MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 2

The angle made by the tangent line at (1, 3) on the curve

$y=4x-{ x }^{ 2 }$ with

$\overset { - }{ OX }$ is

0%
${ tan }^{ -1 }(2)$
0%
${ tan }^{ -1 }(1/3)$
0%
${ tan }^{ -1 }(3)$
0%
$\pi /4$

Explanation

Given,

$y=4x-x^2$

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

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

Therefore, angle made by tangent,

$\tan \theta =2$

$\therefore \theta =\tan ^{-1}2$

equation of tangent at

$(0,0)$ for the equation

$y^2=16x$

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

Explanation

The equation is

$y^2=16x$

The point is

$(0,0)$

The slope of tangent

$2y \dfrac{dy}{dx}=16\\\dfrac{dy}{dx}=\dfrac 8y\\\left.\dfrac{dy}{dx}\right|_{(0,0)}=\infty$

Equation of tangent is

$y-0=\dfrac{8}{0}(x-0)\\x=0$

The intercept on x-axis made by tangent to the curve,

$\displaystyle y=\int _{ 0 }^{ x }{ \left| t \right| } dt,x\in R$ , which are parallel to the line

$y=2x$ , are equal to

0%
$\pm 1$
0%
$\pm 2$
0%
$\pm 3$
0%
$\pm 4$

Explanation

$y = \int_{0}^{x}{\left| t \right| dt}$

$\Rightarrow \cfrac{dy}{dx} = \left| x \right|$

Since tangent to the curve is parallel to the line

$y = 2x$ .

$\therefore \cfrac{dy}{dx} = 2$

$\Rightarrow x = \pm 2$

Therefore,

$y = \int_{0}^{\pm 2}{\left| t \right| dt}$

$\Rightarrow y = \pm 2$

Therefore,

Equation of tangents are-

$y - 2 = 2 \left( x - 2 \right)$

$y + 2 = 2 \left( x + 2 \right)$

For

$x$ -intercept, substitute

$y = 0$ , we have

$0 - 2 = 2 \left( x - 2 \right) \Rightarrow x = +1$

$0 + 2 = 2 \left( x + 2 \right) \Rightarrow x = -1$

Hence the intercept on

$x$ -axis are

$\pm 1$ .

The area of triangle formed by tangent and normal at point

$(\sqrt{3}, 1)$ of the curve

$x^2+y^2=4$ and x-axis is?

0%
$\dfrac{4}{\sqrt{3}}$
0%
$\dfrac{2}{\sqrt{3}}$
0%
$\dfrac{8}{\sqrt{3}}$
0%
$\dfrac{5}{\sqrt{3}}$

Explanation

Slope of OP

$=\dfrac{1}{\sqrt{3}}$ , slope of PQ

$=-\sqrt{3}$

$y-1=-\sqrt{3}(x-\sqrt{3})=-\sqrt{3}x+3$

$\Rightarrow \sqrt{3}x+y=4$ and

$Q\left(\dfrac{4}{\sqrt{3}}, 0\right)$

$\Delta OPQ=\dfrac{2}{\sqrt{3}}$ .
1245724_1611460_ans_76b3a467db73433698dc5776e9a56489.PNG

The function

$f(x)=x^3-6x^2+9x+3$ is decreasing for

0%
$1 < x < 3$
0%
$x > 1$
0%
$x < 1$
0%
$x < 1$ or $x > 3$

A curve

$y=me^{mx}$ where

$m > 0$ intersects y-axis at a point

$P$ .

What is the equation of tangent to the curve at

$P$ ?

0%
$y=mx+m$
0%
$y=-mx+2m$
0%
$y=m^2x+2m$
0%
$y=m^2x+m$

Explanation

$y = m {e}^{mx}, \; m > 0$

Substituting

$x = 0$ , we have

$y = m {e}^{m \cdot 0} = m$

Therefore,

Slope

$= \cfrac{dy}{dx} = {m}^{2} {e}^{mx} = {m}^{2} {e}^{m \cdot 0} = {m}^{2}$

Therefore, equation of tangent at

$P$ is given by-

$y - m = \cfrac{dy}{dx} \left( x - 0 \right)$

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

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

$f(x)=kx^3 -9x^2 +9x+3$ is increasing for every real number

$x$ , then

0%
$k > 3$
0%
$k \ge 3$
0%
$k < 3$
0%
$k \le 3$

The tangent to the curve

$y=e^{kx}$ at a point (0, 1) meets the x-axis at (q, 0) where

$a \epsilon \left [ -2,-1 \right ]$ then

$k \epsilon$

0%
[-1/2,0]
0%
[-1,-1/2]
0%
[0,1]
0%
[1/2, 1]

The function

$f(x)=x^3-27x+8$ is increasing when

0%
$|x| < 3$
0%
$|x| > 3$
0%
$-3 < x < 3$
0%
$none\ of\ these$

Explanation

$f'(x)=3x^2 -27=3(x^2 -9)=3(x+3)(x-3)$

$\therefore \ f'(x)$ is increasing when

$x < -3$ or

$x > 3$ i.e when

$|x| > 3$ .

There are two factors in

$f'(x)$ . so we start with

$+ve$ sign

$\therefore \ f(x)$ is increasing when

$x < -3$ or

$x > 3$ i.e when

$|x| > 3$ .

A stationary point of

$\mathrm{f}(\mathrm{x})=\sqrt{16 -\mathrm{x}^{2}}$ is

0%
(4, 0)
0%
(-4, 0)
0%
(0, 4)
0%
(-4, 4)

Explanation

$f(x)=\sqrt {16-x^2}$

$y=\sqrt {16-x^2}$

$a$ starting point

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

then

$\dfrac {dy}{dx}=\dfrac {1}{2} \dfrac {-2x}{\sqrt {16-x^2}}=\dfrac {x}{\sqrt {16-x^2}}=0$

$x=0$

Now

$y=\sqrt {16-0}$

$=\sqrt {16}=4$

$(0,4)\$ be starting point.

$\mathrm{f}(\mathrm{x})=\mathrm{x}+2$ cosx is increasing in

0%
$(0,\displaystyle \frac{\pi}{2})$
0%
$(\displaystyle \frac{-\pi}{2},\frac{\pi}{6})$
0%
$(\displaystyle \frac{\pi}{2},{\pi})$
0%
$(\displaystyle \frac{-\pi}{2}\frac{\pi}{2})$

Explanation

$f'(x)=1-2 sin x$

$f'(x)>0$

$\forall \ \ x\epsilon (\dfrac {-\pi}{2}, \dfrac {\pi}{6})$

$\displaystyle \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\log \mathrm{x}}-\frac{\log 5}{5}$ is decreasing in

0%
$(\mathrm{e}, \infty)$
0%
$(0, 1)\mathrm{U}(1, \mathrm{e})$
0%
(0, 1)
0%
(1, e)

Explanation

$f\left( x \right) =\cfrac { x }{ \log { x } } -\cfrac { \log { 5 } }{ 5 }$ is decreasing in ....

For decreasing functions

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

$\cfrac { \log { x } \times \cfrac { d }{ dx } \left( x \right) -x\times \cfrac { d }{ dx } \left( \log { x } \right) }{ { \left( \log { x } \right) }^{ 2 } } -0<0$

$\cfrac { \log { x } -1 }{ { \left( \log { x } \right) }^{ 2 } } <0$

$\log _{ e }{ x-1 } <0$

$\log _{ e }{ x-1 } <1$

$x<e$

And also

$x>0$

Now, at

$x=1,f'\left( x \right)$ wont exit

Therefore,

$\left( 0,1 \right) \bigcup { \left( 1,e \right) }$ is the correct answer.

The condition that

$\mathrm{f}(\mathrm{x}) =\mathrm{x}^{3}+\mathrm{a}\mathrm{x}^{2}+\mathrm{b}\mathrm{x}+\mathrm{c}\mathrm \ {i}\mathrm{s}\mathrm \ {a}\mathrm{n}$ increasing function for all real values of

$\mathrm{x}$ is

0%
$\mathrm{a}^{2}<12\mathrm{b}$
0%
$\mathrm{a}^{2}<3\mathrm{b}$
0%
$\mathrm{a}^{2}<4\mathrm{b}$
0%
$\mathrm{a}^{2}<16\mathrm{b}$

Explanation

$f^1(x)=3x^2+2ax+b$

$f^1(x)>0$

$3x^2+2ax+b>0$
Only If

$(2a)^2-4\times 3b<0$ (Property of quadratic

$eq^n$ )

$a^2<3b$

I. lf

$f'(\mathrm{a})>0$ then

$\mathrm{f}$ is increasing at

$\mathrm{x}=\mathrm{a}$
II: If f is increasing at

$\mathrm{x}=\mathrm{a}$ then

$f'(\mathrm{a})$ need not to be positive

0%
only I
0%
only II
0%
both I and II
0%
neither I nor II

Explanation

$f(x)$ is increasing when

$f'(x)>0$

$\mathrm{f}(\mathrm{x})=$ sinx-ax is decreasing in R if

0%
$\mathrm{a}>1$
0%
$\mathrm{a}<13$
0%
$\displaystyle \mathrm{a}>\frac{1}{2}$
0%
$\displaystyle \mathrm{a}<\frac{1}{2}$

Explanation

$f^1(x)=cos x-a$

$f^1(x)<0$

$cos x-a<0$

$0<cos x<1$ for

$\forall x\epsilon R$
So a has to be greater than 1

$\displaystyle$ x

$\epsilon \ (0,\frac{\pi}{2})$ ,

$\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{x}\sin \mathrm{x}+\cos \mathrm{x}+\frac{1}{2}\cos^{2}\mathrm{x}$ is

0%
Increasing
0%
Decreasing
0%
Constant
0%
Nothing can be determi ned

Explanation

$f^1(x)=sin x+x cos x-sin x-cos x sin x$

$=x cos x-\frac {sin2x}{2}$

$=cos x(x-sin x)$
as

$x>sin x$ so

$f^1(x)>0$
Therefore

$f(x)$ is inc.

A stationary value of

$\mathrm{f}(\mathrm{x})=\mathrm{x}(\ln \mathrm{x})^{2}$ is

0%
$2\mathrm{e}^{-2}$
0%
$4\mathrm{e}^{-2}$
0%
$2\mathrm{e}^{2}$
0%
$4\mathrm{e}^{2}$

Explanation

$f(x)=x(\ln x)^2$

$\Rightarrow f'(x)=(\ln x)^2+2x \dfrac {\ln x}{x}$

$\Rightarrow f'(x)=\ln x(\ln(x)+2)$
For stationary points ,

$f'(x)=0$

$\Rightarrow x=1$ or

$e^{-2}$

$f(1)=0$
and

$f(e^{-2})=4e^{-2}$

$\displaystyle \mathrm{f}(\mathrm{x})=\frac{\mathrm{a}\sin \mathrm{x}+\mathrm{b}\cos \mathrm{x}}{\mathrm{a}\cos \mathrm{x}-\mathrm{b}\sin \mathrm{x}} \ \ \ \ \ (\tan \mathrm{x}\neq\frac{\mathrm{a}}{\mathrm{b}})$ is

0%
increasing in domain $\mathrm{f}$
0%
Decreasing in domain $\mathrm{f}$
0%
Constant
0%
Nothing can be determined

Explanation

$f'(x)=\dfrac {(a cos x- b sin x)^2+(a sin x+ b cos x)^2}{(a cos x-b sin x)^2}$

$=\dfrac {a^2+b^2}{(a cos x-b sin x)^2}$

$f'(x)=\dfrac {a^2+b^2}{(a cos x-b sin x)^2}$

$\displaystyle \mathrm{f}(\mathrm{x})=\frac{\mathrm{a}\sin \mathrm{x}+\mathrm{b}\cos \mathrm{x}}{\mathrm{c}\sin \mathrm{x}+\mathrm{d}\cos \mathrm{x}}$ is an increasing funtion if

0%
$ad-bc =0$
0%
$ad -bc < 0$
0%
$ad - bc > 0$
0%
$ab + cd = 0$

Explanation

$f'(x)=\dfrac {(a cos x-b sin x)(c sin x+d cos x)-(a sin x+b cos x)(c cos x-d sin x)}{(c sin x+d cos x)^2}$

$=\dfrac {ad cos^2x-bc sin^2x-bc cos^2x+ad sin^2x}{(c sin x+d cos x)^2}$

$=\dfrac {ad-bc}{(c sin x+d cos x)^2}$

for

$f(x)$ to be increasing function the

$f^{\prime}(x)$ will have to be greater than

$0$

$f'(x)>0$ so

$ad-bc>0$

Assertion A: The curves

$x^{2}=y,\ x^{2}=-y$ touch each other at (0, 0).
Reason R: The slopes of the tangents at (0, 0) for both the curves are equal.

0%
Both A and R are true R is the correct explanation of A
0%
Both A and R are true but R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

$y'=2x$

$y'|_{x=0}=0=m_{1}$

$y'=-2x$

$y'|_{x=0}=0=m_{2}$

$m_{1}=m_{2}$

$\therefore$ both the curve touches each other

If the slope of the tangent to the curve

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

$(1, 1)$ on it is

$2$ , then values of

$a$ and

$b$ are

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

Explanation

$a+b +1=0$

${xy}'+y+a+{b y}'=0$

${y}'=\dfrac{-(y+a)}{(x+b )}$

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

$2+2b +a+1=0$

$2+2 b -b =0$

$b =-2$

$a=1$

If the slope of the tangent to the curve

$y = x^{3}$ at a point on it is equal to the ordinate of the point then the point is

0%
$(27, 3)$
0%
$(3, 27)$
0%
$(3, 3)$
0%
$(1, 1)$

Explanation

Differntiating the given equation

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

$3a^{2}=a^{3}$

$a=3$

$(3,3^{3})$

$(3,27)$

The critical point of

$\mathrm{f}(\mathrm{x})=|2\mathrm{x}+7|$ at

$\mathrm{x}=$

0%
$0$
0%
$7$
0%
$\dfrac{-7}{2}$
0%
$-7$

Explanation

$f(x)=\mid 2x+7\mid$
So critical point is at

$x=\frac {-7}{2}$

P(1, 1) is a point on the parabola

$y=x^{2}$ whose vertex is A. The point on the curve at which the tangent drawn is parallel to the chord

$\overline{AP}$ is

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

Explanation

$tan \theta =\dfrac{(1-0)}{(1-0)}=1$

${y}'=2x$

$2x=1$

$x=\dfrac{1}{2}, y=\dfrac{1}{4}$

The stationary point of

$\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}$ is

0%
(1, 2)
0%
(2, 3)
0%
(1, 3)
0%
(0, 2)

Explanation

$f'(x)=e^x-e^{-x}$

$f'(x)=0$ at

$x=0$

$f(0)=2$

The number of ciritical points of

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

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

Explanation

$\displaystyle \mathrm{f}(\mathrm{x})=\dfrac{|x-1|}{x^{2}}$

$\displaystyle f'(x)=\dfrac{x^{2}(x-1)-2x|x-1|^{2}}{x^{4}|x-1|}$
For maxima or minima,

$f'(x)=0$

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

$\Rightarrow x=0,1,2$
Hence, number of critical points of

$f(x)$ are 3.

For the parabola

$y^{2}=8x$ , tangent and normal are drawn at

$P(2, 4)$ which meet the axis of the parabola in

$A$ and

$B$ , then the length of the diameter of the circle through

$A, P, B$ is

0%
$2$
0%
$4$
0%
$8$
0%
$6$

Explanation

Considering the equation of the parabola

$y^{2}=8x$

The axis of symmetry is the positive

$x-$ axis.

By, differentiating with respect to

$x$ , we get

$2y.y'=8$

$y'=\dfrac{4}{y}$

Hence,

$y'_{(2,4)}=1$

Thus slope of the tangent at the point of contact

$P(2,4)$ is

$1$ while that of normal is

$-1$ .

Hence, equation of the tangent will be

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

$-x+y=2$ ...(i)

Hence, the tangent meets the x axis at

$(-2,0)$

Therefore,

$A=(-2,0)$ .

Equation of normal will be

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

$x+y=6$ ...(ii)

Hence the normal meets the x axis at

$(6,0)$ .

Thus,

$B=(6,0)$

Therefore the three points through which the circle passes are

$(-2,0),(6,0),(2,4)$

Now let the equation of the circle be

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Therefore

$(-2-h)^{2}+k^{2}=r^{2}$

$\rightarrow (2+h)^{2}+k^{2}=r^{2}$

$(6-h)^{2}+k^{2}=r^{2}$

Subtracting {ii} from {i}, we get

$2h-4=0$

$h=2$

Therefore the equation of the circle reduces to

$(x-2)^{2}+(y-k)^{2}=r^{2}$
Now

$(2-2)^{2}+(4-k)^{2}=r^{2}$

$\rightarrow (4-k)^{2}=r^{2}$

$(6-2)^{2}+k^{2}=r^{2}$

$\rightarrow 16+k^{2}=r^{2}$

Subtracting i from ii, we get

$16+k^{2}-(k-4)^{2}=0$

$16+(2k-4)(4)=0$

$4+2k-4=0$

$k=0$

Thus the centre of the circle lies at

$C=(2,0)$ .

Hence the radius of the circle is

$CA=CP=CB$
Now

$CA=\sqrt{(2-(-2))^{2}+0^{2}}$

$=2-(-2)$

$=4$

$=r$

Hence, the diameter of the circle is

$8$ units.

$\mathrm{f}(\mathrm{x})=(\sin^{-1}\mathrm{x})^{2}+(\cos^{-1}\mathrm{x})^{2}$ is stationary at

0%
$\displaystyle \mathrm{x}=\frac{1}{\sqrt{2}}$
0%
$\displaystyle \mathrm{x}=\frac{\pi}{4}$
0%
$\mathrm{x} = 1$
0%
$\mathrm{x} = 0$

Explanation

$f'(x)=2 sin^{-1}(x)\dfrac {1}{\sqrt {1-x^2}}+2 cos^{-1}(x)(\dfrac {-1}{\sqrt {1-x^2}})$

$=2\dfrac {1}{\sqrt{1-x^2}}(sin^{-1}(x)-cos^{-1}(x))$

$f'(x)=0$ when

$x=\dfrac {1}{\sqrt 2}$

The point on the hyperbola

$y = \dfrac {x - 1}{x + 1}$ at which the tangents are parallel to

$y = 2x + 1$ are

0%
$(0, -1)$ only
0%
$(-2, 3)$ only
0%
$(0, -1)$ , $(-2, 3)$
0%
$(-2, 3)$ , $(5, 4)$

Explanation

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

${y}'=2 (\therefore$ parallel to

$y =2x+1)$

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

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

$x=0$ or

$x=-2$
Using the equation of hyperbola, when

$x=0$ ,

$y=-1$ .
when

$x=-2$ ,

$y=3$

The arrangment of the slopes of the normals to the curve

$y=e^{\log(cosx)}$ in the ascending order at the points given below.

$A) \displaystyle x=\frac{\pi}{6}, B) \displaystyle x=\frac{7\pi}{4}, C)x=\frac{11\pi}{6}, D)x=\frac{\pi}{3}$

0%
$C, B, D, A$
0%
$B, C, A, D$
0%
$A, D, C, B$
0%
$D, A, C, B$

Explanation

We have,

$y = e^{\log \cos x} =(\cos x)^{\log e}[\because a^{\log_bc} =c^{\log_ba}] =\cos x$

$\Rightarrow \dfrac{dy}{dx} =-\sin x=m$ (say)
Thus slope of normal to the given curve at any point is,

$m'=- \dfrac{1}{m}=\dfrac{1}{\sin x}$
Now,

$m'_{\frac{\pi}{6}}=2, m'_{\frac{7\pi}{4}}=-\sqrt{2}, m'_{\frac{\pi}{6}}=-2,m'_{\frac{\pi}{6}}=\dfrac{\sqrt{3}}{2}$
Clearly ascending order is,

$C,B,D,A$
Note: Given function is not defined where

$\cos x<0$

Assertion(A): The tangent to the curve

$y=x^{3}-x^{2}-x+2$ at (1, 1) is parallel to the x axis.
Reason(R): The slope of the tangent to the above curve at (1, 1) is zero.

0%
Both A and R are true R is the correct explanation of A
0%
Both A and R are true but R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

$y^{1}=3x^{2}-2x-1$

$y^{1}|_{x=1}=0=m$

$m=0$

$\therefore$ the curve is parallel to x-axis.

The number of tangents to the curve

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

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

Explanation

$\dfrac{dy}{dx}=-1$ (where x is equally inclined)

$\dfrac{3}{2}x^{^{1}/_{2}}+^{3}/_{2}y^{^{1}/_{2}}\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\sqrt{\dfrac{x}{y}}=1$

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

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

$x=y$

$x^{^{3}/_{2}}+x^{^{3}/_{2}}=a^{^{3}/_{2}}$

$2x^{^{3}/_{2}}=a^{^{3}/_{2}}$

$x=\dfrac{a}{2^{^{2}/_{3}}}$

$y=\dfrac{a}{2^{^{2}/_{3}}}$

$\therefore$ only one tangent.

Match the points on the curve

$2y^{2}=x+1$ with the slope of normals at those points and choose

the correct answer.

Point	Slope of normal
I : $(7, 2)$	$a){-4\sqrt{2}}$
II: $(0, \displaystyle \frac{1}{\sqrt{2}})$	$b) -8$
III : $(1, 1)$	$c) -4$
IV: $(3, \sqrt{2})$	$d){-2\sqrt{2}}$

0%
$i-b,ii- d,iii- c,iv- a$
0%
$i-b,ii- a,iii- d,iv- c$
0%
$i-b,ii- c,iii- d,iv- a$
0%
$i-b,ii- d,iii- a,iv- c$

Explanation

$\displaystyle y'=\frac{1}{4y}=m$

Slope of normal

$m'=-4y$

For given points slope is

(1)

$m'_{1}=-4\times 2 ==-8$
(2)

$m'_{2}=-4\times \dfrac{1}{\sqrt 2}=-2\sqrt{2}$
(3)

$m'_{3}=-4\times 1=-4$
(4)

$m'_{4}=-4\times {\sqrt 2}=-4\sqrt{2}$

Option A is correct answer

lf the tangent to the curve

$f(x)=x^{2}$ at any point

$(c, f(c))$ is parallel to the line joining points

$(a, f(a))$ and

$(b,f(b))$ on the curvel then

$a,\ c,\ b$ are in

0%
AP
0%
GP
0%
HP
0%
AGP

Explanation

$\partial^{'}M=2s$ at

$(c,\ \partial(c))=2c$

$\dfrac{\partial(a)-\partial(L)}{a-b}=2c$

$\dfrac{a^{2}=b^{2}}{a-b}=2c$

$a+b=2c$

$\therefore\ a,c,b$ are in AP

lf the parametric equation of a curve given by

$x=e^{t}\cos t,\ y=e^{t}\sin t$ , then the tangent to the curve at the point

$t=\dfrac{\pi}{4}$ makes with axis of

$x$ the angle.

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

Explanation

$at\ t={\pi}/{4}$

$y=e^{t}sint$

$\dfrac{dy}{dl}=e^{t}(sin\ t+cos\ t)$

$\dfrac{dx}{dl}=d^{t}(cos\ t+sin\ t)$

$\dfrac{dy}{dx}=\dfrac{sin \ t+cos\ t}{cos\ t-sin\ t}$

$= \alpha$

$\therefore tan^{-1}(\infty)={\pi}/{2}$

lf the curve

$y=px^{2}+qx+r$ passes through the point (1, 2) and the line

$y=x$ touches it at the origin, then the values of

$p,\ q$ and

$r$ are

0%
$p=1,q=-1,r=0$
0%
$p=1,q=1,r=0$
0%
$p=-1,q=1,r=0$
0%
$p=1,q=2,r=3$

Explanation

$2=p+q+r$

$r=0$

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

$\dfrac{dy}{dx}=2px+q$

$q=1$

$2=p+1+0$

$p=1$

lf the chord joining the points where

$x= p,\ x =q$ on the curve

$y=ax^{2}+bx+c$ is parallel to the tangent drawn to the curve at

$(\alpha, \beta)$ then

$\alpha=$

0%
$2pq$
0%
$\sqrt{pq}$
0%
$\displaystyle \frac{p+q}{2}$
0%
$\displaystyle \frac{p-q}{2}$

Explanation

$y'=2ax+b$

$y'|_{x=b}=(2a\alpha +b)$

$2a\alpha+b=\left ( \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )$

$2a\alpha+b=\dfrac{ap^{2}+bp-aq^{2}-bq}{(p-q)}$

$=\dfrac{a(p-q)(p+q)+b(p-a)}{(p-q)}$

$2a\alpha+b=a(p+q)+b$

$\alpha=\left ( \dfrac{p+q}{2} \right )$

The arrangement of the following curves in the ascending order of slopes of their tangents at the given points.

$A) \displaystyle y=\frac{1}{1+x^{2}}$ at

$x=0$

$B) y=2e^{\frac{-x}{4}},$ where it cuts the y-axis

$C) y= cos(x)$ at

$\displaystyle x=\frac{-\pi}{4}$

$D) y=4x^{2}$ at

$x=-1$

0%
DCBA
0%
ACBD
0%
ABCD
0%
DBAC

Explanation

(A)

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

$y^{1}|_{x=0}=0$

(B)

$y^{1}=-\frac{2x}{(4}$

$y^{1}|_{x=0}=-\frac{1}{2}$

(C)

$y^{1}=-sin\ x$

$y^{1}|_{x=^{-pi}/_4}=\frac{1}{\sqrt{2}}$

(D)

$y^{1}=8x$

$y^{1}|_{x=-1}=-8$

$(C)>(A)>(B)>(D)$

Observe the following lists for the curve

$y=6+x-x^{2}$ with the slopes of tangents at the given points; I, II, III, IV

Point	Tangent slope
I: $(1, 6)$	a) $3$
II: $(2, 4)$	b) $5$
III: $(-1, 4)$	c) $-1$
IV: $(-2, 0)$	d) $-3$

0%
$a ,b, c ,d$
0%
$b, c, d ,a$
0%
$c, d ,b ,a$
0%
$c, d ,a ,b$

Explanation

$y^{'}=\ \ \ 1-2x$
(A)

$m_{1}=-1$
(B)

$m_{2}=-3$
(C)

$m_{3}=3$
(D)

$m_{4}=5$

If the circle

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

$y=x$ at

$P$ such that

$OP=6\sqrt{2},$ then the value of

$c$ is

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

Explanation

Let the point of contact be

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

$y=x$
Hence

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

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

$2x_{1}^2=72$

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

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

$y'=1$ since

$y'$ is the slope of the line

$y=x$ .
By substituting, we get

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

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

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

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

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

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

$c=144-72$

$c=72$

The points on the hyperbola

$x^{2}-y^{2}=2$ closest to the point (0, 1) are

0%
$(\displaystyle \pm\frac{3}{2},\frac{1}{2})$
0%
$(\displaystyle \frac{1}{2},\pm\frac{3}{2})$
0%
$(\displaystyle \frac{1}{2},\frac{1}{2})$
0%
$(\displaystyle \pm\frac{3}{4}\pm\frac{3}{2})$

Explanation

Let point be

$(\sqrt 2 sec\theta,\sqrt 2 tan\theta)$ closest to (0, 1)
So distance

$=\sqrt {2 sec^2\theta+(\sqrt 2 tan\theta-1)^2}$

$=\sqrt {2 sec^2\theta+2tan^2\theta-2\sqrt 2 tan\theta+1}$

$=\sqrt {4 tan^2\theta-2\sqrt 2 tan\theta+3}$

$=2\sqrt {tan^2\theta-\frac {tan\theta}{\sqrt 2}+\frac {3}{4}}$

$=2\sqrt {(tan\theta-\frac {1}{2\sqrt 2})^2+\frac {3}{4}-\frac {1}{8}}$
So distance is min when

$tan\theta=\frac {1}{2\sqrt 2}$

$sec\theta=\pm \frac {3}{2\sqrt 2}$

The number of tangents to the curve

$x^{3/2} + y^{3/2}= 2a^{3/2}$ ,

$a>0$ , which are equally inclined to the axes, is

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

Explanation

For a tangent to be equally inclined to the axes, the slope of the tangent should be 1.

Thus

$y'=1$ .

Now

$x^{\frac{3}{2}}+y^{\frac{3}{2}}=2a^{\frac{3}{2}}$ .

$\dfrac{3}{2}(\sqrt{x})+\dfrac{3}{2}(\sqrt{y})=0$

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

$x=y$

$x_{1}=y_{1}$

Substituting int he above equation, we get

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

$x=y=a$ .

Hence the tangent touches the curve at

$(a,a)$ .

Thus we get only one tangent, satisfying that criteria.

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

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

0%
$\left | m \right | > 1$
0%
$m>1$
0%
$m>-1$
0%
$\left | m \right | \leq 1$

Explanation

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

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

$|m|=|\dfrac{2x}{x^{2}+1}|$

Now, since

$x^{2}\ge 0$

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

$\Rightarrow \dfrac{1}{ x^{2}+1} \le 1$

$\Rightarrow \dfrac{2x}{ x^{2}+1} \le 1$

$|m| \le 1$

If the circle

$x^2 + y^2 + 2gx + 2fy + c =0$ is touched by y = x at P in the first quadrant, such that

$OP = 6 \sqrt2$ , then the value of

$c$ is

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

Explanation

Let the point of contact be

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

$y=x$
Hence

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

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

$2x_{1}^2=72$

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

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

$y'=1$ since

$y'$ is the slope of the line

$y=x$ .
By substituting, we get

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

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

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

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

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

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

$c=144-72$

$c=72$

$\Delta (x)=\begin{vmatrix} \sin x & \cos x &\sin 2x+\cos 2x \\ 0 &1 &1 \\ 1 &0 &-1 \end{vmatrix}$

${\Delta }'(x)$ vanishes at least once in

0%
$\left(0, \dfrac{\pi}2\right)$
0%
$\left(\dfrac{\pi}{2}, \pi \right)$
0%
$\left(0, \dfrac{\pi}4\right)$
0%
$\left(-\dfrac{\pi}{2}, 0\right)$

Explanation

$\Delta' (x)=\begin{vmatrix} \cos x &-\sin x &2\cos 2x-2 \sin 2x \\ 0 &1 &1 \\ 1 &0 &-1 \end{vmatrix}$

$\Rightarrow \Delta' (x)= -\cos x - \sin x -1(2\cos2x - 2\sin2x)$

(Expanded along first column)

$\Rightarrow \Delta' (x)= 2\sin2x -\cos x - \sin x -2\cos2x$

Clearly,

$\Delta' (x)$ is continuous and

$\Delta' (0) =-3 < 0$ and

$\Delta' \left(\dfrac{\pi}{2}\right)= 1 > 0$
Hence,

$\Delta' (x)$ will vanish at least once in

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

Hence, option A.

$\alpha = \cos 10^{\circ} - \sin 10^{\circ}, \beta = \cos 45^{\circ} - \sin 45^{\circ}, \gamma = \cos 70^{\circ} - \sin 70^{\circ}$ then the descending order of

$\alpha, \beta, \gamma$ is

0%
$\alpha, \beta, \gamma$
0%
$\gamma, \beta, \alpha$
0%
$\alpha, \gamma, \beta$
0%
$\beta, \alpha, \gamma$

Explanation

let

$f(x) = \cos {x} - \sin {x}$

$f'(x) = -(\sin {x} + \cos {x}) < 0$ for

$0 < x < \dfrac {\Pi}{2}$
so

$f(x)$ is a decreasing function.

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

Explanation

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

$\dfrac{dy}{d\theta}=d\dfrac{3}{2}(sin2\theta)$

$\dfrac{dy}{d\theta}=d\dfrac{3}{4}(cos2\theta)$ ---- (1)

$\dfrac{dy}{d\theta}=e^{\theta}sin\theta+e^{\theta}cos\theta$ ---- (2)

$\dfrac{dx}{dy}=\dfrac{3}{4}\dfrac{cos2\theta}{e^{\theta}(sin\ \theta+ cos\ \theta)}=0$

$cos2\theta=0$

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

$\theta={\pi}/{4}$

The curve given by the equation

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

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

Explanation

$y-e^{xy}+x=0$

$\therefore \dfrac{dy}{dx}-e^{xy}\left ( x\dfrac{dy}{dx}+y \right )+1=0$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{-1+ye^{xy}}{1-xe^{xy}}$

$\therefore$ if the tangent is vertical at

$(x,y),xe^{xy}=1$ 1 which is possible only if

$x=1,y=0.$

The sum of the intercepts made on the axes of coordinates by any tangent to the curve $\sqrt{x}+\sqrt{y}=2$ is equal to

0%
$4$
0%
$2$
0%
$8$
0%
None of these

Explanation

Curve equation $\sqrt{x}+\sqrt{y}=2 \Rightarrow x > 0, y>0$
$\dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=0$
$\dfrac{dy}{dx}=-\sqrt{\dfrac {y}{x}}$
and since in question its asking in general not for a particular point on curve. Hence it we find it for any point on curve then it will be true for any point.
So as use can see (1, 1) lies on curve
So, $\dfrac{dy}{dx} \int_{1, 1}=-1$
So tangent equation at (1, 1) is $y= (-1)x+c$
and it will pass through (1, 1)
So, $c=2$
tangent equation : $y= -x +2$
$x+y=2$
$\dfrac{x}{2}+\dfrac{y}{2} =1$
So x - intercept $=2$
y- intercept $=2$
So sum $=4$

If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is

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

Explanation

Time period

$\displaystyle T=2\pi \sqrt {\frac{l}{g}}$

$\displaystyle \frac { dT }{ dl } =\frac { \pi }{ \sqrt { gl } }$
Percentage error in measuring l = 2%

$\displaystyle \Rightarrow \frac { \Delta l }{ l } =\frac { 2 }{ 100 }$

$\displaystyle \Rightarrow \Delta l =\frac { 2l }{ 100 }$
Approximate error in

$T \ \displaystyle= dT= \frac { dT }{ dl } \Delta l$

$\displaystyle =\frac{T}{100}=1\% of T$
Percentage error in

$T \ =1 \%$

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.