MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 13

Let

$\emptyset$

$(x)=(f(x))^{3}-3(f(x))^{2}+4f(x)+5x+3 sin x+4 cos x\forall x\in$ R,where

$f(x)$ is a differentiable function

$\forall x\in R$ , then

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$\emptyset$ is increasing whenever f is increasing
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$\emptyset$ is increasing whenever f is decreasing
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$\emptyset$ is decreasing whenever f is decreasing
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$\emptyset$ is decreasing if f'(x)=-11

Let N be the set of positive integers. For all

$n \in N$ , let

$f_n = (n + 1)^{1/3} - n^{1/3}$ and

$A = \left\{n \in N : f_{n +1} < \dfrac{1}{3(n + 1)^{2/3}} < f_n \right\}$
Then

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A = N
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A is a finite set
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the complement of A in N is nonempty, but finite
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A and its complement in N are both infinite

A Stationary point of

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

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

$f:R\rightarrow R$ is the function defined by

$f\left( x \right) =\frac { { e }^{ { x }^{ 2 } }-{ e }^{ -x^{ 2 } } }{ { e }^{ x^{ 2 } }+{ e }^{ { -x }^{ 2 } } } ,$ then

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$f(x)$ is an increasing function
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$f(x)$ is an decreasing function
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$f(x)$ is onto (surjective)
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Into function

$\int { \dfrac { { 2x }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) ^{ 2 } } } dx=f\left( x \right) +c$ where f (0) = 0, then

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f(x) is an increasing function
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x = 0 is point of extremum
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f(x) is concave up for $x\in \left( -\infty ,-1 \right) \cup \left( 0,1 \right)$
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the curve f(x) has 3 points of inflection

$A\;sationary\;po\operatorname{int} \;of\;f\left( x \right) = \sqrt {16 - {x^2}} \;is$

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$\left( {4,0} \right)$
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$\left( { - 4,0} \right)$
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$\left( {0,4} \right)$
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$\left( { - 4,4} \right)$

The slope of the tangent to the curve at a point (x, y) on it is proportional to (x-2). If the slope of the tangent to the curve at (10,-9) on it -The equation of the curve is

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$y=k{ \left( x-2 \right) }^{ 2 }$
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$y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +1$
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$y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +3$
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$y=k{ \left( x+2 \right) }^{ 2 }$

Explanation

${\textbf{Step -1: Find the value of slope}}{\textbf{.}}$

${\text{The slope of the tangent to the curve at a point }}\left( {x,y} \right){\text{ on it is proportional to }}\left( {x - 2} \right).$

$= \dfrac{{dy}}{{dx}} = m\left( {x - 2} \right)$

$= \dfrac{{dy}}{{dx}} = k\left( {x - 2} \right)\left[ {{\text{where }}k{\text{ is a proportionality constant}}} \right]$

${\text{Slope of the tangent to the curve at }}\left( {10, - 9} \right){\text{ is }} - 3.$

$\Rightarrow - 3 = k\left( {10 - 2} \right)$

$\Rightarrow - 3 = 8k$

$\Rightarrow k = - \dfrac{3}{8}$

$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{3}{8}\left( {x - 2} \right)$

${\textbf{Step -2: Integrate the formed equation and solve it further}}{\textbf{.}}$

${\text{Integrating both sides,}}$

$\Rightarrow y = - \dfrac{3}{8}\dfrac{{{{\left( {x - 2} \right)}^2}}}{2} + c{\text{ }}.......\left( i \right){\text{ }}\left[ {{\text{Eqaution of curve}}} \right]$

${\text{Also, }}\left( {10, - 9} \right){\text{ satisfy the equation of curve as it lies on the curve}}{\text{.}}$

$\therefore - 9 = - \dfrac{3}{8}\dfrac{{{{\left( {10 - 2} \right)}^2}}}{2} + c$

$\Rightarrow - 9 = - \dfrac{3}{{16}} \times {8^2} + c$

$\Rightarrow c = 3$

${\textbf{Hence, the equation of the curve is }}\mathbf{y= - \dfrac{3}{{16}}{\left( {x - 2} \right)^2} + 3.}$

At any point on the curve

$2x^{2}y^{2}-x^{4}=c$ , the mean proportional between the abscissa and the difference between the abscissa and the sub-normal drawn to the curve at the same point is equal to

0%
ordinate
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radius vector
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x-intercept of tangent
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sub-tangent

Given

$g(x)= \dfrac{x+2}{x-1}$ and the line 3x + y -10 =0, then the line is

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tangent to g(x)
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normal to g(x)
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chord of g(x)
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none of these

If the 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
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$\dfrac{k}{3}$
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None of these

If the line joining the point (0, 3 ) and (5, -2) is a tangent to the curve

$y= \dfrac{c}{x+1}$ , then the value of c is

0%
1
0%
-2
0%
4
0%
None of these

$f(x) = \left\{\begin{matrix} -x^2, & \text{for} \ x < 0 \\ x^2 + 8, & \text{for} \ x \ge 0 \end{matrix}\right.$

Let . Then x-intercept of the line, thet is , the tangent to the graph of f(x) is

0%
zero
0%
-1
0%
-2
0%
-4

$N$ characters of information are held on magnetic tape, in batches of

$x$ characters each, the batch processing time is

$\alpha +\beta x^2$ seconds.

$\alpha$ and

$\beta$ are constants. The optical value of

$x$ for last processing is,

0%
$\dfrac{\alpha}{\beta}$
0%
$\dfrac{\beta }{\alpha}$
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$\sqrt{\dfrac{\alpha}{\beta}}$
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$\sqrt{\dfrac{\beta }{\alpha}}$

The equation of the curve

$y = be^{-x/a}$ at the point where it crosses the y-axis is

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$\dfrac{x}{a}-\dfrac{y}{b}=1$
0%
$ax = by = 1$
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$ax-by=1$
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$\dfrac{x}{a}+\dfrac{y}{b}=1$

A curve is represented by the equations

$x=sec^{2}t$ and

$y=\cot t,$ where t is a parameter. If the tangent at the point P on the curve, where

$t=\pi /4$ , meets the curve again at the point Q, then

$\left | PQ \right |$ is equal to

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$\dfrac{5\sqrt{3}}{2}$
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$\dfrac{5\sqrt{5}}{2}$
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$\dfrac{2\sqrt{5}}{3}$
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$\dfrac{3\sqrt{5}}{2}$

Let f be a continuous, differetiable and bijective function. If the tangent to y= f (x) at x = a is also the normal to y = f (x) at x = b then there exists at least one

$c \epsilon (a, b)$ such that

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f'(c) = 0
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$f (c)> 0$
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$f (c)< 0$
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None of these

Given the curves

$y=f(x)$ passing through the point

$(0, 1)$ and

$y=\displaystyle \int_{-\infty}^{x}{f(t)}$ passing through the point

$\left( 0, \dfrac{1}{2}\right).$ The tangents drawn to both the curves at the points with equal abscissae intersect on the

$x$ - axis. Then the curve

$y=f(x)$ is

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

If f(x) and g(x) are differentiable function for

$0 \leq x\leq 1$ such that f(0) = 10 , g(0) = 2, f(1) = 2, g(1) =4, then in the interval (0,1)

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f (x) = 0 for all x
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f (x) + 4g' (x) =0for at least one x
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f (x) +2g' (x) for at most one x
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none of these

The angle formed bt the positive y-axis and the tangent to

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

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

0%
$\tan ^{-1}(9)$
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$\dfrac{\pi }{2}\tan ^{-1}(9)$
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$\dfrac{\pi }{2}\tan ^{-1}(9)$
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None of these

The abscissa of a point on the curve

$xy = (a+y)^{2}$ , the normal which cuts off numerically equal intercept from the coordinate axes, is

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$-\dfrac{a}{\sqrt{2}}$
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$\sqrt{2a}$
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$\dfrac{a}{\sqrt{2}}$
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$-\sqrt{2a}$

The co-ordinates of the point (s) on the graph of the function

$f(x)= \dfrac{x^{3}}{3} - \dfrac{5x^{2}}{2} + 7x - 4$ , where the tangent drawn cuts off intercept from the co-ordinate axes which

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(2, 8/3)
0%
(3, 7/2)
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(1, 5/6)
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None of these

The triangle by the tangent to the curve

$f(x) = x^{2} bx -b$ at the point (1, 1) and the co-ordinate axes lies in the first quadrant. If its area is 2, then the value of b is

0%
-1
0%
3
0%
-3
0%
1

If the normal to the curve y = f(x) at the point (3, 4) makes an angle

$\dfrac{3\pi }{4}$ with the positive x-axis, then f'(3) is equal to

0%
-1
0%
$-\dfrac{3 }{4}$
0%
$\dfrac{4}{5}$
0%
1

Let

$f(x, y)$ be a curve in the

$x-y$ plane having the property that distance from the origin of any tangent to the curve is equal to distance of point of contact from the

$y-$ axis. Of

$f(1, 2)=0$ , then all such possible curves are

0%
$x^2+y^2=5x$
0%
$x^2-y^2=5x$
0%
$x^2y^2=5x$
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$All\ of\ these$

The angle between the tangents at ant point P and the line joining P to the original, where P is a point on the curve in

$(x^{2}+y^{2})=c \tan ^{-1}\dfrac{y}{x},c$ is a constnt, is

0%
independent of x
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independent of y
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independent of x but dependent on y
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independent of y but dependent on x

Consider a curve

$y=f(x)$ in

$xy$ - plane. The curve passes through

$(0,0)$ and has the property that a segment of tangent drawn at any point

$P(x, f(x))$ and the line

$y=3$ gets bisected by the line

$x+y=1$ , then the equation of the curve is

0%
$y^2=9(x-y)$
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$(y-3)^2=9(1-x-y)$
0%
$(y+3)^2=9(1-x-y)$
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$(y-3)^2-9(1+x+y)$

The percentage error in the

$11^{th}$ root of the number

$28$ is approximately _______ times the percentage error in

$28$ .

0%
$11$
0%
$28$
0%
$\dfrac{1}{28}$
0%
$\dfrac{1}{11}$

Explanation

Given: To find percentage error in

$11th$ root

$28$ with respect to

$28$

$\therefore$ As we know that if number

$=x^n$

then

$\%$ error in

$x^n$ =n times the (

$\%$ error in

$x$ )

$\therefore$ We can say that percentage error of nth root of any number is approximately

$1/n$ times percentage error in number.

$\therefore 11th$ root of

$28=28^{1/11}$

here

$x=28,n=1/11$

hence ,

$1/11$

The curve possessing the property text the intercept made by the tangent at any point of the curve on the

$y-$ axis is equal to square of the abscissa of the point of tangency, is given by

0%
$y^{2}=x+C$
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$y=2x^{2}+C$
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$y=-x^{2}+cx$
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$None\ of\ these$

The tangent at a point

$P$ of a curve meets the

$y-$ axis at

$A$ and the line parallel to

$y-$ axis at

$A$ , and the line parallel to

$y-$ axis through

$P$ meets the

$x-$ axis at

$B$ . If area of

$\Delta OAB$ is constant (

$O$ being the origin). Then the curve is

0%
$cx^2-xy+k=0$
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$x^{2}+y^{2}=cx$
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$3x^{2}+4y^{2}=k$
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$xy-x^{2}y^{2}+kx=0$

Consider the curved mirror

$y=f(x)$ passing through

$(0, 6)$ having the property that all light rays emerging from origin, after getting reflected from the mirror becomes parallel to

$x-$ axis, then the equation of curve is

0%
$y^2=4(x-y)$ or $y^2=36(9+x)$
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$y^2=4(1-x)$ or $y^2=36(9-x)$
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$y^2=4(1+x)$ or $y^2=36(9-x)$
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$None\ of\ these$

A normal

$P(x,y)$ on a curve meets the

$X-$ axis at

$Q$ and

$N$ is the ordinate at

$P$ .

$NQ=\dfrac {x(1+y^2)}{1+x^2}$ .
Then the equation of curves passing through

$(3,1)$ is

0%
$5(1+y^2)=(1+x^2)$
0%
$5(1+y^2)=5(1+x^2)$
0%
$5(1+x^2)=(1+y^2)$
0%
None of these

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 13 - 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 13 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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