MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 3

If there is an error of

$k%$ in measuring the edge of a cube, then the percent error in estimating its volume is

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

Explanation

Volume of cube

$V=x^{3}$

$\displaystyle \frac{dV}{dx}=3x^{2}$
Percentage error in measuring side

$= k%$

$\displaystyle \Rightarrow \frac{\delta x}{x}=\frac{k}{100}$

$\displaystyle {\delta{x}}=\frac{xk}{100}$
Approximate error in estimating volume

$=dV=(\frac{dV}{dx}){\delta x}=3x^{2}\frac{xk}{100}$
Percentage error in estimating volume

$=\frac{dV}{V}\times 100=3k$

At what points of curve

$y= \displaystyle \frac {2}{3} x^3 + \displaystyle \frac{1}{2} x^2$ , the tangent makes equal angle with the axis

0%
$\left( \displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 5 }{ 24 } \right)$ and $\left( -1,-\displaystyle \frac { 1 }{ 6 } \right)$
0%
$\left(\displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 4 }{ 9 } \right)$ and $\left( -1,0 \right)$
0%
$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 1 }{ 7 } \right)$ and $\left( -3,\displaystyle \frac { 1 }{ 2 } \right)$
0%
$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 4 }{ 47 } \right)$ and $\left( -1,-\displaystyle \frac { 1 }{ 3 } \right)$

Explanation

Given equation of curve is

$y= \dfrac {2}{3} x^3 + \dfrac{1}{2} x^2$

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

Since ,the tangent makes equal angle with x-axis,

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

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

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

$x=-1\Rightarrow y=-\dfrac{1}{6}$

$x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{24}$

If at each point of the curve

$y=x^3-ax^2 +x+1$ , the tangent is inclined at an acute angle with the positive direction of the

$x$ -axis, then

0%
$a>0$
0%
$a\leq \sqrt 3$
0%
$-\sqrt 3 \leq a \leq \sqrt 3$
0%
none of these

Explanation

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

$\dfrac{dy}{dx}=3x^{2}-2ax+1$
Since, the tangent is inclined at an acute angle with the positive direction of x-axis.

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

$\Rightarrow 3x^{2}-2ax+1>0$

$\Rightarrow D<0$

$\Rightarrow 4(a^2-3)<0$

$\Rightarrow (a-\sqrt{3})(a+\sqrt{3})<0$

$\Rightarrow -\sqrt{3}<a<\sqrt{3}$

$x+ 4y=14$ is a normal to the curve

$y^2=\alpha x ^3-\beta$ at

$(2,3)$ , then the value of

$\alpha+\beta$ is

0%
$9$
0%
$-5$
0%
$7$
0%
$-7$

Explanation

Given equation of curve is

$y^2=\alpha x ^3-\beta$
Since, it passes through (2,3)

$4=8\alpha -\beta$

$\displaystyle \frac{dy}{dx}=\frac{3\alpha x^2}{2y}$
Slope of tangent at (2,3) is

$2\alpha$
Slope of normal at

$\displaystyle (2,3)= -\frac{1}{2\alpha}$
Given equation of normal is

$x+4y=14$
Slope of normal is

$-\cfrac{1}{4}$

$\Rightarrow \alpha =2$

$\Rightarrow \beta=7$
Hence

$\alpha+\beta =9$

If there is an error of

$a \%$ in measuring the edge of a cube, then the percentage error in its surface area is

0%
$2a$
0%
$\dfrac {a}{2}$
0%
$3a$
0%
None of the above

Explanation

Surface area of cube

$S=6x^{2}$

$\Rightarrow log S = log 6 + 2log x$
On differentiating we get,

$\dfrac{\Delta S}{S} = 2 \dfrac{\Delta x}{x}$

$\dfrac{\Delta S}{S} = 2 a \%$

The angle formed by the positive

$y-axis$ and the tangent to

$y=x^2+4x-17$ at

$(5/2,-3/4)$ is

0%
${\tan^{-1}(9)}$
0%
$\displaystyle \frac{\pi}{2}-{\tan^{-1}(9)}$
0%
$\displaystyle \frac{\pi}{2}+{\tan^{-1}(9)}$
0%
none of these

Explanation

Slope of tangent to given curve at

$\dfrac{5}{2},-\dfrac{3}{4}$ is 9.

$\theta$ is the angle between the tangent and x-axis , then

$\theta=\pm tan^{-1}{9}$

Angle between y-axis and x-axis is

$\dfrac{\pi}{2}$

So, the angle between y-axis and tangent is

$\dfrac{\pi}{2} \pm tan^{-1}{9}$

The slope of the tangent to the curve

$y= \sqrt {4-x^2}$ at the point where the ordinate and the abscissa are equal is

0%
-1
0%
1
0%
0
0%
none of these

Explanation

Given equation of curve is

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

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

$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{x}{y}$
Let

$P(x_1,y_1)$ be the point on the curve such that

$x_1=y_1$
So, slope of tangent to curve at P is

$-1$ .

The curve given by

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

0%
(0,1)
0%
(1,0)
0%
(1,1)
0%
none of these

Explanation

Given equation of curve is

$x+y=e^{xy}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{y-1}{1-xe^{xy}}$
Since, the tangent is parallel to y-axis,

$\displaystyle\frac{dy}{dx}=\frac{1}{0}$

$\Rightarrow xe^{xy}=1$
Also,

$y=0$ on the line parallel to y-axis

$\Rightarrow x=1$
Hence, the required point is (1,0)

The pressure P and volume V of a gas are connected by the relation

$PV^{1/4}=constant$ . The percentage increase in the pressure corresponding to a deminition of

$\dfrac12 \%$ in the volume is

0%
$\dfrac {1}{2}$ %
0%
$\dfrac {1}{4}$ %
0%
$\dfrac {1}{8}$ %
0%
none of these

Explanation

$PV^{1/4}=constant$

$\displaystyle \Rightarrow P=\dfrac{k}{V^{{1}/{4}}}$

$\displaystyle \Rightarrow \dfrac{dP}{dV}=-\dfrac{k}{4}V^{-{5}/{4}}$

Percentage error in V

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

$\Rightarrow\displaystyle \dfrac{\Delta V}{V} =-\dfrac{1}{200}$

$\Rightarrow\displaystyle {\Delta V}=-\dfrac{V}{200}$

Approximate change in

$P\displaystyle=dP=(\dfrac{dP}{dV}){\Delta V}$

$\displaystyle =\dfrac{1}{800} {kV^{-1/4}}=\dfrac{1}{8}\%$ of P

Percentage increase in

$V \ \displaystyle =\dfrac{1}{8}\%$

If the percentage error in the edge of a cube is 1, then error in its volume is

0%
$1 \%$
0%
$2 \%$
0%
$3 \%$
0%
none of these

Explanation

Volume of cube

$V=x^{3}$

$\Rightarrow \displaystyle \dfrac{dV}{dx}=3x^{2}$

Percentage error in x is 1%.

$\Rightarrow \displaystyle \dfrac{\Delta x}{x}=\dfrac{1}{100}$

$\Rightarrow \displaystyle \Delta x=\dfrac{x}{100}$

Approximate error in V

$\displaystyle=dV=(\dfrac{dV}{dx}) \Delta x$

$\displaystyle = \dfrac{3{x}^{3}}{100}$

Percentage error in V

$\displaystyle= \dfrac{dV}{V}$

$\displaystyle=\dfrac{3}{100}=3\%$

While measuring the side of an equilateral triangle an error of

$k \%$ is marked, the percentage error in its area is

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

Explanation

Area of equilateral triangle

$A=\dfrac { \sqrt { 3 } }{ 4 } { x }^{ 2 }$

$\displaystyle \dfrac{dA}{dx}=\dfrac { \sqrt { 3 } }{ 2 }x$

Given, percentage error in measuring side

$=k\%$

$\Rightarrow \displaystyle \dfrac { \Delta x }{ x } =\dfrac { k }{ 100 }$

$\Rightarrow \displaystyle { \Delta x }=\dfrac {k x }{ 100 }$

Now, approximate error in measuring A

$\displaystyle =dA= (\dfrac{dA}{dx}){ \Delta x}$

$\displaystyle = \dfrac{k }{100} \dfrac{\sqrt{3}}{2}x^{2} =2k\%$ of A

Percentage error in measuring

$A =2k\%$

$y=x^n$ , then the ratio of relative errors in

$y$ and

$x$ is

0%
$1:1$
0%
$2:1$
0%
$1:n$
0%
$n:1$

Explanation

$y=x^{n}$

$\Rightarrow \displaystyle \dfrac{dy}{dx}=nx^{n-1}$

Approximate error in y is

$\displaystyle dy=\left (\dfrac{dy}{dx}\right) \Delta x$

$=nx^{n-1} \Delta x$

Relative error in y is

$\displaystyle \dfrac{dy}{y}=\dfrac{n}{x}\Delta x$

Approximate error in x is

$\displaystyle dx=\left (\dfrac{dx}{dy}\right) \Delta y$

$\displaystyle=\dfrac{1}{nx^{n-1}} \Delta y$

Relative error in x is

$\displaystyle \dfrac{dx}{x}=\dfrac{1}{nx^{n}}\Delta y$

Required ratio

$\displaystyle = \dfrac{\dfrac{n}{x}\Delta x}{\dfrac{1}{nx^{n}}\Delta y}$

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

$\displaystyle =\dfrac{n}{1}$

So, the ratio of relative errors in y and x is

$n:1$ .

If the percentage error in measuring the surface area of a sphere is

$\alpha$ %, then the error in its volume is

0%
$\dfrac {3}{2}\alpha\%$
0%
$\dfrac {2}{3}\alpha\%$
0%
$3\alpha\%$
0%
none of these

Explanation

Volume of sphere

$V=\dfrac{4}{3}\pi r^{3}$

$\displaystyle \dfrac{dV}{dr}=4\pi r^{2}$

Surface area of sphere

$S=4\pi r^{2}$

$\displaystyle \dfrac{dS}{dr}=8\pi r$

$\displaystyle\Rightarrow \dfrac{dV}{dS}=\dfrac{r}{2}$

Also, given percentage error in measuring S is

$\alpha$ %

$\Rightarrow \displaystyle \dfrac{\Delta S}{S}=\dfrac{\alpha}{100}$

$\Rightarrow \displaystyle \Delta S=\dfrac{\alpha S}{100}=\dfrac{4\pi r^{2}\alpha}{100}$

Approximate error in V is

$\displaystyle = dV= (\dfrac{dV}{dS}) \Delta S$

$\displaystyle = \dfrac {\alpha}{100} (2\pi r^{3})$

Percentage error in V=

$\displaystyle \dfrac{dV}{V}$

$\displaystyle=\dfrac{3}{2}\alpha$ %

If there is an error of

$0.01 cm$ in the diameter of a sphere then percentage error in surface area when the radius

$= 5 cm$ , is

0%
$0.005\%$
0%
$0.05\%$
0%
$0.1\%$
0%
$0.2\%$

Explanation

Surface area of sphere

$S=4\pi r^{2}$

$\displaystyle S=\pi D^{2}$

$\Rightarrow S=100\pi$
Also,

$\displaystyle \frac{dS}{dD}=2\pi D=20\pi$ [r=5]
Approximate error in S is

$\displaystyle dS=(\frac{dS}{dD})\Delta D$

$=20\pi (0.01)$

then

$\dfrac{dS}{S}=\dfrac{1}{500}$

$dS=0.2$ % of S
Percentage error in

$S=0.2%$

The circumference of a circle is measured as

$28 cm$ with an error of

$0.01 cm$ . The percentage error in the area is

0%
$\dfrac {1}{14}$
0%
$0.01$
0%
$\dfrac {1}{7}$
0%
none of these

Explanation

Circumference

$C=2\pi r$

$\Rightarrow\displaystyle r=\frac{14}{\pi}$
Also,

$\displaystyle \frac{dC}{dr}=2\pi$
Area of circle

$A=\pi r^{2}$

$\Rightarrow\displaystyle A=\frac{{14}^{2}}{\pi}$
Also,

$\displaystyle \frac{dA}{dr}=2\pi r$

$\displaystyle \Rightarrow \frac{dA}{dC}=r=\frac{14}{pi}$
Approximate error in

$A$ is

$\displaystyle dA=( \frac{dA}{dC}) \Delta C$

$\displaystyle=\frac{14}{\pi}\frac{1}{100}$

$\displaystyle=\frac{1}{1400}$ of A
Percentage error in

$A \ \displaystyle =\frac{1}{14}\%$

The height of a cylinder is equal to the radius. If an error of

$\alpha$ % is made in the height, then percentage error in its volume is

0%
$\alpha$ %
0%
$2\alpha$ %
0%
$3\alpha$ %
0%
none of these

Explanation

Volume of cylinder

$V= \pi { r }^{ 2 }h$

Since,

$h=r$

$V=\pi {h}^{3}$

$\displaystyle \dfrac{dV}{dh}=3\pi h^{2}$

Given, percentage error in measuring height

$=\alpha$ %

$\Rightarrow \displaystyle \dfrac { \Delta h }{ h } =\dfrac { \alpha }{ 100 }$

$\Rightarrow \displaystyle { \Delta h }=\dfrac {\alpha h }{ 100 }$

Now, approximate error in measuring V

$\displaystyle =dV= (\dfrac{dV}{dh}){ \Delta h}$

$\displaystyle = \dfrac{3\alpha }{100} {\pi h^{3}} =3\alpha$ % of V

Percentage error in measuring

$V =3\alpha$ %

If an error of

$k\%$ is made in measuring the radius of a sphere, then percentage error in its volume is

0%
$k\%$
0%
$3k\%$
0%
$2k\%$
0%
$\dfrac23k\%$

Explanation

Volume of sphere

$V=\dfrac { 4 }{ 3 } \pi { r }^{ 3 }$

$\displaystyle \dfrac{dV}{dr}=4\pi r^{2}$

Given, percentage error in measuring radius =k%

$\Rightarrow \displaystyle \dfrac { \Delta r }{ r } =\dfrac { k }{ 100 }$

$\Rightarrow \displaystyle { \Delta r }=\dfrac { kr }{ 100 }$

Now, approximate error in measuring V

$\displaystyle =dV= (\dfrac{dV}{dr}){ \Delta r}$

$\displaystyle = \dfrac{k}{100} {4\pi r^{3}} = \dfrac{3k}{100}V =3k\%$ of V

Percentage error in measuring

$V =3k\%$

If the ratio of base radius and height of a cone is 1:2 and percentage error in radius is

$\lambda$ %, then the error in its volume is

0%
$\lambda$ %
0%
$2\lambda$ %
0%
$3\lambda$ %
0%
none of these

Explanation

Volume of cone

$V=\dfrac { 1 }{ 3 } \pi { r }^{ 2 }h$

Given,

$\displaystyle \dfrac{r}{h}=\dfrac{1}{2}$

$\Rightarrow V=\dfrac { 2 }{ 3 } \pi { r }^{ 3 }$

$\displaystyle \dfrac{dV}{dr}=2\pi r^{2}$

Percentage error in measuring r

$=\lambda$ %

$\Rightarrow \displaystyle \dfrac{\Delta r}{r}=\dfrac{\lambda}{100}$

$\Rightarrow \displaystyle \Delta r =\dfrac{\lambda r}{100}$

Approximate error in V

$\displaystyle=dV=(\dfrac{dV}{dr}) \Delta r$

$\displaystyle=\dfrac{\lambda}{100} (2\pi r^{3})=\lambda\%$ of V

Percentage error in

$V =\lambda\%$

$T=2\pi \sqrt {\dfrac {l}{g}}$ , then relative errors in T and l are in the ratio

0%
$1/2$
0%
$2$
0%
$1/2\pi$
0%
none of these

Explanation

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

$\displaystyle \dfrac{dT}{dl}=\dfrac{\pi}{\sqrt{lg}}$

Approximate error in T

$=\displaystyle dT=(\dfrac{dT}{dl})\Delta l$

$\displaystyle = \dfrac{\pi}{\sqrt{lg}}\Delta l$

Relative error in T

$\displaystyle =\dfrac{dT}{T}=\dfrac{1}{2l} \Delta l$

Approximate error in l

$=\displaystyle dl=(\dfrac{dl}{dT})\Delta T$

$\displaystyle = \dfrac{\sqrt{lg}}{\pi}\Delta T$

Relative error in l i

$\displaystyle =\dfrac{dl}{l}=\dfrac{1}{\pi} \sqrt{\dfrac{g}{l}}\Delta T$

Required ratio

$\displaystyle =\dfrac{\dfrac{1}{2l} \Delta l}{\dfrac{1}{\pi} \sqrt{\dfrac{g}{l}}\Delta T}$

$\displaystyle =\dfrac {\pi}{2\sqrt{lg}} \dfrac{\Delta l}{\Delta T}$

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

In a

$\Delta ABC$ if sides a and b remain constant such that

$\alpha$ is the error in C, then relative error in its area is

0%
$\alpha \cot C$
0%
$\alpha \sin C$
0%
$\alpha\tan C$
0%
$\alpha\cos C$

Explanation

Area of triangle

$S\displaystyle =\dfrac {1}{2}ab \sin C$

$\displaystyle \Rightarrow \dfrac {dS}{dC}=\dfrac {1}{2}ab \cos C$

Now, approximate error in S is

$\Delta S=\dfrac {dS}{dC}\Delta C$

$\displaystyle\Rightarrow \Delta S=\dfrac {1}{2}ab \cos C \alpha [\because \Delta C=\alpha]$

$\displaystyle\Rightarrow \dfrac {\Delta S}{S}=\dfrac {\dfrac {1}{2}ab \cos C}{\dfrac {1}{2}ab \sin C}\alpha=\alpha \cot C$

The circumference of a circle is measured as

$56$ cm with an error

$0.02$ cm. The percentage error in its area is

0%
$\dfrac {1}{7}$
0%
$\dfrac {1}{28}$
0%
$\dfrac {1}{14}$
0%
$\dfrac {1}{56}$

Explanation

Circumference of circle

$C=2\pi r=56cm$

$\Rightarrow \displaystyle r=\frac{28}{\pi}$

Also,

$\displaystyle \frac{dC}{dr}=2\pi$

Area of circle

$A=\pi r^{2}$

$\displaystyle \frac{dA}{dr}=2\pi r$

$\Rightarrow\displaystyle \frac{dA}{dC}=r =\frac{28}{\pi}$

Approximate error in A

$=\displaystyle dA=(\frac{dA}{dC})\Delta C$

$= r (0.02)$

$\Rightarrow\displaystyle \frac{dA}{A}= \frac{0.02}{\pi r}=\frac{1}{1400}$

Percentage error in A is

$\displaystyle\frac{1}{14}$ %

The point(s) at each of which the tangents to the curve

$\displaystyle y = x^3 - 3x^2 - 7x + 6$ cut off on the positive semi axis

$OX$ a line segment half that on the negative semi axis

$OY$ , then the co-ordinates of the point(s) is/are give by:

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

Explanation

$y={ x }^{ 3 }-3{ x }^{ 2 }-7x+6\\ \dfrac { dy }{ dx } =3{ x }^{ 2 }-6x-7$
Let coordinate of X be

$\left( a,0 \right)$ then coordinate of Y is

$\left( 0,-2a \right)$
Therefore slope of XY is

$\cfrac { 0+2a }{ a-0 } =2$
Equating slope

$3{ x }^{ 2 }-6x-7=2\\ 3{ x }^{ 2 }-6x-9=0$
Solving we get

$x=-1,y=9\quad \\ x=3,y=-15$
Hence, option 'B' is correct.

If an error of

$1^o$ is made in measuring the angle of a sector of radius

$30 \ cm$ , then the approximate error in its area is

0%
$450 cm^2$
0%
$25\pi cm^2$
0%
$2.5\pi cm^2$
0%
none of these

Explanation

Area of sector

$\displaystyle A=\dfrac{\pi r^{2}\theta}{360}$

Given

$r=30 cm, d{\theta}=1^{0}$

Approximate error in A is

$\displaystyle=dA=(\dfrac{dA}{d\theta})\Delta \theta$

$\displaystyle = \dfrac{900\pi}{360}$

$\displaystyle \Rightarrow dA =2.5 \pi cm^{2}$

A line L is perpendicular to the curve

$\displaystyle y = \dfrac {x^2}{4} - 2$ at its point P and passes through (10, -1). The coordinates of the point P are

0%
(2, -1)
0%
(6, 7)
0%
(0, -2)
0%
(4, 2)

Explanation

Let point

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ on curve

$y=\cfrac { { x }^{ 2 } }{ 4 } -2$
Slope of tangent is

$\dfrac { dy }{ dx } =\dfrac { { x }_{ 1 } }{ 2 }$
So slope of perpendicular line is

$-\dfrac { 2 }{ { x }_{ 1 } }$ ...(1)
That perpendicular line also passes through

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$\left( 10,-1 \right)$
Slope of line from these point is

$\dfrac { { y }_{ 1 }+1 }{ { x }_{ 1 }-10 }$ ...(2)
Equating (1) and (2) and solving we get,

${ x }_{ 1 }=4$ and

${ y }_{ 1 }=2$
Thus point is

$\left( 4,2 \right)$
Hence, option 'D' is correct.

If the tangent at each point of the curve

$\displaystyle y=\frac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$ makes an acute angle with the positive direction of x-axis, then

0%
$a\ge 1$
0%
$-1\le a\le 1$
0%
$a\le -1$
0%
none of these

Explanation

We have,

$\displaystyle y=\dfrac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$

$\displaystyle \Rightarrow \dfrac { dy }{ dx } =2{ x }^{ 2 }-4ax+2.$

Since, the tangent makes an acute angle with the positive direction of x-axis, therefore,

$\displaystyle \dfrac { dy }{ dx } \ge 0\Rightarrow 2{ x }^{ 2 }-4ax+2\ge 0$ for all

$x$

$\Rightarrow 16{ a }^{ 2 }-16\le 0$

$(\because$ Disc.

$={ \left( 4a \right) }^{ 2 }-4\left( 2 \right) \left( 2 \right) \le 0)$

$\Rightarrow { a }^{ 2 }-1\le 0$ i.e.,

$\left( a-1 \right) \left( a+1 \right) \le 0$

$\Rightarrow -1\le a\le 1.$

Let

$f\left( x \right) =\left\{ \begin{matrix} { x }^{ { 3 }/{ 5 } }\quad \quad \quad x\le 1 \\ -{ \left( x-2 \right) }^{ 3 }\quad x>1 \end{matrix} \right.$
then the number of critical points on the graph of the function is

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

Explanation

$f\left( x \right) =\left\{ \begin{matrix} { x }^{ { 3 }/{ 5 } }\quad \quad \quad x\le 1 \\ -{\left( x-2 \right) }^{ 3 }\quad x>1 \end{matrix} \right.$
Critical points will at
At

$x=2$ , as

${ f }^{ ' }(x)=0$
At

$x=0$ ,

${ f }^{ ' }(x)$ is not defined
At

$x=1$ ,

${ f }^{ ' }(x)$ does not exist , therefore its a critical point.
Thus, there are three critical points.
Hence, option 'C' is correct.

In a

$\Delta ABC$ the sides b and c are given. If there is an error

$\Delta A$ in measuring angle A, then the error

$\Delta a$ in side a is given by

0%
$\dfrac {S}{2a}\Delta A$
0%
$\dfrac {2S}{a}\Delta A$
0%
bc sin A $\Delta A$
0%
none of these

Explanation

$\triangle ABC$ we have

$\Rightarrow d\left( 2bc\cos { A } \right) =d\left( { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } \right) \\ \Rightarrow -2bc\sin { A } dA=-2ada\\ \Rightarrow bc\sin { A } dA=ada$

$\displaystyle \Rightarrow \frac { 2 }{ a } \left( \frac { 1 }{ 2 } bc\sin { B } \right) dA=da$

$\displaystyle \Rightarrow da=\frac { 2S }{ a } dA$

$\displaystyle \Rightarrow \triangle a=\frac { 2S }{ a } dA\\ \left[ \because dx\equiv \triangle a\quad and\quad dA=BA \right]$

If errors of

$1\%$ each are made in the base radius and height of a cylinder, then the percentage error in its volume is

0%
$1\%$
0%
$2\%$
0%
$3\%$
0%
none of these

Explanation

Given, percentage error in r is 1%

$\Rightarrow \displaystyle \frac{\Delta r}{r}=\frac{1}{100}$

$\Rightarrow \displaystyle \Delta r=\frac{r}{100}$
Also given, percentage error in h is 1%

$\Rightarrow \displaystyle \frac{\Delta h}{h}=\frac{1}{100}$

$\Rightarrow \displaystyle \Delta h=\frac{h}{100}$

Now, volume of cylinder

$V=\pi r^{2}h$

$\dfrac{dV}{dr}=\pi2rh+\pi r^2\dfrac{dh}{dr}$

$\Delta V=\pi [r^{2}\Delta h+2rh\Delta r]$

$\displaystyle \Delta V=\pi[r^{2}\frac{h}{100}+2rh\frac{r}{100}]$

$\displaystyle\Delta V=\pi r^{2}h[\frac{3}{100}]$

$\Rightarrow\displaystyle\frac{\Delta V}{V}=\frac{3}{100}$
Percentage error in V is 3%

Let

$\displaystyle g'(x) > 0$ and

$\displaystyle f'(x) < 0 \forall x \epsilon R$ , then

0%
$\displaystyle f(f(x + 1)) > f(f(x - 1))$
0%
$\displaystyle f(g(x - 1)) > f(g(x + 1))$
0%
$\displaystyle g(f(x + 1)) > g(f(x - 1))$
0%
$\displaystyle g(g(x + 1)) > g(g(x - 1))$

Explanation

Given

$g'(x) > 0$ and

$f'(x) < 0$

$\Rightarrow g(x)$ is increasing and

$f(x)$ is decreasing.
Now option A,
Hence,

$f(x+1) < f(x-1)$

$\Rightarrow f(f(x+1)) > f(f(x-1))$
option B,
Hence,

$g(x-1) < g(x+1)$

$\Rightarrow f(g(x-1)) > f(g(x+1))$
option C,
Hence,

$f(x+1) < f(x-1)$

$\Rightarrow g(f(x-1)) > g(f(x+1))$
option D,
Hence,

$g(x-1) < g(x+1)$

$\Rightarrow g(g(x+1)) > g(g(x-1))$
Therefore option A,B,D are correct

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

$\displaystyle y^2 = px^3 + q$ at

$(2, 3)$ , then

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

Explanation

Given equation of curve

$\displaystyle y^2 = px^3 + q\\$

$\displaystyle \dfrac{dy}{dx}=\dfrac{3px^{2}}{y}\\$
Slope of tangent at (2,3)

$= \displaystyle (\dfrac{dy}{dx})_{(2,3)}=2p$

Given equation of tangent is

$y=4x-5$
Slope of tangent =4

$\Rightarrow 2p=4$

$\Rightarrow p=2$

Since, (2,3) lies on

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

$9=16+q$

$\Rightarrow q=-7$

If the tangent to the curve

$xy + ax + by = 0$ at (1, 1) makes an angle

$\displaystyle \tan ^{-1}(2)$ with x-axis, then

$\displaystyle a + 2b$ is equal to

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

Explanation

$xy'+y+a+by'=0$

$(x+b)y'+a+y=0$

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

$y'_{1,1}=\dfrac{-(a+1)}{b+1}$

$=2$
Or

$2b+2=-a-1$

$2b+a=-3$

The set of values of

$a$ for which the function

$\displaystyle f(x) = (4a - 3) (x + \ln5) + 2(a - 7) \cot \frac {x}{2} \sin^2 \frac {x}{2}$ does not posses critical points in its domain is

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

Explanation

$\textbf{Step 1: Simplify the given equation}$

$f(x)=(4a-3)(x+ \ln5) + 2(a-7) \cos {\frac{x}{2}} \times \sin{\frac{x}{2}}$

$f(x)=(4a-3)(x+ \ln5) + (a-7) (2\cos {\frac{x}{2}} \times \sin{\frac{x}{2}})$

$f(x)=(4a-3)(x+ \ln5) + (a-7) (\sin x)$

$\boldsymbol{[\sin2\theta = 2sin \theta \cos \theta]}$

$\textbf{Step 2: Identifying range of a such that function doesn't posses critical points}$

$\text{Differentiate}\ f(x)\ \text{with respect to x}$

$f'(x)=(4a-3)(1) + (a-7) (\cos x)$

$\text{To find critical points, equate}\ f'(x) =0$

$f'(x)=(4a-3)(1) + (a-7) (\cos x)=0$

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

$\text{But the range of} \cos x\ \text{is} [-1,1]$

$\text{Hence,}$

$-1\le-\cfrac{4a-3}{a-7}\le1$

$\text{On simplifying, we get}$

$a\le-\cfrac{4}{3}\ \text{and}\ a\ge2$

$\textbf{Hence, the range where the given function does not posses critical points is}\ \boldsymbol{\left(-\infty,-\frac{4}{3}\right) \cup (2,\infty)}$

$\textbf{Answer: (A)}$

Let

$f$ be a decreasing function in

$(a,b]$ , then which of the following must be true?

0%
$f$ is continuous at $b$
0%
$\displaystyle f'(b)<0$
0%
$\displaystyle \lim_{x\to b}f(x)\leq f(b)$
0%
$\displaystyle \lim_{x\to b}f(x)\geq f(b)$

Explanation

$f$ is a decreasing function in

$(a,b]$
So

$f'(x) \le 0$ not strictly decreasing.
Hence

$f(b) \le f(x)$ in

$(a,b]$
Regarding continuity, only LHS known at b, no information about RHS.

$\displaystyle f:(0, \infty) \rightarrow (-\frac {\pi}{2}, \frac {\pi}{2})$ be defined as,

$\displaystyle f(x) = arc \: \tan( \: x)$

The above function can be classified as

0%
injective but not surjective
0%
surjective but not injective
0%
neither injective nor surjective
0%
both injective as well as surjective

Explanation

$f(x) = \tan^{-1}x$

$f'(x) = \cfrac{1}{1+x^2} > 0 \forall x \in R$
Thus

$f(x)$ is injective.
Also

$f(x)$ is surjective as co-domain and range are same.

If at each point of the curve

$y=x^{3}-ax^{2}+x+1$ the tangent is inclined at an acute angle with the positive direction of the x-axis then

0%
$4a>0$
0%
$a\leq \sqrt{3}$
0%
$-\sqrt{3}\leq a\leq \sqrt{3}$
0%
none of these

The line

$ax + by = 1$ is tangent to the curve

$\displaystyle ax^2 + by^2 = 1$ , if

$(a, b)$ can be equal to

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

Explanation

$\displaystyle ax^2 + by^2 = 1$

$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{ax}{by}$
Slope of tangent at

$(x_1, y_1)$ is

$-\dfrac{ax_1}{by_1}$

Equation of tangent is

$ax+by=1$
Slope of tangent is

$\displaystyle -\frac{a}{b}$

$\Rightarrow \displaystyle -\frac{a{x}_{1}}{b{y}_{1}}=-\frac{a}{b}$

$\Rightarrow x_{1}=y_{1}$

substitute in

$ax^2+by^2=1, ax+by=1$

$\Rightarrow x_1=y_1 = \dfrac1{a+b}$ and we get

$a+b=1$

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

${ e }^{ 2y }=1+4{ x }^{ 2 }$ , then

0%
$m<1$
0%
$\left| m \right| \le 1$
0%
$\left| m \right| >1$
0%
None of these

Explanation

We have,

$\displaystyle { e }^{ 2y }=1+4{ x }^{ 2 }\quad \Rightarrow { e }^{ 2y }.2\dfrac { dy }{ dx } =8x$

$\displaystyle \Rightarrow \dfrac { dy }{ dx } =\dfrac { 4x }{ { e }^{ 2y } } =\dfrac { 4x }{ 1+4{ x }^{ 2 } } .$

$\therefore$ Slope of tangent

$\displaystyle =m=\dfrac { 4x }{ 1+4{ x }^{ 2 } }$

$\displaystyle \Rightarrow \left| m \right| =\dfrac { 4\left| x \right| }{ 1+4{ \left| x \right| }^{ 2 } } \le 1$

$\displaystyle \left[ \because { \left( 1-2\left| x \right| \right) }^{ 2 }\ge 0\quad \quad \Rightarrow 1+4{ \left| x \right| }^{ 2 }-4\left| x \right| \ge 0 \Rightarrow \dfrac { 4\left| x \right| }{ 1+4{ \left| x \right| }^{ 2 } } \le 1 \right]$

$m$ be the slope of tangent to the curve

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

0%
$|m|>1$
0%
$m<1$
0%
$|m|<1$
0%
$|m|\leq 1$

Explanation

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

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

$\dfrac{dy}{dx}$

$=m$

$=\dfrac{2x}{1+x^{2}}$
Now

$(1-x)^{2}\geq0$
Or

$1+x^{2}-2x\geq0$
Or

$1+x^{2}\geq2x$
Or

$\dfrac{2x}{1+x^{2}}\leq 1$
Hence

$|m|\leq 1$

The slope of the tangent to the curve

$y=x^{2}-x$ at the point where the line

$y=2$ cuts the curve in the first quadrant is

0%
$2$
0%
$3$
0%
$-3$
0%
none of these

Explanation

For

$y=2, { x }^{ 2 }-x-2=0$ gives

$x=-1$ and

$x=2$ as point is in first quadrant.
Therefore point is

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

$\cfrac { dy }{ dx } =2x-1=4-1=3$

The slope of the tangent to the locus

$y=\cos^{-1}\left ( \cos x \right )$ at

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

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

Explanation

For

$x=\cfrac { \pi }{ 4 } \quad y=\cos^{ -1 }\cos\left( \cfrac { \pi }{ 4 } \right) =\cfrac { \pi }{ 4 }$

Therefore point is

$\left( \cfrac { \pi }{ 4 } ,\cfrac { \pi }{ 4 } \right)$

Now

$y=\cos^{ -1 }\cos\left( x \right) \Rightarrow \cos\left( y \right) =\cos\left( x \right)$

Slope is

$\cfrac { dy }{ dx } =\cfrac { -\sin\left( x \right) }{ -\sin\left( y \right) } =1$

$P(2, 2)$ and

$Q\left ( \displaystyle \frac{1}{2}, -1 \right )$ are two points on the parabola

$y^{2}=2x$ . The coordinates of the point

$R$ on the parabola, where tangent to the curve is parallel to the chord

$PQ$ is

0%
$\left ( \displaystyle \frac{5}{4}, \sqrt{\frac{5}{2}} \right )$
0%
$(2, -1)$
0%
$\left ( \displaystyle \frac{1}{8}, \frac{1}{2} \right )$
0%
none of these

Explanation

Let point

$\left( 2{ t }^{ 2 },2t \right)$ lies on curve

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

Then slope is

$\displaystyle\dfrac { dy }{ dx } =\dfrac { 2 }{ 2y } =\dfrac { 1 }{ 2t }$

And this slope is equal to the the slope of line

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

Equating slope we get

$\displaystyle t=\dfrac { 1 }{ 4 }$

Hence point is

$\displaystyle\left( \dfrac { 1 }{ 8 } ,\dfrac { 1 }{ 2 } \right)$

The function

$\: f\left( x \right) =x^{ 3 }+\lambda x^{ 2 }+5x+\sin 2x$ will be an invertible function if

$\: \lambda$ belongs to

0%
$\displaystyle \:\left ( -\infty, -3 \right )$
0%
$\displaystyle \:\left ( -3, 3 \right )$
0%
$\displaystyle \:\left ( 3, +\infty \right )$
0%
none of these

Explanation

$\: \mathrm{f}\left( x \right) =x^{ 3 }+\lambda x^{ 2 }+5x+\sin 2x$

$\Rightarrow \mathrm{f'}\left( x \right) =3x^2+2 \lambda x+5+2\cos 2x$
For

$\mathrm{f}(x)$ to be invertible

$\mathrm{f'}(x) > 0\forall x \in R$

$3x^2+2\lambda x+5+2\cos2x> 0\Rightarrow 3x^2+2\lambda x +3> 0$
For this discriminant of above quadratic should be negative,

$\Rightarrow \lambda^2-9< 0$

$\Rightarrow \lambda \in (-3,3)$

The cricital points(s) of f(x)=

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

0%
$x=0$
0%
$x=2$
0%
$x=4$
0%
none of these

Explanation

for

$x\geq 2$

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

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

for

$x<2$

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

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

for critical points:

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

$\Rightarrow x=0,4$
now,

$f'\left( 2+ \right) =1$ &

$f'\left( 2- \right) =-1$
Since,

$f'(x)$ changes sign at

$x=2$
Therefore,

$x=2$ is also a critical point
Thus critical points of

$f(x)$ are

$x=0,2,4$

Ans: A,B,C

The curve given by

$x+y=e^{xy}$ has a tangent as the

$y$ -axis at the point

0%
$(0, 1)$
0%
$(1, 0)$
0%
$(1, 1)$
0%
none of these

Explanation

We have,

$x+y=e^{xy}\Rightarrow (1)$
Differentiate both sides w.r.t.

$x$ we get,

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

$\Rightarrow \dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$
Now for tangent to be y-axis,

$\left|\dfrac{dy}{dx}\right|\to \infty$

$\Rightarrow 1-xe^{xy}=0\Rightarrow (2)$
Solving (1) and (2) we get,

$(x,y)=(1,0)$ , which is the required point

Let

$\displaystyle f(x)=e\:^{x}\sin x$ be the equation of a curve. If at

$\displaystyle x=a,0\leq a\leq 2\pi$ , the slope of the tangent is the maximum then the value of

$a$ is

0%
$\displaystyle \pi /2$
0%
$\displaystyle 3\pi /2$
0%
$\displaystyle \pi$
0%
$\displaystyle \pi /4$

Explanation

Let

$y={ e }^{ x }sin\left( x \right)$
Slope is

$\cfrac { dy }{ dx } ={ e }^{ x }\left( sin\left( x \right) +cos\left( x \right) \right)$
To maximize slope

$\cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ 2e }^{ x }cos\left( x \right) =0$
Critical point is

$x=\cfrac { \pi }{ 2 }$
And at

$x=\cfrac { \pi }{ 2 }$ is the point of maxima

The slope of the tangent to the curve

$y=\sqrt{4-x^{2}}$ at the point where the ordinate and the abscissa are equal, is

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

Explanation

Let abscissa = ordinate =

$a$ , then point is

$\left( a,a \right)$
Now for

$y=\sqrt { 4-{ x }^{ 2 } } \Rightarrow { y }^{ 2 }=4-{ x }^{ 2 }$
Hence slope is

$\cfrac { dy }{ dx } =\cfrac { -2x }{ 2y } =\cfrac { -2a }{ 2a } =-1$

The equation of the curve is given by

$x=e^{t}\sin t$ ,

$y=e^{t}\cos t$ . The inclination of the tangent to the curve at the point

$t=\displaystyle \frac{\pi }{4}$ is

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

Explanation

$dx=e^{t}(\cos t+\sin t).dt$
$dy=e^{t}[\cos t-\sin t).dt$
Hence
$\dfrac{dy}{dx}_{t=\tfrac{\pi}{4}}$

$=\dfrac{\cos t-\sin t}{\cos t+\sin t}_{t=\tfrac{\pi}{4}}$

$=0$

The point on the curve

$\sqrt{x}+\sqrt{y}=2a^{2}$ , where the tangent is equally inclined to the axes, is

0%
$\left ( a^{4}, a^{4} \right )$
0%
$\left ( 0, 4a^{4} \right )$
0%
$\left ( 4a^{4}, 0 \right )$
0%
none of these

Explanation

$y=x$ is the equation of line as it is equally inclined to axes
So let point

$(k,k)$ is the point of tangency
Therefore,

$\sqrt { k } +\sqrt { k } =2{ a }^{ 2 }\\ k={ a }^{ 4 }$
Hence point is

$\left( { a }^{ 4 },{ a }^{ 4 } \right)$

Let

$y=f(x)$ be the equation of a parabola which is touches by the line

$y=x$ at the point where

$x=1$ . Then

0%
$f^{'}\left ( 0 \right )=f^{'}\left ( 1 \right )$
0%
$f^{'}\left ( 1 \right )=1$
0%
$f\left ( 0 \right )+f^{'}\left ( 0 \right )+f^{''}\left ( 0 \right )=1$
0%
$2f\left ( 0 \right )=1-f^{'}\left ( 0 \right )$

Explanation

Solving

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

$y=x$ we get

$f\left( x \right) =x$
Differentiating this w.r.t x we get

${ f }^{ ' }\left( x \right) =1$ ...(1)
At

$x=1$ we get

${ f }^{ ' }\left( 1 \right) =1$
Differentiating (1) w.r.t x we get

${ f }^{ " }\left( x \right) =0$
For

$x=0$

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

${ f }^{ ' }\left( 0 \right) =1$ and

${ f }^{ " }\left( 0 \right) =0$
Thus,

$f\left( 0 \right) +{ f }^{ ' }\left( 0 \right) +{ f }^{ " }\left( 0 \right) =0+1+0=1$

$f'(0)=f'(1)=1$ and

$2f'(0) = 1-f'(0)=0$
Hence, options 'A', 'B', 'C' and 'D' are correct.

The number of tangents to the curve

$y^{2}-2x^{3}-4y+8=0$ that pass through

$(1, 0)$ is

0%
$3$
0%
$1$
0%
$2$
0%
$6$

Explanation

Point

$(0,1)$ doesn't lie on curve, so let

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ is point of contact
Therefore slope of

${ y }^{ 2 }-2{ x }^{ 3 }-4y+8=0$ is

$\cfrac { dy }{ dx } =\cfrac { 6{ x }^{ 2 } }{ 2y-4 } =\cfrac { 6{ x }_{ 1 }^{ 2 } }{ 2{ y }_{ 1 }-2 }$
And slope of line trough

$\left( 1,0 \right)$ and

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ is

$\\ \cfrac { 0-{ y }_{ 1 } }{ 1-{ x }_{ 1 } }$
Equating slopes we get

$6{ x }_{ 1 }^{ 2 }-6{ x }_{ 1 }^{ 3 }=-2{ y }_{ 1 }^{ 2 }-4{ y }_{ 1 }$
And on solving it with equation of curve we get a quadratic equation. That give two values of slope of tangent.
Hence 2 tangents is possible.

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 3 - 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 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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