MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 7

Let

$\quad f(x)+2f(1-x)={ x }^{ 2 }+2\forall x\in R\quad$ then the interval in which

$f(x)$ increases is

0%
$\left( -\infty ,-2 \right) \quad$
0%
$\left( -\infty ,\infty \right)$
0%
$\left( 2,\infty \right)$
0%
None of these

Explanation

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

$\Rightarrow f(1-x)+2f(x)={ (1-x) }^{ 2 }+2...(ii)$
replacing

$x\rightarrow 1-x\quad$
multiplying (ii) by

$2$ then subtracting (1) we get

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

$\Rightarrow f(x)=\cfrac { { (x-2) }^{ 2 } }{ 3 }$
differentiating w.r.to

$x$ both sides we get

$f'(x)=\cfrac { 2(x-2) }{ 3 } >0\forall \left( 2,\infty \right)$

The angle between the curves

$x^{2} + y^{2} = 25$ and

$x^{2} + y^{2} - 2x + 3y - 43 = 0$ at

$(-3, 4)$ is

0%
$\tan^{-1}(1)$
0%
$\tan^{-1}\left (\dfrac {1}{68}\right )$
0%
$\dfrac {\pi}{2}$
0%
$\tan^{-1}\left (\dfrac {3}{4}\right )$

Explanation

slope of first curve at

$(-3,4)$

$m_1 = dy/dx = \dfrac{-x}{y} = \dfrac{3}{4}$

slope of second curve at

$(-3,4)$

$m_2 = \dfrac{-2x+2}{2y+3} = \dfrac{8}{11}$

angle between them

$= tan^{-1} \dfrac{m_1 - m_2}{1+m_1 +m_2}$

$=tan^{-1} \dfrac{1/44}{17/11} =tan^{-1} \dfrac{1}{68}$

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

$3$ is

0%
$y = x^{2} + 3x + 2$
0%
$y = x^{2} + 3x + 1$
0%
$y = x^{3} + 3x + 1$
0%
None of these

Explanation

Given

$\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \left( { x }^{ 2 }+1 \right) =2x\cfrac { dy }{ dx }$

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

Slope,

${ \left( \cfrac { dy }{ dx } \right) }_{ x=0 }={ (2x+3) }_{ x=0 }\Rightarrow 2\times 0+3\Rightarrow 3$

At point

$x=0$

$y={ (0) }^{ 2 }+3\times 0+1$

Thus this curve passes through point

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

$x=0$ as

$3$

$y={ x }^{ 2 }+3x+1$ is the answer

The slope of the tangent at each point of the curve is equal to the sum of the coordinate of the point. Then, the curve that passes through the origin is

0%
$x + y = e^{x} - 1$
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$e^{x} = x + y$
0%
$y = e^{x}$
0%
$y = e^{x} + 1$

Explanation

Given,

$\dfrac {dy}{dx}=x+y\Rightarrow dy=xdx+ydx\Rightarrow dy+dx=xdx+ydx+dx$

So,

$d(x+y)=(x+y+1)dx\Rightarrow dx=\dfrac {d(x+y+1)}{x+y+1}$

Integrating on both sides,

$x=\ln(x+y+1)$

On taking antilog,

$e^x=x+y+1$

Or.

$e^x-1=x+y$

So, Option A is correct answer.

The set of all values of a for which the function

$f(x) = (a^{2} - 3a + 2)(\cos^{2}\dfrac{x} {4} - \sin^{2}\dfrac{x}{4}) + (a - 1)x + \sin 1$ does not possess critical points is

0%
$[1, \infty)$
0%
$(0, 1) \cup (1, 4)$
0%
$(-2, 4)$
0%
$(1, 3)\cup (3, 5)$

Explanation

$f(x) = (a^{2} =- 3a + 2)(\cos^{2}x / 4 - \sin^{2}x / 4) + (a - 1)x + \sin 1$

$\Rightarrow f(x) = (a - 1)(a - 2)\cos x/2 + (a - 1) x + \sin 1$

$\Rightarrow f'(x) = -\dfrac {1}{2} (a - 1)(a - 2) \sin \dfrac {x}{2} + (a - 1)$

$\Rightarrow f'(x) = (a - 1) \left [1 - \dfrac {(a - 2)}{2}\sin \dfrac {x}{2}\right ]$
If

$f(x)$ does not possess critical points, then

$f'(x)\neq 0$ for any

$x \epsilon R$

$\Rightarrow (a - 1)\left [1 -\dfrac {(a - 2)}{2} \sin \dfrac {x}{2}\right ]\neq 0$ for any

$x\epsilon R$

$\Rightarrow a\neq 1$ and

$1 - \left (\dfrac {a - 2}{2}\right )\sin \dfrac {x}{2} = 0$
must not have any solution in

$R$ .

$\Rightarrow a\neq 1$ and

$\sin \dfrac {x}{2} = \dfrac {2}{a - 2}$ is not solvable in

$R$ .

$\Rightarrow a\neq 1$ and

$\left |\dfrac {2}{a - 2}\right | > 1$ [For

$a = 2, f(x) = x + \sin 1\therefore f'(x) = 1 \neq 0]$

$\Rightarrow a \neq 1$ and

$|a - 2| < 2\Rightarrow a\neq 1$ and

$-2 < a - 2 < 2$

$\Rightarrow a\neq 1$ and

$0 < a < 4\Rightarrow a\epsilon (0, 1)\cup (1, 4)$ .

$g\left( x \right) =2f\left( 2{ x }^{ 3 }-3{ x }^{ 2 } \right) +f\left( 6{ x }^{ 2 }-4{ x }^{ 3 }-3 \right)$

$\forall \ x\ \in \ R$ and

$f^{''}\left( x \right) > 0$ ,

$\forall \ x \in \ R$ , then

$g\left ( x \right)$ is increasing in the interval

0%
$\left( \infty ,-\dfrac { 1 }{ 2 } \right)$
0%
$\left( -\dfrac { 1 }{ 2 } ,\ 0 \right) \bigcup \left( 1,\infty \right)$
0%
$\left( 0,\infty \right)$
0%
None of these

The slope of the tangent at the point

$(h, h)$ of the circle

$x^{2} + y^{2} = a^{2}$ is :

0%
$0$
0%
$1$
0%
$-1$
0%
Depends on $h$

Explanation

Slope of the tangent for the cicle

$x^{2}+y^{2}=a^{2}$ can be calculated by differentiating the circle's equation.

On differentiating w.r.t x,

$2x + 2y\dfrac{dy}{dx}=0$

Slope

$m$

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

Point given is

$(h,h)$

Putting point in the slope equation

$m=\dfrac{dy}{dx}=-\dfrac{h}{h}$

Slope

$m=-1$

Hence,

$(C)$

The angle at which the curve

$y={ x }^{ 2 }$ and the curve

$x=\cfrac { 5 }{ 3 } \cos { t } ,y=\cfrac { 5 }{ 4 } \sin { t }$ intersect is

0%
$\tan ^{ -1 }{ \cfrac { 2 }{ 41 } }$
0%
$\tan ^{ -1 }{ \cfrac { 41 }{ 2 } }$
0%
$-\tan ^{ -1 }{ \cfrac { 2 }{ 41 } }$
0%
$2\tan ^{ -1 }{ \cfrac { 41 }{ 2 } }$

Explanation

Given

$y={ x }^{ 2 }...(i)$

$\quad x=\cfrac { 5 }{ 3 } \cos { t } ;y=\cfrac { 5 }{ 4 } \sin { t } ...(ii)$
Which is parametric equation, we change this equation is caresian equation as follows

$\cos { t } =\cfrac { 3 }{ 5 } x;\sin { t } =\cfrac { 4 }{ 5 } y$
On Squaring and adding both i.e.,

$\cos{t}$ and

$\sin {t}$ we get

$\cfrac { 9 }{ 25 } { x }^{ 2 }+\cfrac { 16 }{ 25 } { y }^{ 2 }=\cos ^{ 2 }{ t } +\sin ^{ 2 }{ t }$

$\Rightarrow 9{ x }^{ 2 }+16{ y }^{ 2 }=25....(iii)\quad \quad \left[ \because \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1 \right]$

$\therefore$ The intersection points at Eq.(i) and (iii) are

$(1,1)$ and

$(-1,1)$
Now, slope of tangent of Eq.(i) at point

$(1,1)$ is

${ m }_{ 1 }=\cfrac { dy }{ dx } =2x\quad \therefore { m }_{ 1 }={ \left| \cfrac { dy }{ dx } \right| }_{ (1,1) }=2$
And slope of tangent of Eq. (iii) at point

$(1,1)$ is

${ m }_{ 2 }=\cfrac { dy }{ dx } =-\cfrac { 9 }{ 16 }$

$\therefore$ Angle at point of intersection of Eqs. (i) and (iii) we get

${ \theta }_{ 1 }=\tan ^{ -1 }{ \left| \cfrac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| } =\tan ^{ -1 }{ \cfrac { 41 }{ 2 } }$
similarly, slope of tangent of Eq. (i) at point

$(-1,1)$

${ m }_{ 1 }={ \left| \cfrac { dy }{ dx } \right| }_{ (-1,1) }=-2$
And slope of tangent of Eq. (iii) at point

$(-1,1)$

${ m }_{ 2 }=\cfrac { dy }{ dx } =\cfrac { 9 }{ 16 }$

$\therefore$ Angle at point of intersection of Eqs. (i) and (iii) we get

${ \theta }_{ 2 }=\tan ^{ -1 }{ \left| \cfrac {-2- \cfrac { 9 }{ 16 } }{ 1-\cfrac { 18 }{ 16 } } \right| } =\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } \quad \quad$

The slope of the tangent to the curve

$y=\int _{ 0 }^{ x }{ \frac { dt }{ 1+{ t }^{ 3 } } }$ at the point where

$x=1$ is

0%
$\frac { 1 }{ 4 }$
0%
$\frac { 1 }{ 3 }$
0%
$\frac { 1 }{ 2 }$
0%
$1$

Explanation

$\frac { dy }{ dx } =\frac { 1 }{ 1+{ x }^{ 3 } }$

$m=\left( \frac { dy }{ dx } \right) x=1=\frac { 1 }{ 1+1 } =\frac { 1 }{ 2 }$

A point P moves such that sum of the slopes of the normal drawn from it to the hyperbola

$xy=16$ is equal to the sum of the ordinates of the feet of the normal. Let 'P' lies on the curve C, then.
The equation of 'C' is?

0%
$x^2=4y$
0%
$x^2=16y$
0%
$x^2=12y$
0%
$y^2=8x$

Explanation

Let P(h,k) be the point in the plane of hyperbola

$xy=16=4^2$

The equation of the normal at a point

$(4t, \dfrac {4} {t})$ to the hyperbola

$xy=4^2$ is

$xt^3-yt-4t^4+4=0$

if if passes through

$P(h,k)$ then,

$ht^3-yt-4t^4+4=0$

$\Rightarrow 4t^4-ht^3+kt-4=0$

this is a fourth degree equation in t. So, it gives 4 values of t,

$t_1,t_2,t_3,t_4$ corresponding to each value of t there is a point on hyperbola such that the normal passes through

$P(h,k)$ .

let the 4 points be

$A(ct_1, \dfrac{c}{t_1})$ ,

$B(ct_2, \dfrac{c}{t_2})$ ,

$C(ct_3, \dfrac{c}{t_3})$ and

$D(ct_4, \dfrac{c}{t_4})$

such that normal at these points pass through

$P(h,k)$ .

Since

$t_1,t_2,t_3,t_4$ are roots of equation (1).

Therefore,

$t_1+t_2+t_3+t_4 = \dfrac{h}{c}=\dfrac{h}{4}$ .......(2)

$\sum t_1t_2 =0$ .......(3)

$\sum t_1t_2t_3 = - \dfrac{k}{4}$ .........(4)

$t_1t_2t_3t_4 =-1$ .......(5)

It s given that the sum of the slopes of the normals at A,B,C and D is equal to the sum of the coordinates of these points.

Therefore

$t_1^2+t_2^2+t_3^2+t_4^2 = \dfrac{c}{t_1}+\dfrac{c}{t_3}+\dfrac{c}{t_3}+\dfrac{c}{t_4}=\dfrac{4}{t_1}+\dfrac{4}{t_2}+\dfrac{4}{t_3}+\dfrac{4}{t_4}$

$(t_1+t_2+t_3+t_4 )^2-2\sum t_1t_2=4\left ( \dfrac {\sum t_1t_2t_3}{t_1t_2t_3t_4} \right )$

$\Rightarrow \dfrac{h^2}{4}-2\times 0 = 4 \times \left ( \dfrac{\dfrac {-k}{4}} {-1} \right )$

$\Rightarrow \dfrac{h^2}{4} = k$

$\Rightarrow h^2 = 16k$

so the locus of

$(h,k)$ is

$x^2=16y$ , which is a parabola

A tangent PT is drawn to the circle

$x^2+y^2=4$ at the point

$P(\sqrt{3}, 1)$ . A straight line L, perpendicular to PT is a tangent to the circle

$(x-3)^2+y^2=1$ .

$(1)$ A possible equation of L is?

0%
$x-\sqrt{3}y=1$
0%
$x+\sqrt{3}y=1$
0%
$x-\sqrt{3}y=-1$
0%
$x+\sqrt{3}y=5$

Explanation

Slope of tangent to

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

$P(\sqrt{3},1)$ is given by differentiating given equation w.r.t. x

$\dfrac{dy}{dx}=\dfrac{-x}{y}=-\sqrt{3}$

So slope of line L

$=\dfrac{-1}{-\sqrt{3}}=\dfrac{1}{\sqrt{3}}$

Let equation of L be

$x-\sqrt3y+c=0$

As L is tangent to

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

So perpendicular distance of L from its center should be equal to its radius i.e.

$1$ .

Centre

$(3,0)$

$\dfrac{|3+c|}{\sqrt4}=1$

$\Rightarrow c=-1 or -5$

So equation of L is

$x-\sqrt3y=1$

$x-\sqrt3y=5$

Hence option A is correct.

$f(x) = |x\log_{e}x|$ monotonically decreases in

0%
$(0, 1/e)$
0%
$(1/e, 1)$
0%
$(1, \infty)$
0%
$(1/e, \infty)$

Explanation

Monotomic functions are those functions who either increase or decrease over a given domain.

Example: the function

$x^{3}$ in an always increasing function.

The given equation is:

$f(x) = |x\log_{e}x|$

Differentiating once w.r.t to

$x$ we get,

$\Rightarrow f'(x) = \dfrac {|x\log_{e}x|}{x\log_{e}x}\times (\log_{e}x + 1)$

Now for

$0 < x < 1$ we have

$\Rightarrow f'(x) = \dfrac {-x\log_{e}x}{x\log_{e}x} \times (\log_{e} x + 1)$

$\Rightarrow f'(x) = -(\log_{e}x + 1)$

Now for monotonically decreasing function we have

$f'(x) < 0$

$\Rightarrow -(\log_{e}x + 1) < 0$

$\Rightarrow (\log_{e} x + 1) > 0$

$\Rightarrow x > \dfrac {1}{e}$

Now for

$x > 1$ we have,

$\Rightarrow f'(x) = \dfrac {x\log_{e}x}{x\log_{e}x}\times (\log_{e}x + 1)$

$\Rightarrow f'(x) = (\log_{e}x + 1)$

$\Rightarrow (\log_{e} x + 1) < 0$

This cannot be true as

$\log_{e} > 0$ in this domain hence the inequality is never satisfied.

Hence the function decreases in the range

$\left (\dfrac {1}{e}, 1\right )$ .

1935021_876407_ans_0260cc66d0864b6ba3a9414ce0ba2dfe.PNG

The points of the curve

$y={ x }^{ 3 }+x-2$ at which its tangent are parallel to the straight line

$y=4x-1$ are

0%
$\left( 2,7 \right) ,\left( -2,-11 \right)$
0%
$\left( 0,-2 \right) ,\left( { 2 }^{ 1/3 },{ 2 }^{ 1/3 } \right)$
0%
$\left( { -2 }^{ 1/3 },{ -2 }^{ 1/3 } \right) \left( 0,-4 \right)$
0%
$\left( 1,0 \right) ,\left( -1,-4 \right)$

Explanation

Given

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

$y=4x-1....(ii)$

Slope of tangent to the curve (i)

$\cfrac { dy }{ dx } =3{ x }^{ 2 }+1$

slope of tangent at point

$\left( \alpha ,\beta \right)$ is

$\quad { \left| \cfrac { dy }{ dx } \right| }_{ \left( \alpha ,\beta \right) }^{ }=3{ \alpha }^{ 2 }+1...(iii)$

Given, tangent of curve (i) is parallel to line (ii)

$\therefore$ Slope of line (ii) is

$4$

$\therefore$ From Eq(iii) we get

$3{ \alpha }^{ 2 }+1=4\Rightarrow \alpha =\pm 1$

$\therefore \left( \alpha ,\beta \right)$ lie on curve (i)

$\beta ={ \left( \pm 1 \right) }^{ 2 }+\left( \pm 1 \right) -2\Rightarrow \beta =0,-4\quad$

$\therefore$ Points are

$(1,0)$ and

$(-1,-4)$

If the tangent at

$(1, 7)$ to the curve

$x^{2} = y - 6$ touches the circle

$x^{2} + y^{2} + 16x + 12y + c = 0$ then the value of

$c$ is

0%
$85$
0%
$95$
0%
$195$
0%
$185$

Explanation

Given equation of curve is

$x^{2}=y-6$

$\Rightarrow y=x^{2}+6$

$\Rightarrow \dfrac{dy}{dx}=2x$

Slope of tangent is

$m=\left|\dfrac{dy}{dx}\right|_{(1,7)}=2\times 1=2$

Equation of tangent at the point

$(1,7)$ is given by

$(y-7)=2(x-1)$

$\Rightarrow y-7=2x-2$

$\Rightarrow 2x-y+5=0$

Given equation of circle is

$x^{2} + y^{2} + 16x + 12y + c = 0$

$\Rightarrow (x+8)^{2}+(y+6)^{2}+c-64-36=0$

$\Rightarrow (x+8)^{2}+(y+6)^{2}=100-c$

If the tangent touches the circle, then

Distance from centre

$=$ radius

$\Rightarrow$ Distance of

$(-8,-6)$ from

$2x-y+5=0$ is the radius

$d=\left|\dfrac{2(-8)-(-6)}{\sqrt{4+1}}\right|=\left|\sqrt{5}\right|$

$\Rightarrow Radius=\sqrt{100-c}=\sqrt{5}$

$\Rightarrow c={95}$

The percentage error in the surface area of a cube with edge x cm, when the edge is increased by

$11\%$ is _________.

0%
$11$
0%
$22$
0%
$10$
0%
$44$

Explanation

Surface area of cube

$=$ S

$=6x^2$

$\therefore \dfrac{dS}{dt}=12x\dfrac{dx}{dt}$
Side of cube increase

$11\%$ .

$\therefore \dfrac{dx}{dt}=11\%$ increment in side

$=\dfrac{11x}{100}$

$\therefore \dfrac{dx}{dt}=12\left(\dfrac{11x}{100}\right)$

$=6\left(\dfrac{22}{100}\right)x^2$

$=6x^2\left(\dfrac{22}{100}\right)$

$=22\%$ in surface area

$\therefore$ Surface area increase

$22\%$ .

Consider the following statements:
Statement I:

$x > \sin x$ for all

$x > 0$
Statement II:

$f(x) = x - \sin x$ is an increasing function for all

$x > 0$
Which one of the following is correct in respect of the above statements?

0%
Both Statements I and II are true and Statement II is the correct explanation of Statement I
0%
Both Statements I and II are true and Statement II is the not correct explanation of Statement I
0%
Statement I is true but Statement II is false
0%
Statement I is true but Statement II is true

Explanation

Assume

$f(x)=x-\sin x$

$\therefore f'(x)=1-\cos x\geq 0$ and hence,

$x-\sin x$ is an increasing function and hence,

$x>\sin x$

So, both I and II are true and II is correct explanation for I.

The critical points of the function

$f(x)={ x }^{ 3/5 }\left( 4-x \right) ,x\in { R }^{ + }\cup \left\{ 0 \right\}$ is _______

0%
$0,-\cfrac { 3 }{ 2 }$
0%
$0,\cfrac { 3 }{ 2 }$
0%
$0,\cfrac { 2 }{ 3 }$
0%
$0,-\cfrac { 2 }{ 3 }$

Explanation

$f(x)={ x }^{ \cfrac { 3 }{ 5 } }\left( 4-x \right) x\in { R }^{ + }$

$f(x)=4{ x }^{ \cfrac { 3 }{ 5 } }-{ x }^{ \cfrac { 8 }{ 5 } }$

$\therefore f'(x)=4\left( \cfrac { 3 }{ 5 } \right) { x }^{ -\cfrac { 2 }{ 3 } }=\cfrac { 8 }{ 5 } { x }^{ \cfrac { 3 }{ 5 } }$

$\therefore f'(x)=\cfrac { 4 }{ 5 } \left( \cfrac { 3 }{ { x }^{ \cfrac { 2 }{ 3 } } } -2{ x }^{ \cfrac { 3 }{ 5 } } \right) =\cfrac { 4 }{ 5 } \left( \cfrac { 3-2x }{ { x }^{ \cfrac { 2 }{ 3 } } } \right) \quad$
For

$x=0;f'(x)$ does not exists and for

$x=\cfrac { 2 }{ 3 }$ we have

$f'(x)=0$

$\therefore$ Critical points are

$0$ and

$\cfrac{2}{3}$

The tangent to

$\left( a{ t }^{ 2 },2at \right)$ is perpendicular to X-axis at _____ point

$t\in R$ .

0%
$(4a,4a)$
0%
$(a,2a)$
0%
$(0,0)$
0%
$(a,-2a)$

Explanation

Here

$\left( x,y \right) =\left( a{ t }^{ 2 },2at \right)$

$\therefore x=a{ t }^{ 2 };y=2at$

$\therefore \cfrac { dx }{ dt } =2at;\cfrac { dy }{ dt } =2a$

$\therefore \cfrac { dy }{ dx } \cfrac { dy/dt }{ dx/dt } =\cfrac { 2a }{ 2at } =\cfrac { 1 }{ t }$
Since, tangent is perpendicular to X-axis slope is undefined

$\therefore$

$t=0$

$\therefore t=0$

$\therefore \left( a{ t }^{ 2 },2at \right) =\left( 0,0 \right)$
Thus, tangent is perpendicular to X-axis at

$(0,0)$

Let,

$f:A\rightarrow B$ be an invertible function. If

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

$f'^{-1}(5)$ =

0%
$\dfrac{1}{13}$
0%
$1$
0%
$6$
0%
can not be determined

Explanation

Given

$f$ be an invertible function and

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

Then

$f'^{-1}(5)=\dfrac{1}{f'(x)}\forall x\in f(A)$ such that

$f(x)=5$ .

Now,

$f(x)=5$ given only

$x=1$ a real root, remaining are imaginary.

Now,

$f'(x)=6(x^2+x)+1$ .

$\therefore$

$f'^{-1}(5)=\dfrac{1}{f'(1)}=\dfrac{1}{13}$

Number of critical points of the function

$\displaystyle f\left( x \right) =\dfrac { 2 }{ 3 } \sqrt { { x }^{ 3 } } -\dfrac { x }{ 2 } +\int _{ 1 }^{ x }{ \left( \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \cos { 2t } -\sqrt { t } \right) } dt$ which lie in the interval

$\left[ -2\pi ,2\pi \right]$ is:

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

The point on the curve

$y^{2}=x$ where tangent makes

$45^{o}$ angle with

$x-$ axis ?

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

Explanation

$\dfrac{dy}{dx} =$ Slope of the tangent at that point of the curve.

Given curve is

$x = y^2$

Keep

$y = a, \, then \, x = y^2$

point

$(a^2, a)$

$\displaystyle \dfrac{dy}{dx} (y^2 - x) = 0 \dfrac{dy}{dx} = \dfrac{1}{2y} = \dfrac{1}{2a} = \tan(45^0)$

$\Rightarrow a= \dfrac{1}{2} \quad a^2= \dfrac{1}{4}$

point

$\left(\dfrac{1}{4}, \dfrac{1}{2}\right)$

Line

$y=x$ and curve

$y=x^2+bx+c$ touches at

$(1, 1)$ then __________.

0%
$b=-1, c=1$
0%
$b=1, c=2$
0%
$b=1, c=1$
0%
$b=0, c=1$

Explanation

Given line

$l : y =x$

$l : x-y=0$

$\therefore$ slope of line

$l=1$
Now

$y=x^2+bx+c$

$\therefore \dfrac{dy}{dx}=2x+b$

$\therefore \left(\dfrac{dy}{dx}\right)_{P(1, 1,)}=2(1)+b$

$=2+b$
But slope of tangent of the curve

$=1$ .

$\therefore 2+b=1$

$\therefore b=-1$
Now

$(1, 1)\in y=x^2+bx+c$

$\therefore 1=1+b(1)+c$

$\therefore b+c=0$

$\therefore c=-b$

$c=-(-1)$

$\therefore c=1$

$\therefore$ We have

$b=-1$ and

$c=1$ .

$f(x)=min(|x|^{2}-5|x|,1)$ then

$f(x)$ is non differentiable at

$\lambda$ points, then

$\lambda+13$ equals

0%
$16$
0%
$14$
0%
$13$
0%
$15$

Find the angle between tangent of the curve

$y = (x + 1) (x - 3)$ at the point where it cuts the axis of

$x$ .

0%
$\tan^{-1} \left(\dfrac{8}{15}\right)$
0%
$\tan^{-1} \left(\dfrac{15}{8}\right)$
0%
$\tan^{-1} 4$
0%
$None\ of\ these$

Explanation

The curve

$y=(x+1)(x-3)=x^2-2x-3$ .........(1) cuts the axis of

$x$ at

$(-1,0)$ and

$(3.0)$ . These points are obtained by putting

$y=0$ in the equation (1).

Now the slope of the tangent to the curve (1) at

$(-1,0)$ be

$(m_1)=\left.\dfrac{dy}{dx}\right|_{(-1,0)}=[2x-2]_{(-1,0)}=-4$ .

Again the slope of the tangent to the curve (1) at

$(-1,0)$ be

$(m_2)=\left.\dfrac{dy}{dx}\right|_{(3,0)}=[2x-2]_{(3,0)}=4$ .

Now the angle between the tangents to (1) at those points be

$\tan^{-1}\left(\dfrac{m_1-m_2}{1+m_1.m_2}\right)$

$=\tan^{-1}\left(\dfrac{8}{15}\right)$ .

So the required angle be

$\tan^{-1}\left(\dfrac{8}{15}\right)$ .

If the curves

$y^2 = 4ax$ and

$xy = c^2$ cut orthogonally then

$\dfrac{c^4}{a^4} =$

0%
$4$
0%
$8$
0%
$16$
0%
$32$

Explanation

$y^2=4ax, xy=c^2$ cut orthogonally

Let they intersect at

$(x_1, y_1)$

$y^2=4ax$

$2y\dfrac{dy}{dx}=4a$

$\dfrac{dy}{dx}=\dfrac{2a}{y}$

$\dfrac{dy}{dx}|_{(x_1, y_1)}=\dfrac{2a}{y_1}$ …..

$(1)$

$xy=c^2$

$y+x\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}|_{(x_1, y_1)}=\dfrac{-y_1}{x_1}$ ……

$(2)$

From

$(1), (2)$

$(\because m_1m_2=-1)$

$\dfrac{2a}{\not{y_1}}\times \dfrac{\not{-}\not{y_1}}{x_1}=\not{-}1$

$x_1=2a$

From

$y^2=4ax$

$y_1=\sqrt{8a^2}=2\sqrt{2}a$

$xy=c^2$

$x_1y_1=c^2$

$2a(2\sqrt{2}a)=c^2$

$\dfrac{c^2}{a^2}=4\sqrt{2}$

$\dfrac{c^4}{a^4}=32$ .

An equation of the tangent to the curve

$y=x^{4}$ from the point

$(2,0)$ not on the curve is:

0%
$y=0$
0%
$x=0$
0%
$x+y=0$
0%
$none\ of\ these$

Explanation

let a,b be point on curve (x, y)

$\displaystyle b = a^4$

$\displaystyle \frac{dy}{dx} (x^4) = 4x^3 = 4a^3$

$\displaystyle (y- y_1) = m (x - x_1) \rightarrow$ be tangent equation passes through (2, 0)

$\displaystyle (b-0) = 4a^3 (a-2)$

$\displaystyle a^4 = 4a^3 (a-2)$

$\displaystyle 3a^4 - 8a^3 = 0$

$\displaystyle a^3 (3a - 8) = 0$

$\displaystyle a= 0 \quad or \quad a= \frac{8}{3} \quad (0,0)$ &

$\left[\dfrac{8}{3},\left(\dfrac{8}{3} \right)^4\right]$ points on curve

$\displaystyle b=0 \quad or \quad b= \left(\frac{8}{3}\right)^4$

$\displaystyle (y - 0) = \frac{0-0}{2-0} (x-0)\Rightarrow y = 0 \rightarrow$ tangent (1)

$\displaystyle \left[y-\left(\frac{8}{3}\right)^4\right] = \frac{(\frac{8}{3})^4_{-0}}{(\frac{8}{3})_{-2}} \quad (x - \frac{8}{3}) \rightarrow$ tangent (2)

1058172_1038190_ans_b5c2db2c6f76484f88ed313c09fb571a.jpg

The equation of the tangent to the curve

$y = b{e^{ -\dfrac{x}{a}}}$ at a point , where

$x=0$ is

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

Explanation

$y=b{ e }^{ \cfrac { -x }{ a } }$ when

$x=0\, y=b$

$\dfrac { dy }{ dx } =\cfrac { -b }{ a } { e }^{ \cfrac { -x }{ a } }$ at

$x=0$ ,

$\dfrac { dy }{ dx } =\cfrac { -b }{ a }$

$y=\cfrac { -bx }{ a } +c$

$y=\cfrac { -bx }{ a } +b\\ \cfrac { y }{ b } =\cfrac { -x }{ a } +1\\ \cfrac { x }{ a } +\cfrac { y }{ b } =1$

The function

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

0%
Always increasing
0%
Always decreasing
0%
Neither increasing nor decreasing
0%
None of these

Explanation

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

$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4x \times {x^4} - 4{x^3}\left( {2{x^2} - 1} \right)}}{{{x^8}}}$

$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4{x^3} - 4{x^5}}}{{{x^8}}}$

$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4{x^3}\left( {1 - {x^2}} \right)}}{{{x^8}}}$

$\dfrac{{dy}}{{dx}} \geqslant 0\forall x \in \left[ { - 1,1} \right]$

So,this function is increasing for

$x \in \left[ { - 1,1} \right]$ and decreasing for other values of

$x$

Hence this function is neither increasing nor decreasing

The equation of the curve passing through

$(1,3)$ whose slope at any point

$(x,y)$ on it is

$\dfrac { y }{ { x }^{ 2 } }$ is given by

0%
$y={ 3e }^{ -1/x }$
0%
$y={ 3e }^{ 1-1/x }$
0%
$y={ ce }^{ 1/x }$
0%
$y={ 3e }^{ 1/x }$

Explanation

$\dfrac{{dy}}{{dx}} = \dfrac{y}{{{x^2}}}\,\,\,\,\,\,\,\left( {given} \right)$

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

$\Rightarrow \log y = - \dfrac{1}{x} + c$

$\Rightarrow y = {e^{ - \frac{1}{x} + c}}$

$\Rightarrow y = {e^{ - \frac{1}{x}}},c$

since this curve is passing through point $\left( {1,3} \right)$

so, $3 = {e^{ - \dfrac{1}{t}}}.c$

$\Rightarrow c = 3e$

therefore , eqn of the curve is

$y = {e^{ - \dfrac{1}{x}}}.3e$

$\Rightarrow y = 3{e^{ - \dfrac{1}{x} + 1}}$

$\Rightarrow y = 3{e^{1 - \dfrac{1}{x}}}$

Let

$f:R\rightarrow R$ be a function defined by

$f\left(x\right)=Min \left\{x+1, \left|x\right|+1\right\}$ . Then which of the following is true?

0%
$f(x)\ge 1$ for all $x\in R$
0%
$f(x)\ge 1$ is not differentiable at $x=1$
0%
$f(x)\ge 1$ is differentiable at $x=1$
0%
$f(x)\ge 1$ is not differentiable at $x=0$

Explanation

Function has the minimum value of

$x+1$ and

$\left|x\right|+1$

Hence

$f\left(x\right)$ is differentiable everywhere.

Area of the triangle formed by the tangent at

$x=2$ on the curve

$y= \dfrac{8}{4+x^2}$ with the coordinate axes is (in sq. units)

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

Explanation

$y=\dfrac {8}{4+x^2}$

$At \,\,x=2\ y=1$

$y' =\dfrac {-8(2x)}{(4+x^2)^2}=\dfrac {-16\times 2}{8^2}=\dfrac {-1}{2}$

Tangent is

$y=mx+c$

$y=\dfrac {-1}{2}x+c$

passes through

$(2, 1)\ C=2$

$2y+x=4$

Base of triangle

$=4$

Height of triangle

$=\dfrac {4}{2}=2$

Area of

$\Delta =\dfrac {1}{2}\times 4\times 2=4\ sq.unit$

1416924_1090027_ans_25bd3feb84ce43e795654362a41fb34a.png

Find the points on the ellipse

$\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{9}=1$ , on which the normals are parallel to the line

$3x-y=1$ .

0%
$(\pm\dfrac{6}{\sqrt {10}},\pm\dfrac{3}{\sqrt {10}})$
0%
$(\pm\dfrac{3}{\sqrt {10}},\pm\dfrac{1}{\sqrt {10}})$
0%
$(\pm\dfrac{1}{\sqrt {10}},\pm\dfrac{2}{\sqrt {10}})$
0%
None of these

The entire graph of the equation

$y=x^{2}+kx-x+9$ is strictly above the

$x-$ axis if and only if :

0%
$k<7$
0%
$-5< k <7$
0%
$k>-5$
0%
$none\ of\ these$

Explanation

$y=x^2+kx-x+9$

$a>0$

$b^2-4ac<0$ &

$a>0$

then the graph will always be above x-axis

$a = 1 \therefore a > 0$ .

$(K-1)^2-4\times 9<0$

$(K-1)^2-6^2<0$

$(K-1-6)(K-1+6)<0$

$(K-7)(K+5)<0$

$K>-5,K<7$

$-5<K<7$

Option B.

$y=e^{4x}+2e^{-x}$ satisfied the equation

$\dfrac{d^{3}y}{dx^{3}}+A\dfrac{dy}{dx}+By=0$ then value of

$|A+B|$ is

0%
$36$
0%
$25$
0%
$24$
0%
$15$

Explanation

$\dfrac{d^3y}{dx^3}$ =

$64e^{4x}$ -

$2e^{-x}$

$\dfrac{dy}{dx}$ =

$4e^{4x}$ -

$2e^{-x}$

$\dfrac{d^3y}{dx^3}$ +

$A$

$\dfrac{dy}{dx}$ +

$By$ = 0

$\Rightarrow$

$e^{4x} (64+4A+B)$ +

$2e^{-x} (-1-A+B)$ = 0

64 +4A + B = 0 -(I)

-1 - A + B = 0 -(II)

Solving (i) and (ii),

A = -13 ; B = -12

$\therefore$

$|A + B|$ =

$25$

So, the correct option is 'B'.

Let

$f(x)$ be a differentiate function and

$f(\alpha)=f(\beta)=0(\alpha < \beta)$ , then in the interval

$(\alpha, \beta)$ .

0%
$f(x)+f'(x)=0$ has at least one root
0%
$f(x)-f'(x)=0$ has at least one real root
0%
$f(x)\cdot f'(x)=0$ has at least one real root
0%
None of these

Explanation

$f(x)$ is a differentiable function with

$f(\alpha)=f(\beta)=0(\alpha< \beta)$

then in interval

$(\alpha,\beta)$ for some part

$f'(x)<0$

$A+$ are point

$f(x)=0$

$f(x).f'(x)$ has at least are real root.

1112352_1061786_ans_67cf9ddce449455bb34d434cc938ec04.png

If the tangent to the curve

$y=x\log { x }$ at

$\left( c,f\left( x \right) \right)$ is parallel to the line-segment joining

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

$B\left(e,e\right)$ , then c=...... .

0%
$\dfrac {e-1}{e}$
0%
$\log { \dfrac { e-1 }{ e } }$
0%
${ e }^{ \dfrac { 1 }{ 1-e } }$
0%
${ e }^{ \dfrac { 1 }{ e-1 } }$

Explanation

slope of AB

$\displaystyle =\frac{e}{e-1}$

$\displaystyle \frac{dy}{dx}=1+\log x$

slope at

$(c,f\left(x\right ))$

$\displaystyle 1+\log c=\frac{e}{e-1}\implies=e^{\frac{1}{e-1}}$

$f(x) = g(x) (x - \lambda)^2 \,$ and

$\, g (\lambda)$ , where

$0 < x \le 1$ , then in this interval

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

Let

$h(x) = f(x) - (f (x) )^2 + (f(x))^3$ for every real number

$x$ , then

0%
$h$ is increasing whenever $f$ is increasing and decreasing whenever $f$ is decreasing
0%
$h$ is increasing whenever $f$ is decreasing
0%
$h$ is decreasing whenever $f$ is increasing
0%
Nothing can be said in general

Explanation

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

When $x=2$

$h(x)=2-2^2+2^3=6$

When $x=-2$

$h(x)=-2-(-2)^2+(-2)^3=-14$

Therefore, h is increasing whenever f is increasing and decreasing whenever f is decreasing.

Find the slope of the normal to the curve

$2x^{2} - xy + 3y^{2} = 18$ at

$(3,1)$ .

0%
$\dfrac {-11}{3}$
0%
$\dfrac {3}{11}$
0%
$\dfrac {11}{3}$
0%
$\dfrac {-3}{11}$

Explanation

Equation of curve

$2{ x }^{ 2 }-xy+3{ y }^{ 2 }=18$

Differentiating above equation w.r.t

$x,$

$4x-y-x\dfrac { dy }{ dx }+6y\dfrac { dy }{ dx } =0$

$4x-y+\dfrac { dy }{ dx } (6y-x)=0$

$\dfrac { dy }{ dx } (6y-x)=y-4x$

$\dfrac { dy }{ dx } =\dfrac { y-4x }{ 6y-x }$

$\dfrac { dy }{ dx| } _{ (3,1) }=\dfrac { 1-4\left( 3 \right) }{ 6-3 }$

$\dfrac { dy }{ dx| } _{ (3,1) }=\dfrac { -11 }{ 3 }$

Slope of normal to the curve

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

$=\dfrac { -1 }{ \dfrac { -11 }{ 3 } }$

$=\dfrac { 3 }{ 11 }$

$\boxed { \therefore Ans =\dfrac { 3 }{ 11 } }$

The radius of the sphere is measured as

$\left( {10 \pm 0.02} \right)cm$ . The error in the measurement of its volume is

0%
$25.1 cc$
0%
$25.21 cc$
0%
$2.51 cc$
0%
$251.2 cc$

Explanation

Let

$r$ be the radius of the sphere.

$\Rightarrow$

$r=10$

Error in the measurement of radius

$=\Delta r$

$\therefore$

$\Delta r=0.02\,m$

$\Rightarrow$ Volume of the sphere

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

We need to find error in calculating the volume that is

$\Delta V$

$\Delta V=\dfrac{dv}{dr}\times \Delta r$

$=\dfrac{d\left(\dfrac{4}{3}\pi r^3\right)}{dr}\times \Delta r$

$=\dfrac{4}{3}\pi\dfrac{d(r^3)}{dr}\times \Delta r$

$=\dfrac{4}{3}\pi(3r^2)\times (0.0.2)$

$=4\pi r^2\times 0.02$

$=4\times 3.14\times (10)^3\times 0.02$

$=251.2\,cm^3$ i.e.

$251.2\,cc$

The curve

$y = ax^3 + bx^2 + cx + 8$ touches

$x-$ axis at

$P(-2, 0)$ and cuts

$y-$ axis at a point

$Q$ where its gradient is

$3$ . The values of

$a, b, c$ are respectively ?

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

Explanation

$y=ax^{3}+bx^{2}+cx+8$

$\Rightarrow 0=-8a+4b+c(-2)+8$

$0=-8a+4b-2c+8 ----(1)$

It touches

$y-$ axis at

$Q\Rightarrow x=0$

$y=8$

$Q.(0,8)$

$\dfrac{dy}{dx}=3ax^{2}+2bx+c$

Given gradient is

$3$ at

$G(0,8)$

$3=c$

$\therefore c=3$

$\because$ the curve touches

$x-$ axis

$\therefore$ slope of the tangent at

$P$ is

$0$

$\dfrac{dy}{dx}=3ax^{2}+2bx+3$

$0=12a-4b+3$

$12a-4b=-3 ---- (1)$

eqn

$(1)$

$-8a+4b-6+8=0$

$-8a+4b=-2 --- (ii)$

Adding

$(10$ &

$(2)$ , we get

$12a-4b=-3$

$-8a+4b=-2$

____________

$4a=-5$

$a=-5/4$

$b=-3$

$\boxed{\therefore a=-5/4, b=-3, c=3}$

The curve satisfying D.E

$y dx - (x+3{y}^{2})dy=0$ and passing through the point

$(1,1)$ also passes through the point:

0%
$\left( \cfrac { 1 }{ 4 } ,-\cfrac { 1 }{ 2 } \right)$
0%
$\left( -\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } \right)$
0%
$\left( \cfrac { 1 }{ 3 } ,-\cfrac { 1 }{ 3 } \right)$
0%
$\left(- \cfrac { 1 }{ 4 } ,-\cfrac { 1 }{ 2 } \right)$

Slope of the line

$\sqrt { { x }^{ 2 }+{ 4y }^{ 2 }-4xy+4 } +x-2y=1$ equals to

0%
$\dfrac { 1 }{ 2 }$
0%
$2$
0%
$-\dfrac { 1 }{ 2 }$
0%
$None\ of\ these$

$f(x)=x^{3/2}(3x-10), x\ge-0)$ , then

$f(x)$ is decreasing in

0%
$(-\infty,0)\cup(0,\infty)$
0%
$(2,\infty)$
0%
$(-\infty,-1]\cup[1,\infty)$
0%
$(-\infty,0]\cup[2,\infty)$

Explanation

R.E.F image

$f'(x) = 3x^{3/2}+(3x-10)\dfrac{3}{2}x^{1/2}$

Putting

$f'(x) = 0 \Rightarrow 3x^{3/2}+(3x-10)\dfrac{3}{2}x^{1/2} = 0$

$\Rightarrow 2x^{3/2} = (10-3x)x^{1/2}$

$\Rightarrow 2x = 10-3x$

$\Rightarrow 5x = 10$

$\Rightarrow x = 2$

value of

$x$ Intervals sign of

$f'(x)$

$x > 2$

$x\epsilon (2,\infty )$

$< 0$ decreasing

$x< 2$

$x\epsilon (-\infty ,2)$

$> 0$ Increasing

The numbers of tangent to the curve

$y - 2 = {x^5}$ which are drawn
from point

$\left( {2,2} \right)$ is/are

0%
$30$
0%
$80$
0%
$20$
0%
$50$

Explanation

$y-2=x^5 +2$

$y=x^5+2$

$y^1=5x^4$

$y^1=5(2)^4$

$y^1=5\times 16=80$

If the normal to the curve

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

${(3,4)}$ makes angle

$\dfrac {3\pi}{4}$ with

$\bar {OX}$ then

$f^{ 1 }\left( 3 \right)=$

0%
$-1$
0%
$1$
0%
${(-3/4)}$
0%
${(4/3)}$

Explanation

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

${y}^{‘}={f}^{‘}\left(x\right)$

Slope of tangent

$={f}^{‘}\left(x\right)$

Slope of normal

$=-\dfrac{1}{{f}^{‘}\left(x\right)}$

$\Rightarrow \tan{\dfrac{3\pi}{4}}=-\dfrac{1}{{f}^{‘}\left(3\right)}$

$\Rightarrow -1=-\dfrac{1}{{f}^{‘}\left(3\right)}$

$\Rightarrow \boxed{{f}^{‘}\left(3\right)}=1$

If the curves

$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{4}=1$ and

$y^{2}=16x$ intersect at right angles then value of

$a^{2}$ is

0%
$2$
0%
$4$
0%
$\dfrac{8}{3}$
0%
$\dfrac{16}{3}$

Tangents are drawn from a point on the circle

$x^2+y^2=25$ to the ellipse

$9x^2+16y^2-144=0$ then the angle between the tangents is

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

Area of the triangle formed by the tangent, normal to the curve

$x^{2}/a^{2}+y^{2}/b^{2}=1$ at the point

$(a/\sqrt{2} , b/\sqrt{2})$ and the

$x-$ axis is

0%
$\dfrac{ab}{4}\sqrt{a^{2}+b^{2}}$
0%
$4ab$
0%
$\dfrac{b}{4a}({a^{2}+b^{2}})$
0%
$none$

The slope of normal to the curve y= log (logx) at x = e is

0%
e
0%
-e
0%
$\frac{1}{e}$
0%
- $\frac{1}{e}$

Explanation

$\begin{array}{l} \frac { d }{ { dx } } \log \left( { \log x } \right) \\ Substitute\, u=\log x \\ =\dfrac { d }{ { du } } \log u\frac { d }{ { dx } } \log x \\ =\dfrac { 1 }{ u } \left( { \frac { 1 }{ x } } \right) \\ =\dfrac { 1 }{ { \log x } } \left( { \frac { 1 }{ x } } \right) \\ Substitute\, x=e \\ =\dfrac { 1 }{ { \log e } } \left( { \dfrac { 1 }{ e } } \right) \\ =\dfrac { 1 }{ e } \end{array}$

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 7 - 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 7 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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