MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 12

Let f be a function which is continuous and differentiable for all real x. If

$f(2)=-4$ and

$f'(x)\geq 6$ for all

$x\epsilon [2, 4]$ , then.

0%
$f(4)<8$
0%
$f(4)\geq 8$
0%
$f(4) > 12$
0%
$f(4) >8$

Explanation

$f(2)=-4$ and

$f'(x) \geq 6 \: \forall x\in [2,4]$

Applying Lagrange's mean value theorem,

$\dfrac{f(4)-f(2)}{4-2}=f'(c) \geq 6$

$\Rightarrow \dfrac{f(4)-f(2)}{2}\geq 6$

$\Rightarrow f(4)-f(2)\geq 12$

$\Rightarrow f(4)\geq 12+f(2)$

$\Rightarrow f(4)\geq 12-4$

$\Rightarrow f(4)\geq 8$

Consider

$f(x)=\lim _{ n\rightarrow \infty }{ \cfrac { { x }^{ n }-\sin { { x }^{ n } } }{ { x }^{ n }+\sin { { x }^{ n } } } }$ for

$x>0,x\neq 1,f(1)=0$ then

0%
$f$ is continuous at $x=1$
0%
$f$ has a discontinuity at $x=1$
0%
$f$ has an infinite or oscillatory discontinuity at $x=1$
0%
$f$ has a removal type of discontinuity at $x=1$

A point where function

$f(x)$ is not continuous where

$f(x)=\left[ \sin { \left[ x \right] } \right]$ in

$\left( 0,2\pi \right)$ ; is (

$\left[ \ast \right]$ denotes greatest integer

$\le x$ )

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

$u=f(x,y)$ is a differentiable function of

$x$ and

$y$ , where

$x$ and

$y$ are differentiable functions of

$t$ then:

0%
$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { \partial x }{ \partial t } +\cfrac { \partial f }{ \partial y } .\cfrac { \partial y }{ \partial t }$
0%
$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { dx }{ dt } -\cfrac { \partial f }{ \partial y } .\cfrac { \partial y }{ \partial t }$
0%
$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { dx }{ dt } +\cfrac { \partial f }{ \partial y } .\cfrac { dy }{ dt }$
0%
$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { \partial x }{ \partial t } -\cfrac { \partial f }{ \partial y } .\cfrac { \partial y }{ \partial t }$

Explanation

Given

$u=f(x,y)$ is a a differentiable function of

$x$ and

$y$ , where

$x, y$ are differentiable functions of

$t$ .

We have to find

$\dfrac{du}{dt}$

Since

$x, y$ are differentiable functions of

$t$

Let

$x=g(t), y=h(t)$

Thus by chain rule we get

$\dfrac{du}{dt}=\dfrac{\partial f}{\partial x}\cdot \dfrac{dx}{dt}+\dfrac{\partial f}{\partial y}\cdot \dfrac{dy}{dt}$

Let

$f:\left[ {0,2} \right] \to R$ a function which is continuous on

$\left[ {0,2} \right]$ and is differentiable on

$\left( {0,2} \right)$ with

$f\left( 0 \right) = 1$ . Let

$F\left( x \right) = \int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)} dt,$ for

$x \in \left[ {0,2} \right]$ , if

$F'\left( x \right) = f'\left( x \right),\forall x \in \left( {0,2} \right),$ then

$F(2)$ equals to

0%
${e^2} - 1$
0%
${e^4} - 1$
0%
$e-1$
0%
${e^4}$

Explanation

From Newton-Leibnitz's formula,

$\dfrac{d}{dx}\left[ \displaystyle \int_{\phi(x)}^{\psi(x)} f(t)dt\right]=f\left\{\psi(x)\right\} \left\{ \dfrac{d}{dx}\psi(x)\right\}-f\left\{ \phi(x) \right\} \left\{ \dfrac{d}{dx}\phi(x)\right\}$

Given that,

$F(x)=\displaystyle \int_{0}^{x^2}f(\sqrt{t})dt$

$\therefore F'(x)=f(x)\cdot \dfrac d{dx}(x^2)-f(0)\cdot \dfrac d{dx}(0)$

$\Rightarrow F'(x)=2xf(x)\qquad...(i)$

Also given that,

$F'(x)=f'(x)$

$\Rightarrow 2xf(x)=f'(x)$ [ from

$(i)$ ]

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

Integrating both sides w.r.t.

$x$ ,

$\Rightarrow \displaystyle \int \dfrac{f'(x)}{f(x)}dx=\displaystyle \int 2xdx$

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

$\Rightarrow f(x)=e^{{x^2}+c}$

$\Rightarrow f(x)=ke^{x^2}$

Now,

$f(0)=1$

$\Rightarrow k=1$

Hence,

$f(x)=e^{x^2}$

$\therefore F(2)=\displaystyle \int_{0}^{4}e^tdt=\big[e^t\big]_{0}^{4}=e^4-1$

$\int_{1}^{3} F'(x)dx=-12$ and

$\int_{1}^{3} x^{3} F' '(x) dx=40$ , then the correct expression(s) is/are

0%
$9f'(3)+f'(1)-32=0$
0%
$\int_{1}^{3} f(x)dx=12$
0%
$9f' (3)-f'(1)+32=0$
0%
$\int_{1}^{3} f(x) dx=-12$

Which of the following function is differentiable at x=0

0%
$\cos \left( {|x|} \right) + |x|$
0%
$\cos \left( {|x|} \right) - |x|$
0%
$\sin \left( {|x|} \right) + |x|$
0%
$\sin \left( {|x|} \right) - |x|$

Explanation

A function is differentiable at 0 only if it's L.H.L &R.H.L is equal.

$f(x)=\sin\mid x \mid-\mid x \mid$ then given that

$\sin(-x)=-\sin x$ it can be expressed as,

$\therefore f(x)=\begin{Bmatrix}\sin x-x\quad if \,x\ge 0\\ -\sin x+x\quad if \,x<0\\ \end{Bmatrix}$

$\Rightarrow f^{'}(x)=\begin{Bmatrix}\cos x-1 \quad if\, x\ge 0 \\ -\cos x+1 \quad if\, x<0\\ \end{Bmatrix}$

Given that

$\cos 0=1$ we have

$f_{+}^{'}(0)=2$ &

$f_{-}^{'}(0)=-2$

$\therefore f_{+}^{'}(0) \ne f_{-}^{'}(0)$

$f(x)=\cos \mid x \mid-\mid x\mid$ , then given that

$\cos(-x)=\cos(x)$ can be written as:

$\therefore f(x)=\begin{Bmatrix}\cos x-x\quad if \,x\ge 0\\ \cos x+x\quad if \,x<0\\ \end{Bmatrix}$

$\Rightarrow f^{'}(x)=\begin{Bmatrix}-\sin x+1 \quad if\, x\ge 0 \\ -\sin x+1 \quad if\, x<0\\ \end{Bmatrix}$

Given that

$\sin 0=0$ we have

$f_{+}^{'}(0)=-1$ &

$f_{-}^{'}(0)=1$

$\therefore f_{+}^{'}(0) \ne f_{-}^{'}(0)$

$f(x)=\cos \mid x \mid+\mid x \mid$ , then

$\therefore f(x)=\begin{Bmatrix}\cos x+x\quad if \,x\ge 0\\ \cos x-x\quad if \,x<0\\ \end{Bmatrix}$

$\Rightarrow f^{'}(x)=\begin{Bmatrix}-\sin x+1 \quad if\, x\ge 0 \\ -\sin x-1 \quad if\, x<0\\ \end{Bmatrix}$

Given that

$\sin 0=0$ we have

$f_{+}^{'}(0)=1$ &

$f_{-}^{'}(0)=-1$

$\therefore A\Rightarrow \cos(\mid x \mid)+\mid x \mid$

$f : R \rightarrow R$ is defined by

$f(x) = \left \{\begin{matrix} \dfrac{x + 2}{x^2 + 3x + 2} & if & x \in R - \{-1, -1\} \\ -1 & if & x = -2 \\ 0 & if &x = -1\end{matrix} \right.$ then

$f(x)$ continuous on the set

0%
$R$
0%
$R - \{-2\}$
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$R - \{-1, -2\}$
0%
$R-\{-1\}$

$f\left( x+y \right) =f\left( x \right) +f\left( y \right)$ then

$f\left( x \right)$ may be

0%
$x$
0%
$x+1$
0%
${ x }^{ 2 }+1$
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$\log { x }$

The set of points, where the function

$f(x) = x|x|$ , is differentiable, is given by

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$(-\infty, \infty)$
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$(-\infty, 0) \cup (0, \infty)$
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$(0, \infty)$
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$[0, \infty)$

The function

$\displaystyle f(x) = (x^2 - 1) |x^2 - 3x + 2| + cos (|x|)$ , is not differentiable at x=

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

The function

$f\left( x \right) = \,{\sin ^{ - 1}}\left( {\cos \,x} \right)\,is\,: -$

0%
discontinuous at $=0$
0%
continuous at $=0$
0%
differentiable $=0$
0%
none of these

Explanation

Given:

$f\left(x\right)={\sin}^{-1}{\left(\cos{x}\right)}$

Continuity at

$x=0$

We have,LHL at

$x=0$

$\displaystyle \lim_{x\rightarrow\,{0}^{-}}{f\left(x\right)}=\displaystyle \lim_{h\rightarrow\,0}{\sin}^{-1}{\left(\cos{\left(0-h\right)}\right)}$

$=\displaystyle \lim_{h\rightarrow\,0}{\sin}^{-1}{\left(\cos{h}\right)}$

$={\sin}^{-1}{1}=\dfrac{\pi}{2}$

RHL at

$x=0$

$\displaystyle \lim_{x\rightarrow\,{0}^{+}}{f\left(x\right)}=\displaystyle \lim_{h\rightarrow\,0}{\sin}^{-1}{\left(\cos{\left(0+h\right)}\right)}$

$=\displaystyle \lim_{h\rightarrow\,0}{\sin}^{-1}{\left(\cos{h}\right)}$

$={\sin}^{-1}{1}=\dfrac{\pi}{2}$

$f\left(0\right)={\sin}^{-1}{\left(\cos{0}\right)}={\sin}^{-1}{1}=\dfrac{\pi}{2}$

Differentiability at

$x = 0:$

LHD at

$x = 0=\displaystyle \lim_{x\rightarrow\,{0}^{-}}{\dfrac{f\left(x\right)-f\left(0\right)}{x-0}}$

$=\displaystyle \lim_{x\rightarrow\,{0}^{-}}{\dfrac{{\sin}^{-1}{\left(\cos{\left(0-h\right)}\right)}-\dfrac{\pi}{2}}{-h}}$

$=\displaystyle \lim_{x\rightarrow\,{0}^{-}}{\dfrac{{\sin}^{-1}{\left(\cos{\left(-h\right)}\right)}-\dfrac{\pi}{2}}{-h}}$

$=\displaystyle \lim_{x\rightarrow\,{0}^{-}}{\dfrac{{\sin}^{-1}{\left(\cos{h}\right)}-\dfrac{\pi}{2}}{-h}}$

$=\displaystyle \lim_{x\rightarrow\,{0}^{-}}{\dfrac{{\sin}^{-1}{\left(\sin{\left(\dfrac{\pi}{2}-h\right)}\right)}-\dfrac{\pi}{2}}{-h}}$

$=\displaystyle \lim_{h\rightarrow\,0}\dfrac{-h}{-h}=1$

RHD at

$x = 0=\displaystyle \lim_{x\rightarrow\,{0}^{+}}{\dfrac{f\left(x\right)-f\left(0\right)}{x-0}}$

$=\displaystyle \lim_{x\rightarrow\,{0}^{+}}{\dfrac{{\sin}^{-1}{\left(\cos{\left(0+h\right)}\right)}-\dfrac{\pi}{2}}{h}}$

$=\displaystyle \lim_{x\rightarrow\,{0}^{+}}{\dfrac{{\sin}^{-1}{\left(\cos{\left(h\right)}\right)}-\dfrac{\pi}{2}}{h}}$

$=\displaystyle \lim_{x\rightarrow\,{0}^{+}}{\dfrac{{\sin}^{-1}{\left(\cos{h}\right)}-\dfrac{\pi}{2}}{h}}$

$=\displaystyle \lim_{x\rightarrow\,{0}^{+}}{\dfrac{{\sin}^{-1}{\left(\sin{\left(\dfrac{\pi}{2}-h\right)}\right)}-\dfrac{\pi}{2}}{h}}$

$=\displaystyle \lim_{h\rightarrow\,0}\dfrac{-h}{h}=-1\\$

$\therefore\,LHD\neq\,RHD$

Hence, the function is not differentiable at

$x = 0$ but is continuous at

$x = 0$

Hence option

$(b)$ is correct.

Given that

$f(x)$ is a differentiable function of

$x$ and that

$f\left( x \right).f\left( y \right) = f\left( x \right) + f\left( y \right) + f\left( {xy} \right) - 2$ and that

$f\left( 2 \right) = 5.$ Then

$f'\left( 3 \right)$ is equal to

0%
$6$
0%
$24$
0%
$15$
0%
$19$

$y^{y^{y^{....{^\infty}}}} = \log_e(x+\log_e(x+....))$ , then

$\dfrac{dy}{dx}$ at

$(x= e^2-2, y= \sqrt2)$ is

0%
$\dfrac{\log\left(\dfrac{e}{2}\right)}{2\sqrt2(e^2-1)}$
0%
$\dfrac{\log2}{2\sqrt2(e^2-1)}$
0%
$\dfrac{\sqrt2 \log\dfrac{e}{2}}{(e^2-1)}$
0%
None of these

Let

$y=x^{x^x}$ , then differentiate

$y$ w.r.t

$x$ .

0%
$x^{x^x}\left(\dfrac{1}{x}+\log x+(\log x)^2\right)$
0%
$x^{x^x}(x^x)\left(\dfrac{1}{x}+\log x-(\log x)^2\right)$
0%
$x^{x^x}(x^x)\left(\dfrac{1}{x}+\log x+(\log x)^2\right)$
0%
$x^{x^x}(x^x)\left(\dfrac{1}{x}-\log x-(\log x)^2\right)$

Explanation

$y=x^{x^x}$

diffrentiate it w.r.t.

$x$

$\dfrac{dy}{dx}=(e^{lnx^{x^x}})^{'}$

$\dfrac{dy}{dx}=(e^{x^x lnx})^{'}$

$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})^{'}$

$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}lnx)^{'}$

$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}(xlnx)^{'}ln x$ +

$\dfrac{e^{xlnx}}{x})$

$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}(xlnx)^{'}ln x$ +

$\dfrac{e^{xlnx}}{x})$

$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}(lnx+\dfrac{x}{x})ln x$ +

$\frac{e^{xlnx}}{x}$

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

$(x^x(lnx+1)lnx+\dfrac{x^x}{x}$

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

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

so option C is correct.

$x = 2\left( {\theta + \sin \theta } \right)and\,y = 2\left( {1 - \cos \theta } \right),then\;value\;of\frac{{dy}}{{dx}}\;is\;$

0%
$\tan \left( {\frac{\theta }{2}} \right)$
0%
$\cot \left( {\frac{\theta }{2}} \right)$
0%
$\sin \left( {\frac{\theta }{2}} \right)$
0%
$\cos \left( {\frac{\theta }{2}} \right)$

$f\left( x \right) = x + \left| x \right| + {\kern 1pt} \,\cos \left( {\left[ {{\pi ^2}} \right]x} \right)\,\,and\,\,g\left( x \right) = \,\sin \,x\,where\,\left[ . \right]$ denotes the greatest integer function) then :-

$(1)$

$f\left( x \right) + g\left( x \right)$ is discontinuous

$(2)$

$f\left( x \right) + g\left( x \right)$ is differentiable everywhere

$(3)$

$f\left( x \right) \times g\left( x \right)$ is differentiable everywhere

$(4)$

$f\left( x \right) \times g\left( x \right)$ is countimuos but not diffrentiable at

$x=0$

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

$f\left( x \right) = \left\{ \begin{array}{l}\frac{{1 - \left| x \right|}}{{1 + x}},{\rm{ }}x \ne - 1\\1,{\rm{ }}x = - 1{\rm{ }}\end{array} \right.$ then

$f\left( {\left[ {2x} \right]} \right),$ where

$\left[ {} \right]$ represents the greatest integer function , is

0%
discontinuous at $x = - 1$
0%
continuous at $x = 0$
0%
continuous at $x = \frac{1}{2}$
0%
continuous at $x = 1$

$x=a\cos^{3}\theta, y=a\sin^{3}\theta$ , then

$1+ {(\dfrac {dy}{dx})}^{2}$ is

0%
$\sec^{2}\theta$
0%
$\tan \theta$
0%
$1$
0%
$\tan^{2}\theta$

Explanation

Differentiate $x = a{\cos ^3}\theta$ with respect to $\theta$ ,

$\frac{{dx}}{{d\theta }} = - 3a{\cos ^2}\theta \sin \theta$

Differentiate $y = a{\sin ^3}\theta$ with respect to $\theta$ ,

$\frac{{dy}}{{d\theta }} = 3a{\sin ^2}\theta \cos \theta$

Now,

$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}$

$= - \frac{{3a{{\sin }^2}\theta \cos \theta }}{{3a{{\cos }^2}\theta \sin \theta }}$

$= - \tan \theta$

Then,

$1 + {\left( {\frac{{dy}}{{dx}}} \right)^2} = 1 + {\left( { - \tan \theta } \right)^2}$

$= 1 + {\tan ^2}\theta$

$= {\sec ^2}\theta$

Let f:R

$\rightarrow$ (0,1) be a continuous function.. Then, which of the following function(s) has (have) the value zero at some point in the interval (0,1)?

0%
$e ^ { x } - \int _ { 0 } ^ { 1 } f ( t ) \sin t d t$
0%
$f ( x ) + \int _ { 0 } ^ { 1 } f ( t ) \sin t d t$
0%
$x - \int _ { 0 } ^ { \frac { \pi } { 2 } - x } f ( t ) \cos t d t$
0%
$x ^ { 3 } - f ( x )$

$f:\left[ {0,1} \right] \to \left[ {0,1} \right]$ be definded by f(x) =

$\begin{cases} x,\quad \quad \quad \quad \quad \quad if\quad x\quad is\quad rational\quad \\ 1-x,\quad \quad \quad \quad if\quad x\quad is\quad irrational\quad \quad \quad \quad \quad \end{cases}$ then

$\left( {f \circ f} \right)x$ ______________.

0%
constant
0%
1+x
0%
x
0%
None of these

$\left( \frac { 1-x }{ 1+x } \right) =x$ and

$g\left( x \right) =\int { f\left( x \right) } dx$ then

0%
$g\left( x \right)$ is continuous in domain
0%
$g\left( x \right)$ is discontinuous st two points in its domain
0%
$\lim _{ x\rightarrow \infty }{ g\left( x \right)=-1 }$
0%
$\int { g\left( x \right) dx=-\frac { { x }^{ 2 } }{ 2 } +\left( 2x+1 \right) \lambda n\left( \frac { 1+x }{ e } \right) +C }$

$f(x)=\begin{cases} \dfrac { 1-\sqrt { 2 } \sin { x } }{ \pi -4x } ,\quad \quad ifx\neq \dfrac { \pi }{ 4 } \\ a\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ,\quad \quad ifx=\dfrac { \pi }{ 4 } \end{cases}$ is continous at

$x=\dfrac {\pi}{4}$ then

$a=$

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

A function

$f$ satisfies the relation

$f(x)=f'(x)+f"(x)+.....\infty$ terms, where

$f(x)$ is differentiable indefinitely, If

$f'(-1)=1$ then

$f(-1)$ is equal to

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

$f(x)= x\sin\dfrac{1}{x} , \ for x\neq 0$

$= 0,\ for x=0$

Then.

0%
$f'(0^+) exit\ but \ f'(0^-)$ does not exit
0%
$f'(0^+) \ and \ f'(0^-)$ do not exit
0%
$f'(0^+) = f'(0^-)$
0%
none of these

$f(x) = \left\{\begin{matrix}(3/x^{2})\sin 2x^{2} & if x M 0 \\\dfrac {x^{2} + 2x + c}{1 - 3x^{2}} & if\ x \geq 0, x \neq \dfrac {1}{\sqrt {3}}\\ 0 & x = 1/ \sqrt {3}\end{matrix}\right.$ then in order that

$f$ be continuous at

$x = 0$ , the value of

$c$ is

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

Let

$g(x)$ be a continuous function for all

$x$ , and

$f(x)=f(\alpha)+(x-\alpha).g(x)\ \forall \ x\ =\epsilon \ R$ . Then;

0%
$f(x)$ is necessarily differentiable at $x=\alpha$
0%
$f(x)$ is not necessarily differentiable at $x=\alpha$
0%
$f(x)$ is not necessarily continuous at $x=\alpha$
0%
$None\ of\ these$

Let a function

$f: R\rightarrow R$ be given by

$f(x+y)=f(x)f(y)$ for all

$x, y \in R$ and

$f(x)\neq 0$ for any

$x_{1}$ function

$f(x)$ is differentiable at

$x=0$ . Find

$f(x) gi ven f(0)=1$ .

0%
$e^{x}$
0%
$x.{f'(0)}$
0%
$\dfrac{x^{2}}{2}f'(x)$
0%
$None\ of\ these$

Define

$f\left( x \right) = \left\{ \begin{array}{l}{x^2} + bx + c\,\,\,\,\,,x < 1\\x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ge 1\end{array} \right.$ . If

$f(x)$ is differentiable at

$x=1$ then

$(b-c)=$

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

$\dfrac { d }{ dx } \left[ { cos }^{ -1 }\left( x\sqrt { x } -\sqrt { \left( 1-x \right) \left( 1-{ x }^{ 2 } \right) } \right) \right] =$

0%
$\dfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } -\dfrac { 1 }{ 2\sqrt { { x-x }^{ 2 } } }$
0%
$\dfrac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } -\dfrac { 1 }{ 2\sqrt { { x-x }^{ 2 } } }$
0%
$\dfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } +\dfrac { 1 }{ 2\sqrt { { x-x }^{ 2 } } }$
0%
$\dfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 12 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 12 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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