MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 15

if the function

$f(x)=\begin{cases} a+{\sin}^{-1}(x+b),\,x\geq 1 \\ x,\,x<1 \end{cases}$ is differentiable at

$x=1$ , then

$\displaystyle \frac{a}{b}$ is equal to

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

The function

$f(x) = \dfrac {x}{1 + |x|}$ is differentiable at which of the following?

0%
Every where
0%
Everywhere except at $x = 1$
0%
Everywhere except at $x = 0$
0%
Everywhere except at $x = 0$ or $1$

$y=e^{\sin^{-1}(t^{2}-1)}$ &

$x=e^{\sec^{-1}\left (\frac{1}{t^{2}-1}\right)}$ , then

$\dfrac{dy}{dx}$ is equal to

0%
$\frac{x}{y}$
0%
$\dfrac{-y}{x}$
0%
$\dfrac{y}{x}$
0%
$\dfrac{-x}{y}$

$f(x)={ sin }^{ -1 }\left[ \dfrac { 2x }{ 1+{ x }^{ 2 } } \right]$ ,then

$f(x)$ is differentiable on

0%
[-1,1]
0%
R-{-1,1}
0%
R-(-1,1)
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None of these

f(X)=|x|+|x-1| is continuous at

0%
'0' only
0%
0,1 only
0%
Every where
0%
No where

Explanation

$\textbf{Step-1: Finding values of}$

$\mathbf{f(x)}$

$f(x)=|x|+|x-1|$

$f(x)=-x-|x-1|$

$\text{at}$

$x\leq{0}$

$=x-(x-1)$

$\text{at}$

$0\leq{x}<1$

$=x+(x-1)$

$\text{at}$

$x\geq{1}$

$\text{Hence, we get}$

$f(x)=1-2x$

$\text{at}$

$x\leq{0}$

$=1$

$\text{at}$

$0\leq{x}<1$

$=2x-1$

$\text{at}$

$x\geq{1}$

$\textbf{Step-2: Finding limits of the function at constraint values}$

$\text{At}$

$x=0,$

$\lim_{x\to 0} f(x)=1$

$\text{Hence,}$

$f(x)$

$\text{is continuous at}$

$x=0$

$\text{At}$

$x=1,$

$\lim_{x\to 1} f(x)=1$

$\text{Hence,}$

$f(x)$

$\text{is continuous at}$

$x=1$

$\text{Hence,}$

$f(x)$

$\text{is continuous everywhere}$

$\textbf{Hence,Correct option is (C)}$

$\displaystyle f_{n}(x) = e^{f_{n-1}(x)}$ for all

$\displaystyle n \,\epsilon \,N$ and

$f_{0} (x) = x$ then

$\dfrac{d}{dx} \left \{f_{n} (x)\right \}$ is equal to

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$\displaystyle f_{n}(x) \,.\dfrac{d}{dx}\left \{f_{n - 1} (x)\right \}$
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$\displaystyle f_{n}(x) \,.f_{n - 1}(x)$
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$\displaystyle f_{n}(x) \,.f_{n - 1}(x) ..... f_{2}(x)\,.f_{1} (x)$
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$\displaystyle \prod_{i = 1}^{n} f_{i}(x)$

The order of the differential equation of all circles whose radius is

$4$ , is?

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

Explanation

Circle with radius

$4$ is represented as follows:-

$(x - h)^2 + (y - k)^2 = 4^2$ .....

$(1)$

Differentiating w.r.t

$x$

$2(x - h) + 2(y - k) \dfrac{dy}{dx} = 0$

$\Rightarrow\ (x - h) + (y - k) \dfrac{dy}{dx} = 0$ .....

$(2)$

Again differentiating w.r.t

$x$

$1 + (y - k)\dfrac{d^2y}{dx^2} + \left (\dfrac{dy}{dx} \right )^2 = 0$

$\Rightarrow\ (y - k) = \dfrac{-\left (\dfrac{dy}{dx} \right )^2 - 1}{\dfrac{d^2y}{dx^2}}$ ....

$(3)$

From

$(2)$ we get

$\Rightarrow\ (x - h) = - (y - k) \dfrac{dy}{dx}$

$\Rightarrow\ (x - h) = - \left \{ \dfrac{\left (\dfrac{dy}{dx} \right )^2 - 1 }{\dfrac{d^2y}{dx^2}} \right \} \dfrac{dy}{dx}$

$\therefore\ (x - h) = \dfrac{\left (\dfrac{dy}{dx} \right )^3 + \left (\dfrac{dy}{dx} \right )}{\dfrac{d^2 y}{dx^2}}$ ....

$(4)$

Substituting

$(4)$ and

$(3)$ in eq.

$(1)$

$\left (\dfrac{\left(\dfrac{dy}{dx} \right )^3 + \dfrac{dy}{dx}}{\dfrac{d^2y}{dx^2}} \right )^2 + \left (\dfrac{-\left (\dfrac{dy}{dx} \right )^2 - 1}{\dfrac{d^2y}{dx^2}} \right )^2 = 4^2$

$\left [\left (\dfrac{dy}{dx} \right )^3 + \dfrac{dy}{dx} \right ]^2 + \left [\left (\dfrac{dy}{dx} \right )^2 + 1 \right ]^2 = 16 \dfrac{d^2y}{dx^2}$

$\therefore$ Order of D.E of circle whose radius is

$4$ is

$2$ .

$f\left( x \right) = {\left| x \right|^{\left| {\sin x} \right|}}$ , then

${f'}\left( { - \dfrac{\pi }{4}} \right)$ is equals

0%
${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( { - \dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } - \dfrac{{2\sqrt 2 }}{\pi }} \right)$
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${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } + \dfrac{{2\sqrt 2 }}{\pi }} \right)$
0%
${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{\pi }{4} + \dfrac{{2\sqrt 2 }}{\pi }} \right)$
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None of these.

Let

$f:(-1,1)\rightarrow R$ be a differentiable function satisfying

$(f'(x))^4=16(f(x))^2$ for all

$x\in (-1,1)$

$f(0)=0$
The number of such functions is

0%
$2$
0%
$3$
0%
$4$
0%
more than $4$

Explanation

Given:

${\left({f}^{\prime}{\left(x\right)}\right)}^{4}=16{\left(f{\left(x\right)}\right)}^{2}$ for all

$x\in\,\left(–1 , 1\right)$

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

$\Rightarrow\, {\left({f}^{\prime}{\left(x\right)}\right)}^{2}=\pm\,4f\left(x\right)$

$\Rightarrow\, {f}^{\prime}{\left(x\right)}=\pm\,2\sqrt{\pm\,f\left(x\right)}$

$(i)$ Case

$1$

${f}^{\prime}{\left(x\right)}=2\sqrt{\pm\,f\left(x\right)}$

$\displaystyle\int{\dfrac{d\left(f\left(x\right)\right)}{f\left(x\right)}}=\displaystyle\int{2\,dx}$

$\Rightarrow\,2\sqrt{f\left(x\right)}=2x+c$

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

$\Rightarrow\,\sqrt{f\left(x\right)}=x$

$\Rightarrow\,x\ge 0$

$f\left(x\right)={x}^{2},\,\,0\le x<1$

$(ii)$ Case

$2$

${f}^{\prime}{\left(x\right)}=-2\sqrt{f\left(x\right)}$

$\Rightarrow\,\sqrt{f\left(x\right)}=-x\Rightarrow\,x\le 0$

$f\left(x\right)={x}^{2};\,\,\,-1<x\le 0$

$(iii)$ Case

$3$

${f}^{\prime}{\left(x\right)}=2\sqrt{-f\left(x\right)}$

$\sqrt{-f\left(x\right)}=x$

$-f\left(x\right)={x}^{2}$

$f\left(x\right)=-{x}^{2};\,\,\,0\le x< 1$

$(iv)$ Case

$4$

${f}^{\prime}{\left(x\right)}=-2\sqrt{-f\left(x\right)}$

$\sqrt{-f\left(x\right)}=-x$

$f\left(x\right)=-{x}^{2};\,\,\,-1< x\le 0$

$(v)$ Also, one singular solution of given differential equation is

$f\left(x\right)=0,\,\,-1<x<1$

Hence, there are more than

$4$ function possible

${f}_{1}\left( x \right) =\begin{cases} {x}^{2};0\le x< 1 \\ {-x}^{2}; -1<x<0 \end{cases}$

${f}_{2}\left( x \right) =\begin{cases} -{x}^{2};0\le x< 1 \\ {x}^{2}; -1<x<0 \end{cases}$

${f}_{3}\left( x \right) = {x}^{2},\,-1<x<1$

${f}_{4}\left( x \right) =-{x}^{2},\,-1<x<1$

${f}_{5}\left( x \right) =0,\,-1<x<1$ ...

Hence there are more than

$4$ solutions

$\displaystyle y = \dfrac{\sqrt{(1 + t^{2})} - \sqrt{(1 - t^{2})}}{\sqrt{(1 + t^{2})} +\sqrt{(1 - t^{2})}}$ and

$\displaystyle x = \sqrt {(1 - t^{4})}$ , then

$\dfrac{dy}{dx}$

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$\displaystyle \dfrac{-1}{t^{2}\left \{1 +\sqrt{1 - t^{4}} \right \}}$
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$\displaystyle \dfrac{\left \{\sqrt{(1 - t^{4})} - 1\right \}}{t^{6}}$
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$\displaystyle \dfrac{1}{t^{2}\left \{1 + \sqrt{(1 - t^{4})}\right \}}$
0%
$\displaystyle \dfrac{1 - \sqrt{( 1 - t^{4})}}{t^{6}}$

f(x) is diffrentiable function and (f(x). g(x)) is differentiable a x=a , then

0%
g(x) must be differentiable at x=a
0%
if g(x) is discontinuous , then f(a) =0
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f(a) $\neq$ 0 then g(x) must be differentiable
0%
nothing can be said

Give that f(x) =xg(x) /

$\left | x \right |$ , g(0) = 0 and f(x) is continous at x=Then the value of f' (0)

0%
Does not exist
0%
is -1
0%
is 1
0%
is 0

Let f(x)=

0%
f' is differentiable
0%
f is differentiable
0%
f' is continuous
0%
f is continuous

$f(x) = \left\{\begin{matrix} \dfrac{x\log \cos x}{\log(1+x^2)}, & x \neq 0\\ 0, & x=0\end{matrix}\right.$ then

0%
f(x) is not continuous at x=0.
0%
f(x) is continuous at x=0.
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f(x) is continuous at x=0 but not differentiable at x=0.
0%
f(x) is differentiable at x=0.

Let f:

$R\rightarrow R$ be a function such that f(x+y)= f(x)+f(y),

$\forall$ x,y

$\epsilon R$ . If f(x) is differentiable at x=0, then

0%
f(x) is differentiable only in a finite interval containing zero
0%
f(x) is continuous $\forall x\epsilon R$
0%
f(x) is constant $\forall x\epsilon R$
0%
f(x) is differentiable except at finitely many points

Let

$f(x)$ be a function satisfying

$f(x+y) = f(x)+f(y)$ and

$f(x) = xg(x)$

$\forall x, ~y \in$ R, where

$g(x)$ is a continuous function then, which of the following is true?

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$f'(x)=g'(0)$
0%
$f'(x)=g (0)$
0%
$f(x)=g (0)$
0%
$f(x)=g'(0)$

Which of the following is differentiable at x= 0

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

$\cos |x|$ is differentiable everywhere.

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True
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False

If the function

$f:[0,8] \rightarrow R$ is differentiable, then for

$0<a, b<2, \int_{0}^{8} f(t) d t$ is equal to

0%
$3\left[\alpha^{3} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]$
0%
$3\left[\alpha^{3} f(\alpha)+\beta^{3} f(\beta)\right]$
0%
${3}\left[\alpha^{2} f\left(\alpha^{3}\right)+\beta^{2} f\left(\beta^{3}\right)\right]$
0%
$3\left[\alpha^{2} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]$

Explanation

$\operatorname{Let} g(x)=\int_{0}^{x^{3}} f(t) d t\\$

$\text { Now } \int_{0}^{8} f(t) d t=g(2) =\dfrac{g(2)-g(1)}{2-1}+\dfrac{g(1)-g(0)}{1-0} \\$

$=g^{\prime}(\alpha)+g^{\prime}(\beta) \\$

$=3\left[\alpha^{2} f\left(\alpha^{3}\right)+\beta^{2} f\left(\beta^{3}\right)\right]$

Let f(x) and g(x) be differentiable for

$0\leq x\leq 1$ , such that f(0) such that f'(c)=2g'(c), then the value of g(1) must be

0%
1
0%
3
0%
-2
0%
-1

$f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2)$ and

$g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)$

The value of

$g(0)$ is

0%
0
0%
-3
0%
2
0%
None of these

Explanation

$f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2)$ and

$g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)$

$\text { Here put } g^{\prime}(1)=a, g^{\prime \prime}(2)=b \quad\quad\quad(1)$

$\text { Then } f(x)=x^{2}+a x+b, f(1)=1+a+b \Rightarrow f^{\prime}(x)=2 x+a$

$\quad f^{\prime \prime}(x)=2$

$\therefore g(x)=(1+a+b) x^{2}+(2 x+a) x+2=x^{2}(3+a+b)+a x\quad+2$

$\Rightarrow g^{\prime}(x)=2 x(3+a+b)+a \text { and } g^{\prime \prime}(x)=2(3+a+b)$

$\text { Hence, } g^{\prime}(1)=2(3+a+b)+a \quad\quad\quad(2)$

$g^{\prime \prime} (2)=2(3+a+b)\quad\quad\quad(3)$

From (1),(2) and (3) we have

$a=2(3+a+b)+a$

$\text { and } b=2(3+a+b)$

$\Rightarrow 3+a+b=0 \text { and } b+2 a+6=0$

Hence,

$b=0$ and

$a=-3 .$ So,

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

$g(x)=-3 x+2$

$g(x)=-3 x+2 \Rightarrow g(0)=2$

$f(x)$ is not invertible for

0%
$x \in\left[-\dfrac{\pi}{2}-\tan ^{-1} 2, \dfrac{\pi}{2}-\tan ^{-1} 2\right]$
0%
$x \in\left[\tan ^{-1} \dfrac{1}{2}, \pi+\tan ^{-1} \dfrac{1}{2}\right]$
0%
$x \in\left[\pi+\cot ^{-1} 2,2 \pi+\cot ^{-1} 2\right]$
0%
None of these

Explanation

$\begin{array}{l}f(x)=\sin x+\sin x \int_{-\pi / 2}^{\pi / 2} f(t) d t+\cos x \int_{-\pi / 2}^{\pi / 2} t f(t) d t \\ \qquad \begin{aligned}=\sin x\left(1+\int_{-\pi / 2}^{\pi / 2} f(t) d t\right)+\cos x \int_{-\pi / 2}^{\pi / 2} t(t) d t & \\=& A \sin x+B \cos x \\ \text { Thus, } A=& 1+\int_{-\pi / 2}^{\pi / 2} f(t) d t \\=& 1+\int_{-\pi / 2}^{\pi / 2}(A \sin t+B \cos t) d t \\=& 1+2 B \int_{0}^{\pi / 2} \cos t d t \\ &=\int_{-\pi / 2}^{\pi / 2} t(A \sin t+B \cos t) d t \\ &=1+2 B \\ = & 2 A \int_{0}^{\pi / 2} t f(t) d t \end{aligned}\end{array} \\$

$\begin{aligned}&=2 A[-t \cos t+\sin t]_{0}^{\pi / 2} \\\Rightarrow \quad B &=2 A\end{aligned}\\$
From equations (1) and

$(2),$ we get

$A=-1 / 3, B=-2 / 3 \\$

$\Rightarrow f(x)=-\dfrac{1}{3}(\sin x+2 \cos x)\\$
Thus, the range of

$f(x)$ is

$\left[-\dfrac{\sqrt{5}}{3}, \dfrac{\sqrt{5}}{3}\right]\\$

$\begin{aligned}f(x) &=-\dfrac{1}{3}(\sin x+2 \cos x) \\&=-\dfrac{\sqrt{5}}{3} \sin \left(x+\tan ^{-1} 2\right) \\&=-\dfrac{\sqrt{5}}{3} \cos \left(x-\tan ^{-1} \dfrac{1}{2}\right)\end{aligned}\\$

$f(x)$ is invertible if

$-\dfrac{\pi}{2} \leq x+\tan ^{-1} 2 \leq \dfrac{\pi}{2}\\$

$\Rightarrow-\dfrac{\pi}{2}-\tan ^{-1} 2 \leq x \leq \dfrac{\pi}{2}-\tan ^{-1} 2\\$
or

$0 \leq x-\tan ^{-1} \dfrac{1}{2} \leq \pi\\$

$\Rightarrow \tan ^{-1} \dfrac{1}{2} \leq x \leq \pi+\tan ^{-1} \dfrac{1}{2}\\$
or

$\pi \leq x-\tan ^{-1} \dfrac{1}{2} \leq 2 \pi\\$

$\Rightarrow x \in\left[\pi+\cot ^{-1} 2,2 \pi+\cot ^{-1} 2\right]\\$

Let

$f(0,\infty)\rightarrow R$ be a differentiable function such that

$f'(x)=2-\dfrac{f(x)}{x}$ for all

$x\epsilon (0,\infty)$ and

$f(1)\neq 1$ Then

0%
$\underset{x\rightarrow 0+}{lim}f'(\dfrac{1}{x})=1$
0%
$\underset{x\rightarrow 0+}{lim}xf'(\dfrac{1}{x})=2$
0%
$\underset{x\rightarrow 0+}{lim}x^2f'(\dfrac{1}{x})=0$
0%
$|f(x)|\le 2$ for all $x\epsilon(0,2)$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 15 - 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 15 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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