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

Given

$P(x)=x^4+ax^3+bx^2+cx+d$ such that

$x=0$ is the only real root of

$P(x)=0$ . If

$P(-1) < P(1)$ , then in the interval

$[-1, 1]$ .

0%
$P(-1)$ is the minimum and $P(1)$ is the maximum of P
0%
$P(-1)$ is not minimum but $P(1)$ is the maximum of P
0%
$P(-1)$ is the minimum and $P(1)$ is not the maximum of P
0%
Neither $P(-1)$ is the minimum not $P(1)$ is the maximum of P

Find the slope of tangent of the curve

$x = a\,{\sin ^3}t,y = b\,\,{\cos ^3}t$ at

$t = \frac{\pi }{2}$

0%
$cott$
0%
$-tant$
0%
$-cott$
0%
$\text{not defined at}$ $\frac{\pi}{2}$

Explanation

$x=asin^3t$

$dx=a3sin^2t(cost)dt$ …(1)

$y=bcos^3t$

$dy=b3cos^2t(-sint)dt$ ..(2)

$t=\dfrac{\Pi}{2}$

$\dfrac{dx}{dt}=0$ and hence

$\dfrac{dy}{dx}=\text{not defined}$

The interval in which

$y = x^{2} e^{-x}$ is increasing is

0%
$(-\infty, \infty)$
0%
$(-2, 0)$
0%
$(2, \infty)$
0%
$(0, 2)$

Explanation

$\textbf{Step -1: Find values of x at which the given function changes its nature.}$

$\text{The given function is }y(x)=x^2e^{-x}$

$\dfrac{dy}{dx}=x^2\dfrac{d}{dx}e^{-x}+e^{-x}\dfrac{d}{dx}x^2$

$=-x^2e^{-x}+2xe^{-x}$

$\text{Put }\dfrac{dy}{dx}=0.$

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

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

$\because e^{-x}>0$

$\therefore\text{Either }x=0\text{ or }2-x=0$

$\Rightarrow x=0\text{ or }2$

$\textbf{Step -2: Find the interval in which the given function is increasing.}$

$\text{The points }x=0\text{ and }x=2\text{ divide the real line into three disjoint intervals,}$

$\text{i.e., }(-\infty,0),(0,2)\text{ and }(2,\infty)$

$\because\text{In the interval }(0,2),\dfrac{dy}{dx}>0.$

$\therefore \text{In }(0,2),\text{ the given function is increasing.}$

$\textbf{Hence, The correct option is D.}$

The set of all values of

$a$ for which

$f\left( x \right) =\left( { a }^{ 2 }-3a+2 \right) \left( \cos ^{ 2 }{ \dfrac { x }{ 4 } -\sin ^{ 2 }{ \dfrac { x }{ 4 } } } \right) +\left( a-1 \right) x+\sin { 1 }$ does not possess critical points is

0%
$(1, \infty)$
0%
$(-2, 4)$
0%
$(1, 3)\cup (3, 5)$
0%
$(-\infty, 1)\cup (1, 4)$

Let

$f(x) = \underset{x}{\overset{x + \dfrac{\pi}{3}}{\int}} |\sin \, \theta | \, d \theta \, \, (x \in [0, \pi])$

0%
$f(x)$ is strictly increasing in this interval
0%
$f(x)$ is differentiable in this interval
0%
Range of $f(x)$ is $[2 - \sqrt{3} , 1]$
0%
$f(x)$ has a maxima at $x = \dfrac{\pi}{3}$

The slope of the tangent to the curve

${r^2} = {a^2}\cos 2\theta$ , where

$x = r\cos \theta ,y = r\sin \theta$ , at the point

$\theta=\frac{\pi}{6}$ is

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

The line

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

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

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

$m$ is the slope of a tangent to the curve

$e^{y}=1+x^{2}$ , then

$m$ belongs to the interval

0%
$[-1, 1]$
0%
$[-2, -1]$
0%
$[1, 2]$
0%
$[1, 3]$

$f(x)=\dfrac{x^2}{2-2cos x} ; \ g(x)=\dfrac{x^2}{6x-6sin x}$ where

$0< \times < 1$ , then

0%
both $'f'$ and $'g'$ are increasing functions
0%
$'f'$ is decreasing & $'g'$ is increasing function
0%
$'f'$ increasing functions & $'g'$ is decreasing function
0%
both $'f'$ & $'g'$ are decreasing function

A tangent drawn to the curve

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

$P\left( {x,y} \right)$
cuts the x and y axes at A and B, respectively, such that

$AP:PB = 1:3$ . If

$f\left( 1 \right) = 1$ then the curve passes through

$\left( {k,\frac{1}{8}} \right)$ where

$k$ is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Equation of tangent at

$P(x,y)$ is

$Y - y = \dfrac{dy}{dt}(X-x)$

It meets x - axis at

$A$

$A = \left(x-\dfrac{dx}{dy}y, 0\right)$

and y -axis at

$B$

$B = (0, y - \dfrac{dy}{dt})$

And also given

$\dfrac{AP}{PB} = 1 : 3$

Using section formula

$\left[\dfrac{mx_1+m_2x_2}{m_1m+m_2}, \dfrac{m_1y_1+m_2y_2}{m_1+m_2}\right]$

$\dfrac{3}{4} \left(x - \dfrac{dx}{dy} y\right)v = \dfrac{1}{4} \left(y - x \dfrac{dy}{dx}\right) = (x,y)$

$\dfrac{3}{4} \left( x-y \dfrac{dx}{dy}\right) = x$ ;

$\dfrac{1}{4} \left(y - x \dfrac{dy}{x} \right) = y$

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

$\log(x^3y) =$ const (on integrative)

Given

$f(1) = 1$ , so here

$cons \, t = 1$

$x^3y = 1$ (k, 1/8)

$k^3 = 8$

$k = 2$

1331071_1107146_ans_c4d4706b3f1e499a93282a153729ec90.png

On the curve

${x}^{3} = 12y$ , then the interval at which the abscissa changes at a faster rate than the ordinate ?

0%
$x\in \left( -2,2 \right)$
0%
$x\in \left( -2,2 \right) -\left\{ 0 \right\}$
0%
$x\in \left( -3,3 \right) -\left\{ 0 \right\}$
0%
None of these

Explanation

$\left| dy \right| >\left| dx \right|$

$\left| \dfrac { dy }{ dx } \right| >1$

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

diff wrto y

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

$\dfrac { dx }{ dy } =\dfrac { 12 }{ { 3x }^{ 2 } } =\dfrac { 4 }{ { x }^{ 2 } }$

$\left| \dfrac { dy }{ dx } \right| >1$

$\left| \dfrac { y }{ { x }^{ 2 } } \right| >1$

$\left| { x }^{ 2 } \right| >4$

$x>\pm 2$

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

The point on the curve

$y = b e^{\dfrac {-x}{a}}$ at which the tangent drawn is

$\dfrac {x}{a} + \dfrac {y}{b} = 1$ is

0%
$(0, b)$
0%
$\left (a, \dfrac {1}{e}\right )$
0%
$(0, 1)$
0%
$(1, 0)$

Let

$f ( x ) = \frac { 1 } { 1 + x ^ { 2 } }$ . Let

$m$ be the slope,

$'a'$ be the

$x$ -intercept and

$'b'$ be the

$y$ -intercept of tangent to

$y = f ( x )$ .Abscissa of point of contact of the tangent for which

$'m'$ is greatest is:

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

$V$ is the set of points on the curve

$y^{3} - 3xy +2 = 0$ where the tangent is vertical then

$V =$ .

0%
$\phi$
0%
$\left \{(1 , 0)\right \}$
0%
$\left \{(1, 1)\right \}$
0%
$\left \{(0, 0), (1, 1)\right \}$

Paraboals

$(y-\alpha )^{2}=4a(x-\beta )and (y-\alpha )^{2}=4a'(x-\beta ')$ will have a common normal (other than the normal passing through vertex ofparabola)if:

0%
$\frac{2(a-a')}{\beta '-\beta }< 1$
0%
$\frac{2(a-a')}{\beta -\beta' }< 1$
0%
$\frac{2(a'-a)}{\beta -\beta' }< 1$
0%
$\frac{2(a'a)}{\beta -\beta' }> 1$

The curve

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

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

The slope of the curve

$y=\sin { x } +\cos ^{ 2 }{ x }$ is zero at a point , whose x-coordinate can be ?

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

If the curves

$\dfrac {x^{2}}{a^{2}} + \dfrac {y^{2}}{4} = 1$ and

$y^{3} = 16x$ intersect at right angles, then

$a^{2}$ is equal to

0%
$5/3$
0%
$4/3$
0%
$6/11$
0%
$3/2$

Explanation

$y^{3}=16x$

$3y^{2}\dfrac{dy}{dx}=16$

$\dfrac{dy}{dx}=m_{1}=\dfrac{16}{3y^{2}}$

$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{4}=1$

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

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

$m_{1}.m_{2}=-1$

$\dfrac{16}{3y^{2}}\left(\dfrac{-4x}{a^{2}y}\right)=-1$

$\dfrac{16\times 4\times x}{3a^{2}y^{3}}=1\Rightarrow 64x=3a^{2}y^{3}$

$a^{2}=\dfrac{64 x}{3y^{3}}$ Now

$y^{3}=16 x$

$a^{2}=\dfrac{64 x}{3\times 16 x}\Rightarrow a^{2}=\dfrac{4}{3}$

If the slope of one of the lines given by

${a^2}{x^2} + 2hxy+by^2 = 0$ be three times of the other , then h is equal to

0%
$2\sqrt 3 ab$
0%
$-2\sqrt 3 ab$
0%
$\frac{2}{{\sqrt 3 }}ab$
0%
$-\frac{2}{{\sqrt 3 }}ab$

If for a curve represented parametrically by

$x={ sec }^{ 2 }t,\quad y=cot\quad t\quad$ , the tangent at a point

$P(t=\frac { \pi }{ 4 } )$ meets the curve again at the point Q, then

$\begin{vmatrix} PQ \end{vmatrix}$ is equal to

0%
$\frac { 2\sqrt { 5 } }{ 3 }$
0%
$\frac { 3\sqrt { 5 } }{ 2 }$
0%
$\frac { 5\sqrt { 3 } }{ 3 }$
0%
$\frac { 5\sqrt { 5 } }{ 4 }$

$\dfrac { d } { d x } \left( \sin ^ { 5 } x \cdot \sin 5 x \right) =$

0%
$\sin ^ { 4 } x \cdot \sin 5 x$
0%
$5\sin ^ { 4 } x \cdot \sin 6 x$
0%
$5\sin ^ { 4 } x \cdot \sin 5 x$
0%
$- 5 \sin ^ { 4 } x \cdot \sin 6 x$

Explanation

$\frac { d }{ dx } \left( { sin }^{ 5 }x.sin5x \right)$

$={ sin }^{ 5 }x\frac { d }{ dx } \left( sin5x \right) +sin5x\frac { d }{ dx } \left( { sin }^{ 5 }x \right)$

$={ sin }^{ 5 }x\left[ 5cos5x \right] +sin5x\left[ 5{ sin }^{ 4 }x.cosx \right]$

$=5{ sin }^{ 5 }xcos5x+5sin5x{ sin }^{ 4 }x.cosx$

$=5{ sin }^{ 4 }x\left[ sinx.cos5x+cosx.sin5x \right]$

$=5{ sin }^{ 4 }x\left[ sin\left( x+5x \right) \right]$

$=5{ sin }^{ 4 }xsin\left( 6x \right)$

Let f and g be non-increasing and non-decreasing functions respectively from

$[0,\infty ]$ vto

$[0,\infty ]$ and

$h(x)=f(g(x)),h(0)=0$ , then in

$[0,\infty ]$ ,

$h(x)-h(1)$ is

0%
<0
0%
>0
0%
=0
0%
none of these

Let

$f$ and

$g$ be differentiable function satisfying

$g'(a)=2,g(a)=b$ and

$fog=I$ (identity function). Then

$f'(b)$ is equal to

0%
$\frac{1}{2}$
0%
$2$
0%
$\frac{2}{3}$
0%
None of these

The function

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

0%
One-One in $\left( { - \infty ,1 }\right)$
0%
$\left( {0,\infty} \right)$
0%
$\left( {0,1} \right)$
0%
$\left( { - \infty ,0 }\right)$

$y = 6\tan \,x\left( {{e^x} - x - 1} \right) - 3{x^3} - {x^4} - \frac{5}{4}{x^5},\,$ if

${n^{th}}$ derivative at x=0 is non zero then least value of n is

0%
3
0%
4
0%
5
0%
6

$(x+{ y }^{ 3 })\dfrac { dy }{ dx }$ =y and y(0)=then sum of all possible value(s) of y(1) is ________________.

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

The function

$f(x)=x-ln|2x+1|,x\epsilon \left(-100,\dfrac{-1}{2}\right)\cup \left(\dfrac{-1}{2},\dfrac{1}{2}\right)$ is decreasing in interval

0%
$\left(\dfrac{-1}{2},\dfrac{1}{2}\right)$
0%
$\left(-100,\dfrac{-1}{2}\right)$
0%
$\left(\dfrac{-1}{2},0\right)$ only
0%
$\left(0,\dfrac{1}{2}\right)$ only

Explanation

Given

$f\left( x \right) =x-log\left| 2x+1 \right|$

Differentiate function w.r.t. x, we get,

${ f }^{ \prime }\left( x \right) =\frac { d }{ dx } \left[ x-log\left| 2x+1 \right| \right]$

$\therefore { f }^{ \prime }\left( x \right) =1-\frac { 1 }{ 2x+1 } \times \frac { d }{ dx } \left( 2x+1 \right)$

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

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

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

For function to be decreasing,

${ f }^{ \prime }\left( x \right) <0$

$\therefore \frac { 2x-1 }{ 2x+1 } <0$

Case 1)

$2x-1<0$ and

$2x+1>0$

$\therefore 2x<1$ and

$2x>-1$

$\therefore x<\frac { 1 }{ 2 }$ and

$x>-\frac { 1 }{ 2 }$

Thus, function is decreasing in the interval

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

Case 2)

$2x-1>0$ and

$2x+1<0$

$\therefore 2x>1$ and

$2x<-1$

$\therefore x>\frac { 1 }{ 2 }$ and

$x<-\frac { 1 }{ 2 }$

Satisfying both the conditions simultaneously is not possible.

For what values of a ,

$f(x) = -x^3 +4ax^2 +2x-5$ is decreasing

$\forall$ x

0%
$(1, 2)$
0%
$(3, 4)$
0%
R
0%
no value of a

Explanation

Given

$y=f\left( x \right) =-{ x }^{ 3 }+4a{ x }^{ 2 }+2x-5$

For continuously decreasing function,

$\frac { dy }{ dx } <0$

$\therefore \frac { d }{ dx } \left[ -{ x }^{ 3 }+4a{ x }^{ 2 }+2x-5 \right] <0$

$\therefore -3{ x }^{ 2 }+4a\left( 2x \right) +2<0$

$\therefore -3{ x }^{ 2 }+8ax+2<0$

$\therefore 3{ x }^{ 2 }-8ax-2>0$

$\therefore \frac { -\left( -8a \right) \pm \sqrt { { \left( -8a \right) }^{ 2 }-4\times 3\times \left( -2 \right) } }{ 2\times 3 } >0$

$\therefore \frac { 8a\pm \sqrt { 64{ a }^{ 2 }+24 } }{ 6 } >0$

$\therefore 8a\pm \sqrt { 64{ a }^{ 2 }+24 } >0$

Consider first equation

$8a+\sqrt { 64{ a }^{ 2 }+24 } >0$

$\therefore \sqrt { 64{ a }^{ 2 }+24 } >-8a$

Squaring both sides, we get,

$\therefore 64{ a }^{ 2 }+24>64{ a }^{ 2 }$

$\therefore 24>0$

We don't get value of a in this condition.

Consider second equation

$8a-\sqrt { 64{ a }^{ 2 }+24 } >0$

$\therefore 8a>\sqrt { 64{ a }^{ 2 }+24 }$

Squaring both sides, we get,

$\therefore 64{ a }^{ 2 }>64{ a }^{ 2 }+24$

$\therefore 0>24$

This condition is not possible.

Thus, option (D) is correct.

Equation of the tangent line at

$y=\dfrac{a}{4}$ to the curve

$y\left( { x }^{ 2 }+{ a }^{ 2 } \right) ={ ax }^{ 2 }$ .

0%
$8y=3\sqrt { 3 }x -a$
0%
$8y=3\sqrt { 3 }x -5a$
0%
$8y=3\sqrt { 3 }x +a$
0%
None

Which of the following is not always correct for the function

$f(x)$ and

$g(x)$ these are inverse to each other.

0%
If $f(x)$ is an increasing function, then $g(x)$ will also be increasing.
0%
If $f(x)$ is an odd function, then $g(x)$ will also be an odd function.
0%
Tangent at $(\alpha,\ \beta )$ to $f(x)$ is parallel to tangent at $(\beta,\ \alpha)$ to $g(x)$
0%
Tangent at $(\alpha,\ \beta )$ to $f(x)$ and tangent at is parallel to $(\beta,\ \alpha)$ to $g(x)$ forms complementary angles with x-axis

The increasing function in

$(0,\ \pi /4)$ is

0%
$\cos x+\sin x$
0%
$\cos x-\sin x$
0%
$\dfrac {\sin x}{x}$
0%
$\dfrac {x}{\sin x}$

Let

$f :\left[ 2,4 \right] \rightarrow \left[ 3,5 \right]$ be a bijective decreasing function, then find

$\int _{ 2 }^{ 4 }{ f(t)dt- } \int _{ 3 }^{ 5 }{ { f }^{ -1 }(t)dt. }$

0%
2
0%
14
0%
0
0%
10/3

The increasing function in

$\left(0,\pi/4\right)$ is

0%
$\cos x+\sin x$
0%
$\cos x-\sin x$
0%
$\dfrac{\sin x}{x}$
0%
$\dfrac{x}{\sin x}$

$f(x)=\cos x+a^{2}x+b$ is an increasing function for all values of

$x$ , then the value which

$'a'$ can take.

0%
$a\in [-1,1]$
0%
$a\in(-\infty,-1]\cup [1,\infty)$
0%
$a\in [-1,\infty)$
0%
$a\in(-\infty,1]$

Which of the following statements is/are correct ?

0%
x + sinx is increasing function
0%
tanx is an increasing function
0%
x + sinx is decreasing function
0%
sec x is an increasing function

$y = \log _ { \sin x } ( \tan x ) ,$ then

$\frac { d y } { d x }$ at

$x = \frac { \pi } { 4 }$ is:

0%
$\frac { 4 } { \log 2 }$
0%
$-\frac { 4 } { \log 2 }$
0%
$\frac { 1 } { \log 2 }$
0%
none of these

$f(x)$ is differentiable function satisfying the relation

$f(x)=x^{2}+\displaystyle \int^{x}_{0}e^{-t}f(x-t)dt$ , then

$\displaystyle \sum^{9}_{k=1}f(k)$ equals

0%
$1060$
0%
$1260$
0%
$960$
0%
$1224$

$f(x)=\frac{x}{log x}-\frac{log}{x}$ is increasing in

0%
$(e,\infty )$
0%
$(0,1)\epsilon (1,e)$
0%
(0,1)
0%
(1,e)

Solve it:-

$y = x + \dfrac{1}{x},$

0%
$x=1$ is a point of local maximum
0%
$x=-1$ is a point of local minimum
0%
Local maximum value > Local minimum value
0%
Local maximum value < Local minimum value

In the interval

$\left( {7,\infty } \right),f(x) = \left| {x - 5} \right| + 2\left| {x - 7} \right|$ is

0%
Increasing
0%
Decreasing
0%
Constant
0%
Cannot be estimated

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

$g(x)=\dfrac {x}{\tan x}$ where

$0 < x < 1$ , then in this interval

0%
$g(x)$ are decreasing 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

$\theta$ is angle of intersection between

$y=10-x^{2}$ and

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

$|\tan \theta|$ is-

0%
$\dfrac {5\sqrt {3}}{11}$
0%
$\dfrac {7\sqrt {3}}{15}$
0%
$\dfrac {4\sqrt {3}}{11}$
0%
$none$

Explanation

Given:

$y=10-{x}^{2}$ ......

$(1)$

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

$(2)$

On solving,

$10-{x}^{2}=4+{x}^{2}$

$\Rightarrow 2{x}^{2}=6$

$\Rightarrow {x}^{2}=\dfrac{6}{2}=3$

$\therefore x=\pm \sqrt{3}$ from

$(1)$

$\Rightarrow y=10-{\left(\pm\sqrt{3}\right)}^{2}=10-3=7$

Consider

$y=10-{x}^{2}$

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

${m}_{1}=\left[\dfrac{dy}{dx}\right]_{\left(\sqrt{3},7\right)}=-2\times\sqrt{3}=-2\sqrt{3}$

Consider

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

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

${m}_{2}=\left[\dfrac{dy}{dx}\right]_{\left(\sqrt{3},7\right)}=2\times\sqrt{3}=2\sqrt{3}$

$=\left|\dfrac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}\right|$

$\tan{\theta}=\left|\dfrac{-2\sqrt{3}-2\sqrt{3}}{1+\left(-2\sqrt{3}\right)\left(2\sqrt{3}\right)}\right|$

$=\left|\dfrac{-4\sqrt{3}}{1-12}\right|$

$=\dfrac{4\sqrt{3}}{11}$

$\therefore \tan{\theta}=\dfrac{4\sqrt{3}}{11}$

The function

$f(x)=\sqrt{25-4x^{2}}$ is increasing in

0%
(-3,0)
0%
(4,0)
0%
(-5/20)
0%
R

The function log (log x) increases in

0%
$(1,\infty )$
0%
$(0,\infty )$
0%
$\infty$
0%
R

The function f defined by

$f(x)=(x+2)e^{-x}$ si

0%
decreasing for all x
0%
decreasing in $(-\infty ,-1)$ and increasing in $(1,\infty )$
0%
increasing for all x
0%
decreasing in $(-1,-\infty )$ and increasing in $(-\infty ,-1)$

The function

$\frac{ln(1+x)}{x}$ in

$(0,\infty )$ is

0%
increasing
0%
decreasing
0%
not decreasing
0%
not increasing

Let

$f(x)=\left\{\begin{matrix} max \{|x|, x^2\}, & |x|\leq 2\\ 8-2|x|, & 2 < |x|\leq 4\end{matrix}\right.$
Let S be the set of points in the interval

$(-4, 4)$ at which f is not differentiable. Then S?

0%
Is an empty set
0%
Equals $\{-2, -1, 1, 2\}$
0%
Equals $\{-2, -1, 0, 1, 2\}$
0%
Equals $\{-2, 2\}$

If the subnormal to the curve

${ x }^{ 2 }.{ y }^{ n }={ a }^{ 2 }$ is a constant then n=

0%
$-4$
0%
$-3$
0%
$-2$
0%
$-1$

Explanation

Given,

$x^2y^n=a^2$

$\Rightarrow x=\sqrt{\dfrac{a^2}{y^n}}$

$\Rightarrow 2xy^n+x^2ny^{n-1}\dfrac{dy}{dx}=0$

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{2y}{nx}$

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{2y}{n\sqrt{\frac{a^2}{y^n}}}$

$\therefore \dfrac{dy}{dx}=-\dfrac{2}{na}y^{\frac{n+2}{2}}$

Subnormal is given by,

$SN=y \dfrac{dy}{dx}$

$=y\left ( -\dfrac{2}{na}y^{\frac{n+2}{2}} \right )$

$\therefore SN=-\dfrac{2}{na}y^{\frac{n+4}{2}}$

$\therefore n=-4$

Let

$f(x)$ be a function satisfying

$f'(x)=f(x)$ with

$f(0)=1$ and g be the function satisfying

$f(x)+g(x)=x^2$ the value of the integral

$\displaystyle\int^1_0f(x)g(x)dx$ is?

0%
$\dfrac{1}{4}(e-7)$
0%
$\dfrac{1}{4}(e-2)$
0%
$\dfrac{1}{2}(e-3)$
0%
None of these

Explanation

Given:

$f'(x)=f(x)$ , with

$f(0)=1$ and

$f(x)+g(x)=x^2$

To find: Value of

$\displaystyle \int'_0 f(x)g(x)dx$ .

Solution :

$f'(x)=f^0(x)=\dfrac{f'(x)}{f(x)}=1$

$\displaystyle \therefore \int \dfrac{f'(x)}{f(x)}dx=\int dx\Rightarrow \log |f(x)|=x+c$

$\Rightarrow f(x)=e^{x+c}$

$\therefore f(0)=e^e=1$

$\Rightarrow e=0$

$\therefore f(x)=e^x$

Now,

$f(x)+g(x)=x^2$

$\Rightarrow g(x)=x^2-f(x)=x^2-e^x$

$\displaystyle \therefore \int^1_0 f(x) g(x)dx=\int^1_0 e^x(x^2-e^x)dx$

$\displaystyle =\int^1_0 e^x x^2 dx-\int^1_0 e^{2x}dx=I$ (say)

Let

$\displaystyle I_1=\int e^x x^2 dx=x^2 \int e^x dx-\int [\dfrac{d}{dx}x^2\int e^x dx]dx$

$\displaystyle =x^2e^x-2\int xe^x dx+c_1$

$\displaystyle =x^2e^x-2x\int e^x dx+2\int [\dfrac{d}{dx} x.\int e^x dx]dx+c_1$

$\displaystyle =x^2e^x-2xe^x+2\int e^x dx+c_2$

$=x^2e^x-2xe^x+2e^x+c_3$

$=e^x[x^2-2x+2]+c_3$

$\therefore I=\left[e^x(x^2-2x+2)-\dfrac{e^{2x}}{2}\right]^1_0$

$=(e^1(1-2+2)-\dfrac{e^2}{2})-(e^0(0-0+2-\dfrac{1}{2}))$

$=e-\dfrac{e^2}{2}-\dfrac{3}{2}=\dfrac{1}{4}(4e-e^2-3)$

Hence the answer is (D) None of these.

Define

$f(x) = \dfrac{1}{2} [ |\sin x| + \sin x], 0 < x \le 2\pi$ . The

$f$ is

0%
increasing in $\left(\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right)$
0%
decreasing in $\left(0, \dfrac{\pi}{2}\right)$ and increasing in $\left(\dfrac{\pi}{2}, \pi\right)$
0%
increasing in $\left(0, \dfrac{\pi}{2}\right)$ and decreasing in $\left(\dfrac{\pi}{2}, \pi\right)$
0%
increasing in $\left(0, \dfrac{\pi}{4}\right)$ and decreasing in $\left(\dfrac{\pi}{4}, \pi\right)$

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

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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