MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 10

The equation of the curves through the point

$(1,0)$ and whose slope is

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

0%
$(y - 1)(x + 1) + 2x = 0$
0%
$2x(y - 1) + x + 1 = 0$
0%
$x(y - 1)(x + 1) + 2 = 0$
0%
None of these

Explanation

Slope

$= \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = \dfrac{y - 1}{x^{2} + x}$

$\implies \dfrac{dy}{y-1} = \dfrac{dx}{x^{2} + x}$

$\implies \int\dfrac{1}{y-1}dy = \int(\dfrac{1}{x} - \dfrac{1}{x+1})dx + C$

$\implies \log(y-1)=\log(\dfrac{x}{x+1})+\log{c}$

$\implies \dfrac{(y-1)(x+1)}{x} = k$
Putting

$x=1, y=0$ , we get

$k=-2$
The equation is

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

The function f(x) is

0%
increasing for all x
0%
non-monotonic
0%
decreasing for all x
0%
None of these

Explanation

We have the equations of the tangents of the curve

$y=\displaystyle \int_{-∞}^{x} f(t)dt$ and

$y=f(x)$ at arbitrary points on them are

$Y-\displaystyle \int_{-∞}^{x} f(t)dt=f(x)(X-x)$

and

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

As Eqs.(i) and (ii) intersect at the same point on the X-axis

Putting Y=0 and equating x-coordinates, we have

$x-\dfrac{f(x)}{f'(x)}=x-\dfrac{\displaystyle \int_{-∞}^{x}f(t)dt}{f(x)}$

$\Rightarrow \dfrac{f(x)}{\displaystyle \int_{-∞}^{x}f(t)dt}=\dfrac{f'(x)}{f(x)}$

$\Rightarrow \displaystyle \int_{-∞}^{x}f(t)dt=cf(x)$

As, f(0)=1

$\Rightarrow\displaystyle \int_{-∞}^{x}f(t)dt=c\times1\Rightarrow c=\dfrac{1}{2}$

$\Rightarrow \displaystyle \int_{-∞}^{x}f(t)dt =\dfrac{1}{2}f(x);$ differentiating both the sides and integrating and using boundary conditions, we get

$f(x)=e^{2x}; y=2ex$ is tangent to

$y=e^{2x}$

$\therefore$ Number of solutions=1

Clearly, f(x) is increasing for all x.

$\therefore lim_{x→∞}(e^{2x})^{e^{-2x}}=1$

$f(x) = \displaystyle \int_{1}^{x} e^{t^2/2}(1-t^2)dt,$ then

$\dfrac{d}{dx} f(x)$ at x=1 is

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

Explanation

We have,

$f(x)=\displaystyle \int_{1}^{x}e^{t^2/2}(1-t^2)dt$

$f'(x)=[e^{x^2/2}(1-x^2)]^2$

$f'(1)=e^{1/2}.0=0$

The curve for which the ratio of the length of the segment by any tangent on the

$Y-$ axis to the length of the radius vector is constant

$(K)$ , is

0%
$(y+\sqrt {x^2 -y^2})x^{k-1}=c$
0%
$(y+\sqrt {x^2 +y^2})x^{k-1}=c$
0%
$(y-\sqrt {x^2 -y^2})x^{k-1}=c$
0%
$(y+\sqrt {x^2 +y^2})x^{k-1}=c$

Number of critical point for

$y=f(x)$ for

$x \in [0,2]$

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

Explanation

Integrating,

$\dfrac{d^2y}{dx^2}=6x-4,$
we get

$\dfrac{dy}{dx}=3x^2-4x+A$
When

$x=1, \dfrac{dy}{dx}=0$ so that

$A=1$ .
Hence,

$\dfrac{dy}{dx}=3x^2-4x+1$
Integrating we get

$y=x^3-2x^2+x+B$ .
When

$x=1, y=5$ so that

$B=5$
Thus, we have

$y=x^3-2x^2+x+5$
From Eq. (i) we get the critical points

$x=\dfrac{1}{3}, x=1$
At the critical point

$x=\dfrac{1}{3}, \dfrac{d^2y}{dx^2}$ is negative.
Therefore, at

$x=\dfrac{1}{3},y$ has a local maximum.
At

$x=1, \dfrac{d^2y}{dx^2}$ is positive. THerefore, at

$x=1,y$ has a local
minimum.
Also,

$f(1)=5,f\left(\dfrac{1}{3}\right)=\dfrac{157}{27}, f(0)=5, f(2)=7$
Hence, the global value

$=7$
And the global minimum value

$=5$

The point of the curve

$y^2 = x$ where the tangent makes an angle of

$\frac { \pi}{4}$ with x-axis is

0%
$( \frac {1}{2}, \frac {1}{4} )$
0%
$( \frac {1}{4}, \frac {1}{2} )$
0%
$(4,2 )$
0%
$(1,1)$

Explanation

Given curve is

$y^2 = x$

Therefore as per given condition

$\dfrac{dy}{dx}=\dfrac{1}{2y} = tan \dfrac{\pi}{4}=1 \Rightarrow y=\dfrac{1}{2} \\ \Rightarrow x= \dfrac{1}{4}$

Therefore, correct answer is

$B$ .

The abscissa of the point on the curve

$3y=6x- 5x^3$ the normal at which passes through origin is :

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

Explanation

Let

$(x_1,y_1)$ be the point on the given curve

$3y=6x-5x^3$ at which the normal passes through the origin. Then we have

$\left( \frac { dy }{ dx } \right) _{ (x_{ 1 },y_{ 1 }) } =2-5x_1^2 .$

Again the equation of the normal at

$(x_1,y_1)$ passing through the origin gives

$2-5x_1^2= \dfrac{-x_1}{y_1}=\dfrac{-3}{6-5x_1^2}$

Since.

$x_1=1$ satisfies the equation, therefore, correct answer is

$(A) .$

The curve

$y=x^{\frac{1}{5}}$ has at

$(0,0)$

0%
a vertical tangent (parallel to y-axis)
0%
a horizontal tangent (parallel to x-axis)
0%
an oblique tangent
0%
no tangent

Explanation

We have,

$y=x^{1/5}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{5}x^{\tfrac{1}{5}-1}=\frac{1}{5}x^{-4/5}$

$\therefore \left( \dfrac { dy }{ dx } \right) _{ (0,0) }= \tfrac{1}{5} \times (0)^{-4/5}= \infty$

So, the curve

$y=x^{1/5}$ has vertical tangent at

$(0,0)$ , which is parallel to y-axis.

The equation of the curve satisfying the differential equation

$y_2(x^2 + 1) = 2xy_1$ passing through the point

$(0, 1)$ and having slope of tangent at

$x = 0$ as

$3$ (where

$y_2$ and

$y_1$ represents 2nd and 1st order derivative), then

0%
$y=f(x)$ is a strictly increasing function
0%
$y=f(x)$ is non-monotomic finction
0%
$y=f(x)$ has three distinct real roots
0%
$y=f(x)$ has only one negative root

Explanation

The given differential equation is

$y_2 (x^2 + 1) = 2xy_1$

or,

$\dfrac{y_2}{y_1} = \dfrac{2x}{x^2 + 1}$
Integrating both sides, we get

$\log y_1 = \log(x^2 + 1) + \log C$

$y_1 = C(x^2 + 1)$ (1)
It is given that

$y_1 = 3$ at

$x = 0$
Putting

$x = 0, y_1 = 3$ in equation (1), we get

$3=C$
Substituting the value of

$C$ in (1), we obtain

$y_1 = 3 (x^2 + 1)$ (2)
Integrating both sides w.r.t. to

$x$ , we get

$y = x^3 + 3x + C_2$
This passes through the point

$(0,1)$ . Therefore,

$1 = C_2$
Hence, the requiredequation of the curve is

$y = x^3 + 3x + 1$
Obviously it is strictly increasing from equation (2)
Also

$f(0) = 1 > 0$ , then the only root is negative.

The tangent to the curve

$y=e^{2x}$ at the point

$(0,1)$ meets x-axis at:

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

Explanation

The equation of curve is

$y=e^{2x}$

Since, it passes through the point

$(0.1).$

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

$\Rightarrow \left( \frac { dy }{ dx } \right) _{ x=0 }=2e^{2\times0} =2=$ slope of tangent to the curve

$\therefore$ Equation of tangent is

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

$\Rightarrow y=2x+1$

Since, tangent to curv

$y=e^{2x}$ at the point

$(0,1)$ meets X-axis i.e.,

$y=0$ .

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

So, the required point is

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

The slope of tangent to the curve

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

$(2,-1)$ is:

0%
$\frac{22}{7}$
0%
$\frac{6}{7}$
0%
$\frac{-6}{7}$
0%
$-6$

Explanation

Equation of curve is given is given by

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

$y=2t^2-2t-5.$

$\therefore \dfrac{dx}{dt}= 2t+3$ and

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

$\Rightarrow\dfrac{dy}{dx}=\dfrac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } =\dfrac { 4t-2 }{ 2t+3 } ....(i)$

Since, the curve passes through the point

$(2,-1).$

$\therefore 2=t^2+3t-8$

and

$-1=2t^2-2t-5$

$\Rightarrow t^2+3t-10=0$

and

$2t^2-2t-4=0$

$\Rightarrow t^2+5t-2t-10=0$

and

$2t^2+2t-4t-4=0$

$\Rightarrow t(t+5)-2(t+5)=0$

and

$2t(t+1)-4(t+1)=0$

$\Rightarrow (t-2)(t+5)=0$

and

$(2t-4)(t+1)=0$

$\Rightarrow t=2,-5$ and

$t=-1,2$

$\Rightarrow t=2$

$\therefore$ Slope of tangent,

$\left( \frac { dy }{ dx } \right) _{ at\ t-2 }=\dfrac { 4\times 2-2 }{ 2\times 2+3 } =\dfrac { 6 }{ 7 }$ [using Eq.(i)]

The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is :

0%
an ellipse
0%
parabola
0%
circle
0%
rectangular hyperbola

Explanation

${\textbf{Step -1: Differentiate slope}}{\textbf{.}}$

${\text{Given: slope of tangent is equal to ratio of abscissa to the ordinate of the point}}{\text{.}}$

${\text{Let slope of the tangent be }}\dfrac{{dy}}{{dx}}.$

$\text{According to the question}$

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

${\textbf{Step -2: Solve further to find the equation}}{\textbf{.}}$

${\text{Separating the variables}}{\text{.}}$

$ydy = xdx$

${\text{Upon integration we get,}}$

$\int {ydy = \int {xdx} }$

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

$\Rightarrow{y^2} = {x^2} + 2C$

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

$\Rightarrow {x^2} - {y^2} = k$

$\textbf{[ k = - 2 C]}$

${\textbf{Hence, the given equation is of a rectangular hyperbola}}{\textbf{.}}$

The line

$5x-2y+4k=0$ is tangent to

$4x^{2}-y^{2}=36$ , then k is:

0%
$\dfrac{9}{4}$
0%
$\dfrac{81}{16}$
0%
$\dfrac{4}{9}$
0%
$\dfrac{2}{3}$

Explanation

The equation of the hyperbola is

$4x^2-y^2=36$

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

$\dfrac{4x^2}{9}-\dfrac{y^2}{36}=1$ ............(1)

$a^2=9;$

$b^2=36$

The equation of the line is

$5x-2y+4k=0$

$2y=5x+4k$

$y=\dfrac{5}{2}x+2k$ ............(2)

From equation (2)

$m=\dfrac{5}{2},\,c=2k$

$c^2=a^2m^2-b^2$

$(2k)^2=9(\dfrac{5}{2})^2-36$

$4k^2=9\times \dfrac{25}{4}-36$

$4k^2=\dfrac{225-144}{4}$

$k^2=\dfrac{81}{16}\Rightarrow k=\dfrac{9}{4}$

The function

$f\left( x \right) =tan x-x$

0%
always increases
0%
always decreases
0%
never increases
0%
sometime increase and sometimes decreases

Which of the following function is decreasing on

$\left( 0,\frac { \pi }{ 2 } \right)$

0%
$sin 2 x$
0%
$tan x$
0%
$cos x$
0%
$cos 3 x$

If the tangent at

$(1,1)$ on

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

$P$ , then

$P$ is

0%
$(4,4)$
0%
$(-1,2)$
0%
$(3,6)$
0%
$\left(\dfrac{9}{4}, \dfrac{3}{8}\right)$

Explanation

$\left(\dfrac{9}{4}, \dfrac{3}{8}\right)$
[Hint:

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

$\therefore 2y\dfrac{dy}{dx}=3x^{2}-8x+4$

$\therefore \dfrac{dy}{dx}=\dfrac{3x^{2}-8x+4}{2y}$

$\therefore \left(\dfrac{dy}{dx}\right)_{at\ (1,1)}=\dfrac{3(1)^{2}-8(1)+4}{2(1)}=\dfrac{1}{2}$
= slope of the tangent at

$(1,1)$

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

$\therefore 2y-12=-x+1$

$\therefore x+2y=3$
Only the coordinates

$\left(\dfrac{9}{4}, \dfrac{3}{8}\right)$ satisfy both the
equation

$y^{2}=x(2-x)^{2}$ and

$x+2y=3$

$\therefore P$ is

$\left(\dfrac{9}{4}, \dfrac{3}{8}\right)$ ].

The slope of the tangent to the curve

$x = t^{2} + 3 t - 8, y = 2t^{2} - 2t - 5$ at the point

$(2, -1)$ is

0%
$\dfrac{22}{7}$
0%
$\dfrac{6}{7}$
0%
$\dfrac{7}{6}$
0%
$\dfrac{-6}{7}$

Explanation

$Slope \space of \space tangent \space to \space curve \space y=f(x) \space is \dfrac{dy}{dx}$

$Using\space chain \space rule$

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

$Given\space x=t^2+3t-8 \implies \dfrac{dx}{dt}=2t+3$

$Given\space y=2t^2-2t-5 \implies \dfrac{dy}{dt}=4t-2$

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

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

$\therefore$ Answer (B)

$\dfrac{6}{7}$

The line

$y = mx + 1$ is a tangent to the curve

$y^{2} = 4x$ if the value of m is .......

0%
$1$
0%
$2$
0%
$3$
0%
$\dfrac{1}{2}$

Explanation

Given,

$y^{2} = 4x \Rightarrow 2y \dfrac{dy}{dx} = 4 \Rightarrow \dfrac{dy}{dx} = \dfrac{2}{y}$

but slope of tangent is

$m = \dfrac{2}{y} \Rightarrow y = \dfrac{2}{m}$

When

$y= \dfrac{2}{m}, x = \dfrac{y^{2}}{4} = 4/m^{2}_{/4} = \dfrac{1}{m^{2}}$

$x = \dfrac{1}{m^{2}}, y = \dfrac{2}{m}$ lie on the line

$y = mx + 1$

$\dfrac{2}{m} = m \times \dfrac{1}{m^{2}} + 1 \Rightarrow \dfrac{2}{m} = \dfrac{1}{m} + 1$

$\dfrac{2}{m} = 1 \Rightarrow m = 1$

Correct answer (A).

The normal at the point

$(1, 1)$ on the curve

$2y + x^{2} - 3$ is .............

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

Explanation

Given,

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

$2 \dfrac{dy}{dx} + 2x = 0 \\ \Rightarrow \dfrac{dy}{dx} = -x \\ \therefore \dfrac{dy}{dx} = -1$

Slope of tangent =

$-1$

$\therefore$ equation of normal is

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

$\Rightarrow y - x = 0$

$\Rightarrow x-y = 0$

$\therefore$ Answer (B).)

The slope of the normal to the curve

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

$x = 0$ is

0%
$3$
0%
$1/3$
0%
$-3$
0%
$-1 /3$

Explanation

Given,

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

slope

$= \dfrac{dy}{dx} = 4x + 3 \cos x$

$\dfrac{dy}{dy}|_{x=0} = 4 ( 0 ) = 3$

slope of normal =

$-1/ \dfrac{dy}{dx} = -1/3$

The normal to the curve

$x^{2} = 4y$ passing

$(1, 2)$ is

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

Explanation

Given,

$4y = x^{2} \Rightarrow 4 \dfrac{dy}{dx} = 2x \Rightarrow \dfrac{dy}{dx}|_{x = x1} = \dfrac{x_{1}}{2}$

slope of tangent is

$= \dfrac{x_{1}}{2}$

slope of normal is

$= \dfrac{-2}{x_{1}}$

equation of normal

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

but this passes they

$(1, 2)$

$2 - y_{1} = \dfrac{-2}{x} (1 - x) \Rightarrow 2 - y_{1} = \dfrac{-2}{x_{1}} + 2$

$y_{1} = \dfrac{2}{x_{1}}$

but

$x^{2} = 4y$

$x^{2}_{1} = 4y_{1}$

$x^{2}_{1} = 4 \times \dfrac{2}{x_{1}} = 8 \Rightarrow x_{1} = 2$

when

$x_{1} = 2 y_{1} = \dfrac{2}{2} = 1$

$\therefore$ equation of normal is

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

$x + y = 3$

The line

$y = x + 1$ is a tangent to the curve

$y^{2} = 4 x$ at the point

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

Explanation

Given,

$y^{2} = 4 x \\ \Rightarrow 2y \dfrac{dx}{dx} = 4 \\ \Rightarrow \dfrac{dx}{dy} = 2/y$

but given that

$y = x + 1$ is a tangent to the curve its slope =

$1 ( y = mx +c )$

$\dfrac 2y =1\implies y=2$

$x=2-1=1$

The line

$y = x + 1$ is a tangent to the curve

$y^{2} = 4 x$ at the point

$(1,2)$ .

The points on the curve

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

0%
$\left ( 4, \pm \dfrac{8}{3} \right )$
0%
$\left ( 4, \dfrac{-8}{3} \right )$
0%
$\left ( 4, + \dfrac{8}{3} \right )$
0%
$\left (\pm 4, \dfrac{8}{3} \right )$

For

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

$n\in I$ , the critical points of

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

0%
$x=n\pi$
0%
$x=2n\pi$
0%
$x=(2n+1)\pi$
0%
no critical points

Explanation

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

$f'(x)=\sec^{2}x (\sin a \tan^{2}x+\sin a -1)$
For critical points,

$\>f'(x)=0$

$\Rightarrow \sec^{2}x (\sin a \tan^{2}x+\sin a -1)=0$

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

$\because \sec^{2}x=0$ (not possible)]
Since,

$a\in [\pi,2\pi]$

$\Rightarrow \displaystyle \frac{1-\sin a}{\sin a} <0$

$\Rightarrow \tan^{2}x <0$ which is false.
Hence, no critical points.

Let

$\mathrm{f}(\mathrm{x})=\mathrm{a}\mathrm{x}^{3}+\mathrm{b}\mathrm{x}^{2}+ cx + \mathrm{d}$ , where

$a,b,c,d$ are real and

$3\mathrm{b}^{2}<\mathrm{c}^{2}$ , is an increasing function and

$\mathrm{g}(\mathrm{x})=\mathrm{a}\mathrm{f}'(\mathrm{x})+\mathrm{b}\mathrm{f}''(\mathrm{x})+\mathrm{c}^{2}$ . lf

$\displaystyle \mathrm{G}(\mathrm{x})=\int_{\alpha}^{\mathrm{x}}\mathrm{g}(\mathrm{t})\mathrm{d}\mathrm{t},\alpha \in \mathrm{R}$ , then for

$\alpha < x < \alpha +1$ ,

0%
G(x) is a decreasing function
0%
G(x) is an increasing function
0%
G(x) is neither increasing nor decreasing
0%
G(x) is a one-one function

Explanation

${f}({x})={a}{x}^{3}+{b}{x}^{2}+ cx + {d}$

$f'(x)=3ax^{2}+2bx+c$
Since,

$f(x)$ is increasing.

$\Rightarrow f'(x)=3ax^{2}+2bx+c >0$

$\Rightarrow a>0$ and

$4b^{2}-12ac<0$

$\Rightarrow a>0$ and

$b^{2}<3ac$
Also,

$g(x)=af'(x)+bf''(x)+c^{2}$

$\Rightarrow g(x)=a(3ax^{2}+2bx+c)+b(6ax+2b)+c^{2}$

$\Rightarrow g(x)=3a^{2}x^{2}+8abx+2b^{2}+c^{2}+ac$

$D=64a^{2}b^{2}-4(3a^2)(2b^{2}+c^{2}+ac)$

$\Rightarrow D=4a^{2}(10b^{2}-3c^{2}-3ac)<4a^{2}(10b^{2}-3c^{2}-b^{2})$

$\Rightarrow D<4a^{2}(9b^{2}-3c^{2})$

$\Rightarrow D<12a^{2}(3b^{2}-c^{2})$

Since

$\Rightarrow 3b^{2}-c^{2}<0$ ,

$\Rightarrow D<0$

$\Rightarrow g(x)>0$ for all

$x\in R$ .

$\displaystyle {G}({x})=\int_{\alpha}^{x}{g}({t}){d}{t}$ is a strictly increasing function.
Hence,

$G(x)$ is one-one in the given interval.

Let

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

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

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

$g(x)$ is decreasing in

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

Explanation

Given

$g\left( x \right) =f\left( \sin { x } \right) +f\left( \cos { x } \right)$

Differentiating w.r.t

$x$ , we get

$g'\left( x \right) =f'\left( \sin { x } \right) .\cos { x } -f'\left( \cos { x } \right) .\sin { x }$

Again differentiating w.r.t

$x$ , we get

$g''\left( x \right) =-f'\left( \sin { x } \right) .\sin { x } +f''\left( \sin { x } \right) .\cos ^{ 2 }{ x } -f'\left( \cos { x } \right) .\cos { x } +f''\left( \cos { x } \right) .\sin ^{ 2 }{ x }$ ...(2)

Now, given

$\displaystyle f'\left( \sin { x } \right) <0\Rightarrow f'\left( \sin { \left( \frac { \pi }{ 2 } -x \right) } \right) <0\quad \quad \left( \because x\in \left( 0,\frac { \pi }{ 2 } \right) \Rightarrow \frac { \pi }{ 2 } -x\in \left( 0,\frac { \pi }{ 2 } \right) \right)$

$\Rightarrow f'\left( \cos { x } \right) <0$ ...(2)

Similarly

$\displaystyle f''\left( \sin { x } \right) >0\Rightarrow f''\left( \sin { \left( \frac { \pi }{ 2 } -x \right) } \right) >0\Rightarrow f''\left( \cos { x } \right) >0$ ...(3)

$\therefore g''(x)>0$

$($ Using (1),(2) and (3)

$)$

$\Rightarrow g'(x)$ is increasing in

$\displaystyle \left( 0,\frac { \pi }{ 2 } \right)$ .

Also

$\displaystyle g'\left( \frac { \pi }{ 4 } \right) =0$

$\Rightarrow \displaystyle g'\left( x \right) >0\quad \forall x\in \left( \frac { \pi }{ 4 } ,\frac { \pi }{ 2 } \right)$ and

$\displaystyle g'\left( x \right) <0\quad \forall x\in \left( 0,\frac { \pi }{ 4 } \right)$

Thus

$g(x)$ is decreasing in

$\displaystyle \left( 0,\frac { \pi }{ 4 } \right)$

The point of contact of vertical tangent to the curve given by the equations

$\mathrm{x}=3-2\cos\theta, \mathrm{y}=2+3\sin\theta$ is

0%
(1, 5)
0%
(1, 2)
0%
(5, 2)
0%
(2, 5)

Explanation

$x=3-2cos\theta ,y=2+3sin\, \theta$
The slope of the tangent at

$'\theta '$ is

$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta }}{\dfrac{dx}{d\theta }}$

$\therefore$ for vertical tangent

$\dfrac{dx}{d\theta }=0i.e.,2sin\, \theta =0$

$\Rightarrow \theta =0\,$ or

$\pi or2\pi$
Then

$x=1,y=2;x=5 ,y=2;x=1,y=2$

$\therefore$ the points of contact are

$(1,2) ,(5,2)$

The value of a for which the function

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

0%
$(-\displaystyle \infty,-\frac{4}{3})\mathrm{\cup}(2,\infty)$
0%
$(-\infty, -1)$
0%
$[1, \infty)$
0%
$(-2,\infty)$

Explanation

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

$f'\left( x \right) =\left( 4a-3 \right) \left( 1+0 \right) +2\left( a-7 \right) \left[ \cot { \cfrac { x }{ 2 } } \times 2\sin { \cfrac { x }{ 2 } } \cos { \cfrac { x }{ 2 } } \times \cfrac { 1 }{ 2 } +\sin ^{ 2 }{ \cfrac { x }{ 2 } } \times \left( -\csc ^{ 2 }{ \cfrac { x }{ 2 } } \right) \times \cfrac { 1 }{ 2 } \right]$

$f'\left( x \right) =\left( 4a-3 \right) +2\left( a-7 \right) \left[ \cos ^{ 2 }{ \cfrac { x }{ 2 } } -\cfrac { 1 }{ 2 } \right]$

$f'\left( x \right) =\left( 4a-3 \right) +2\left( a-7 \right) \times \cfrac { \left( \cos { x } \right) }{ 2 } \quad \left[ \because 1+\cos { x } =2\cos ^{ 2 }{ \cfrac { x }{ 2 } } \right]$

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

Now, for no critical points

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

$\left( 4a-3 \right) \neq \left( 7-a \right) \cos { x }$

$\cos { x } \neq \cfrac { 4a-3 }{ 7-a }$

$\therefore \cfrac { 4a-3 }{ 7-a } >1$

$4a-3>7-a$

$5a>10$

$a>2$

and

$\cfrac { 4a-3 }{ 7-a } <-1$

$4a-3<a-7$

$3a<-4$

$a<-{ 4 }/{ 3 }$

$\therefore a\epsilon \left( -\infty ,{ -4 }/{ 3 } \right) \bigcup { (2,\infty ) }$ is correct.

The greatest inclination between the tangents is

0%
$\displaystyle \tan^{-1} \left ( \dfrac{\mathrm{a}+\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$
0%
$\displaystyle \tan^{-1} \left (\dfrac{\mathrm{a}-\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$
0%
$\tan^{-1}\sqrt{\dfrac{\mathrm{a}}{\mathrm{b}}}$
0%
$\tan^{-1}\sqrt{\dfrac{\mathrm{b}}{\mathrm{a}}}$

Explanation

The slope of the tangent to the ellipse

$=-\displaystyle \dfrac{\mathrm{b}}{\mathrm{a}}\cot\theta$
The slope of tangent to the circle is

$-\cot\theta$

$\therefore$ the angle

$\phi$ between the tangents is given by

$\tan\phi=\dfrac{(-\dfrac{\mathrm{b}}{\mathrm{a}}+1)\cot\theta}{1+\dfrac{\mathrm{b}}{\mathrm{a}}\cot^{2}\theta}=\dfrac{(\mathrm{a}-\mathrm{b})}{\mathrm{a}\tan\theta+\mathrm{b}\cot\theta}=\dfrac{\mathrm{a}-\mathrm{b}}{(\sqrt{\mathrm{a}\tan\theta}-\sqrt{\mathrm{b}\cot\theta})^{2}+2\sqrt{\mathrm{a}\mathrm{b}}}$
This is maximum when

$\sqrt{\mathrm{a}\tan\theta}-\sqrt{\mathrm{b}\cot\theta}=0$

$\therefore$ the maximum angle

$=\displaystyle \tan^{-1}(\dfrac{\mathrm{a}-\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}})$

A function

$y=f(x)$ has a second order derivative

$f''(x)=6(x-1)$ .
If its graph passes through the point

$(2,1)$ and at that point the tangent to the graph is

$y=3x-5$ , then the function is

0%
$(x-1)^{2}$
0%
$(x+1)^{2}$
0%
$(x+1)^{3}$
0%
$(x-1)^{3}$

Explanation

$f''(x)=6(x-1)$

$\Rightarrow f'(x) = 3x^2 -6x+k$ ...(1)
Slope of tangent at

$(2,1)$ is

$3$ .

$\Rightarrow 3=3 \times 2^2 -6 \times 2 +k$

$\Rightarrow k=3$ ...(2)
From (1) and (2)

$f(x)=x^3-3x^2+3x+c$
Using

$f(x)=y=1$ when

$x=2$ , we get

$1=2+c \Rightarrow c=-1$
Hence, the function is

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

$\Rightarrow f(x)=(x-1)^3$

$f(x) = \displaystyle \frac{x}{{\sin x}}$ and

$g(x) = \displaystyle \frac{x}{{\tan x}}$ where

$0 < x \leq 1$ then in the interval

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

Explanation

We know that if

$f'(x)>0$ then

$f$ is increasing on that interval and if

$f'(x)<0$ then

$f$ is decreasing on the interval.

Given

$f(x)=\dfrac{x}{sinx}$

$\implies f'(x)=\dfrac{sinx-xcosx}{sin^2x}=0$

$\implies sinx-xcosx=0$

$\implies x=\dfrac{sinx}{cosx}=tanx$

In the interval

$(0,1)$ there is no solution for

$x=tanx$

Therefore,

$f'(x)>0$

Hence,

$f(x)$ is an increasing function.

Given

$g(x)=\dfrac{x}{tanx}$

$\implies g'(x)=\dfrac{tanx-xsec^2x}{tan^2x}=0$

$\implies tanx-xsec^2x=0$

$\implies x=\dfrac{sec^2x}{tanx}=\dfrac{1}{sinxcosx}$

In the interval

$(0,1)$ ,

$g(x)$ is not an increasing function.

A function

$y = f (x)$ is given by

$x = \cos^2\theta$ &

$y =\dfrac{\cot\, \theta }{\sec^2\, \theta }$ for all

$\theta >0$ , then

$f$ is :

0%
increasing in $x \in \left (0, \dfrac {3}{2}\right)$ & decreasing in $x \in \left ( \dfrac {3}{2}, \infty\right)$
0%
increasing in $x \in (0, 1)$
0%
increasing in $x \in (0, 2)$
0%
decreasing in $x \in ( 2, \infty)$

Suppose

$a,b,c$ are such that the curve

$y = ax^2 + bx + c$ is tangent to

$y = 3x -3$ at

$(1, 0)$ and is also tangent to

$y = x + 1$ at

$(3, 4)$ then the value of

$(2a -b -4c)$ equals

0%
$7$
0%
$8$
0%
$9$
0%
$10$

Explanation

$y = ax^2 + bx + c$
Since

$(1,0)$ lies on the curve

$\Rightarrow a + b + c = 0$ ...... (i)
Slope of curve

$\displaystyle =\frac {dy}{dx} = 2ax + b$
Slope of curve at (1,0)

$=2a+b$
Slope of curve at (3,4)

$=6a+b$
Slope of tangent

$y=3x-3$ is 3
Slope of tangent

$y=x+1$ is 1

$(\dfrac {dy}{dx})_{x=1} = 3$ and

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

$2a + b = 3$ .... (ii)

$6a + b = 1$ .... (iii)
on solving we get a

$= -\dfrac {1}{2}, b = 4, c = -\dfrac {7}{2}$

$\therefore 2a - b - 4c = -1 -4 + 14 = 9$

For the curve

$y=3 \sin \theta \cos \theta, x= e^{\theta} \sin \theta, 0 \leq \theta \leq \pi$ , the tangent is parallel to x-axis when

$\theta$ is :

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

Explanation

$y=3\sin { \theta } \cos { \theta }$

$\Rightarrow \displaystyle\frac { dy }{ d\theta } =3(\cos { \theta } -\sin { \theta } )(\cos { \theta } +\sin { \theta } )$

$x={ e }^{ \theta }\sin { \theta }$

$\displaystyle\frac { dx }{ d\theta } =3{ e }^{ \theta }(\cos { \theta } -\sin { \theta } )$

$\displaystyle\frac { dy }{ dx} = 3{ e }^{ -\theta }(\cos { \theta } -\sin { \theta })$

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

$\Rightarrow \sin { \theta } =\cos { \theta }$

$\Rightarrow \displaystyle \theta = \frac{\pi}{4}$

$f(x) = x^{2/3}$ then

0%
$(0,0)$ is a point of maxima
0%
$(0,0)$ is a point of minima
0%
$(0,0)$ is a critical point
0%
There is no critical point

Explanation

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

$f'(x)=\dfrac{2}{3x^{1/3}}$
Here,

$f'(x)$ is undefined at

$x=0$ . So,

$f'(x)$ has singularity at

$x=0$ .

$f(x)=\left\{\begin{matrix}x^2+2 & x<0\\ 3 & x = 0\\ x+2 & x>0\end{matrix}\right.$ , then which of the following statement(s) is/are false ?

0%
$f(x)$ has a local maximum at $x=0$
0%
$f(x)$ is strictly decreasing on the left of $x=0$
0%
$f'(x)$ is strictly increasing on the left of $x=0$
0%
$f'(x)$ is strictly increasing on the right of $x=0$

Explanation

$f(0)=3$

$\displaystyle \lim_{x \rightarrow 0^{-}}f(x)=2$

$\displaystyle \lim_{x \rightarrow 0^{+}}f(x)=2$
Therefore there is a local maximum at

$x=0$

$f'(x)=2x$ for

$x<0$ therefore

$f(x)$ is decreasing for

$x<0$

$f''(x) =2$ for

$x<0$ , hence

$f'(x)$ is increasing for

$x<0$

$f'(x)=1 >0 \ for\ x>0$ therefore

$f(x)$ is increasing for

$x>0$

$f'(x)$ is constant for

$x>0$ , hence option D is false

For the curve represented parametrically by the equations,

$x = 2 ln \cot( t) + 1$ &

$y = \tan( t) + \cot( t)$

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

Explanation

$x=2ln cot(t)+1$

$\displaystyle \frac{dx}{dt}=\frac{2}{cot(t)}(-cosec^2 t)=\frac{-2}{sin^2( t) {cot}(t)}$

$\displaystyle \left(\frac{dx}{dt}\right)_{ t=\frac{\pi}{4}} =-4$

$y=tan(t) +cot(t)$

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

$\displaystyle \left (\dfrac{dy}{dt}\right )_{ t=\frac{\pi}{4} }=0$

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

Slope of tangent

$=0$

$\Rightarrow$ Tangent is parallel to x-axis.

Slope of normal

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

$\Rightarrow$ Normal is parallel to y-axis.

A curve passes through

$(2, 0)$ and the slope of the tangent at any point

$(x, y)$ is

$x^2 -2x$ for all values of

$x$ . The point of minimum ordinate on the curve where

$x > 0$ is

$(a, b)$ '

Then find the value of

$a + 6b$ .

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

Explanation

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

$y = \dfrac {x^3}{3} - x^2 + C$
Passing through (2, 0)

$\Rightarrow C = \dfrac {4}{3}$

$\therefore y = \dfrac {x^3}{3} - x^2 + \dfrac {4}{3}$

$y' = x^2 - 2x = x(x - 2)$
For maxima or minima,

$y'=0$

$\Rightarrow x= 0,2$

$y''>0$ at

$x=2$
At x = 2, y takes the minimum value.

$\therefore$ minimum value of y is

$= \dfrac {8}{3} - 4 +\dfrac {4}{3} = 0$

$\therefore a = 2, b = 0$

$a + 6b = 2$

The value of

$x$ at which tangent to the curve

$y=x^3-6x^2+9x+4, 0\leq x \leq 5$ has maximum slope is

0%
$0$
0%
$2$
0%
$\dfrac{5}{2}$
0%
$5$

Explanation

$y=x^3-6x^2+9x+4$

$\dfrac{dy}{dx}=3x^2-12x+9$

Slope of tangent,

$m=f(x)=3x^2-12x+9$

For maxima or minima,

$f'(x)= 0$

$\Rightarrow 6x-12=0$

$\Rightarrow x=2$ .

Now,

$m(0)=9, m(2)=-3$ and

$m(5)=24$
Hence,

The maximum value is attained at

$x = 5$ .

Hence, option D.

The point on the curve

$y^{2} = x ,$ the tangent at which makes an angle of

$45^{0}$ with positive direction of

$x -$ axis will be given by

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

Explanation

$\textbf{Step 1: Find the slope of tangent to given curve at any point}$

$\text{ Given equation of curve is } y^{2} = x$ ...(1)

$\text{Differentiating w.r.t. x we get }$

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

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

$\text{Hence slope of tangent to given curve at any point } (x, y) \text{ is }\dfrac{dy}{dx}=\dfrac{1}{2y}$

$\textbf{Step 2: Find the required point}$

$\text{Given slope of tangent}= \tan 45^{\circ}$

$\text{Hence }\displaystyle\frac{1}{2y}=1\Rightarrow y=\frac{1}{2}$

$\displaystyle \Rightarrow x=\frac{1}{4}$ [From eq.(1)]

$\text{Hence, the required point is } \left ( \dfrac{1}{4},\dfrac{1}{2} \right )$

$\textbf{Hence, Option 'D' is correct}$

A function

$y=f(x)$ has a second-order derivative

$f''(x)=6(x-1)$ . If its graph passes through the point

$(2,1)$ and at the point tangent to the graph is

$y=3x-5$ , then the value of

$f(0)$ is

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

Explanation

$f''(x)=6x-6$

$\int { f''(x)dx } =\int { (6x-6)dx }$

$\Rightarrow f'(x)=3x^{2}-6x+C$
Since, it passes through (2,1) and tangent is

$y=3x-5$

$\Rightarrow C=3$
So,

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

$\int { f'(x)dx } =\int { (3{ x }^{ 2 }-6x+3)dx }$

$\Rightarrow f(x)=x^{3}-3x^{2}+3x+C_1$
Since, it passes through

$(2,1)$

$\Rightarrow C_1=-1$
Hence,

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

$f(0)=-1$

The period of oscillation

$T$ of a pendulum of length

$l$ at a place of acceleration due to gravity

$g$ is given by

$T=2\pi \sqrt {\dfrac {l}{g}}$ . If the calculated length is

$0.992$ times the actual length and if the value assumed for

$g$ is

$1.002$ times its actual value, the relative error in the computed value of

$T$ is

0%
$0.005$
0%
$-0.005$
0%
$0.003$
0%
$-0.003$

Explanation

Relative error will calculate by

$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{\Delta L}{L}-\dfrac{\Delta g}{g}\right]$

$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{0.992L}{L}-\dfrac{1.002 g}{g}\right]$

$\Delta T=-0.005T$

The focal length of a mirror is given by

$\dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$ . If equal errors (

$\alpha$ ) are made in measuring

$u$ and

$v$ , then the relative error in

$f$ is

0%
$\dfrac {2}{\alpha}$
0%
$\alpha \left (\dfrac {1}{u}+\dfrac {1}{v}\right )$
0%
$\alpha \left (\dfrac {1}{u}-\dfrac {1}{v}\right )$
0%
none of these

Explanation

Given,

$\displaystyle \dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$

$\Rightarrow \displaystyle -\dfrac {\Delta v}{v^2}+\dfrac {\Delta u}{u^2}=-\dfrac {2\Delta f}{f^2}$
Since, given equal errors in measuring u and v i.e.

$\Delta u=\Delta v=\alpha$

$\Rightarrow {\alpha}\left(\dfrac{1}{u}-\dfrac{1}{v}\right)\left(\dfrac{1}{u}+\dfrac{1}{v}\right)=-\dfrac {2\Delta f}{f^2}$
But,

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{2}{f}$

$\Rightarrow\displaystyle \frac{ \Delta f}{f}={\alpha}\left(\dfrac{1}{u}+\dfrac{1}{v}\right)$
Hence, correct option is B

The tangent of the acute angle between the curves

$y=|x^2-1|$ and

$y=\sqrt {7-x^2}$ at their points of intersection is

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

Explanation

The given curves are

$y=|x^{2}-1|$ and

$y=\sqrt{7-x^2}$

To find the point of intersection, let us equate both the curves.

$|x^{2}-1|=\sqrt{7-x^2}$

Squaring on both sides gives us

$x^{4}-x^{2}-6=0$

$(x^2-3)(x^2+2) =0$

On solving above quadratic equation, we get

$x^{2}=3$

The point of intersection is

$\left(\pm \sqrt{3},2\right)$

The slope of the tangent to

$y=|x^{2}-1|$ at the point

$\left( \sqrt{3},2\right)$ is

$m_{1}=\displaystyle\dfrac{dy}{dx}=2x=2\sqrt{3}$

The slope of tangent to

$y=\sqrt{7-x^2}$ at the point

$\left(\sqrt{3},2\right)$ is

$m_{2}=\displaystyle\dfrac{dy}{dx}=\dfrac{-x}{\sqrt{7-x^2}}=\dfrac{-\sqrt{3}}{2}$

Let

$\theta$ be the acute angle between the two tangents.

$\displaystyle\tan\theta = \dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}}$

$\Rightarrow \displaystyle\tan\theta = \dfrac{-5\sqrt{3}}{4}$

$\therefore$ The tangent of the acute angle between the curves is

$\displaystyle\dfrac{5\sqrt{3}}{4}$ .

Hence, option C is correct.

The angle made by the tangent of the curve

$x=a (t+\sin t \cos t)$ ,

$y=a(1+sint)^2$ with the

$x- axis$ at any point on it is

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

Explanation

$x=a (t+sin t cos t)$

$\dfrac{dx}{dt}=2acos^{2}t$

$y=a(1+sint)^2$

$\dfrac{dy}{dt}=2a(cost+costsint)$

$\Rightarrow \displaystyle \dfrac{dy}{dx}=\dfrac{1+sint}{cost}$

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

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

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

Slope of tangent to the curve

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

So, the angle made by the tangent to the curve with the x-axis

$=\dfrac{\pi}{4}+\dfrac{t}{2}$ or

$\dfrac{1}{4}(\pi+2t)$

Consider the function

$f(x)= \begin{cases} x \sin \displaystyle \frac {\pi}{x}, for \ x>0\\ 0, for \ x=0 \end{cases}$ . Then, the number of points in

$(0,1)$ where the derivative

$f'(x)$ vanishes is

0%
$0$
0%
$1$
0%
$2$
0%
infinite

The abscissas of points

$P$ and

$Q$ on the curve

$y=e^x+e^{-x}$ such that tangents at

$P$ and

$Q$ make

$60^{\circ}$ with the

$x$ -axis are

0%
$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {7}\right )$ and $\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 5} {2}\right )$
0%
$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$
0%
$\ln \left (\displaystyle \frac {\sqrt 7-\sqrt 3} {7}\right )$
0%
$\pm\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$

Explanation

$y=e^{x}+e^{-x}$

$\displaystyle\frac{dy}{dx}=e^{x}-e^{-x}=\frac{e^{2x}-1}{e^{x}}$
Let P

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

$(x_2,y_2)$ be the points on the given curve.
Slope of tangent at P is

$\displaystyle m_1=\frac{e^{2x_{1}}-1}{e^{x_{1}}}$

$\Rightarrow \displaystyle \sqrt{3}=\frac{e^{2x_{1}}-1}{e^{x_{1}}}$

$\displaystyle\Rightarrow e^{2x_1}-\sqrt{3}e^{x}-1=0$

$\displaystyle\Rightarrow (e^{x_1}-\frac{\sqrt{3}}{2})^{2}=\frac{7}{4}$

$\displaystyle\Rightarrow e^{x_1}=\pm\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}$

$\displaystyle\Rightarrow {x_1}=ln{(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})}$ as logarithm of negative numbers is not defined.
Similarly,

$\displaystyle\Rightarrow {x_2}=ln{(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})}$ as it makes same angle,

$60^{0}$ with x-axis.

The graphs

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

$y=x^3+2x-1$ intersect at exactly 3 distinct points. The slope of the line passing through two of these points

0%
is equal to 4
0%
is equal to 6
0%
is equal to 8
0%
is not unique

Explanation

Given equation of curves

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

$y=x^3+2x-1$
Let

$(x_{1},y_{1})$ be a point of intersection of the curves.

$\Rightarrow y_{1}=2x_{1}^3-4x_{1}+2$
And

$y_{1}=x_{1}^3+2x_{1}-1$

$\Rightarrow y_1=8x_{1}-4$
Let

$(x_{2},y_{2})$ be another point of intersection of the curves.

$\Rightarrow y_{2}=2x_{2}^3-4x_{2}+2$
And

$y_{2}=x_{2}^3+2x_{2}-1$

$\Rightarrow y_2=8x_{2}-4$
Now, slope

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

$\Rightarrow m =\displaystyle \dfrac{8(x_2-x_1)}{x_2-x_1}$

$\Rightarrow m=8$

If the curve represented parametrically by the equations

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

$y=\tan t+ \cot t$

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

Explanation

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

$\displaystyle \frac{dx}{dt}=-\frac{2cosec^{2}t}{\cot t}=-\frac{2}{\sin t\cos t}$

$y=\tan t+ \cot t$

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

$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{sin^{2}t-cos^{2}t}{sin2t}$

$\Rightarrow \displaystyle \frac{dy}{dx}=\cot 2t$
Slope of tangent at

$t=\dfrac{\pi}{4}$ is 0.
Hence, tangent is parallel to x-axis.

Let

$S$ be a square with sides of length

$x$ . If we approximate the change in size of the area of

$S$ by

$\displaystyle h.\frac{dA}{dx}|_{x=x_0}$ , when the sides are changed from

$x_0$ to

$x_o+h$ , then the absolute value of the error in our approximation, is

0%
$h^2$
0%
$2hx_0$
0%
$x_0^2$
0%
$h$

Explanation

We have S be a square with side length x

Since, area of square

$A=x^{2}$

Differentiate both sides with respect to x

$\displaystyle \dfrac{dA}{dx}=2x$

Given , approximate change in size of area

$=\Delta A = \displaystyle h.\dfrac{dA}{dx}|_{x=x_0}$

$\Rightarrow \Delta A=2x_{0}h$

Also,

$(A+\Delta A)-A=(x_{0}+h)^{2}-{x_0}^{2}$

$=2x_{0}h+h^{2}$

Hence, absolute value of error in our approximation is

$h^{2}$

Hence, Option (A) is correct.

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 10 - 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 10 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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