If $$\displaystyle f(x) = sin^{1}\left (\dfrac{2x}{1 + x^{2}} \right ) $$ , then

$$ f $$ is derivative for all $$ x $$ with ,$$ \left |x \right | > 1 $$

$$ f $$ is derivative for all $$ x $$ with ,$$ \left |x \right | < 1 $$

$$ f $$ is not derivative at $$ x = 1 $$

$$ f $$ is not derivable at $$ x = -1 $$

If $$y=\cos^{-1}x^{3}$$ then $$\dfrac{dy}{dx}=?$$

$$\dfrac{-3x^{2}}{\sqrt{1-x^{6}}}$$

$$\dfrac{-3}{x^{2}\sqrt{1-x^{6}}}$$

MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 9

$f(x)$ is a differentiable function

$\forall \ x\ \epsilon \ R$ so that,

$f(2)=4,\ f'(x)\ge 5\ \forall \ x\ \epsilon \ [2,6]$ , then,

$f(6)$ is :

0%
$\ge 24$
0%
$\le 24$
0%
$\ge 9$
0%
$\le 9$

$x=\sin^ {-1}\left( \dfrac { 2\theta }{ 1+{ \theta }^{ 2 } } \right),y=\sec^ {-1}\sqrt {1+\theta^ {2}}$ , then

$\dfrac {dy}{dx}=$

0%
$\dfrac {-1}{2}$
0%
$\dfrac {1}{2}$
0%
$-2$
0%
$2$

$f(x)=0$ for

$x < 0$ and

$f(x)$ is differential at

$x=0$ then for

$x\ge 0, f(x)$ may be

0%
$x^{2}$
0%
$x$
0%
$-x$
0%
$-x^{3/2}$

$f(x)=\begin{cases} mx-1,\quad x\le 5 \\ 3x-5,\quad x>5 \end{cases}$ is continuous then value of m is:

0%
$\dfrac{11}{5}$
0%
$\dfrac{5}{11}$
0%
$\dfrac{5}{3}$
0%
$\dfrac{3}{5}$

Explanation

$f{\left( x \right)} \begin{cases} mx-1 & \text{ if } x \le 5 \\ 3x - 5 & \text{ if } x > 5 \end{cases}$

Since the function is continuous at

$x = 5$ .

Therefore,

$\lim_{x \rightarrow {5}^{-}}{f{\left( x \right)}} = \lim_{x \rightarrow {5}^{+}}{f{\left( x \right)}} = f{\left( 5 \right)}$

$\lim_{x \rightarrow {5}^{-}}{\left( mx - 1 \right)} = \lim_{x \rightarrow {5}^{+}}{\left( 3x - 5 \right)}$

$5m - 1 = 15- 5$

$5m = 10 + 1$

$\Rightarrow m = \cfrac{11}{5}$

Hence the value of

$m$ is

$\cfrac{11}{5}$ .

$y={ tan }^{ -1 }\left( \frac { ax-b }{ bx+a } \right)$ , the value of

$\frac { dy }{ dx }$ is

0%
$\frac{ a}{1+x^{2}}$
0%
$\frac{1}{1+x^{2}}$
0%
$\frac{-b}{1+x^{2}}$
0%
$\frac{b}{1+x^{2}}$

Let

$f:[0,2]\rightarrow R$ be a twice differentiable function such that

$f"(x)>0$ , for all

$x\in (0,2)$ If

$\phi (x)=f(x)+f(2-x)$ , then

$\phi$ is:

0%
decreasing on $(0,2)$
0%
decreasing on $(0,1)$ and increasing on $(1,2)$
0%
increasing on $(0,2)$
0%
increasing on $(0,1)$ and decreasing on $(1,2)$

Explanation

$\phi(x)=f(x)+f(2-x)$

$\phi '(x)=f'(x)-f'(2-x)$ ...(1)
Since

$f"(x)>0$

$\Rightarrow f'(x) is increasing \forall x\in (0,2)$
case I when

$x>2-x \Rightarrow x>1$

$\Rightarrow \phi'(x) >0\forall x \in (1,2)$

$\therefore \phi (x) is increasing on (1,2)$
case II: when

$x<2-x \Rightarrow x<1$

$\Rightarrow \phi '(x) <0\forall x\in (0,1)$

$\therefore \phi (x)$ is decreasing on

$(0,1)$

$f(x)=|\cos x-\sin x|$ , then

$f'\left(\dfrac{\pi}{6}\right)$ equal to?

0%
$-\dfrac{1}{2}(1+\sqrt{3})$
0%
$\dfrac{1}{2}(1+\sqrt{3})$
0%
$-\dfrac{1}{2}(1-\sqrt{3})$
0%
$\dfrac{1}{2}(1-\sqrt{3})$

$f(x)=\left\{\begin{matrix} \dfrac{log_ex}{x-1} & x\neq 1\\ k & x=1\end{matrix}\right.$ continuous at

$x=1$ , then the value of k is?

0%
$e$
0%
$1$
0%
$-1$
0%
$0$

Explanation

Given

$f(x)=\left\{\begin{matrix} \dfrac{log_ex}{x-1} & x\neq 1\\ k & x=1\end{matrix}\right.$

we know that at

$x=1$ is continuous so

$k=\displaystyle\lim_{x\rightarrow 1}\dfrac{log_ex}{x-1}$

It is

$\dfrac 00$ form so by applying L-hospital rule

$=\displaystyle\lim_{x\rightarrow 1}\dfrac{1/x}{1}=1$ .

Let

$f(x)=15-|x-10|; x\in R$ . Then the set of all values of x, at which the function,

$g(x)=f(f(x))$ is not differentiable, is?

0%
$\{5, 10, 15, 20\}$
0%
$\{10, 15\}$
0%
$\{5, 10, 15\}$
0%
$\{10\}$

Explanation

$f(x)=15-|x-10|, x\in R$

$f(f(x))=15-|f(x)-10|$

$=15-|15-|x-10|-10|$

$=15-|5-|x-10||$

$x=5, 10, 15$ are points of non differentiability

Aliter:

$x=10$

$f(x)$ is no differentiable

also, when

$15-|x-10|=10$

$\Rightarrow x=5, 15$

$\therefore$ non differentiability points are

$\{5, 10, 15\}$ .

$f(x)=\begin{vmatrix} \sqrt{1+kx}-\sqrt{1-kx} & if & -1\leq x < 0\\ \begin{matrix} 2x+1\\ x-1\end{matrix} & if & 0\leq x\leq 1\end{vmatrix}$ is continuous at

$x=0$ , then the value of k is?

0%
$k=1$
0%
$k=-1$
0%
$k=0$
0%
$k=2$

Explanation

$\displaystyle\lim_{x\rightarrow 0^-}\dfrac{\sqrt{1+Kx}-\sqrt{1-Kx}}{x}=\displaystyle\lim_{x\rightarrow 0^+}\dfrac{mKx}{x(\sqrt{1-Kx}+\sqrt{x})}$

$\displaystyle\lim_{x\rightarrow 0^+}\dfrac{2x+1}{x-1}=-1=K$ (after rationlation)

$\therefore K=-1$ .

$f(x) = \begin{cases} \dfrac{\sin(p+1)x + \sin x}{x}& , &x < 0\\ \quad \quad \quad q&,& x = 0\\ \dfrac{\sqrt{x^2 + x}- \sqrt{x}}{x^{3/2}}&,& x > 0 \end{cases}$ is continuous at

$x = 0$ the

$(p, q)$ is

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

$x=3\sin {t} ,\ y=3\cos {t} ,$ find

$\dfrac {dy}{dx}$ at

$t=\dfrac { \pi }{ 3 }$

0%
$3$
0%
$0$
0%
$-\sqrt{3}$
0%
$1$

Explanation

$x=3\sin t$

$\dfrac{dx}{dt}=3\cos t$

$y=3\cos t$

$\dfrac{dy}{dt}=-3\sin t$

Thus,

$\dfrac{dy}{dx}=-\tan t$

$t=\dfrac{\pi}{3}$ ,

$\dfrac{dy}{dx}=-\sqrt3$

Let

$f(x)=(x+|x|)|x|$ . Then, for all x.

0%
$f$ is continuous
0%
$f$ is differentiable for some x
0%
$f'$ is continuous
0%
$f''$ is continuous

Explanation

Consider the given function.

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

$f(x)=\left\{\begin{matrix} (x+x)x, & x>0\\ (x-x)x, & x<0 \end{matrix}\right.$

$f(x)=\left\{\begin{matrix} 2x^2, & x>0\\ 0, & x<0 \end{matrix}\right.$

$f'(x)=\left\{\begin{matrix} 4x, & x>0\\ 0, & x<0 \end{matrix}\right.$

Now,

$\lim_{x\to\ 0^{+}}\ 2x^2=\lim_{x\to 0^{-}}0=0$

Since, it is polynomial function.

Hence, it is continuous and differentiable at

$0$ .

The set of points where the function

$f(x)=x|x|$ is differentiable is?

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

Explanation

Consider the given function.

$f(x)=x|x|$

$f(x)=\left\{\begin{matrix} x^2, & x>0\\ -x^2, & x<0 \\ 0 & x =0\end{matrix}\right.$

Since, it is polynomial function.

Hence, it is differentiable at

$(-\infty, \infty)$ .

Differentiate the following function with respect to x.
If for

$f(x)=\lambda x^2+\mu x+12$ ,

$f'(14)=15$ and

$f'(2)=11$ , then find

$\lambda$ and

$\mu$ .

0%
$\lambda =1, \mu =6$ .
0%
$\lambda =1, \mu =7$ .
0%
$\lambda =6, \mu =7$ .
0%
$\lambda =6, \mu =1$ .

Explanation

Given function

$f(x)=\lambda x^2 +\mu x +12$

$\implies f^{\prime}(x)=\dfrac d{dx}(\lambda x^2+\mu x+12)$

$f^{\prime}(x)=2\lambda x+\mu$

$f^{\prime}(2)=2\lambda (2)+\mu$

$\implies 4\lambda +\mu=11\cdots (1)$

$f^{\prime}(4)=2\lambda (4)+\mu$

$\implies 8\lambda+\mu=15 \cdots (2)$

$(2)-(1)$

$8\lambda-4\lambda+\mu-\mu=15-11$

$4\lambda =4$

$\implies \lambda=1$

Put it in

$(1)$

$\implies 4(4)+\mu =11$

$\mu=11-4$

$\mu =7$

$f(9) = 9, f'(9) = 0$ , then

$\underset{x\to 9}{\lim} \dfrac{\sqrt{f(x)}-3}{\sqrt{x}-3}$ is equal to

0%
$0$
0%
$f(0)$
0%
$f'(3)$
0%
$f(9)$
0%
$1$

Explanation

$\underset{x\to 9}{\lim} \dfrac{\sqrt{f(x)} -3}{\sqrt{x} -3}$

$\left[\dfrac{0}{0}from\right]$

$= \underset{x\to 9}{\lim} \dfrac{\dfrac{f'(x)}{2\sqrt{f(x)}}}{\dfrac{1}{2\sqrt{x}}}$

$= \dfrac{\sqrt{x}f'(x)}{\sqrt{f(x)}}$

$= \dfrac{\sqrt{9}f'(9)}{\sqrt{f(9)}}$

$= \dfrac{3\times 0}{3} = 0$

Let

$f : R \rightarrow R$ be a continuous function such that

$f(x^2) = f(x^3)$ for all

$x \in R$ . Consider the following statements.
I. f is an odd function.
II. f is an even function.
III. f is differentiable everywhere

0%
I is true and III is false
0%
II is true and III is false
0%
both I and III are true
0%
both II and III are true

Explanation

$f:R\rightarrow\,R$ be a continuous function such that

$f\left({x}^{2}\right)=f\left({x}^{3}\right)$ ..........

$(1)$ for all

$x\in R$

Put

$x=-x$

$f\left({x}^{2}\right)=f\left(-{x}^{3}\right)$

From

$(1)$ we have

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

Put

${x}^{3}=t$ we have

$f\left(t\right)=f\left(-t\right)$

$\Rightarrow\,f\left(x\right)$ is an even function.

$(ii)$ Now take

${x}^{3}=t$

$\Rightarrow\,f\left({t}^{\frac{2}{3}}\right)=f\left(t\right)$

Put

$t={t}^{\frac{2}{3}}\Rightarrow\,f\left({\left({t}^{\frac{2}{3}}\right)}^{2}\right)=f\left(t\right)$

$\Rightarrow\,f\left(t\right)=f\left({t}^{\frac{2}{3}}\right)=f\left({\left({t}^{\frac{2}{3}}\right)}^{2}\right)=f\left({\left({t}^{\frac{2}{3}}\right)}^{3}\right)=....=f\left({\left({t}^{\frac{2}{3}}\right)}^{n}\right)$

This is true for all

$t\in R$ and any

$n\in I$

Hence if we take

$n\rightarrow\,\infty,\,{\left(\dfrac{2}{3}\right)}^{n}\rightarrow\,0$

Then

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

$\Rightarrow\,f\left(x\right)$ is a constant function, hence it is differentiable everywhere.

Let

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

$x$ and

$y$ . If

$f(0) = 1, f(3) = 3$ and

$f'(0) = 11$ , then

$f'(3)$ is equal to

0%
$11$
0%
$22$
0%
$33$
0%
$44$

Explanation

We have,

$f'(x) = \underset{h\to 0}{\lim} \dfrac{f(x+h)-f(x)}{h}$

$\Rightarrow f'(3) = \underset{h\to 0}{\lim} \dfrac{f(3+h)-f(3)}{h}$

$=\underset{h\to 0}{\lim} \dfrac{f(3)f(h) - f(3+0)}{h}$

$=\underset{h\to 0}{\lim} \dfrac{f(3)f(h) - f(3)f(0)}{h}$

$=f(3) \underset{h\to 0}{\lim} \dfrac{f(h)-f(0)}{h}$

$=f(3) \underset{h\to 0}{\lim} \dfrac{f(0+h)-f(0)}{h}$

$= f(3)f'(0)$

$=3\times 11$

$=33$

Let

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

$f(x) = 1+\sin(3x)g(x)$ , where

$g$ is differentiable. The

$f'(x)$ is equal to

0%
$3f(x)$
0%
$g(0)$
0%
$3f(x) g(0)$
0%
$3g(x)$

Explanation

$f'(x) = \underset{h\to 0}{\lim} \dfrac{f(x+h) - f(x)}{h}$

$=\underset{h\to 0}{\lim} \dfrac{f(x)f(h) - f(x)}{h}$

$= f(x) \underset{h \to 0}{\lim} \left(\dfrac{1+\sin 3h(g(h))-1}{h}\right)$

$3f(x) \underset{h\to 0}{\lim} \dfrac{\sin 3h}{3h} \underset{h\to 0}{\lim} g(h)$

$=3 f(x) \times 1 \times g(0) = 3f(x) g(0)$

$y=\tan^{-1}\left(\dfrac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)$ then

$\dfrac{dy}{dx}=?$

0%
$\dfrac{a}{b}$
0%
$\dfrac{-b}{a}$
0%
$1$
0%
$-1$

Explanation

Let,

$\dfrac{a}{b}=\tan\theta$

$y=\tan^{-1}\left(\dfrac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)\\\,\,=\tan^{-1}\left(\dfrac{\dfrac{a}{b}-\tan x}{1+\dfrac{a}{b}\tan x}\right)\\\,\,=\tan^{-1}\left(\dfrac{\tan \theta-\tan x}{1+\tan \theta \tan x}\right)$

$=\tan^{-1}\tan (\theta-x)\\=\theta-x\\=\left(\tan^{-1}\dfrac{a}{b}-x\right)$ .

$\therefore \dfrac{dy}{dx}=-1$ .

$y=\tan^{-1}\left\{\dfrac{\cos x+\sin x}{\cos x-\sin x}\right\}$ then

$\dfrac{dy}{dx}=?$

0%
$1$
0%
$-1$
0%
$\dfrac{1}{2}$
0%
$\dfrac{-1}{2}$

$y=\sin^{-1}(3x-4x^3)$ then

$\dfrac{dy}{dx}=?$

0%
$\dfrac{3}{\sqrt {1-x^2}}$
0%
$\dfrac{-4}{\sqrt {1-x^2}}$
0%
$\dfrac{3}{\sqrt {1+x^2}}$
0%
$none\ of\ these$

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

0%
$f$ is derivative for all $x$ with , $\left |x \right | < 1$
0%
$f$ is not derivative at $x = 1$
0%
$f$ is not derivable at $x = -1$
0%
$f$ is derivative for all $x$ with , $\left |x \right | > 1$

$f(x) = \left\{\begin{matrix} x^2 (sgn [x]) + \{x\}), & 0 \le x < 2 \\ \sin x + | x - 3|, & 2 \le x < 4\end{matrix}\right.$

were [ ] and { } represent the greatest integer and the fractional part function, respectively.

0%
f (x) is differentiable at x=1.
0%
f(x) is continuous but non-differentiable at x=1.
0%
f(x) is non-differentiable at x=2.
0%
f(x) is non-differentiable at x=2.

$y=\cos^{-1}x^{3}$ then

$\dfrac{dy}{dx}=?$

0%
$\dfrac{-1}{\sqrt{1-x^{6}}}$
0%
$\dfrac{-3x^{2}}{\sqrt{1-x^{6}}}$
0%
$\dfrac{-3}{x^{2}\sqrt{1-x^{6}}}$
0%
$none\ of\ these$

$y=\cos^{-1}(4x^{3}-3x)$ then

$\dfrac{dy}{dx}=?$

0%
$\dfrac{3}{\sqrt{1-x^{2}}}$
0%
$\dfrac{-3}{\sqrt{1-x^{2}}}$
0%
$\dfrac{4}{\sqrt{1-x^{2}}}$
0%
$\dfrac{-4}{(3x^{2}-1)}$

Which of the following statement is always true ([ .] represents the greatest integer function)

0%
If f(x) is discontinuous, then $\left | f(x) \right |$ is discontinuous
0%
If f(x) is discontinuous , then f $\left ( \left | x \right | \right )$ is discontinuous
0%
f(x) = $\left [g \left (x \right ) \right ]$ is discontinuous when g(x) is an integer
0%
None of these

The largest value of c such that there exists a differential function

$h(x) for -c < x < c$ that is a solution of

$y_1 = 1 +y^2$ with h(0) = 0 is

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

Which of the following function is thrice differentiable at x=0?

0%
$f\left ( x \right )=\left | x^{3} \right |$
0%
$f\left ( x \right )=x^{3}\left | x \right |$
0%
$f\left ( x \right )=\left | x \right |\sin ^{3}x$
0%
$f\left ( x \right )=x\left | \tan ^{3}x \right |$

$y=\tan^{-1}(\sec x+6\tan x)$ then

$\dfrac{dy}{dx}=?$

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

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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