Limits, Continuity and Differentiability

(A)

Let

$f(x)=\begin{cases} \dfrac{1}{x}\sin\left(\dfrac{1}{x}\right), & x\neq 0\\\\ 0, & x=0 \end{cases}$

Now,

$g(x)=x f(x)=\begin{cases} \sin\left(\dfrac{1}{x}\right), & x\neq 0\\\\ 0, & x=0 \end{cases}$

When $x \neq 0$, the function $g(x)$ is $\sin\left(\dfrac{1}{x}\right)$, which keeps oscillating between $-1$ and $1$ as $x \to 0$. So, $\lim\limits_{x \to 0} g(x)$ does not exist. Therefore, $g(x)$ is not continuous at $x=0$.

(B)

Assume that $f(x)$ is continuous at $x=0$. This means $f(0^-) = f(0^+) = f(0)$.

Now consider the derivative of $g(x)$ at $x=0$:

$g'(0)=\lim\limits_{h\to 0}\dfrac{g(h)-g(0)}{h}$

Since $g(0)=0$, this becomes

$g'(0)=\lim\limits_{h\to 0}\dfrac{g(h)}{h}=\lim\limits_{h\to 0}\dfrac{h f(h)}{h}$

$g'(0)=\lim\limits_{h\to 0} f(h)=f(0)$

This limit exists and is finite because $f(x)$ is continuous at $x=0$. Hence, $g(x)$ is differentiable at $x=0$.

(C)

From the above, if $g(x)$ is differentiable at $x=0$, then

$g'(0)=\lim\limits_{h\to 0} f(h)$

This shows that the limit of $f(h)$ as $h \to 0$ exists. However, we cannot say anything about the value of $f(0)$. So, even if $g(x)$ is differentiable at $x=0$, we cannot conclude that $f(x)$ is continuous at $x=0$.

(D)

This statement is correct and follows from the reasoning in option (C).

$ \mathrm{e}^{\mathrm{x}_0}+\mathrm{x}_0=0 $

$ \begin{aligned} & g(x)=\frac{3 x e^x+3 x-\alpha e^x-\alpha x}{3\left(e^x+1\right)} \\ & g(x)=x-\frac{\alpha\left(e^x+x\right)}{3\left(e^x+1\right)}, g(x)+e^{x_0}=x+e^{x_0}-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) \end{aligned} $

$ g(x)+e^{x_0}=\left(x-x_0\right)-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) $

$\Rightarrow$ As very clear $\mathrm{g}\left(\mathrm{x}_0\right)+\mathrm{x}_0 \Rightarrow 0$

$ \lim \limits_{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right| \frac{0 \uparrow}{0 \uparrow}=\lim \limits_{x \rightarrow x_0}\left|\frac{g^{\prime}(x)}{1}\right|=\left|g^{\prime}\left(x_0\right)\right| $

$ \mathrm{g}^{\prime}(\mathrm{x})=1-\frac{\alpha}{3}\left(\frac{\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{e}^{\mathrm{x}}+1\right)-\left(\mathrm{e}^{\mathrm{x}}+\mathrm{x}\right) \mathrm{e}^{\mathrm{x}}}{\left(\mathrm{e}^{\mathrm{x}}+1\right)^2}\right) $

$ \mathrm{g}^{\prime}\left(\mathrm{x}_0\right)=1-\frac{\alpha}{3}\left(\frac{\left(\mathrm{e}^{\mathrm{x}_0}+1\right)^2}{\left(\mathrm{e}^{\mathrm{x}_0}+1\right)^2}\right)=1-\frac{\alpha}{3} $

$ \Rightarrow \lim \limits_{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=\left|g^{\prime}\left(x_0\right)\right|=\left|1-\frac{\alpha}{3}\right| $

The function $ f : \mathbb{R} \to \mathbb{R} $ is defined by:

$ f(x) = \begin{cases} 2 - 2x^2 - x^2 \sin \frac{1}{x}, & \text{if } x \neq 0, \\ 2, & \text{if } x = 0. \end{cases} $

Explanation:

(A) To determine differentiability at $ x = 0 $, consider the right-hand derivative (RHD):

$ \lim\limits_{h \rightarrow 0} \frac{(2 - 2h^2 - h^2 \sin \frac{1}{h}) - 2}{h} = 0 $

Similarly, for the left-hand derivative (LHD) at $ x = 0 $, we also find it equals 0. Therefore, $ f $ is differentiable at $ x = 0 $.

(B) For $ x \in (0, \delta) $, the derivative of $ f $ is:

$ f^{\prime}(x) = -\left(4x + 2x \sin \frac{1}{x}\right) + \cos \frac{1}{x} $

Thus:

$ f^{\prime}(x) = -\left(2x\left(4 - \sin \frac{1}{x}\right)\right) + \cos \frac{1}{x} $

Since $\cos \frac{1}{x}$ oscillates, we cannot assert that $ f(x) $ is strictly decreasing on $ (0, \delta) $.

(C) For $ x \in (-\delta, 0) $ and any $ \delta > 0 $, $ f(x) $ is not increasing since $\cos \frac{1}{x}$ oscillates between -1 and 1.

(D) Evaluating at $ x = 0 $:

$ f(0) = 2 $

For small $ h $:

$ f(0 + h) < 2 \quad \text{and} \quad f(0 - h) < 2 $

Therefore, $ x = 0 $ is a local maximum, not a local minimum.

$\begin{aligned} & \text { (P) Let } g(x)=10 x^3-45 x^2+60 x+35 \\\\ & g^{\prime}(x)=30\left(x^2-3 x+2\right)=30(x-1)(x-2)\end{aligned}$

$\Rightarrow g(x)$ is decreasing in $[1,2]$

Now, $f(1)=\left[\frac{g(1)}{n}\right]=\left[\frac{60}{n}\right], f(2)=\left[\frac{55}{n}\right]$

For $f(x)$ to be continuous in $[1,2]$ its integral value should remain same in whole interval for $n=9, f(1)=6, f(2)=6$

$(P) \rightarrow(2)$

(Q)

$ \begin{aligned} & g^{\prime}(x)=\left(2 n^2-13 n-15\right)\left(3 x^2+3\right) \\ & =(2 n-15)(n+1)\left(3 x^2+3\right)>0 \end{aligned} $

for $g(x)$ to be increasing $\min n=8$

$\Rightarrow Q \rightarrow 1$ matching

(R) $h(x)=\left(x^2-9\right)^n\left(x^2+2 x+3\right)$

$h(3)=0$ for $n>5$

at $n=6 \Rightarrow n(3+\delta)>h(3)$

$ n(3-\delta)>n(3) $

$\Rightarrow h(x)$ has local minima at $x=3$ for $n=6$

$R \rightarrow 4$ matching

(S) $\quad I(x)=\sin |x|+\cos \left|x+\frac{1}{2}\right|+\sin |x-1|+\cos \left|x-\frac{1}{2}\right|$ $+\ldots \ldots+\sin |x-4|+\cos \left|x-\frac{7}{2}\right|$

as $\sin |x-a|$ is non differentiable at $x=a$

but $\cos |x-a|$ remains differentiable at $x=a$

$\Rightarrow$ given $I(x)$ is non-differentiable at

$\begin{aligned} & x_0=0,1,2,3,4(5 \text { points }) \\\\ & (S) \rightarrow(3) \text { matching } \\\\ & (P) \rightarrow(2)(Q) \rightarrow(1)(R) \rightarrow(4)(S) \rightarrow(3)\end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{2}{x}(\sin (\sin k x)+\cos x+x-1)=6 \\ & \lim _{x \rightarrow 0} \frac{\sin (\sin k x) \cdot \sin k x}{(\sin k x) k x} \cdot k+1-\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2} \cdot x=3 \\ & k+1=3 \Rightarrow k=2 \end{aligned}$

Option-A : $\mathrm{f}(\mathrm{x})=\mathrm{x}^2 \sin \frac{\pi}{\mathrm{x}^2}$

$\mathrm{f}(\mathrm{x})=0 \Rightarrow \sin \frac{\pi}{\mathrm{x}^2}=0 \Rightarrow \frac{\pi}{\mathrm{x}^2}=\mathrm{n} \pi, \mathrm{n} \in \mathrm{N}$

$\mathrm{x}^2=\frac{1}{\mathrm{n}} \Rightarrow \mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}$

$\frac{1}{\sqrt{\mathrm{n}}} \geq \frac{1}{10^{10}} \Rightarrow 10^{10} \geq \sqrt{\mathrm{n}} \Rightarrow \mathrm{n} \leq 10^{20}$, finite number of solutions

Option-B : $\mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}} \Rightarrow \frac{1}{\sqrt{\mathrm{n}}}>\frac{1}{\pi} \Rightarrow \pi>\sqrt{\mathrm{n}} \Rightarrow \mathrm{n}<\pi^2$, Number of solutions is 9

Option-C : $\mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}, \frac{1}{\sqrt{\mathrm{n}}}<\frac{1}{10^{10}} \Rightarrow \sqrt{\mathrm{n}}>10^{10} \Rightarrow \mathrm{n}>10^{20}$, Infinite number of solutions

Option-D : $\frac{1}{\pi^2}<\frac{1}{\sqrt{\mathrm{n}}}<\frac{1}{\pi} \Rightarrow \sqrt{\mathrm{n}} \in\left(\pi, \pi^2\right) \Rightarrow \mathrm{n} \in\left(\pi^2, \pi^4\right)$, Definitely more than 25 solutions

$f(x)=\left\{\begin{array}{cc} x|x| \sin \frac{1}{x} & ; x \neq 0 \\ 0 & ; \quad x=0 \end{array} \quad g(x)=\left\{\begin{array}{cc} 1-2 x & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right.\right.$

$g\left(\frac{1}{2}-x\right)=\left\{\begin{array}{cc} 2 x & ; \quad 0 \leq \frac{1}{2}-x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}=\left\{\begin{array}{cc} 2 x ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right\}\right.$

$g(x)+g\left(\frac{1}{2}-x\right)=\left\{\begin{array}{ll} 1 ; & 0 \leq x \leq \frac{1}{2} \\ 0 & ; \text { otherwise } \end{array}\right\}$

(P) $\quad$ Now $a=0, b=1, c=0, d=0$

$\because h(x)=g(x)+g\left(\frac{1}{2}-x\right)= \begin{cases}1 ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise }\end{cases}$

Hence Range of h(x) is {0, 1}

(Q) $\mathrm{a}=1, \mathrm{~b}=0, \mathrm{c}=0, \mathrm{~d}=0$

$h(x)=f(x)=\left\{\begin{array}{cc} x|x| \sin \frac{1}{x} & ; x \neq 0 \\ 0 & ; x=0 \end{array}\right.$

$\begin{aligned} & R H D=\lim _{x \rightarrow 0} \frac{x^2 \sin \frac{1}{x}-0}{x}=0 \\ & \text { LHD }=\lim _{x \rightarrow 0} \frac{-x^2 \sin \frac{1}{x}-0}{x}=0 \end{aligned}$

Hence $h(x)$ is differentiable on $R$

(R) $\mathrm{a}=0, \mathrm{~b}=0, \mathrm{c}=1, \mathrm{~d}=0$

$h(x)=x-g(x)=\left\{\begin{array}{cl} 3 x-1 & ; 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right.$

$\therefore \mathrm{h}(\mathrm{x}) \text { is ONTO }$

(S) $a=0, b=0, c=0, d=1$

$h(x)=g(x)=\left\{\begin{array}{cl} 1-2 x & ; 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right.$

Range of h(x) is [0, 1]

$\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\sin x^2 \cdot\left(\log _e x\right)^\alpha \cdot \sin \frac{1}{x^2}}{x^{\alpha \beta} \cdot\left(\log _c(1+x)\right)^\beta}=0 \\ & \lim _{x \rightarrow \infty} \frac{\left(\log _e x\right)^\alpha}{\left(\log _e(x+1)\right)^\beta \cdot x^{\alpha \beta+2}}=0 \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{\log _c x}{\log _e(x+1)}\right)^\beta \cdot \frac{\left(\log _e x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \\ & \lim _{x \rightarrow \infty} \frac{\left(\log _e x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \quad \text { Put } \log _e x=t \\ & \lim _{t \rightarrow \infty} \frac{t^{\alpha-\beta}}{\left(e^t\right)^{\alpha \beta+2}}=0 \end{aligned}$

As we know $\lim _\limits{x \rightarrow \infty} \frac{x}{e^x}=0$

$\alpha \beta+2>0 \Rightarrow \alpha \beta>-2$

$ f(x)=\left\{\begin{array}{cc} 0 & ; 0 < x <\frac{1}{4} \\\\ \left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{1}{4} \leq x<\frac{1}{2} \\\\ 2\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{1}{2} \leq x<\frac{3}{4} \\\\ 3\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{3}{4} \leq x<1 \end{array}\right. $

$ f(x) \text { is discontinuous at } x=\frac{3}{4} \text { only } $

$ f^{\prime}(x)=\left\{\begin{array}{cc} 0 & ; 0 < x <\frac{1}{4} \\\\ 2\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+\left(x-\frac{1}{4}\right)^2 & ; \quad \frac{1}{4}< x <\frac{1}{2} \\\\ 4\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+2\left(x-\frac{1}{4}\right)^2 & ; \frac{1}{2} < x <\frac{3}{4} \\\\ 6\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+3\left(x-\frac{1}{4}\right)^2 & ; \frac{3}{4} < x < 1 \end{array}\right. $

$f(x)$ is non-differentiable at $x=\frac{1}{2}$ and $\frac{3}{4}$

Minimum values of $f(x)$ occur at $x=\frac{5}{12}$ whose value is $-\frac{1}{432}$

$ \begin{aligned} & f(n)=n+\frac{16+5 n-3 n^{2}}{4 n+3 n^{2}}+\frac{32+n-3 n^{2}}{8 n+3 n^{2}}+\ldots .+\frac{25 n-7 n^{2}}{7 n^{2}} \\\\ & =\left(\frac{16+5 n-3 n^{2}}{4 n+3 n^{2}}+1\right)+\left(\frac{32+n-3 n^{2}}{8 n+3 n^{2}}+1\right)+\ldots \ldots+\left(\frac{25 n-7 n^{2}}{7 n^{2}}+1\right) \\\\ & f(n)=\frac{9 n+16}{4 n+3 n^{2}}+\frac{9 n+32}{8 n+3 n^{2}}+\ldots . .+\frac{25 n}{7 n^{2}} \\\\ & =\sum_{r=1}^{n} \frac{9 n+16 r}{4 r n+3 n^{2}}=\frac{1}{n} \sum_{r=1}^{n} \frac{9+16\left(\frac{r}{n}\right)}{4\left(\frac{r}{n}\right)+3} \\\\ & \lim _{n \rightarrow \infty} f(n)=\int_{0}^{1} \frac{9+16 x}{4 x+3} d x \\\\ & =\int_{0}^{1} \frac{(16 x+12)-3}{4 x+3} d x \\\\ & =\left.\left(4 x-\frac{3}{4} \ln |4 x+3|\right)\right|_{0} ^{1} \\\\ & =4-\frac{3}{4} \ln \frac{7}{3} \end{aligned} $

Given,

$f(x) = {{{x^2} - 3x - 6} \over {{x^2} + 2x + 4}}$ .... (i)

$ \Rightarrow f'(x) = {{({x^2} + 2x + 4)(2x - 3) - ({x^2} - 3x - 6)(2x + 2)} \over {{{({x^2} + 2x + 4)}^2}}}$

$ \Rightarrow f'(x) = {{5x(x + 4)} \over {{{({x^2} + 2x + 4)}^2}}}$


Sign scheme for f'(x)

Here, f is decreasing in the interval ($-$2, $-$1) and f is increasing in the interval (1, 2).

Now, $f( - 4) = {{11} \over 6},f(0) = {{ - 3} \over 2}$ [from Eq. (i)]

and $\mathop {\lim }\limits_{x \to \pm \,\infty } f(x) = 1$

$\therefore$ Range $ = \left[ {{{ - 3} \over 2},{{11} \over 6}} \right]$

Hence, f(x) is into.


f(x) has local maxima at x = $-$4

and local minima at x = 0.

The given function f : R $ \to $ R is satisfying as

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

So, $f'(x) = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(x) + f(h) + f(x)f(h) - f(x)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(h)} \over h}(1 + f(x))$

$ \because $ $f(x) = xg(x) \Rightarrow g(x) = {{f(x)} \over x}$

$ \therefore $ $\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = \mathop {\lim }\limits_{x \to 0} g(x) = 1$ (given)

$ \therefore $ $f'(x) = 1 + f(x)) \Rightarrow {{f'(x)} \over {1 + f(x)}} = 1$

$ \Rightarrow {\log _e}(1 + f(x)) = x + c$

$ \Rightarrow 1 + f(x) = {e^{x + c}}$

$ \Rightarrow f(x) = {e^{x + c}} - 1$

$ \because $ $f(0) = 0 \Rightarrow c = 0$

Therefore, $f(x) = {e^x} - 1$ is differentiable at every $x \in R$.

And $f'(x) = {e^x} \Rightarrow f'(0) = 1$

Now, $g(x) = f{{(x)} \over x} = {{{e^x} - 1} \over x}$ and if g(0) = 1

LHD (at x = 0) of

$g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 - h) - g(0)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{{e^{ - h}} - 1} \over { - h}} - 1} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{e^{ - h}} - 1 + h} \over {{h^2}}} = {1 \over 2}$

and, RHD (at x = 0) of

$g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 + h) - g(0)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{e^h} - 1 - h} \over {{h^2}}} = {1 \over 2}$

So, if g(0) = 1, then g is differentiable at every x$ \in $R.

Given functions f : R $ \to $ R be defined

by f(x) = x³ $-$ x² + (x + 1) sin x and g : R $ \to $ R be an arbitrary function.

Now, let g is continuous at x = 1, then

$\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$

$ = \mathop {\lim }\limits_{h \to 0} {{(fg)(1 - h) - (fg)(1)} \over {1 - h - 1}}$

$ \because $ $(fg)(x) = f(x)\,.\,g(x)$ (given)

$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1 - h) - f(1)\,.\,g(1)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over { - h}}$

{$ \because $ f(1) = 0 and g is continuous at x = 1, so g(1 $-$ h) = g(1)}

$ = g(1)\mathop {\lim }\limits_{h \to 0} {{{{(1 - h)}^2}( - h) + ( - h)\sin (1 - h)} \over { - h}}$

$ = (1 + \sin 1)g(1)$

Similarly, $\mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over h}$

$ = g(1)\mathop {\lim }\limits_{h \to 0} {{{{(1 + h)}^2}(h) + h\sin (1 + h)} \over h}$

$ = (1 + \sin 1)g(1)$

$ \because $ RHD and LHD of function fg at x = 1 is finitely exists and equal, so fg is differentiable at x = 1

Now, let (fg)(x) is differentiable at x = 1, so

$\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$

$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}}$

$ \Rightarrow $ $\mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x)} \over {x - 1}}$ {$ \because $ f(1) = 0}

$ \Rightarrow $ $\mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,g(1 - h)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} {{f(1 + h)\,g(1 + h)} \over h}$

$ \Rightarrow $ $\mathop {\lim }\limits_{h \to 0} {{[{{(1 - h)}^2}( - h) + ( - h)\sin (1 + h)]g(1 + h)} \over { - h}}$

= $\mathop {\lim }\limits_{h \to 0} {{[{{(1 + h)}^2}(h) + (h)\sin (1 + h)]g(1 + h)} \over h}$

$\mathop {\lim }\limits_{h \to 0} \,[{(1 - h)^2} + \sin (1 - h)]g(1 - h) $

$= \mathop {\lim }\limits_{h \to 0} \,[{(1 + h)^2} + \sin (1 + h)]g(1 + h)$

It does not mean that g(x) is continuous or differentiable at x = 1.

But if g is differentiable at x = 1, then it must be continuous at x = 1 and so fg is differentiable at x = 1.

$\mathop {\lim }\limits_{n \to \infty } \left[ {{{1 + \root 3 \of 2 + \root 3 \of 3 + ...\root 3 \of n } \over {{n^{7/3}}\left( {{1 \over {{{(an + 1)}^2}}} + {1 \over {{{(an + 2)}^2}}} + ... + {1 \over {{{(an + n)}^2}}}} \right)}}} \right]$, $a \in R,\,|a|\, > 1$

$ = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {({r^{1/3}})} } \over {{n^{7/3}}\sum\limits_{r = 1}^n {{1 \over {{{(an + r)}^2}}}} }}$

$ = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^{1/3}}{1 \over n}} } \over {\sum\limits_{r = 1}^n {{1 \over {{{(a + {r \over n})}^2}}}{1 \over n}} }}$

$ = {{\int\limits_0^1 {{x^{1/3}}dx} } \over {\int\limits_0^1 {{{dx} \over {{{(a + x)}^2}}}} }} = 54$, (given)

$ \Rightarrow {{{3 \over 4}[{x^{4/3}}]_0^1} \over {\left[ { - {1 \over {x + a}}} \right]_0^1}} = 54$

$ \Rightarrow {{{3 \over 4}} \over { - {1 \over {a + 1}} + {1 \over a}}} = 54$

$ \Rightarrow {3 \over {4 \times 54}} = {1 \over {a(a + 1)}} $

$\Rightarrow {a^2} + a = 72$

$ \Rightarrow {a^2} + 9a - 8a - 72 = 0$

$ \Rightarrow a(a + 9) - 8(a + 9) = 0$

$ \Rightarrow (a - 8)(a + 9) = 0$

$ \Rightarrow a = 8$ or $ - 9$

Hence, options (c) and (d) are correct.

It is given, that f : R $ \to $ R and

Property 1 : $\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {\sqrt {|h|} }}$ exists and finite, and


Property 2 : $\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {{h^2}}}$ exists and finite.

Option A,

P2 : $\mathop {\lim }\limits_{h \to 0} {{\sin h - \sin 0} \over {{h^2}}} = \mathop {\lim }\limits_{h \to 0} {1 \over h}\left( {{{\sin \,h} \over h}} \right)$ = doesn't exist.

Option B,

P1 : $\mathop {\lim }\limits_{h \to 0} {{{h^{2/3}} - 0} \over {\sqrt {|h|} }} = \mathop {\lim }\limits_{h \to 0} {h^{2/3 - 1/2}} = \mathop {\lim }\limits_{h \to 0} {h^{1/6}} = 0$ exists and finite.

Option C,

P1 : $\mathop {\lim }\limits_{h \to 0} {{|h| - 0} \over {\sqrt {|h|} }} = \mathop {\lim }\limits_{h \to 0} \sqrt {|h|} = 0$, exists and finite.

Option D,

P2 : $\mathop {\lim }\limits_{h \to 0} {{h|h| - 0} \over {{h^2}}} = \mathop {\lim }\limits_{h \to 0} {{|h|} \over h} = \left\{ {\matrix{ 1 & {if\,h \to {0^ + }} \cr { - 1} & {if\,h \to {0^ - }} \cr } } \right.$

So, $\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {{h^2}}}$ does not exist.

Hence, options (b) and (c) are correct.

Given function f : R $ \to $ R is

$f(x) = \left\{ {\matrix{ {{x^5} + 5{x^4} + 10{x^3} + 10{x^2} + 3x + 1,} & {x < 0;} \cr {{x^2} - x + 1,} & {0 \le x < 1;} \cr {{2 \over 3}{x^3} - 4{x^2} + 7x - {8 \over 3},} & {1 \le x < 3;} \cr {(x - 2){{\log }_e}(x - 2) - x + {{10} \over 3},} & {x \ge 3;} \cr } } \right\}$

So,

$f'(x) = \left\{ {\matrix{ {{5x^4} + 20{x^3} + 30{x^2} + 20x + 3,} & {x < 0;} \cr {2x - 1,} & {0 \le x < 1;} \cr {2{x^2} + 8x + 7,} & {1 \le x < 3;} \cr {{{\log }_e}(x - 2),} & {x \ge 3;} \cr } } \right\}$

At x = 1, f"(1^-) = 2 > 0 and f"(1⁺) = 4$ - $8 = $ - $4 < 0

$ \therefore $ f'(x) is not differentiable at x = 1 and f'(x) has a local maximum at x = 1.

For x $ \in $ ($ - $$\infty $, 0)

f'(x) = 5x⁴ + 20x³ + 30x² + 20x + 3

and since

f'($ - $1) = 5$ - $20 + 20 + 30 $ - $ 20 + 3 = $ - $2 < 0

So, f(x) is not increasing on x $ \in $($ - $$\infty $, 0).

Now, as the range of function f(x) is R, so f is onto function.

Hence, options (b), (c) and (d) are correct.

(i) Given,

f₁ : R $ \to $ R and f₁(x) = sin$(\sqrt {1 - {e^{ - {x^2}}}} )$

$ \therefore $ f₁(x) is continuous at x = 0

Now,

${f_1}'(x) = \cos \sqrt {1 - {e^{ - {x^2}}}} .{1 \over {2\sqrt {1 - {e^{ - {x^2}}}} }}(2x{e^{ - {x^2}}})$

At x = 0

f₁'(x) does not exists.

$ \therefore $ f₁(x) is not differential at x = 0

Hence, option (2) for P.

(ii) Given, ${f_2}(x) = \left\{ \matrix{ {{|\sin x|} \over {{{\tan }^{ - 1}}x'}}\,if\,x \ne 0 \hfill \cr 1,\,if\,x = 0 \hfill \cr} \right\}$

$ \Rightarrow {f_2}(x) = \left\{ \matrix{ {{ - \sin x} \over {{{\tan }^{ - 1}}x}}x < 0 \hfill \cr {{\sin x} \over {{{\tan }^{ - 1}}x}}x > 0 \hfill \cr 1\,x = 0 \hfill \cr} \right.$

Clearly, f₂(x) is not continuous at x = 0.

$ \therefore $ Option (1) for Q.

(iii) Given, f₃(x) = [sin(log_e(x + 2))], where [ ] is G.I.F.

and f₃ : ($-$1, e^{$\pi $/2} $-$ 2) $ \to $ R

It is given,

$ - 1 < x < {e^{\pi /2}} - 2$

$ \Rightarrow - 1 + 2 < x + 2 < {e^{\pi /2}} - 2 + 2$

$ \Rightarrow 1 < x + 2 < {e^{\pi /2}}$

$ \Rightarrow {\log _e}1 < {\log _e}(x + 2) < {\log _e}{e^{\pi /2}}$

$ \Rightarrow 0 < {\log _e}(x + 2) < {\pi \over 2}$

$ \Rightarrow \sin 0 < \sin {\log _e}(x + 2) < \sin {\pi \over 2}$

$ \Rightarrow 0 < \sin {\log _e}(x + 2) < 1$

$ \therefore $ $[\sin {\log _e}(x + 2)] = 0$

$ \therefore $ ${f_3}(x) = 0,\,f{'_3}(x) = {f_3}''(x) = 0$

It is differentiable and continuous at x = 0.

$ \therefore $ Option (4) for R

(iv) Given, ${f_4}(x) = \left\{ \matrix{ {x^2}\sin \left( {{1 \over x}} \right),\,if\,x \ne 0 \hfill \cr 0,\,if\,x = 0 \hfill \cr} \right.$

Now, $\mathop {\lim }\limits_{x \to 0} {f_4}(x) = \mathop {\lim }\limits_{x \to 0} {x^2}\sin \left( {{1 \over x}} \right) = 0$

${f_4}'(x) = 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)$

For $x = 0,\,{f_4}'(x) = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h}$

$ \Rightarrow {f_4}'(x) = \mathop {\lim }\limits_{h \to 0} {{{h^2}\sin \left( {{1 \over h}} \right) - 0} \over h}$

$ \Rightarrow {f_4}'(x) = \mathop {\lim }\limits_{h \to 0} h\sin \left( {{1 \over h}} \right) = 0$

Thus,

${f_4}'(x) = \left\{ \matrix{ 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right),\,x \ne 0 \hfill \cr 0,\,x = 0 \hfill \cr} \right.$

Again, $\mathop {\lim }\limits_{x \to 0} $

$f'(x) = \mathop {\lim }\limits_{x \to 0} \left( {2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)} \right)$

does not exists.

Since, $\mathop {\lim }\limits_{x \to 0} $ cos${\left( {{1 \over x}} \right)}$ does not exists.

Hence, f'(x) is not continuous at x = 0.

$ \therefore $ Option (3) for S.

Given,

$\mathop {\lim }\limits_{t \to x} {{f(x)\sin t - f(t)\sin x} \over {t - x}} = {\sin ^2}x$

Using L' Hospital rules

$\mathop {\lim }\limits_{t \to x} {{f(x)\cos t - f'(t)\sin x} \over 1} = {\sin ^2}x$

$ \Rightarrow f(x)\cos x - f'(x)\sin x = {\sin ^2}x$

$ \Rightarrow f'(x)\sin x - f(x)\cos x = - {\sin ^2}x$

$ \Rightarrow {{f'(x)\sin x - f(x)\cos x} \over {{{\sin }^2}x}} = - 1$

$ \Rightarrow d\left( {{{f(x)} \over {\sin x}}} \right) = - 1$

On integrating, we get

${{f(x)} \over {\sin x}} = - x + C$

$ \Rightarrow f(x) = - x\sin x + C\sin x$

It is given that $x = {\pi \over 6},\,f\left( {{\pi \over 6}} \right) = - {\pi \over {12}}$

$ \therefore $ $f\left( {{\pi \over 6}} \right) = - {\pi \over 6}\sin {\pi \over 6} + C\sin {\pi \over 6}$

$ = - {\pi \over {12}} = - {\pi \over {12}} + {1 \over 2}C$

$ \Rightarrow C = 0$

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

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

$f\left( {{\pi \over 4}} \right) = - {\pi \over 4}\sin {\pi \over 4} = - {\pi \over {4\sqrt 2 }}$ false

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

$\sin x > x - {{{x^3}} \over 6},\,\forall x \in (0,\pi )$

$ \Rightarrow - x\sin x < - {x^2} + {{{x^4}} \over 6}$

$ \Rightarrow f(x) < {{{x^4}} \over 6} - {x^2},\,\forall x \in (0,\,\pi )$

It is true.

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

f'(x) = $-$ sin x $-$ x cos x

f'(x) = 0

$ \Rightarrow $ $-$ sin x $-$ x cos x = 0

tan x = $-$ x



$ \Rightarrow $ Their exists $\alpha $$ \in $(0, $\pi $) for which f'($\alpha $) = 0

It is true.

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

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

$f''(x) = - 2\cos x + x\sin x$

$f''\left( {{\pi \over 2}} \right) = {\pi \over 2},\,f\left( {{\pi \over 2}} \right) = - {\pi \over 2}$

$ \therefore $ $f''\left( {{\pi \over 2}} \right) + f\left( {{\pi \over 2}} \right) = 0$

It is true.

We have,

${(f(0))^2} + {(f'(0))^2} = 85$

and $f:R \to [ - 2,2]$

(a) Since, f is twice differentiable function, so f is continuous function.

$ \therefore $ This is true for every continuous function.

Hence, we can always find x $ \in $ (r, s), where f(x) is one-one.

$ \therefore $ This statement is true.

(b) By L.M.V.T.

$f'(c) = {{f(b) - f(a)} \over {b - a}}$

$ \Rightarrow |f'(c)| = \left| {{{f(b) - f(a)} \over {b - a}}} \right|$

$ \Rightarrow |f'({x_0})| = \left| {{{f(0) - f( - 4)} \over {0 + 4}}} \right|$

$ = \left| {{{f(0) - f( - 4)} \over 4}} \right|$

Range of f is [$-$2, 2]

$ \therefore $ $ - 4 \le f(0) - f( - 4) \le 4$

$ \Rightarrow 0 \le \left| {{{f(0) - f( - 4)} \over 4}} \right| \le 1$

Hence, |f'(x₀)| = 1.

Hence, statement is true.

(c) As no function is given, then we assume

$f(x) = 2\sin \left( {{{\sqrt {85} x} \over 2}} \right)$

$ \therefore $ $f'(x) = \sqrt {85} \cos \left( {{{\sqrt {85} x} \over 2}} \right)$

Now, ${(f(0))^2} + {(f'(0))^2} = {(2\sin 0)^2} + {(\sqrt {85} \cos 0)^2}$

${(f(0))^2} + {(f'(0))^2} = 85$

and $\mathop {\lim }\limits_{x \to \infty } f(x)$ does not exists.

Hence, statement is false.

(d) From option b, $|f'({x_0})|\, \le 1$

${x_0} \in ( - 4,0)$

$ \therefore $ ${(f'({x_0}))^2}\, \le 1$

Hence, $g({x_0}) = {(f({x_0}))^2} + {(f'({x_0}))^2}\, \le 4 + 1$

[$ \because $ f(x₀) $ \in $ [$-$2, 2]

$ \Rightarrow g({x_0})\, \le 5$

Now, let p $ \in $ ($-$4, 0) for which g(p) = 5

Similarly, let q be smallest positive number q $ \in $ (0, 4)

such that g(q) = 5

Hence, by Rolle's theorem is (p, q)

g'(c) = 0 for $\alpha $ $ \in $ ($-$4, 4) and since g(x) is greater than 5 as we move from x = p to x = q

and f(x))² $ \le $ 4

$ \Rightarrow $ (f'(x))² $ \ge $ 1 in (p, q)

Thus, g'(c) = 0

$ \Rightarrow $ f'f + f'f" = 0

So, f($\alpha $) + f"($\alpha $) = 0 and f'($\alpha $) $ \ne $ 0

Hence, statement is true.

We have,

$f'(x) = {e^{(f(x) - g(x))}}g'(x)\forall x \in R$

$ \Rightarrow f'(x) = {{{e^{f(x)}}} \over {{e^{g(x)}}}}g'(x)$

$ \Rightarrow {{f'(x)} \over {{e^{f(x)}}}} = {{g'(x)} \over {{e^{g(x)}}}}$

$ \Rightarrow {e^{ - f(x)}}f'(x) = {e^{ - g(x)}}g'(x)$

On integrating both side, we get

${e^{ - f(x)}}$ = ${e^{ - g(x)}}$ + C

At x = 1

${e^{ - f(1)}}$ = ${e^{ - g(1)}}$ + C

${e^{ - 1}} = {e^{ - g(1)}} + C$ [$ \because $ f(1) = 1] ...(i)

At x = 2

${e^{ - f(2)}}$ = ${e^{ - g(2)}}$ + C

$ \Rightarrow $ ${e^{ - f(2)}}$ = ${e^{ - 1}}$ + C [$ \because $ g(2) = 1] .... (ii)

From Eqs. (i) and (ii)

${e^{ - f(2)}}$ = $2{e^{ - 1}}$ $-$ ${{e^{ - g(1)}}}$ ...(iii)

$ \Rightarrow $ ${e^{ - f(2)}}$ > $2{e^{ - 1}}$

We know that, e^$-$x is decreasing

$ \therefore $ $-$f(2) < log_e 2$-$1

f(2) > 1 $-$ log_e 2

$ \Rightarrow $ ${{e^{ - g(1)}}}$ + ${e^{ - f(2)}}$ = $2{e^{ - 1}}$ [from Eq. (iii)]

$ \Rightarrow $ ${{e^{ - g(1)}}}$ < $2{e^{ - 1}}$ $-$g(1) < log_e 2 $-$1

$ \Rightarrow $ g(1) >$-$log_e 2

f'(x) is increasing

For some x in $\left( {{1 \over 2},\,1} \right)$

f'(x) = 1 [LMVT]

$ \therefore $ f'(1) > 1

$f(x) = {{1 - x(1 + |1 - x|)} \over {|1 - x|}}\cos \left( {{1 \over {1 - x}}} \right)$

Now, $\mathop {\lim }\limits_{x \to {1^ - }} f(x)$

$ = \mathop {\lim }\limits_{x \to {1^ - }} {{1 - x(1 + 1 - x)} \over {1 - x}}\cos \left( {{1 \over {1 - x}}} \right)$

$ = \mathop {\lim }\limits_{x \to {1^ - }} (1 - x)\cos \left( {{1 \over {1 - x}}} \right) = 0$

and $\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} $

${{1 - x(1 + 1 - x)} \over {x - 1}}\cos \left( {{1 \over {1 - x}}} \right)$

$ = \mathop {\lim }\limits_{x \to {1^ + }} - (x + 1).cos\left( {{1 \over {x + 1}}} \right)$, which does not exist.

(a) $ \because $ ${e^x} \in (1,e)$ in (0, 1) and $\int_0^x {f(t)\sin t\,dt \in } $ (0, 1) in (0, 1)

$ \therefore $ ${e^x} - \int_0^x {f(t)\sin t\,dt} $ cannot be zero.

So, option (a) is incorrect.

(b) $f(x) + \int_0^{{\pi \over 2}} {f(t)\sin t\,dt} $ always positive

$ \therefore $ Option (b) is incorrect.

(c) Let $h(x) = x - \int_0^{{\pi \over 2} - x} {f(t)\cos t\,dt} $,

$h(0) = - \int_0^{{\pi \over 2}} {f(t)\cos t\,dt} < 0$

$h(1) = 1 - \int_0^{{\pi \over 2} - 1} {f(t)\cos t\,dt} > 0$

$ \therefore $ Option (c) is correct.

(d) Let $g(x) = {x^9} - f(x)$

$g(0) = - f(0) < 0$, $g(1) = 1 - f(1) > 0$

$ \therefore $ Option (d) is correct.

$f(x) = x\cos (\pi (x + [x]))$

At x = 0

$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} x\cos (\pi (x + [x]) = 0$

and f(x) = 0

$ \therefore $ It is continuous at x = 0 and clearly discontinuous at other integer points.

$f(x) = a\cos (|{x^3} - x|) + b|x|\sin (|{x^3} + x|)$

If ${x^3} - x \ge 0$

$ \Rightarrow \cos \left| {{x^3} - x} \right| = \cos ({x^3} - x)$

and if ${x^3} - x \le 0$

$ \Rightarrow \cos |{x^3} - x| = \cos ({x^3} - x)$

$\therefore$ $\cos (|{x^3} - x|) = \cos ({x^3} - x),\,\forall x \in R$ ..... (i)

Again, if ${x^3} + x \ge 0$

$ \Rightarrow |x|\sin (|{x^3} + x|) = x\sin ({x^3} + x)$

and if ${x^3} + x \le 0$

$ \Rightarrow |x|\sin (|{x^3} + x|) = - x\sin \{ - ({x^3} + x)\} $

$\therefore$ $|x|\sin (|{x^3} + x|) = x\sin ({x^3} + x),\,\forall x \in R$ ...... (ii)

$ \Rightarrow f(x) = a\cos (|{x^3} - x|) + b|x|\sin (|{x^3} + x|)$

$\therefore$ $f(x) = a\cos ({x^3} - x) + bx\sin ({x^3} + x)$ ..... (iii)

For a function to be differentiable at x = 0, the function must be continuous.

$f(0) = a\cos (0 - 0) + b(0)\sin (0) = a$

Therefore,

$f({0^ + }) = \mathop {\lim }\limits_{h \to \infty } [a\cos ({h^3} - h) + bh\sin ({h^3} + h)] = 0$

$f({0^ - }) = \mathop {\lim }\limits_{h \to \infty } [a(\cos ( - {h^3} + h) + b( - h)\sin ( - {h^3} - h)]$

$ = \mathop {\lim }\limits_{h \to \infty } [a\cos ({h^3} - h) + bh\sin ({h^3} + h)] = 0$

which is continuous at x = 0; hence, f(x) is differentiable for all values of a and b.

Therefore,

$f(1) = a\cos (1 - 1) + 1\sin (1 + 1) = a + b\sin 2$

$f({1^ + }) = \mathop {\lim }\limits_{h \to 0} a\cos {(1 + h)^3} - (1 + h) + b(1 + h)\sin {(1 + h)^3} + (1 + h)$

$ = f(1)$

$f({1^ - }) = \mathop {\lim }\limits_{h \to 0} a\cos {(1 - h)^3} - (1 - h) + b(1 - h)\sin {(1 - h)^3} + (1 - h)$

$ = f(1)$

Thus, f(x) is continuous and we can also see that f is differentiable at x = 0 and x = 1.

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

$ = \left\{ {\matrix{ { - 3,} & { - 1/2 \le x < 1} \cr { - 2,} & {1 \le x < \sqrt 2 } \cr { - 1,} & {\sqrt 2 \le x < \sqrt 3 } \cr {0,} & {\sqrt 3 \le x < 2} \cr {1,} & {x = 2} \cr } } \right.$

and $g(x) = |x|f(x) + |4x - 7|f(x)$

$ = (|x| + |4x - 7|)f(x)$

$ = (|x| + |4x - 7|)[{x^2} - 3]$

$ = \left\{ {\matrix{ {( - x - 4x - 7)( - 3),} & { - 1/2 \le x < 0} \cr {(x - 4x + 7)( - 3),} & {0 \le x < 1} \cr {(x - 4x + 7)( - 2),} & {1 \le x < \sqrt 2 } \cr {(x - 4x + 7)( - 1),} & {\sqrt 2 \le x < \sqrt 3 } \cr {(x - 4x + 7)(0),} & {\sqrt 3 \le x < 7/4} \cr {(x + 4x - 7)(0),} & {7/4 \le x < 2} \cr {(x + 4x - 7)(1),} & {x = 2} \cr } } \right.$

$\therefore$ $g(x) = \left\{ {\matrix{ {15x + 21,} & { - 1/2 \le x < 0} \cr {9x - 21,} & {0 \le x < 1} \cr {6x - 14,} & {1 \le x < \sqrt 2 } \cr {3x - 7,} & {\sqrt 2 \le x\sqrt 3 } \cr {0,} & {\sqrt 3 \le x < 2} \cr {5x - 7,} & {x = 2} \cr } } \right.$

Now, the graphs of f(x) and g(x) are shown below.

Graph for f(x)

Clearly, f(x) is discontinuous at 4 points.

$\therefore$ Option (b) is correct.

Graph for g(x)

Clearly, g(x) is not differentiable at 4 points, when

x $\in$ ($-$1 / 2, 2).

$\therefore$ Option (c) is correct.

Rewrite f as

$f(x) = \left\{ {\matrix{ {g(x),} & {x > 0} \cr 0 & {x = 0} \cr { - g(x),} & {x < 0} \cr } } \right.$

We have, $f'(x) = \left\{ {\matrix{ {g'(x),} & {x > 0} \cr { - g'(x),} & {x < 0} \cr } } \right.$

At x = 0

$f'(0) = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{g(h) - 0} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{g(h) - g(0)} \over h} = g'(0)$

$f'(x) = \left\{ {\matrix{ {g'(x),} & {x \ge 0} \cr { - g'(x),} & {x < 0} \cr } } \right.$

f is differentiable at x = 0.

$\therefore$ Option (a) is correct.

(b) $h(x) = {e^{|x|}} = \left\{ {\matrix{ {{e^x},} & {x \ge 0} \cr {{e^{ - x}},} & {x < 0} \cr } } \right.$

$ \Rightarrow h'(x) = \left\{ {\matrix{ {{e^x},} & {x \ge 0} \cr { - {e^{ - x}},} & {x < 0} \cr } } \right.$

$ \Rightarrow h'({0^ + }) = 1$

and $h'({0^ - }) = - 1$

So, h(x) is not differentiable at x = 0.

$\therefore$ Option (b) is not correct.

(c) $(foh)(x) = f\{ h(x)\} $, as $h(x) > 0$

$ = \left\{ {\matrix{ {g({e^x}),} & {x \ge 0} \cr {g({e^{ - x}}),} & {x < 0} \cr } } \right.$

$ \Rightarrow (foh)'(x) = \left\{ {\matrix{ {{e^x}g'({e^x}),} & {x \ge 0} \cr { - {e^x}g'({e^{ - x}}),} & {x < 0} \cr } } \right.$

$ \Rightarrow (foh)'({0^ + }) = g'(1),(foh)'({0^ - })$

$ = - g'(1)$

So, $(hof)(x)$ is not differentiable at

$x = 0$.

$\therefore$ Option (c) is not correct.

(d) $(hof)(x) = {e^{\left| {f(x)} \right|}} = \left\{ {\matrix{ {{e^{\left| {g(x)} \right|}},} & {x \ne 0} \cr {{e^0} = 1,} & {x = 0} \cr } } \right.$

Now, $(hof)'(0) = \mathop {\lim }\limits_{h \to 0} {{{e^{\left| {g(x)} \right|}} - 1} \over x}$

$ = \mathop {\lim }\limits_{h \to 0} {{{e^{\left| {g(x)} \right|}} - 1} \over {|g(x)|}}\,.\,{{|g(x)|} \over x}$

$ = \mathop {\lim }\limits_{h \to 0} {{{e^{\left| {g(x)} \right|}} - 1} \over {|g(x)|}}\,.\,\mathop {\lim }\limits_{h \to 0} {{\left| {g(x)| - 0} \right|} \over {|x|}}\,.\mathop {\lim }\limits_{h \to 0} {{|x|} \over x}$

$ = 1\,.\,g'(0)\,.\,\mathop {\lim }\limits_{h \to 0} {{|x|} \over x}$

= 0, as g'(0) = 0

$\therefore$ Option (d) is correct.

Given that $f:(a,b) \to [1,\infty )$

and $g(x) = \left\{ {\matrix{ 0 & , & {x < a} \cr {\int_a^x {f(t)dt} } & , & {a \le x \le b} \cr {\int_a^b {f(t)dt} } & , & {x > b} \cr } } \right.$

Now, $g({a^ - }) = 0 = g({a^ + }) = g(a)$

[as $g({a^ + }) = \mathop {\lim }\limits_{x \to {a^ + }} \int_a^x {f(t)dt = 0} $ and $g(a) = \int_a^a {f(t)dt = 0} $]

$g({b^ - }) = g({b^ + }) = g(b) = \int_a^b {f(t)dt} $

$\Rightarrow$ g is continuous, $\forall$x$\in$R.

Now, $g'(x) = \left\{ {\matrix{ 0 & , & {x < a} \cr {f(x)} & , & {a < x < b} \cr 0 & , & {x > b} \cr } } \right.$

g'(a^$-$) = 0 but g'(a⁺) = f(a) $\ge$ 1

[$\because$ Range of f(x) is [1, $\infty$), $\forall$ x $\in$[a, b]]

$\Rightarrow$ g is non-differentiable at x = a

and g'(b⁺) = 0

but g'(b^$-$) = f(b) $\ge$ 1

$\Rightarrow$ g is not differentiable at x = b.

$\mathop {\lim }\limits_{n \to \infty } {{({1^a} + {2^a} + ... + {n^a})} \over {{{(n + 1)}^{a - 1}}[(na + 1) + (na + 2) + ... + (na + n)]}}$

$ = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {{r^a}} } \over {{{(n + 1)}^{a - 1}}\left\{ {{n^2}a + {{n(n + 1)} \over 2}} \right\}}}$

$ = \mathop {\lim }\limits_{n \to \infty } {{{1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^a}} } \over {{{\left( {1 + {1 \over n}} \right)}^{a - 1}}\left\{ {a + {1 \over 2} + {1 \over {2n}}} \right\}}}$

$ = {{\int\limits_a^1 {{x^a}dx} } \over {a + {1 \over 2}}} = {1 \over {(a + 1)\left( {a + {1 \over 2}} \right)}} = {1 \over {60}}$ given.

$ \Rightarrow (2a + 1)(a + 1) = 120 \Rightarrow 2{a^2} + 3a - 119 = 0$

$ \Rightarrow 2{a^2} - 14a + 17a - 119 = 0$

$ \Rightarrow 2a(a - 7) + 17(a - 7) = 0 \Rightarrow (2a + 17)(a - 7) = 0$.

$\therefore$ $a = 7,\, - {{17} \over 2}$

But $a = - {{17} \over 2}$ has to be discarded because integral $\int\limits_a^1 {{x^a}dx} $ converges if a > $-$1

$\mathop {\lim }\limits_{x \to \infty } \left( {{{{x^2} + x + 1} \over {x + 1}} - ax - b} \right) = 4$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{{x^2} + x + 1 - a{x^2} - ax - bx - b} \over {x + 1}} = 4$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{{x^2}(1 - a) + x(1 - a - b) + (1 - b)} \over {x + 1}} = 4$

Here, we make degree of N^r = degree of D^r

$\therefore$ $1 - a = 0$

and $\mathop {\lim }\limits_{x \to \infty } {{x(1 - a - b) + (1 - b)} \over {x + 1}} = 4$

$ \Rightarrow 1 - a - b = 4$

$ \Rightarrow b = - 4$

To check differentiable at x = 0

$R\{ f'(0)\} = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{h^2}\left| {\cos {\pi \over h}} \right| - 0} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} h.\,\left| {\cos {\pi \over h}} \right| = 0$

$L\{ f'(0)\} = \mathop {\lim }\limits_{h \to 0} {{f(0 - h) - f(0)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{h^2}\left| {\cos \left( { - {\pi \over h}} \right)} \right| - 0} \over { - h}}$

$ = 0$

So, f(x) is differentiable at x = 0

To check differentiability at x = 2

$R\{ f'(2)\} = \mathop {\lim }\limits_{h \to 0} {{f(2 + h) - f(2)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}\left| {\cos \left( {{\pi \over {2 + h}}} \right)} \right| - 0} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\cos \left( {{\pi \over {2 + h}}} \right)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\sin \left( {{\pi \over 2} - {\pi \over {2 + h}}} \right)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\sin \left( {{{\pi h} \over {2(2 + h)}}} \right)} \over {h.{\pi \over {2(2 + h)}}.{\pi \over {2(2 + h)}}}}$

$ \Rightarrow R\left( {f'(2)} \right) = \pi $

$L\left( {f'(2)} \right) = \mathop {\lim }\limits_{h \to 0} {{f(2 - h) - f(2)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2}.\left| {\cos {\pi \over {2 - h}}} \right| - {2^2}.\left| {\cos {\pi \over 2}} \right|} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2} - \left( { - \cos {\pi \over {2 - h}}} \right) - 0} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{ - {{(2 - h)}^2}.\sin \left( {{\pi \over 2} - {\pi \over {2 - h}}} \right)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2}.\sin \left( { - {{\pi h} \over {2(2 - h)}}} \right)} \over {h \times {{ - \pi } \over {2(2 - h)}} \times {{ - \pi } \over {2(2 - h)}}}}$

$ \Rightarrow L(f'(2)) = - \pi $

Thus, f(x) is differentiable at x = 0 but not at x = 2.

We have at the points x = 2n

f(2n) = a_n + sin2n$\pi$ = a_n

Also for the L.H.L., we have

L.H.L. = $\mathop {\lim }\limits_{h \to 0} ({b_n} + \cos \pi (2n - h)) = {b_{n + 1}}$

R.H.L. = $\mathop {\lim }\limits_{h \to 0} ({a_n} + \sin \pi (2n + h)) = {a_n}$

For continuity ${b_{n + 1}} = {a_n}$

Again at $x = 2n + 1$

L.H.L. = $\mathop {\lim }\limits_{h \to 0} ({a_n} + \sin (\pi (2n + 1 - h))) = {a_n}$

R.H.L. = $\mathop {\lim }\limits_{h \to 0} ({b_{n + 1}} + \cos (\pi (2n + 1) - h)) = {b_{n + 1}} - 1$

Also, $f(2n + 1) = {a_n}$

For continuity we require ${b_n} + 1 = {a_n}$ $\therefore$ ${a_n} - {b_n} = 1$

Also, ${a_n} = {b_{n + 1}} - 1$

$\therefore$ ${a_{n - 1}} - {b_n} = - 1$

Here, $\mathop {\lim }\limits_{x \to 0} {\{ 1 + x\log (1 + {b^2})\} ^{1/x}}$ [1^$\infty$ from]

$ \Rightarrow {e^{\mathop {\lim }\limits_{x \to 0} \{ x\log (1 + {b^2})\} \,.\,{1 \over x}}}$

$ \Rightarrow {e^{\log (1 + {b^2})}} = {(1 + b)^2}$ ..... (i)

Given,

$\mathop {\lim }\limits_{x \to 0} {\{ 1 + x\log (1 + {b^2})\} ^{1/x}} = 2b{\sin ^2}\theta $

$ \Rightarrow (1 + {b^2}) = 2b{\sin ^2}\theta $

$\therefore$ ${\sin ^2}\theta = {{1 + {b^2}} \over {2{b^2}}}$ ..... (ii)

By $AM \ge GM$

${{b + {1 \over b}} \over 2} \ge {\left( {b\,.\,{1 \over b}} \right)^{1/2}} \Rightarrow {{{b^2} + 1} \over {2b}} \ge 1$ ...... (iii)

From Eqs. (ii) and (iii), we get ${\sin ^2}\theta = 1$

$ \Rightarrow \theta = \pm {\pi \over 2}$ as $\theta \in ( - \pi ,\pi ]$

Set x = 0 in the functional equation to obtain

$f(0) = f(0) + f(0)$ $\therefore$ $f(0) = 0$

$f'(x) = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x + 0)} \over h} = \mathop {\lim }\limits_{h \to 0} {{f(x) + f(h) - f(x) - f(0)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over h} = f'(0)$

Thus, $f'(x) = \lambda $ (say). Also $f(x) = \lambda x + \mu $

As $f(x) = 0$ we have $\mu = 0$ $\therefore$ $f(x) = \lambda x$.

$\mathop {\lim }\limits_{x \to {{{\pi ^ - }} \over 2}} f(x) = 0 = f( - \pi /2)$

$\mathop {\lim }\limits_{x \to {{{\pi ^ + }} \over 2}} f(x) = \cos \left( { - {\pi \over 2}} \right) = 0$

$f'(x) = \left\{ {\matrix{ { - 1,} & {x \le - \pi /2} \cr {\sin x,} & { - \pi /2 < x \le 0} \cr {1,} & {0 < x \le 1} \cr {1/x,} & {x > 1} \cr } } \right.$

Clearly, f(x) is not differentiable at x = 0 as f'(0^$-$) = 0 and f'(0⁺) = 1.

f(x) is differentiable at x = 1 as $f'({1^{ - 1}}) = f'({1^ + }) = 1$.

The given limit is

$L = \mathop {\lim }\limits_{x \to 0} {{a - \sqrt {{a^2} - {x^2}} - {{{x^2}} \over 4}} \over {{x^4}}}$

$ = \mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}(a + \sqrt {{a^2} - {x^2}} )}} - {1 \over {4{x^2}}}$

$ = \mathop {\lim }\limits_{x \to 0} {{(4 - a) - \sqrt {{a^2} - {x^2}} } \over {4{x^2}(a + \sqrt {{a^2} - {x^2}} )}}$

The numerator $\to$ 0 if $a = 2$; therefore, $L = {1 \over {64}}$.