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CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 13 - MCQExams.com
CBSE
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
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(4,0)
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(-4,0)
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(0,4)
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(-4,4)
If $$f:R\rightarrow R$$ 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
If $$\int { \dfrac { { 2x }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) ^{ 2 } } } dx=f\left( x \right) +c$$ 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
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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
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k
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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
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0%
1
0%
-2
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4
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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
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zero
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-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,
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$$\dfrac{\alpha}{\beta}$$
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$$\dfrac{\beta }{\alpha}$$
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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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$$\dfrac{x}{a}-\dfrac{y}{b}=1$$
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$$ax = by = 1$$
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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
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$$f(x)=x^2+x+1$$
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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
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$$\tan ^{-1}(9)$$
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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)
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(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
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-1
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3
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-3
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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
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-1
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$$-\dfrac{3 }{4}$$
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$$\dfrac{4}{5}$$
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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
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$$x^2+y^2=5x$$
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$$x^2-y^2=5x$$
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$$x^2y^2=5x$$
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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
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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
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$$y^2=9(x-y)$$
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$$(y-3)^2=9(1-x-y)$$
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$$(y+3)^2=9(1-x-y)$$
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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 $$28$$.
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$$11$$
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$$28$$
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$$\dfrac{1}{28}$$
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$$\dfrac{1}{11}$$
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
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$$y^{2}=x+C$$
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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
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$$cx^2-xy+k=0$$
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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
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$$y^2=4(x-y)$$ 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 $$P$$.
If $$NQ=\dfrac {x(1+y^2)}{1+x^2} $$.
Then the equation of curves passing through $$(3,1)$$ is
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$$5(1+y^2)=(1+x^2)$$
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$$5(1+y^2)=5(1+x^2)$$
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$$5(1+x^2)=(1+y^2)$$
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None of these
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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