \alpha } } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) \left( \dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha } } -{ a }^{ 2 } \right) \\ \Rightarrow { a }^{ 2 }{ p }^{ 2 }\cot ^{ 2 }{ \alpha } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) \left( { p }^{ 2 }-\sin ^{ 2 }{ \alpha } \right) \\ \Rightarrow { p }^{ 2 }\left( { a }^{ 2 }\cot ^{ 2 }{ \alpha } -1-{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) =\sin ^{ 2 }{ \alpha } -{ a }^{ 2 }\cos ^{ 2 }{ \alpha } \\ \Rightarrow { p }^{ 2 }={ a }^{ 2 }\cos ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \alpha }$

If the line

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

$xy=1$ , then

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

Explanation

Given equation of curve

$xy=1$

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

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{y}{x}$
Slope of tangent at

$P(x_1,y_1)=-\dfrac{1}{x_1^2}$
Slope of normal at P is

$=x^2$ ....(1)

Given equation of normal

$ax+by+c=0$
Slope of the normal at P is

$-\dfrac{a}{b}$ .....(2)

From (1) and (2),

$x^2=-\dfrac{a}{b}$
Since,

$x^2 >0$

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

$\Rightarrow \dfrac{a}{b}<0$
i.e.

$\dfrac{a}{b}$ should be negative.

$\Rightarrow a<0, b>0$ or

$a>0 , b<0$

Find the co-ordinates of the points on the curve

$\displaystyle y= x/\left ( 1+x^{2} \right )$ where the tangent to the curve has greatest slope.

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

Explanation

Given curve

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

Here slope

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

$\displaystyle \Rightarrow S =\dfrac{\left ( 1-x^{2} \right )}{\left ( 1+x^{2} \right )^2}$

Now,

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

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

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

For maximum or minimum of S,

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

$\displaystyle \Rightarrow x= -\sqrt{3}, 0, \sqrt{3}$ .

Now, at

$x=0$ ,

$\dfrac{ds}{dx}$ changes from +ive to -ive

$\displaystyle x= \pm \sqrt{3}$ . it changes from -ive to +ive .

Hence slope S is maximum when x=0 and min. when

$\displaystyle x= \pm \sqrt{3}$ , thus for greatest slope, we have x=0 and y=0.

Hence the required point is (0, 0), that is, the origin.

The line

$y=x$ is a tangent to the parabola

$\displaystyle y= ax^{2}+bx+c$ at the point

$x=1$ .If the parabola passes through the point

$(-1,0)$ , then determine

$a, b, c.$

0%
$\displaystyle a= \frac{1}{2}, b= \frac{1}{4}, c= \frac{1}{3}.$
0%
$\displaystyle a= \frac{1}{4}, b= \frac{1}{2}, c= \frac{1}{4}.$
0%
$\displaystyle a= 2, b= 1, c= 4.$
0%
$\displaystyle a= 4, b= 2, c= 4.$

Explanation

Given equation of parabola is

$\displaystyle y= ax^{2}+bx+c$

$\displaystyle \frac{dy}{dx}= 2ax+b$
Slope of tangent to the curve at

$x=1$ is

$2a+b$
Given tangent is

$y=x$ . Slope of this tangent is

$1.$
So,

$2a+b=1$ ...(1)

Since, the parabola passes through

$(-1,0)$

$\displaystyle \therefore a-b+c= = 0$ ...(2)

Given

$y=x$ is a tangent at

$x=1$

$\displaystyle \therefore y= 1.$
Hence

$(1,1)$ lies both on tangent and parabola

$\displaystyle \therefore a+b+c= 1$ ...(3)

Solving (1), (2) and (3), we get

$\displaystyle a= \frac{1}{4}, b= \frac{1}{2}, c= \frac{1}{4}.$

Find

$\displaystyle \frac{dy}{dx}$ if

$\:y= \left [ x+\sqrt{x+} \sqrt{x}\right ]^{1/2}$ , at

$x=1$

0%
$\dfrac{3+4\sqrt{2}}{8\sqrt{2}(\sqrt{1+\sqrt{2}})}$
0%
Not defined
0%
$0$
0%
$e$

Explanation

$y=[x+\sqrt{x+\sqrt{x}}]^{\frac{1}{2}}$

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

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

$y^{4}-2xy^{2}+x^{2}=x+\sqrt{x}$

Differentiating with respect to x gives us

$4y^{3}y'-2y^{2}-4xyy'+2x=1+\dfrac{1}{2\sqrt{x}}$

At

$x=1$

$4y^{3}_{x=1}y'-2y^{2}_{x=1}-4yy'_{x=1}+2=1+\dfrac{1}{2}$

$4y^{3}_{x=1}y'-2y^{2}_{x=1}-4yy'_{x=1}=\dfrac{-1}{2}$

Now at

$x=1$

$y=\sqrt{1+\sqrt{1+\sqrt{1}}}$

$=\sqrt{1+\sqrt{2}}$

Substituting in the above equation gives us

$4y^{3}_{x=1}y'-2y^{2}_{x=1}-4yy'_{x=1}=\dfrac{-1}{2}$

$4(\sqrt{1+\sqrt{2}})(1+\sqrt{2})y'-2(1+\sqrt{2})-4(\sqrt{1+\sqrt{2}})y'=\dfrac{-1}{2}$

$4(\sqrt{1+\sqrt{2}})y'(1+\sqrt{2}-1)=\dfrac{-1}{2}+2(1+\sqrt{2})$

$4\sqrt{2}(\sqrt{1+\sqrt{2}})y'=\dfrac{3}{2}+2\sqrt{2}$

$8\sqrt{2}(\sqrt{1+\sqrt{2}})y'=3+4\sqrt{2}$

$y'=\dfrac{3+4\sqrt{2}}{8\sqrt{2}(\sqrt{1+\sqrt{2}})}$
Hence

$\dfrac{dy}{dx}_{x=1}=\dfrac{3+4\sqrt{2}}{8\sqrt{2}(\sqrt{1+\sqrt{2}})}$

A and B are points

$(-2,0)$ and

$(1,3)$ on the curve

$\displaystyle y=4-x^{2}$ . If the tangent at P on the curve be parallel to chord AB, then co-ordinates of point P are

0%
$\displaystyle \left ( -\frac{1}{3}, \frac{5}{3} \right )$
0%
$\displaystyle \left ( \frac{1}{2}, -\frac{15}{4} \right )$
0%
$\displaystyle \left ( -\frac{1}{2}, \frac{15}{4} \right )$
0%
$\displaystyle \left ( -\frac{1}{3}, \frac{1}{5} \right )$

Explanation

Given equation of curve

$\displaystyle y=4-x^{2}$ ...(i)

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

Given points

$A(-2,0)$ and

$(1,3)$
Slope of AB

$=\dfrac{3}{3}=1$

$\Rightarrow -2x=1$

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

So, by equation (i), we get

$y=\dfrac{15}{4}$
Hence, the point is

$(-\dfrac{1}{2},\dfrac{15}{4})$

The line

$\dfrac xa+\dfrac yb=1$ touches the curve

$\displaystyle y=be^{-x/a}$ at the point

0%
$(a,b/a)$
0%
$(-a,b/a)$
0%
$(a,a/b)$
0%
None of these

Explanation

Simplifying the equation of the line we get

$bx+ay=ab$

$ay=-bx+ab$
Or

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

$\dfrac{dy}{dx}=\dfrac{-b}{a}$
Or

$-\dfrac{b}{a}.e^{-x/a}.=\dfrac{-b}{a}$
Or

$e^{-x/a}=1$
Or

$x=0$
Hence

$y=b$ .
Therefore the point is

$(0,b)$

If the line,

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

$xy=2,$ then

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

Explanation

$xy'+y=0$
Or

$y'=\dfrac{-y}{x}$
Hence slope of normal

$=\dfrac{x}{y}$

$=\dfrac{2}{y^{2}}$
Hence the slope is always positive.
Now
slope of the line is

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

$a<0$ and

$b>0$
Or

$a>0$ and

$b<0$ .

The function

$\displaystyle f\left ( x \right )=2\log \left ( x-2 \right )-x^{2}+4x+1$ increases in the interval

0%
$\displaystyle \left ( 1, 2 \right )$
0%
$\displaystyle \left (2, 3 \right )$
0%
$\displaystyle \left ( 5/2, 3 \right )$
0%
$\displaystyle \left ( 2, 4 \right )$

Explanation

$f(x) = 2\log(x-2)-x^2+4x+1$

$f'(x) = 2/(x-2)-2x+4 = 2\cdot \dfrac{1-(x-2)^2}{x-2}$

$f'(x) = -2 \cdot \dfrac{(x-1)(x-3)}{x-2}$
For

$f(x)$ to be increasing,

$f'(x) > 0$

$-2 \cdot \dfrac{(x-1)(x-3)}{x-2} > 0$

$\Rightarrow x \epsilon (2,3)$ (since domain of f is

$(2, \infty)$
Hence option B, C are correct.

The critical points of the function

$\displaystyle f\left ( x \right )= \frac{\left | x-1 \right |}{x^{2}}$ are

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

Explanation

for

$x\geq 1$

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

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

for

$x<1$

$f(x)=\dfrac { -x+1 }{ { x }^{ 2 } }$
&

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

for critical points:

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

$\Rightarrow x=0,2$
now,

$f'\left( 1+ \right) =1$ &

$f'\left( 1- \right) =-1$
Since,

$f'(x)$ changes sign at

$x=1$
Therefore,

$x=1$ is also a critical point
Thus critical points of

$f(x)$ are

$x=0,1,2$

Ans: A,B,C

$\displaystyle f\left ( 0 \right )=0$ and

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

$x > 0$ , then

$\displaystyle \frac{f(x)}{x}$

0%
decreases on $\displaystyle \left ( 0, \infty \right )$
0%
increases on $\displaystyle \left ( 0, \infty \right )$
0%
decreases on $\displaystyle \left ( 1, \infty \right )$
0%
neither increases nor decreases on $\displaystyle \left ( 0, \infty \right )$

Explanation

Given,

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

$f''(x)>0$ , it means

$f'(x)$ is also increasing

Let

$g(x)=\cfrac { f\left( x \right) }{ x }$

$g'\left( x \right) =\cfrac { xf'\left( x \right) -f\left( x \right) }{ { x }^{ 2 } }$

$f'\left( x \right)$ is increasing and

$x\epsilon \left( 0,\infty \right)$ thus

$f'\left( x \right) =$ positive

$f(x)=$ positive

${ x }^{ 2 }=$ positive

Therefore,

$g\left( x \right) =$ positive

$>0$

Thus,

$\cfrac { f\left( x \right) }{ x }$ increases for

$x\epsilon \left( 0,\infty \right)$

The interval(s) of decrease of of the function

$\displaystyle f\left ( x \right )= x^{2}\log 27-6x\log 27+\left ( 3x^{2}-18x+24 \right )\log \left ( x^{2}-6x+8 \right )$ is

0%
$\displaystyle \left ( 3-\sqrt{1+1/3e}, 2\right )$
0%
$\displaystyle \left ( 4, 3+\sqrt{1+1/3e}\right )$
0%
$\displaystyle \left ( 3, 4 +\sqrt{1+1/3e}\right )$
0%
none of these

Explanation

$f'\left( x \right) =2x\log27+\left( 6x-18 \right) \log\left( { x }^{ 2 }-6x+8 \right)$

$\displaystyle +\dfrac { \left( { 3x }^{ 2 }-18+24 \right) \left( 2x-6 \right) }{ { x }^{ 2 }-6x+8 }$

$=6\left( x-3 \right) \log\left( 3\left( { x }^{ 2 }-6x+8 \right) e \right)$

For

$f\left( x \right)$ to be define

${ x }^{ 2 }-6x+8>0$

$\Rightarrow x>4$ or

$x<2$

$x>4$ then

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

$\log3\left( { x }^{ 2 }-6x+8 \right) e<0$

$i.e.3\left( { x }^{ 2 }-6x+8 \right) e<1\quad i.e.{ x }^{ 2 }-6x+\left( 8-1/3e \right) <0\\ i.e.\left( x-\left( 3+\sqrt { 1+1/3e } \right) \right) \left( x-\left( 3-\sqrt { 1+1/3e } \right) \right) <0\\ \Leftrightarrow 3-\sqrt { 1+1/3e } <x<3+\sqrt { 1+1/3e }$

Hence

$x\epsilon \left( 4,3+\sqrt { 1+1/3e } \right)$

Similarly if

$x<2$ then

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

$\log3\left( { x }^{ 2 }-6x+8 \right) e>0$

$i.e\quad x<3-\sqrt { 1+1/3e }$ or

$x>3+\sqrt { 1+1/3e }$

Hence

$x\epsilon \left( 3-\sqrt { 1+1/3e } ,2 \right)$

The slope of the tangent to the curve represented by

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

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

$M\left ( 2,-1 \right )$ is

0%
7/6
0%
2/3
0%
3/2
0%
6/7

Explanation

We first determine the value of

$t$ corresponding to the given values ofx and

$y$ . From

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

$t = 2, -5$ , and from

$2t^{2}-2t-5= 2$ we get

$t = 2, -1$ . Hence to the given point there corresponds the value

$t = 2$ . Therefore, the slope of the tangent at

$\left ( 2,-1 \right )$ is

$\displaystyle \left | y' \right |_{t=2}=\left | \dfrac{dy/dt}{dx/dt} \right |_{t=2}\:=\left | \dfrac{4t-2}{2t+3} \right |_{t=2}=\:\dfrac{6}{7}$

The number of critical points of the fuction

$\displaystyle f'\left ( x \right ),$ where

$\displaystyle f'\left ( x \right )= \frac{\left | x-2 \right |}{x^{3}}$ are

0%
$0$
0%
$1$
0%
$3$
0%
$4$

Explanation

$\displaystyle f\left( x \right) =\begin{cases} \dfrac { x-2 }{ { x }^{ 3 } } ,\quad \quad x>1 \\ \dfrac { 2-x }{ { x }^{ 3 } } ,\quad \quad x<1,x\neq 0 \end{cases}$

$\displaystyle f'\left( x \right) =\begin{cases} \dfrac { 2\left( 3-x \right) }{ { x }^{ 4 } } ,\quad x\in \left( -\infty ,0 \right) \cup \left( 0,1 \right) \\ \dfrac { 2\left( x-3 \right) }{ { x }^{ 4 } } ,\quad \quad x\in \left( 1,\infty \right) \end{cases}$

$\displaystyle f''\left( x \right) =\begin{cases} \dfrac { 6\left( x-4 \right) }{ { x }^{ 5 } } ,\quad x\in \left( -\infty ,0 \right) \cup \left( 0,1 \right) \\ \dfrac { 6\left( 4-x \right) }{ { x }^{ 5 } } ,\quad \quad x\in \left( 1,3 \right) \cup \left( 3,\infty \right) \end{cases}$

$f''\left( x \right)$ doesn't exits at

$x=3$

Thus critical point of

$f'\left( x \right)$ is 3.

The value of a for which the function

$\displaystyle f\left ( x \right )= \left ( 4a-3 \right )\left ( x+\log 5 \right )+2\left ( a-7 \right )\cot\left ( x/2 \right )\sin ^{2}\left ( x/2 \right )$ does not possess critical points is

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

Explanation

$f\left( x \right) =\left( 4a-3 \right) \left( x+\log 5 \right) +2\left( a-7 \right) \cot \left( x/2 \right) \sin ^{ 2 } \left( x/2 \right)$

$\Rightarrow f\left( x \right) =\left( 4a-3 \right) \left( x+\log 5 \right) +\left( a-7 \right) \sin { x }$

$f(x)$ posses critical points when

$f'\left( x \right) =\left( 4a-3 \right) +\left( a-7 \right) \cos { x } =0$

$\Rightarrow \cos { x } =-\dfrac { 4a-3 }{ a-7 }$

$\Rightarrow -1\le -\dfrac { 4a-3 }{ a-7 } \le 1$

$\Rightarrow -\dfrac { 4 }{ 3 } \le a\le 2$
Therefore,

$f(x)$ does not have critical points when

$a\in (-\infty,-4/3)\cup (2,\infty)$

Ans: A,D

The critical points of the function

$\displaystyle f\left ( x\right )=\left ( x-2 \right )^{2/3}\left ( 2x+1\right )$ are

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

Explanation

Given,

$f(x)=(x-2)^{2/3}(3x+1)$

$\displaystyle f'\left( x \right) =\dfrac { 2 }{ 3 } { \left( x-2 \right) }^{ \frac { -1 }{ 3 } }\left( 2x+1 \right) +2{ \left( x-2 \right) }^{ \frac { 2 }{ 3 } }=10\left( x-1 \right) { \left( x-2 \right) }^{ \frac { -1 }{ 3 } }$

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

$\Rightarrow x=1$
Also

$f'\left( x \right)$ does not exits at

$x=2$
Hence the critical points are

$x=1,2$

The coordinates of the point on the curve

$\displaystyle \left ( x^{2}+1 \right )\left ( y-3 \right )=x$ where a tangent to the curve has the greatest slope are given by

0%
$\displaystyle \left ( \sqrt{3}, 3+\sqrt{3}/4 \right )$
0%
$\displaystyle \left ( -\sqrt{3}, 3-\sqrt{3}/4 \right )$
0%
$\displaystyle \left ( 0, 3 \right )$
0%
none of these

Explanation

Solving for y the given equation we have

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

Now

$\displaystyle f'\left( x \right) =\dfrac { -2x\left( 3-{ x }^{ 2 } \right) }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } }$

For extremum of

$f\left( x \right)$ , we have

$f'\left( x \right) =0\Rightarrow x=0,x=\pm \sqrt { 3 }$

$x=0,f'\left( x \right)$ changes sign from positive to negative

$\Rightarrow f\left( x \right)$ has maxima at

$x=0$ .

Thus the required point is

$\left( 0,3 \right)$

The angle at which the curve

$y=ke^{kx}$ intersects the

$y$ -axis is

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

Explanation

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

The curve intersects

$y$ -axis at

$\left ( 0,k \right )$

So,

$\displaystyle \dfrac{dy}{dx}|_{\left ( 0,k \right )}=k^{2}$ .

$\theta$ is the angle at which the given

curve intersects the

$y$ -axis then

$\displaystyle \tan \left ( \pi /2-\theta \right )=\dfrac{k^{2}-0}{1+0.k^{2}}=k^{2}$ .

Hence

$\theta =\cot ^{-1}k^{2}$

The lines tangent to the curves

$\displaystyle y^{3}-x^{2}y+5y-2x=0$ and

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

$\displaystyle \theta$ equal to

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

Let

$\displaystyle f\left ( x \right )=x^{3}+ax+b$ with

$\displaystyle a\neq b$ and suppose the tangent lines to the graph of

$f$ at

$x = a$ and

$x = b$ have the same gradient Then the value of

$f (1)$ is equal to

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

Explanation

$f(x)=x^3+ax+b$

$f'(x)=3x^2+a$
Given gradient at

$x=a$ and at

$x=b$ are same

$\Rightarrow 3a^2 +a=3b^2+a\Rightarrow b^2=a^2$
But given

$a\neq b$

$\Rightarrow a+b=0 ..(1)$
Hence

$f(1)=1+a+b=1$ using (1)

A curve with equation of the form

$\displaystyle y=ax^{4}+bx^{3}+cx+d$ has zero gradient at the point (0, 1) and also touches the x-axis at the point (-1, 0) then the values of x for which the curve has a negative gradient are

0%
x > -1
0%
x < 1
0%
x < -1
0%
$\displaystyle -1\leq \times \leq 1$

Explanation

Given,

$\displaystyle y=ax^{4}+bx^{3}+cx+d$

$\Rightarrow y'=4ax^3+3bx^2+c$
Using given conditions,

$y(0)=1\Rightarrow d=1$

$y'(0)=0\Rightarrow c=0$

$y(-1)=0\Rightarrow a-b=-1 ..(1)$
and

$y'(-1)=0\Rightarrow 4a-3b=0 ..(2)$
Solving equation (1) and (2) we get,

$a=3,b=4$
Hence the polynomial is,

$y=3x^4+4x^3+1$

$y'=12x^2(1+x)$
Now for negative gradient

$y' < 0\Rightarrow 12x^2(1+x)< 0$

$\Rightarrow x< -1$

$f'(x) = g(x)\left ( x-a \right )^{2}$ , where

$g(a)\neq 0$ and

$g$ is continuous at

$x = a$ then

0%
$f$ is increasing near a if $g(a) > 0$
0%
$f$ is increasing near a if $g(a) < 0$
0%
$f$ is decreasing near a if $g(a) > 0$
0%
$f$ is decreasing near a if $g(a) < 0$

Explanation

Since

$g$ is continuous at

$x = a$
if

$g(a) > 0$ , there exist an open interval

$I$ containing

$a$ so that

$g(x) > 0 \forall x\in I$

$\Rightarrow {f}'(x)\geq 0\forall x\in I$ . Therefore,

$f$ is increasing near

$a$ .
Similarly

$f$ is decreasing near

$a$ if

$g(a) < 0$ .

Ans: A,D

The curve

$y= ax^{3}+bx^{2}+cx+8$ touches

$x$ -axis at

$P\left ( -2,0 \right )$ and cuts the

$y$ -axis at a point

$Q(0,8)$ where its gradient isThe values of

$a$ ,

$b$ ,

$c$ are respectively

0%
$-\displaystyle \frac{1}{2},-\frac{3}{4},3$
0%
$\displaystyle 3, -\frac{1}{2},-4$
0%
$\displaystyle -\frac{1}{2},-\frac{7}{4},2$
0%
none of these

Explanation

Given,

$y= a x^3+b x^2+cx+8$

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

Since the curve touches

$x$ -axis at

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

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

$\left ( i \right )$

The curve cut the

$y$ -axis at

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

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

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

$\left ( ii \right )$

$a = -1/4$ ,

$b =0$

Suppose

$f'(x)$ exists for each

$x$ and

$h(x) = f(x) - (f(x))^{2}+(f(x))^{3}$ for every real number

$x$ . Then

0%
$h$ is increasing whenever f is increasing
0%
$h$ is increasing whenever f is decreasing
0%
$h$ is decreasing whenever f is decreasing
0%
nothing can be said in general.

Explanation

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

${h}'(x)={f}'(x)-2{f}'(x)f(x)+3{f}'(x)f(x)^{2}$

$\displaystyle =3{f}'(x)\left [ f(x)^{2}-\frac{2}{3}f(x)+\frac{1}{3} \right ]$

$=3{f}'(x)\left [ \left ( f(x)-1/3 \right )^{2}+2/9 \right ]$

Thus,

${h}'(x)>0$ if

${f}'(x)>0$
and

${h}'(x)<0$ if

${f}'(x)<0$
Therefore,

$h$ increases whenever

$f$ increases
and

$h$ decreases whenever

$f$ decreases

Ans: A,C

The critical points of the function

$f\left( x \right)={ \left( x-2 \right) }^{ 2/3 }\left( 2x+1 \right)$ are

0%
$1$ and $2$
0%
$1$ and $\displaystyle-\frac{1}{2}$
0%
$-1$ and $2$
0%
$1$

Explanation

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

$\displaystyle f'\left( x \right) =\frac { 2 }{ 3 } { \left( x-2 \right) }^{ -1/3 }\left( 2x+1 \right) { \left( x-2 \right) }^{ 2/3 }=0.2$

Clearly

$f'(x)$ is not defined at

$x=2$ .

$\therefore x=2$ is a critical point.

Another critical point is given by

$f'(x)=0,$

i.e.,

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

$\displaystyle \Rightarrow \frac { 2 }{ 3 } \left( 2x+1 \right) +2\left( x-2 \right) =0$

$\Rightarrow 4x+2+6x-12=0$

$\Rightarrow x=1$

The graph a function

$f$ is given. On what interval is

$f$ increasing ?

0%
$(-1, 3]$
0%
$(-3,1)$
0%
$(-3,1]$
0%
none of these

Explanation

A function

$f(x)$ is said to be increasing if as

$x$ increases

$f(x)$ increases as well
It can be Clearly observed from the graph that for the interval

$(-1,3]$ function f is increasing

The points of contact of the vertical tangents

$x= 2-3\sin \theta$ ,

$y= 3+2\cos \theta$ are

0%
$\left ( 2,5 \right ),\left ( 2,1 \right )$
0%
$\left ( -1,3 \right ),\left ( 5,3 \right )$
0%
$\left ( 2,5 \right ),\left ( 5,3 \right )$
0%
$\left ( -1,3 \right ),\left ( 2,1 \right )$

Explanation

For the tangents to be vertical,

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

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

$\dfrac{dx}{d\theta}=0$
Or

$-3\cos\theta=0$
Or

$\theta=\dfrac{2n-1}{2}\pi$
Hence

$\theta=\dfrac{\pi}{2},\dfrac{3\pi}{2}$ .
Now

$x_{\tfrac{\pi}{2}}$

$=-1$

$y_{\tfrac{\pi}{2}}$

$=3$ .
Similarly

$x_{\tfrac{3\pi}{2}}=5$

$y_{\tfrac{3\pi}{2}}=3$ .
Hence the points are

$(-1,3)$ and

$(3,5)$ .

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 4 - 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 4 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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