If $$y=\tan ^{-1} \sqrt{\dfrac{x+1}{x-1}}, $$ then $$ \dfrac{d y}{d x} $$ is

$$\dfrac{-1}{2|x| \sqrt{x^{2}-1}} $$

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

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

If $$P(1)=0$$ and $$\dfrac {dP(x)}{dx} > P(x)$$ for all $$x \ge 1$$, then

$$P(x)$$ is a constant function

neither injective nor surjective

If $$y = \sin(2 \sin^{-1} x)$$, then dx = .......

$$\frac{4x^2 - 1}{\sqrt{1 - x^2}}$$

If $$y=x^{\left(x^{ x}\right)}$$, then $$\displaystyle\frac{dy}{dx}$$ is

$$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x^{\displaystyle x-1}\right]$$

$$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x^{\displaystyle x}\right]$$

$$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x\right]$$

$$y\left[x^{\displaystyle x}\left(\log_e{x}\right)\log{x}+x^{\displaystyle x-1}\right]$$

MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 10

$y=\tan^{-1}\left(\dfrac{\sqrt{a+\sqrt{x}}}{1-\sqrt{ax}}\right)$ then

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

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

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

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

0%
$\dfrac{-2}{(1+x^{2})}$
0%
$\dfrac{2}{(1+x^{2})}$
0%
$\dfrac{-1}{(1-x^{2})}$
0%
$none\ of\ these$

Explanation

$y=\sec^{-1}\left(\dfrac{x^{2}+1}{x^{2}-1}\right)$

Put

$x=\cot\theta$ .

As we know,

$\cos 2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$

Then

$y=\sec^{-1}\left(\dfrac{1+\tan^{2}\theta}{1-\tan^{2}\theta}\right)=\sec^{-1}\left(\dfrac{1}{\cos 2\theta}\right)=\sec^{-1}(\sec 2\theta)=2\theta=2\cot^{-1}x$ .

$\therefore \dfrac{dy}{dx}=\dfrac{-2}{1+x^2}$

$y=\tan ^{-1} \sqrt{\dfrac{x+1}{x-1}},$ then

$\dfrac{d y}{d x}$ is

0%
$\dfrac{-1}{2|x| \sqrt{x^{2}-1}}$
0%
$\dfrac{-1}{2 x \sqrt{x^{2}-1}}$
0%
$\dfrac{1}{2 x \sqrt{x^{2}-1}}$
0%
$None \ of \ these$

Explanation

Let

$x=\sec \theta$

$\text { Then } y=\tan ^{-1} \sqrt{\dfrac{\sec \theta+1}{\sec \theta-1}}$

$\quad=\tan ^{-1} \sqrt{\dfrac{1+\cos \theta}{1-\cos \theta}}=\tan ^{-1}\left(\cot \dfrac{\theta}{2}\right)$

$\Rightarrow y=\tan ^{-1}\left\{\tan \left(\dfrac{\pi}{2}-\frac{\theta}{2}\right)\right\}=\dfrac{\pi}{2}-\dfrac{1}{2} \sec ^{-1} x$

$\Rightarrow \dfrac{d y}{d x}=-\dfrac{1}{2} \times \dfrac{1}{|x| \sqrt{x^{2}-1}}$

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

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

$f(3)$ is

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

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

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

$f(x)=x^{2}-3 x \Rightarrow f(3)=9-9=0$

The function given by y=

$\left | x-1 \right |$ is differentiable function and

$f(1/n)$ = 0

$\forall n\geq 1$ and n\epsilon I $$, then

0%
f(x)=0, $x\epsilon (0,1]$
0%
f(0)=0, f'(0) =0
0%
f(0)= 0=f'(0), $x\epsilon (0,1]$
0%
f(0)=0 and f'(0) need not to be zero

Explanation

Given that f(x) is a continuous and differentiable function and

$f(\dfrac{1}{x})=0,x=n,n\epsilon l$

$\therefore f(0^{+})=f(\dfrac{1}{\infty })=0$
since R.H.L=0,

$\therefore f(0)=0$ for f(x)to be continuous Alsio f'(0)

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

$[Usingf(0)=0][\because f(0^{+})=0]$ Hence f(0)=0 f(0)

$P(1)=0$ and

$\dfrac {dP(x)}{dx} > P(x)$ for all

$x \ge 1$ , then

0%
$P(x) > 0\forall x > 1$
0%
$P(x)$ is a constant function
0%
$P(x) < 0\forall x > 1$
0%
None of these

Explanation

$\dfrac {dP(x)}{dx} > P(x)$

$\Rightarrow \ e^{-x}\dfrac {dP(x)}{dx}-e^{-x}P(x) > 0$

$\Rightarrow \dfrac {d}{dx}(P(x)e^{-x}) > 0$

$\Rightarrow P(x)e^{-x}$ is an increasing function.

$\Rightarrow P(x)e^{-x} > P(1)e^{-1} \forall a \ge 1$

$\Rightarrow P(x) e^{-x} > 0 \forall x > 1\Rightarrow P(x) > 0 \forall x > 1$ .

The function

$f(x)=\dfrac{x}{1+\left | x \right |}$ is differentiable

0%
only at non-integer points
0%
everywhere
0%
only at $x=0$
0%
none of these

Explanation

Thus

$f(x)$ is differentiable at

$x=0$

$f(x)=\dfrac{x}{1+\left | x \right |}$ is everywhere differentiable

1599570_1752603_ans_7786c109f25d40c6a4bd51e80124c9f0.png

$H(x_0)=0$ for some

$x=x_0$ and

$\dfrac {d}{dx}H(x) > 2cxH(x)$ for all

$x \le x_0$ , where

$c > 0$ , then

0%
$H(x)=0$ has root for $x > x_0$
0%
$H(x)=0$ has no root for $x > x_0$
0%
$H(x)=0$ is constant function
0%
None of these

Explanation

Given that

$\dfrac {d}{dx}H(x) > 2cxH(x)$

$\Rightarrow \ e^{-cx^2}\dfrac {d}{dx}H(x)-e^{-cx^2}2cxH(x) > 0$

$\Rightarrow \ \dfrac {d}{dx}(H(x)e^{-cx^2}) > 0$

$\Rightarrow \ H(x)e^{-cx^2} > 0$ is an increasing function.
But

$H(x_0)=0$ and

$e^{-cx^2}$ is always positive.

$\Rightarrow \ H(x) > 0$ for all

$x > x_0$

$\Rightarrow \ H(x)$ cannot be zero for any

$x > x_0$ .

Given a function

$f:[0,4] \rightarrow \mathrm{R}$ is differentiable, then for

$\displaystyle \operatorname{some} \alpha, \beta \in(0,2), \int_{0}^{4} f(t) d t$ equals to

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

Explanation

$\displaystyle I=\int_{0}^{4} f(t) dt$

Put

$t=x^2$

$\Rightarrow d t=2 x dx$

$\displaystyle \therefore I=2 \int_{0}^{2} x f\left(x^{2}\right) dx$

From Lagrange's Mean Value Theorem

$\dfrac{\displaystyle \int_{0}^{2} 2 x f\left(x^{2}\right) d x-\int_{0}^{0} 2 x f\left(x^{2}\right) d x}{2-0}=2 y f\left(y^{2}\right)$

$\Rightarrow \displaystyle \int_{0}^{2} 2 x f\left(x^{2}\right) d x=2 \times 2 y f\left(y^{2}\right) \\$

$=2\left\{\dfrac{2 \alpha f\left(\alpha^{2}\right)+2 \beta f\left(\beta^{2}\right)}{2}\right\}\\$

(where

$0<\beta<y<\alpha<2$ , and using intermediate Mean Value Theorem.)

If f(x)=

$[sin^{-1}(sin2x)]$ (where, [] denotes the greatest integer function), then

0%
$\displaystyle \int_{0}^{\pi/2}f(x)dx=\dfrac{\pi}{2}-sin^{-1}(sin1)$
0%
f(x) is periodic with period
0%
$\displaystyle \lim_{x\dfrac{\pi}{2}}f(x)=-1$
0%
None of the above

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

$f(x)$ satisfying

$f(x+ y) - f(x - y) = f(x) \{ f(y) - f(y) - y \}, \forall x, y \epsilon R, (f(y) \neq f(-y)$ for all

$y \epsilon R)$ and

$f' (0) = 2010$ .
Now answer the following questions

Which of the following is true for

$f(x)$

0%
$f(x)$ is one-one and into
0%
$\{ f(x) \}$ is non-periodic, where {.} denotes fractional part of x.
0%
$f(x) = 4$ has only two solutions.
0%
$f(x) = f^{-1} x$ has only one solution

Explanation

Here,

$2 f' (x) = \displaystyle \lim_{h \rightarrow 0} \left ( \frac{f(x + h) - f(x)}{h} + \frac{f(x x - h) - f(x)}{-h} \right )$

$= \displaystyle \lim_{h \rightarrow 0} \left ( \frac{f(x + h) - f(x - h)}{h} \right )$ (i)

$\therefore 2f'(0) \displaystyle \lim_{h \rightarrow 0} \left ( \frac{f(h) - f (-h)}{h} + \frac{f(-h) - f(0)}{-h} \right )$

$= \displaystyle \lim_{h \rightarrow 0}\frac{f(h) - f(-h)}{h}$ ... ..(ii)
Now by given relation, we have

$f(h) - f(-h) = \frac{f(x + h) - f(x - h)}{-h}$ and

$f(0) = 1$
From Eqs. (i) and (ii), we have

$\frac{f'(x)}{f(x)} = 2010$

$\Rightarrow$

$f(x) = e^{2010e}, f(0) = 1$

$\therefore \{ f(x) \}$ is non-periodic.

If f(x) =

$\displaystyle \int_{0}^{x}(f(t))^2 dt f:R→R$ be differentiable function and f(g(x)) is differentiable function at x=a, then

0%
g(x) must be differentiable at x=a
0%
g(x) may be non-differentiable at x=a
0%
g(x) must be discontinuous at x=a
0%
None of the above

Explanation

Here,

$f'(x)=(f(x))^2>0;\dfrac{d}{dx} f(g(x)) |_{x=a}$

$=f'(g(x))\displaystyle \lim_{x→a}\dfrac{g(x)-g(a)}{x-a}$

It implies that g(x) must be differentiable at x=a.

$y=f(x)$ is

0%
injective but not surjective
0%
surjective but not injective
0%
bijective
0%
neither injective nor surjective

Explanation

$\begin{aligned}f(x)=& x^{2}+\int_{0}^{x} e^{-t} f(x-t) d t \\=& x^{2}+\int_{0}^{x} e^{-(x-t)} f(x-(x-t)) d t \\&=x^{2}+e^{-x} \int_{0}^{x} e^{t} f(t) d t\end{aligned}$
Differentiating w.r.t.

$x,$ we get

$\begin{aligned}\Rightarrow f^{\prime}(x)=& 2 x-e^{-x} \int_{0}^{x} e^{t} f(t) d t+e^{-x} \cdot e^{x} f(x) \\&=2 x-e^{-x} \int_{0}^{x} e^{t} f(t) d t+f(x)\end{aligned}$

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

$\Rightarrow f(x)=\dfrac{x^{3}}{3}+x^{2}+c\\$
Also

$f(0)=0\\$ [using equation from equation]

$\Rightarrow f(x)=\dfrac{x^{3}}{3}+x^{2} \\$

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

$\Rightarrow f^{\prime}(x)=0$ has real roots, hence

$f^{\prime}(x)$ is non-monotonic
Hence,

$f(x)$ is many-one, but range is

$R$ , hence surjective.

The current statement(s) is/are

0%
$f'(1)<0$
0%
$f(2)<0$
0%
$f'(x)\neq 0$ for any $x \in (1,3)$
0%
$f'(x) = 0$ for some $x \in (1,3)$

$u=\sin^{-1}\Bigg(\dfrac{2x}{1+x^2}\Bigg)$ and

$v=\tan^{-1}\Bigg(\dfrac{2x}{1-x^2}\Bigg)$ , then

$\dfrac{du}{dv}$ is

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

The function

$f(x)=\Bigg\{\dfrac{\sin x}{x}+\cos x,if\,x\neq 0\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k,if\,x=0$ is continuous at

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

0%
$3$
0%
$2$
0%
$1$
0%
$1.5$

$|\sin x|$ is a differentiable function for every value of

$x$ .

0%
True
0%
False

The function

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

0%
continuous at $x = 0$ as well as at $x = 1$ .
0%
continuous at $x = 1$ but not at $x = 0$ .
0%
discontinuous at $x = 0$ as well as at $x = 1$ .
0%
continuous at $x = 0$ but not at $x = 1$ .

$f(x) = \sin^{-1} \left ( \frac{4^{x + \frac{1}{2}}}{1 + 2^{4x}} \right )$ , which of the following is not the derivative of f(x)?

0%
$\frac{2.4^{x} \cdot \log 4}{1 + 4^{2x}}$
0%
$\frac{4^{x + 1} \cdot \log 2}{1 + 4^{2x}}$
0%
$\frac{4^{x + 1} \cdot \log 4}{1 + 4^{4x}}$
0%
$\frac{2^{2^(x + 1)} \cdot \log 2}{1 + 2^{4x}}$

Explanation

Put

$4^x = \tan \theta$ . Then

$\theta = \tan^{-1}(4^x)$

$\therefore f(x) = \sin^{-1} \left ( \frac{2 \tan \theta}{1 + \tan \theta} \right )$

$= \sin^{-1} (\sin 2 \theta)$

$= 2 \theta$

$= 2 \tan^{-1} (4^x)$

$\therefore f'(x) = 2 \times \frac{1}{1 + (4^x)^2} \times 4^x \log 4$

$= \frac{2.4^x \cdot \log 4}{1 + 4^{2x}}$ ....(a)

$= \frac{2.4^x \cdot 2 \log 2}{1 + 4^{2x}}$

$= \frac{4^{x + 1} \cdot \log 2}{1 + 4^{2x}}$ ...(b)

$= \frac{(2^{2})^{x + 1} \cdot \log 2}{1 + 2^{4x}}$

$= \frac{2^{2^(x + 1)} \cdot \log 2}{1 + 2^{4x}}$ ....(d)

$y \tan^{-1} \left ( \sqrt{\frac{a - x}{a + x}} \right )$ , where -a < x < a, then

$\frac{dy}{dx} =$ .....

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

Explanation

Put

$x = a \cos \theta$

$y = \tan^{-1} \left ( \frac{x}{1 + \sqrt{1 - x^2}} \right ) + \sin \left [ 2 \tan^{-1} \left ( \sqrt{\frac{1 - x}{1 + x}} \right ) \right ]$ then

$\frac{dy}{dx} =$ ...........

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

Explanation

$y = \tan^{-1} \left ( \frac{x}{1 + \sqrt{1 - x^2}} \right ) + \sin \left [ 2 \tan^{-1} \left ( \sqrt{\frac{1 - x}{1 + x}} \right ) \right ]$
Put

$x = \cos \theta.$ Then

$\theta = \cos^{-1} x$

$\therefore y = \tan^{-1} \left ( \frac{\cos \theta}{1 + \sqrt{1 - \cos \theta^2}} \right ) + \sin \left [ 2 \tan^{-1} \left ( \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \right ) \right ]$

$= \tan^{-1} \left ( \frac{\cos \theta}{1 + 1 + \tan \theta} \right ) + \sin \left [ 2 \tan^{-1} \left ( \sqrt{\frac{2 \sin^2 (\frac{\theta}{2})}{2 \cos^2(\frac{\theta}{2})}} \right ) \right ]$

$= \tan^{-1} \left [ \frac{\sin (\frac{\pi}{2} - \theta)}{1 + \cos (\frac{\pi}{2} - \theta)} \right ] + \sin [2 \tan^{-1} (\tan \frac{\theta}{2})]$

$= \tan^{-1} \left [ \frac{2 \sin (\frac{\pi}{4} - \frac{\theta}{2}) \cdot \cos (\frac{\pi}{4} - \frac{\theta}{2})}{2 \cos^2 (\frac{\pi}{4} - \frac{\theta}{2})} \right ] + \sin (2 \times \frac{\theta}{2})$

$= \tan^{-1} [\tan (\frac{\pi}{4} - \frac{\theta}{2})] + \sin \theta$

$= \frac{\pi}{4} - \frac{\theta}{2} + \sqrt{1 - \cos^2 \theta}$

$= \frac{\pi}{4} - \frac{2}{2} \cos^{-1} x + \sqrt{1 - x^2}$

$\frac{dy}{dx} = 0 - \frac{1}{2} \times \frac{1}{\sqrt{1 - x^2}} + \frac{1}{2 \sqrt{1 - x^2}} \times (-2x)$

$= \frac{1}{2 \sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}}$

$= \frac{1 - 2x}{2 \sqrt{1 - x^2}}$

$y = \sin(2 \sin^{-1} x)$ , then dx = .......

0%
$\frac{2 - 4x^2}{\sqrt{1 - x^2}}$
0%
$\frac{2 + 4x^2}{\sqrt{1 - x^2}}$
0%
$\frac{4x^2 - 1}{\sqrt{1 - x^2}}$
0%
$\frac{1 - 2x^2}{\sqrt{1 - x^2}}$

Explanation

$y = \sin(2 \sin^{-1} x)$
Put

$x = \sin \theta$ . Then

$\theta = \sin^{-1} x$

$\therefore y = \sin 2 \theta$

$\therefore \frac{dy}{dx} = \frac{dy}{d \theta} \times \frac{d \theta}{dx} = 2 \cos 2 \theta \times \frac{1}{\sqrt{1 - x^2}}$

$= \frac{2(1 - 2 \sin^2 \theta)}{\sqrt{1 - x^2}} = \frac{2 - 4x^2}{\sqrt{1 - x^2}}$

If function

$f(x)=\dfrac{x^2-9}{x-3}$ is continuous at

$x=3$ , then value of

$(3)$ will be:

0%
$6$
0%
$3$
0%
$1$
0%
$0$

Explanation

$f(x)=\dfrac{x^2-9}{x-3}$

Left hand limit

$f(3+0)=\displaystyle \lim_{h \rightarrow 0}f(3+h)$

$=\displaystyle \lim_{h \rightarrow 0} \dfrac{(3+h)^2-9}{3+h-3}$

$=\displaystyle \lim_{h \rightarrow 0} \dfrac{h(6+h)}{h}$

$=\displaystyle \lim_{h \rightarrow 0} (6+h)$

$=6$

Function is continous at

$x=3$ , so

$f(3)=f(3+0)$

$f(3)=6$

Hence, option

$(a)$ is correct.

$f(x)=\begin{cases} \begin{matrix} \dfrac{\log (1+mx)- \log (1-nx)}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} k; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$
is continuous at

$x=0$ then the value of

$k$ will be:

0%
$0$
0%
$m+n$
0%
$m-n$
0%
$m.n$

Explanation

$\therefore$ Function is continous at

$x=0$

$\therefore \displaystyle \lim_{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{\log (1+mx)- \log (1-nx)}{x}=k$

$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{ [mx-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3} - ...]- [nx-\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} - ...]}{x}=k$

$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{x \left[ m-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3}-...+n+\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} +... \right]}{x}=k$

$\Rightarrow \displaystyle \lim_{x \rightarrow 0} m-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3}-...+n+\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} +... =k$

$\Rightarrow m+n=k$

$\Rightarrow k=m+n$

Hence, option

$(b)$ is correct.

If function

$f(x)=\begin{cases} \begin{matrix} \dfrac{\sin 3x}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} m; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$
is continuous at

$x=2$ then value of

$m$ will be:

0%
$3$
0%
$1/3$
0%
$1$
0%
$0$

Explanation

$f(x)=\begin{cases} \begin{matrix} \dfrac{\sin 3x}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} m; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$

$f(0)=m$

$f(0+0)=\displaystyle \lim_{h \rightarrow 0}f(0+h)$

$\displaystyle \lim_{h \rightarrow 0} \dfrac{\sin 3 (0+h)}{0+h}$

$\displaystyle \lim_{h \rightarrow 0}3 \dfrac{\sin 3 h}{3h}$

$=3 \times 1$

$=3$

$x=a$ function is continuous.

So,

$f(0)=f(0+0)$

$m=3$

Hence, option

$(a)$ is correct.

Let

$x={f}''(t) cost +{f}'(t) sint$ and

$y={-f}''(t) sint+{f}'(t) cost.$ Then

$\displaystyle \int \left [ \left(\frac{dx}{dt} \right)^2 + \left(\frac{dy}{dt} \right)^2 \right ]^{\frac{1}{2}} dt$ equals
(Note :

$f(x), f'(x), f''(x) , f'''(x) >0$ )

0%
${f}'(t)+{f}''(t)+c$
0%
${f}''(t)+{f}'''(t)+c$
0%
$f(t)+{f}''(t)+c$
0%
${f}'(t)-{f}''(t)+c$

Explanation

$x=f''(t)\cos { t } +f'(t)\sin { t } \\ \dfrac { dx }{ dt } =f'''(t)\cos { t } -f''(t)\sin { t } +f''(t)\sin { t } +f'(t)\cos { t } \\ \dfrac { dx }{ dt } =f'''(t)\cos { t } +f'(t)\cos { t } \\ \dfrac { dx }{ dt } =(f'''(t)+f'(t))\cos { t } ..........(1)\\ y=-f''(t)\sin { t } +f'(t)\cos { t } \\ \dfrac { dy }{ dt } =-f'''(t)\sin { t } -f''(t)\cos { t } +f''(t)\cos { t } -f'(t)\sin { t } \\ \dfrac { dy }{ dt } =-f'''(t)\sin { t } -f'(t)\sin { t } \\ \dfrac { dy }{ dt } =-\sin { t } (f'''(t)+f'(t))........(2)\\ add\quad squares\quad of\quad eq.1\quad and\quad 2\quad \\ { \left( \dfrac { dx }{ dt } \right) }^{ 2 }+{ \left( \dfrac { dy }{ dt } \right) }^{ 2 }={ (f'''(t)+f'(t)) }^{ 2 }\\ { \left( { \left( \dfrac { dx }{ dt } \right) }^{ 2 }+{ \left( \dfrac { dy }{ dt } \right) }^{ 2 } \right) }^{ \dfrac { 1 }{ 2 } }=f'''(t)+f'(t)\\ integrate\quad both\quad sides\quad \\ \int { { \left( { \left( \dfrac { dx }{ dt } \right) }^{ 2 }+{ \left( \dfrac { dy }{ dt } \right) }^{ 2 } \right) }^{ \dfrac { 1 }{ 2 } }dt } =f''(t)+f(t)+c$

The set of all points where the function

$f(\displaystyle \mathrm{x})=\frac{x}{1+|x|}$ is differentiable is

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

Explanation

$f(x) = \dfrac{x}{1+|x|}$

For

$x\geq 0$ ,

$f(x) = \dfrac{x}{1+x}$

For

$x<0$ ,

$f(x) = \dfrac{x}{1-x}$

$f(x)$ is continuous and differentiable at all points except perhaps at

$x=0$ .

Hence, we will check the differentiability at

$x=0$ .

$x \rightarrow 0$ ,

$f(x) \rightarrow 0$ .

$f(0) = 0$ .

Hence,

$f(x)$ is continuous at

$x=0$ .

$RHD = \dfrac{ 1+x-x}{(1+x)^2} = 1$ {see that

$f(x)=\infty$ at

$x=-1$ but

$-1 \notin \ [0,\infty)$ }

$LHD = \dfrac {1-x+x}{(1-x)^2} = 1$

Hence,

$f(x)$ is differentiable at

$x=0$ .

Hence, option A is correct.

$y=x^{\displaystyle x^{\displaystyle x^{\displaystyle \dots^{\displaystyle\infty}}}}$ , find

$\displaystyle\frac{dy}{dx}$ .

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$\displaystyle\frac{y^2}{x(1-y\log{x})}$
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$\displaystyle\frac{y}{x(1-\log{x})}$
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$\displaystyle\frac{y^2}{x(y-\log{x})}$
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None of these

Explanation

The given function may be written as,

$y=x^{\displaystyle y}$
Taking

$\log$ on both sides,

$\log{y}=\log { { x }^{ y } }$

$\log{y}=y\log{x}$
Differentiate w.r. to

$x$

$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =y\frac { d }{ dx } \left( \log { x } \right) +\log { x } \frac { d }{ dx } y$

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

$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } -\log { x } .\frac { dy }{ dx } =\frac { y }{ x }$

$\displaystyle \frac { dy }{ dx } \left( \frac { 1 }{ y } -\log { x } \right) =\frac { y }{ x }$

$\displaystyle \frac { dy }{ dx } \left( \frac { 1-y\log { x } }{ y } \right) =\frac { y }{ x }$

$\therefore \displaystyle \frac { dy }{ dx } =\frac { { y }^{ 2 } }{ x\left( 1-y\log { x } \right) }$

Derivative of

$({\log{x}})^{\displaystyle\cos{x}}$ with respect to

$x$ is

0%
$\displaystyle({\log{x}})^{\displaystyle\cos{x}}\left[\frac{\cos{x}}{x\log{x}}-\sin{x}\log{(\log{x})}\right]$
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$\displaystyle({\log{x}})^{\displaystyle\cos{x}}\left[\frac{\cos{x}}{x\log{x}}-\cos{x}\log{(\log{x})}\right]$
0%
$\displaystyle({\log{x}})^{\displaystyle\sin{x}}\left[\frac{\sin{x}}{x\log{x}}-\cos{x}\log{(\log{x})}\right]$
0%
None of these

Explanation

$y={ \log { x } }^{ \cos { x } }$
Taking log on both sides, we get

$\log { y } =\log { \left( { \log { x } }^{ \cos { x } } \right) }$

$\log { y } =\cos { x } .\left[ \log { \left( \log { x } \right) } \right]$
Differentiate w.r. to

$x$

$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =\cos { x } \frac { d }{ dx } \left[ \log { \left( \log { x } \right) } \right] +\left[ \log { \left( \log { x } \right) } \right] \frac { d }{ dx } \cos { x }$

$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =\cos { x } .\frac { 1 }{ \log { x } } \frac { d }{ dx } \left( \log { x } \right) -\left[ \log { \left( \log { x } \right) } \right] \sin { x }$

$\displaystyle \frac { dy }{ dx } =y\left[ \cos { x.\frac { 1 }{ x\log { x } } - } \left[ \log { \left( \log { x } \right) } \right] \sin { x } \right]$

$\therefore \displaystyle \frac { dy }{ dx } ={ \log { x } }^{ \cos { x } }\left[ \cos { x.\frac { 1 }{ x\log { x } } - } \left[ \log { \left( \log { x } \right) } \right] \sin { x } \right]$

$y=x^{\left(x^{ x}\right)}$ , then

$\displaystyle\frac{dy}{dx}$ is

0%
$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x^{\displaystyle x}\right]$
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$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x\right]$
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$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x^{\displaystyle x-1}\right]$
0%
$y\left[x^{\displaystyle x}\left(\log_e{x}\right)\log{x}+x^{\displaystyle x-1}\right]$

Explanation

$y={ x }^{ \left( { x }^{ x } \right) }$
Take log on both sides

$\log y=\log { { x }^{ \left( { x }^{ x } \right) } }$

$\log { y } ={ x }^{ x }\log { x }$ ....

$(i)$

Let

${ x }^{ x }=z$

$\therefore x\log { x } =\log { z }$
Differentiate both sides with respect to

$x$

$\cfrac { 1 }{ z } \cfrac { dz }{ dx } =\log { x } +\cfrac { x }{ x }$

$\cfrac { dz }{ dx } =z\left[ 1+\log { x } \right]$

$\cfrac { dz }{ dx } ={ x }^{ x }\left[ \log { e } +\log { x } \right]$

$\cfrac { dz }{ dx } ={ x }^{ x }\log { ex }$

Equation

$(i)\Longrightarrow \log { y } =z\log { x }$
Differentiate both sides with respect to

$x$

$\cfrac { 1 }{ y } \cfrac { dy }{ dx } =\log { x\cdot \cfrac { dz }{ dx } } +\cfrac { z }{ x }$

$\therefore \cfrac { dy }{ dx } =y\left[ { x }^{ x }\left( \log { ex } \right) \log { x+{ x }^{ x-1 } } \right]$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 10 - 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 10 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

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