Limits, Continuity and Differentiability

$\lim _\limits{x \rightarrow 0} \frac{\frac{\tan (\tan x)-\tan x}{\tan ^3 x} \frac{\tan ^3 x}{x^3}+\frac{\tan x-\sin x}{x^3}+\frac{\sin x-\sin (\sin x)}{\sin ^3 x} \frac{\sin ^3 x}{x^3}}{\frac{\tan x-\sin x}{x^3}}$

$=\frac{\frac{1}{3}+\frac{1}{2}+\frac{1}{6}}{\frac{1}{2}}=2$

$\begin{aligned} &a+b=\lim _{t \rightarrow-1^{+}}(\alpha+\beta)=\lim _{t \rightarrow-1^{+}}-\frac{(t+2)^{\frac{1}{6}}-1}{(t+2)^{\frac{1}{7}}-1}\\ &\begin{aligned} & \text { let } t+2=y \\ & a+b=\lim _{y \rightarrow 1^{+}} \frac{y^{1 / 6}-1}{y^{1 / 7}-1}=\frac{7}{6} \\ & 72(a+b)^2=72 \frac{49}{36}=98 \end{aligned} \end{aligned}$

To determine the points of discontinuity of the function $ f(x) = \left[\frac{x^2}{2}\right] - [\sqrt{x}] $, where $[\cdot]$ denotes the greatest integer function, we need to identify possible values of $ x $ where discontinuities might occur within the interval $[0,4]$.

Discontinuity Analysis

For the term $\left[\frac{x^2}{2}\right]$:

The probable values of $ x $ that could cause discontinuities are the roots or specific values where the integer part changes between consecutive integers. The transitions happen when:

$ \begin{aligned} & = 1, 2, 3, 4, 5, 6, 7, 8 \\ & \implies x = \sqrt{2}, 2, \sqrt{6}, 2\sqrt{2}, \sqrt{10}, 2\sqrt{3}, \sqrt{14}, 4 \end{aligned} $

For the term $[\sqrt{x}]$:

The values of $ x $ where $[\sqrt{x}]$ changes are straightforward. They occur at:

$ x = 1, 2 $

Discontinuity Check

By evaluating $ f(x) $ at all these potential points, we find the function is indeed discontinuous at:

$ x = 1, \sqrt{2}, 2, \sqrt{6}, 2\sqrt{2}, \sqrt{10}, 2\sqrt{3}, \sqrt{14} $

Thus, the function $ f(x) $ has 8 discontinuities on the interval $[0,4]$.

$\begin{aligned} &\text { for } x \geq 1, x^{21} \geq x^{19} \geq \ldots \geq x \text {. }\\ &f(x)=\left\{\begin{array}{lr} x & x<-1 \\ x^{21} & -1 \leq x \leq 0 \\ x & 0< x<1 \\ x^{21} & x \geq 1 \end{array}\right. \end{aligned}$

$\begin{aligned} &\text { Clearly, } f(x) \text { is continuous everywhere. }\\ &\begin{aligned} & \Rightarrow \alpha=0 \\ & f^{\prime}(x)=\left\{\begin{array}{cl} 1 & ; x<-1 \\ 21 x^{20} & ;-1 \leq x \leq 0 \\ 1 & ; 0< x<1 \\ 21 \cdot x^{20} & ; x \geq 1 \end{array}\right. \\ & \Rightarrow \beta=3 \\ & \Rightarrow \alpha+\beta=3 \end{aligned} \end{aligned}$

To solve the given limit problem, we start by analyzing the expression:

$ \lim\limits_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}} = p $

This limit exhibits the indeterminate form $1^{\infty}$. To handle this form, we use the transformation:

$ p = e^{\lim_{x \to 0} \left( \frac{\tan x}{x} - 1 \right) \frac{1}{x^2}} $

Expanding $\tan x$ using its Taylor series near $x = 0$, we have:

$ \tan x = x + \frac{x^3}{3} + \frac{2}{15}x^5 + \ldots $

Substituting this expansion into the limit, we get:

$ \frac{\tan x - x}{x^3} = \frac{\left(x + \frac{x^3}{3} + \frac{2}{15}x^5 + \ldots - x\right)}{x^3} = \frac{\frac{x^3}{3} + \frac{2}{15}x^5 + \ldots}{x^3} $

This simplifies to:

$ \frac{x^3}{3x^3} = \frac{1}{3} $

Thus, the limit becomes:

$ p = e^{\frac{1}{3}} $

Therefore, the expression for $\log_e p$ is:

$ \log_e p = \frac{1}{3} $

Finally, computing $96 \log_e p$:

$ 96 \log_e p = 96 \cdot \frac{1}{3} = 32 $

To find the least natural number $ p $ for which the following inequality holds:

$ \lim \limits_{x \rightarrow 0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{p}{x}\right]\right)-x^2\left(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+\ldots+\left[\frac{9^2}{x^2}\right]\right)\right) \geq 1 $

we simplify the expression inside the limit.

As $ x \to 0^+ $, $ \left[\frac{n}{x}\right] $ approximates to $ \frac{n}{x} $. Thus, the problem becomes finding:

$ \left(1 + 2 + \ldots + p\right) - \left(1^2 + 2^2 + \ldots + 9^2\right) \geq 1 $

The sum of the first $ p $ natural numbers is given by:

$ \frac{p(p+1)}{2} $

And the sum of the squares of the first 9 natural numbers is:

$ 1^2 + 2^2 + \ldots + 9^2 = \frac{9 \cdot 10 \cdot 19}{6} $

Thus, the inequality becomes:

$ \frac{p(p+1)}{2} - \frac{9 \cdot 10 \cdot 19}{6} \geq 1 $

Solving this, we rewrite:

$ p(p+1) \geq 572 $

The least natural number $ p $ satisfying this condition is $ 24 $.

$\begin{aligned} & f(x)=\lim _{n \rightarrow \infty} \sum_{r=0}^n\left(\tan \frac{x}{2^r}-\tan \frac{x}{2^{r+1}}\right)=\tan x \\ & \lim _{x \rightarrow 0}\left(\frac{e^x-e^{\tan x}}{x-\tan x}\right)=\lim _{x \rightarrow 0} e^{\tan x} \frac{\left(e^{x-\tan x}-1\right)}{(x-\tan x)} \\ & =1 \end{aligned}$

$\begin{aligned} & f(x)=\left\{\begin{array}{ccc} 3 x & ; & x<0 \\ \min \{1+x, x\} & ; & 0 \leq x<1 \\ \min \{2+x, x+2\} & ; & 1 \leq x<2 \\ 5 & ; & x>2 \end{array}\right. \\ & f(x)=\left\{\begin{array}{ccl} 3 \mathrm{x} & ; & x<0 \\ \mathrm{x} & ; & 0 \leq x<1 \\ \mathrm{x}+2 & ; & 1 \leq x<2 \\ 5 & ; & x>2 \end{array}\right. \end{aligned}$

Not continuous at $\mathrm{x} \in\{1,2\} \Rightarrow \alpha=2$

Not diff. at $\mathrm{x} \in\{0,1,2\} \Rightarrow \beta=3$

$\alpha+\beta=5$

$\mathrm{f}(\mathrm{x})$ is continuous and differentiable

$ \begin{array}{ll} \text { at } x=1 ; & \text { LHL }=\text { RHL, LHD }=\text { RHD } \\ & -3 a-2=a^2+b,-6 a=b \\ & a=2,1 ; b=-12 \end{array} $

$f(x)=\left\{\begin{array}{cc}-6 x^2-2, & x<1 \\ 4-12 x, & x \geq 1\end{array}\right.$

$\begin{aligned} & \left.\text { Area }=\int_{-\sqrt{3}}^1\left(-6 x^2-2+20\right) d x+\int_1^2(4-12 x+20) d x\right] \\\\ & =16+12 \sqrt{3}+6=22+12 \sqrt{3} \\\\ & \therefore \quad \alpha+\beta=34\end{aligned}$

$\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} f(x)=f\left(\frac{\pi}{2}\right)=\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}} f(x) \text { for continuity at } x=\frac{\pi}{2}$

$\begin{aligned} & \Rightarrow \quad \lim _{x \rightarrow \frac{\pi^{-}}{2}}\left(\frac{8}{7}\right)^{\left(\frac{\tan 8 x}{\tan 7 x}\right)} \quad \text { Let } x=\frac{\pi}{2}-h \\ & \Rightarrow \quad \lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (4 \pi-8 h)}{\tan \left(3 \pi+\frac{\pi}{2}-7 h\right)}}=\lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (-8 h)}{\cot (7 h)}}=\left(\frac{8}{7}\right)^0=1 \\ & \Rightarrow \quad a-8=1 \Rightarrow a=9 \end{aligned}$

$\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}}(1+|\cot x|)^{\frac{b}{a}|\tan x|}, x=\frac{\pi}{2}+h$

$\lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h}=\lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h}$

$\begin{aligned} & =\lim _\limits{h \rightarrow 0}(1-\tan h)^{\left(\frac{-1}{\tan h}\right) \cdot(-\tan h) \cdot\left(\frac{-b}{9} \cot h\right)} \\ & =e^{\frac{b}{9}}=1 \quad \Rightarrow b=0 \\ & \Rightarrow a^2+b^2=81+0=81 \end{aligned}$

$\begin{aligned} \alpha & =\lim _{x \rightarrow 0^{+}} \frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{(\sqrt{\tan x}-\sqrt{x})} \\ & =\lim _{x \rightarrow 0} \frac{e^{\sqrt{x}}\left(e^{\sqrt{\tan x}-\sqrt{x}}-1\right)}{(\sqrt{\tan x}-\sqrt{x})}=1 \\ \beta & =\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}=\lim _{x \rightarrow 0} e^{(\sin x)\left(\frac{1}{2} \cot x\right)} \\ & =\lim _{x \rightarrow 0} e^{\frac{1}{2} \cos x}=e^{1 / 2} \end{aligned}$

$\begin{aligned} & \text { Product of roots }=\sqrt{e}=\frac{-\sqrt{e}}{a} \Rightarrow a=-1 \\ & \text { Sum of roots }=\frac{-b}{a}=1+\sqrt{e} \\ & \qquad=b=\sqrt{e}+1 \\ & \Rightarrow 12 \ln (a+b)=12 \ln (\sqrt{e}+1-1)=12 \ln \left(e^{1 / 2}\right)=6 \end{aligned}$

$\mathop {\lim }\limits_{x \to 0} 2\left( {{{1 - \cos x{{(\cos 2x)}^{{1 \over 2}}}{{(\cos 3x)}^{{1 \over 3}}}\,...\,{{(\cos 10x)}^{{1 \over {10}}}}} \over {{x^2}}}} \right)$ $\left(\frac{0}{0} \text { form }\right)$

Using L' hospital

$2 \lim _\limits{x \rightarrow 0} \frac{\sin x(\cos 2 x)^{\frac{1}{2}} \ldots(\cos 10 x)^{\frac{1}{10}} \ldots(\sin 2 x)(\cos x)(\cos 3 x)^{\frac{1}{3}}+\ldots}{2 x}$

$\begin{aligned} \Rightarrow & \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}+\frac{\sin 2 x}{x}+\ldots+\frac{\sin 10 x}{x}\right) \\ \quad & =1+2+\ldots+10=55 \end{aligned}$

$\begin{aligned} f(x) & =\left[\frac{x}{3}+3\right]-[\sqrt{x}] \\ & =\left[\frac{x}{2}\right]-[\sqrt{x}]+3 \end{aligned}$

Critical points where $f(x)$ might change behaviours when $\frac{x}{2} \in$ integer and $\sqrt{x} \in$ integer

$\Rightarrow$ Critical points,

$\begin{aligned} & f(0)=3 \\ & f\left(0^{+}\right)=3 \\ & f\left(1^{-}\right)=3 \\ & f\left(1^{+}\right)=2 \\ & f\left(2^{-}\right)=2 \\ & f\left(2^{+}\right)=3 \\ & f\left(3^{+}\right)=f\left(3^{-}\right)=3=f\left(4^{+}\right)=f\left(4^{-}\right)=f\left(5^{+}\right)=f\left(5^{-}\right) \\ & f\left(6^{-}\right)=3 \\ & f\left(6^{+}\right)=4 \\ & f\left(7^{-}\right)=f\left(7^{+}\right)=4 \\ & f\left(8^{-}\right)=4 \\ & f(8)=5 \end{aligned}$

$\begin{aligned} & \Rightarrow f(x) \text { is not continuous at } \\ & x=1,2,6,8 \\ & \Rightarrow \sum_{a \in s} a=1+2+6+8 \\ & \quad=17 \end{aligned}$

$\because 2 x^2+x-2=0$ has two roots where $a=\frac{\sqrt{17}-1}{4}$ and another root is $\frac{-\sqrt{17}-1}{4}$

And $2+x-2 x^2=-2\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)$

Now $\lim _\limits{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-a x)^2}$

$\begin{aligned} & =\lim _{x \rightarrow \frac{1}{a}} \frac{32 \sin ^2\left(\frac{2+x-2 x^2}{2}\right)}{a^2\left(\frac{1}{a}-x\right)^2} \\ & =\lim _{x \rightarrow \frac{1}{a}} \frac{\left(x+\frac{4}{\sqrt{17}+1}\right)^2 32 \cdot\left(\sin \left(\frac{1}{2} \cdot(-2)\right)\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2}{a^2\left(\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2} \\ & =2 \cdot\left(\frac{1}{a}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2 \\ & =2\left(\frac{4}{\sqrt{17}-1}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2 \\ & =\frac{17 \times 4}{18-2 \sqrt{17}}=\frac{68}{9-\sqrt{17}} \\ & =17(9+\sqrt{17}) \\ & \alpha+\beta=170 \end{aligned}$

$\lim _\limits{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{(t-x)}=1 \quad\left(\frac{0}{0} \text { form }\right)$

$\begin{aligned} & \lim _{t \rightarrow x} \frac{2 \operatorname{tf}(x)-x^2 f^{\prime}(t)}{1}=1 \\ & \Rightarrow 2 x f(x)-x^2 f'(x)=1 \\ & \frac{d y}{d x}-\frac{2 x y}{x^2}=\frac{-1}{x^2} \\ & \Rightarrow \frac{d y}{d x}-\left(\frac{2}{x}\right) y=\frac{-1}{x^2} \\ & \Rightarrow \text { I.F. }=e^{\int \frac{-2}{x} d x}=e^{-2 \ln x}=x^{-2}=\frac{1}{x^2} \end{aligned}$

$\begin{aligned} \Rightarrow & y\left(\frac{1}{x^2}\right)=\int\left(\frac{-1}{x^2}\right)\left(\frac{1}{x^2}\right) d x+C \\ & \frac{y}{x^2}=\frac{1}{3 x^3}+C \text { at } x=1, y=1 \\ \Rightarrow & 1=\frac{1}{3}+C \Rightarrow C=\frac{2}{3} \\ \Rightarrow & y=\frac{1}{3 x}+\frac{2}{3} x^2=f(x) \\ \Rightarrow & 2 f(2)+3 f(3)=24 \end{aligned}$

$I=\lim _\limits{x \rightarrow 1} \frac{(5 x+1)^{1 / 3}-(+5)^{1 /}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}$

From: $\frac{0}{0}$, using $\mathrm{L}-\mathrm{H}$ rule

$\begin{aligned} & I=\lim _{x \rightarrow 1} \frac{\frac{1}{3} \times 5(5 x+1)^{-2 / 3}--(+5)}{\frac{1}{2} \times 2(2 x+3)^{-1 / 2}-\frac{1}{2}(x+4)^{-1 / 2}} \\ & =\frac{\left(\left.\frac{5}{3}-\frac{1}{3} \right\rvert\, 6^{-2 / 3}\right.}{\frac{1}{2} 5^{-1 / 2}}=\frac{8}{3} \times \frac{5^{1 / 2}}{6^{2 / 3}}=\frac{m \sqrt{5}}{n(2 n)^{2 / 3}} \\ & \Rightarrow m=8, n=3 \\ & \Rightarrow 8 m+12 n=100 \end{aligned}$

Finding right hand limit

R = $ \begin{gathered} \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) =\lim _{h \rightarrow 0} f(h) \end{gathered} $

$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right) \sin ^{-1}(1-h)}{h\left(1-h^2\right)} \\\\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{\cos ^{-1}\left(1-\mathrm{h}^2\right)}{\mathrm{h}}\left(\frac{\sin ^{-1} 1}{1}\right)\end{aligned}$

Let $\cos ^{-1}\left(1-h^2\right)=\theta \Rightarrow \cos \theta=1-h^2$

$ \begin{aligned} & =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{\theta}{\sqrt{1-\cos \theta}} \\\\ & =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{1}{\sqrt{\frac{1-\cos \theta}{\theta^2}}} \\\\ & =\frac{\pi}{2} \frac{1}{\sqrt{1 / 2}} \\\\ & \therefore \mathrm{R}=\frac{\pi}{\sqrt{2}} \end{aligned} $

Now finding left hand limit

$ \begin{aligned} & L=\lim _{x \rightarrow 0^{-}} f(x) =\lim _{h \rightarrow 0} f(-h) \end{aligned} $

$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-\{-h\}^2\right) \sin ^{-1}(1-\{-h\})}{\{-h\}-\{-h\}^3} \\\\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-(-h+1)^2\right) \sin ^{-1}(1-(-h+1))}{(-h+1)-(-h+1)^3} \\\\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(-h^2+2 h\right) \sin ^{-1} h}{(1-h)\left(1-(1-h)^2\right)}\end{aligned}$

$\begin{aligned} & =\lim _{h \rightarrow 0}\left(\frac{\pi}{2}\right) \frac{\sin ^{-1} h}{\left(1-(1-h)^2\right)} \\\\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{-h^2+2 h}\right) \\\\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{h}\right)\left(\frac{1}{-h+2}\right) \\\\ & \quad L=\frac{\pi}{4}\end{aligned}$

$\begin{aligned} & \frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)=\frac{32}{\pi^2}\left(\frac{\pi^2}{2}+\frac{\pi^2}{16}\right) \\\\ & =16+2 \\\\ & =18\end{aligned}$

$\mathop {\lim }\limits_{x \to 0} {{a{x^2}\left( {1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ....} \right) - b\left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - .....} \right) + cx\left( {1 - x + {{{x^2}} \over {x!}} - {{{x^3}} \over {3!}} + .....} \right)} \over {{x^3}\,.\,{{\sin x} \over x}}}$

$=\lim _\limits{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^3+\ldots \ldots}{x^3}=1$

$\begin{aligned} & \mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0 \\ & \mathrm{a}-\frac{\mathrm{b}}{3}+\frac{\mathrm{c}}{2}=1 \quad \mathrm{a}=\frac{3}{4} \quad \mathrm{~b}=\mathrm{c}=\frac{3}{2} \\ & \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=\frac{9}{16}+\frac{9}{4}+\frac{9}{4} \\ & 16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81 \end{aligned}$

$\mathrm{f}(\mathrm{x})\left\{\begin{array}{c} \frac{1}{\mathrm{x}} ; \mathrm{x} \geq 2 \\ \mathrm{ax}^2+2 \mathrm{~b} ;-2<\mathrm{x}<2 \\ -\frac{1}{\mathrm{x}} ; \mathrm{x} \leq-2 \end{array}\right.$

Continuous at $\mathrm{x}=2 \quad \Rightarrow \frac{1}{2}=\frac{\mathrm{a}}{4}+2 \mathrm{~b}$

Continuous at $\mathrm{x}=-2 \quad \Rightarrow \frac{1}{2}=\frac{\mathrm{a}}{4}+2 \mathrm{~b}$

Since, it is differentiable at $\mathrm{x}=2$

$-\frac{1}{x^2}=2 \mathrm{ax}$

Differentiable at $x=2 \quad \Rightarrow \frac{-1}{4}=4 a \Rightarrow a=\frac{-1}{16}, b =\frac{3}{8}$

$\begin{aligned} & f(1)=1, f(a)=0 \\ & {f^2}(x) = \mathop {\lim }\limits_{r \to x} \left( {{{2{r^2}({f^2}(r) - f(x)f(r))} \over {{r^2} - {x^2}}} - {r^3}{e^{{{f(r)} \over r}}}} \right) \\ & = \mathop {\lim }\limits_{r \to x} \left( {{{2{r^2}f(r)} \over {r + x}}{{(f(r) - f(x))} \over {r - x}} - {r^3}{e^{{{f(r)} \over r}}}} \right) \\ & f^2(x)=\frac{2 x^2 f(x)}{2 x} f^{\prime}(x)-x^3 e^{\frac{f(x)}{x}} \\ & y^2=x y \frac{d y}{d x}-x^3 e^{\frac{y}{x}} \\ & \frac{y}{x}=\frac{d y}{d x}-\frac{x^2}{y} e^{\frac{y}{x}} \end{aligned}$

Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

$\begin{aligned} & v=v+x \frac{d v}{d x}-\frac{x}{v} e^v \\ & \frac{d v}{d x}=\frac{e^v}{v} \Rightarrow e^{-v} v d v=d x \end{aligned}$

Integrating both side

$\begin{aligned} & \mathrm{e}^{\mathrm{v}}(\mathrm{x}+\mathrm{c})+1+\mathrm{v}=0 \\ & \mathrm{f}(1)=1 \Rightarrow \mathrm{x}=1, \mathrm{y}=1 \end{aligned}$

$\begin{aligned} & \Rightarrow c=-1-\frac{2}{e} \\ & e^v\left(-1-\frac{2}{e}+x\right)+1+v=0 \\ & e^{\frac{y}{x}}\left(-1-\frac{2}{e}+x\right)+1+\frac{y}{x}=0 \\ & x=a, y=0 \Rightarrow a=\frac{2}{e} \\ & a e=2 \end{aligned}$

The function $f(x) = |[x]| + \sqrt{x-[x]}$ is composed of two parts : the greatest integer function $[x]$, and the fractional part function $x-[x]$.

1. The greatest integer function $[x]$ is discontinuous at every integer, since it jumps from one integer to the next without taking any values in between. The absolute value does not affect where this function is continuous or discontinuous. So within the interval $(-2,1)$, $[x]$ is discontinuous at $-2, -1, 0$ and $1$. However, because the interval is open $(-2,1)$, the endpoints $-2$ and $1$ are not included.

2. The function $\sqrt{x-[x]}$ is the square root of the fractional part of $x$. The fractional part function $x-[x]$ is continuous everywhere, as it always takes a value between $0$ and $1$ (inclusive of $0$, exclusive of $1$). However, the square root function $\sqrt{x}$ is only defined for $x \geq 0$, and so $\sqrt{x-[x]}$ is discontinuous wherever $x-[x] < 0$. This happens exactly at the points where $x$ is a negative integer, as the fractional part of a negative integer is $1$ (considering that the "fractional part" is defined as the part of the number to the right of the decimal point, which for negative numbers works a bit differently). Within the interval $(-2,1)$, this is the case for $-2$ and $-1$. However, since the interval is open $(-2,1)$, the endpoint $-2$ is not included.

Now, let's analyze the discontinuities within the given interval $(-2,1)$:

At $x=-1$: $f(-1^{+})=1+0=1$ and $f(-1^{-})=2+1=3$. Since these two values are different, $f(x)$ is discontinuous at $x=-1$.

At $x=0$: $f(0^{+})=0+0=0$ and $f(0^{-})=1+1=2$. Again, these two values are different, so $f(x)$ is discontinuous at $x=0$.

At $x=1$: $f(1^{+})=1+0=1$ and $f(1^{-})=0+1=1$. These two values are the same, so $f(x)$ is continuous at $x=1$. However, this point is not within the open interval $(-2,1)$.

So within the interval $(-2,1)$, the function $f(x) = |[x]| + \sqrt{x-[x]}$ is discontinuous at the points $-1$ and $0$ (2 points in total).

Given function is $f(x)=\left\{\begin{array}{cc}x[x], & -2 < x < 0 \\ (x-1)[x], & 0 \leq x<2\end{array}\right.$

When $[x]$ is denotes greatest integer function



Clearly, $|f(x)|$ remains same.

Given that, $m$ and $n$ respectively are the number points in $(-2,2)$ at which $y=|f(x)|$ is not continuous and not differentiable

So, $m=1$ where $y=|f(x)|$ not continuous

and $n=3$ where $|f(x)|$ is not differentiable.

Thus, $m+n=4$

Here, $f(x)=\left\{\begin{array}{cc}3 x^{2}+k \sqrt{x+1}, & 0 < x < 1 \\ m x^{2}+k^{2}, & x \geq 1\end{array}\right.$

$\because f(x)$ is differentiable at $x>0$

So, $f(x)$ is differentiable at $x=1$

$ \begin{gathered} f\left(1^{-}\right)=f(1)=f\left(1^{+}\right) \\\\ 3+k \sqrt{2}=m+k^2 ......(i) \end{gathered} $

$ \begin{aligned} & f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right) \\\\ & 6(1)+\frac{k}{2 \sqrt{1+1}}=2 m(1) \\\\ & \Rightarrow 6+\frac{k}{2 \sqrt{2}}=2 m ......(ii) \end{aligned} $

Using (i) and (ii),

$ \begin{aligned} & 3+k \sqrt{2}=3+\frac{k}{4 \sqrt{2}}+k^2 \\\\ & \Rightarrow k^2+k\left[\frac{1}{4 \sqrt{2}}-\sqrt{2}\right]=0 \end{aligned} $

$ \Rightarrow k\left[k+\frac{1-8}{4 \sqrt{2}}\right]=0 \Rightarrow k=0, \frac{7}{4 \sqrt{2}} $

As the problem specifies k to be a positive real number, we can rule out k = 0. Hence, k = $\frac{7}{4 \sqrt{2}}$

$ \begin{aligned} & \text { for } k=\frac{7}{4 \sqrt{2}}, m=3+\frac{\frac{7}{4 \sqrt{2}}}{4 \sqrt{2}} \\\\ & =3+\frac{7}{32}=\frac{96+7}{32}=\frac{103}{32} \end{aligned} $

$ \text { So, } \frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}=\frac{8 \times\left[2 \times \frac{103}{32} \times 8\right]}{6 \times \frac{1}{8}+\frac{7}{4 \sqrt{2}} \times 2 \sqrt{918}} $

$ =\frac{412}{\frac{3}{4}+\frac{7}{12}}=\frac{412}{\frac{9+7}{12}}=\frac{412 \times 12}{16}=309 $

Concept :

$f(x)$ is differentiable at $x=a$, if $f^{\prime}\left(a^{-}\right)=f'\left(a^{+}\right)$

Given that $ f(x) = [a + 13\sin(x)] $, where $[t]$ is the greatest integer function and $ x \in (0, \pi) $.

The function $[t]$ is not differentiable wherever $ t $ is an integer because at these points, the function has a jump discontinuity.

For $ f(x) $ to have a point of non-differentiability, the value inside the greatest integer function, i.e., $ a + 13\sin(x) $, should be an integer.

So, we need to find the values of $ x $ in the interval $ (0, \pi) $ for which $ a + 13\sin(x) $ is an integer.

Now, $ \sin(x) $ varies from 0 to 1 in the interval $ (0, \pi) $, so the maximum value of $ 13\sin(x) $ in this interval is 13.

For each integer value of $ 13\sin(x) $ between 0 and 13, we'll have a corresponding value of $ x $. There will be two such values of $ x $ for each value of $ 13\sin(x) $ (because of the periodic and symmetric nature of sine function over the interval $ (0, \pi) $), except for the maximum value, 13, which will have only one corresponding value of $ x $ (namely $ x = \frac{\pi}{2} $).

Thus, the total number of integer values of $ 13\sin(x) $ between 0 and 13 is 13. Excluding the maximum value, we have 12 integer values, each giving rise to two values of $ x $. Including the maximum value, which gives one value of $ x $, we have :

$ 12 \times 2 + 1 = 25 $

Thus, $ f(x) $ is not differentiable at 25 points in the interval $ (0, \pi) $.

$f(x)=4|2 x+3|+9\left[x+\frac{1}{2}\right]-12[x+20]$

$ =4|2 x+3|+9\left[x+\frac{1}{2}\right]-12[x]-240 $

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

and $f(x)$ is discontinuous at $\{-19,-18, \ldots ., 18,19\}$

as well as $\left\{-\frac{39}{2},-\frac{37}{2}, \ldots,-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}, \ldots, \frac{39}{2}\right\}$,

at same point they are also non differentiable

$ \begin{aligned} \therefore & \text { Total number of points of non differentiability } \\ &=39+40 \\ &=79 \end{aligned} $

If a graph cuts $y = 2x + 5$ in (0, 1) twice then its concavity changes twice.

$\therefore$ $f'(x) = 0$ at at least two points.

Let $x + 2\cos x = a$

$x + 2 = b$

as $x \to 0$, $a \to 2$ and $b \to 2$

$\mathop {\lim }\limits_{x \to 0} {\left( {{{{a^3} + 2{a^2} + 3\sin a} \over {{b^3} + 2{b^2} + 3\sin b}}} \right)^{{{100} \over x}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \,.\,{{100} \over x}\,.\,{{({a^3} - {b^3}) + 2({a^2} - {b^2}) + 3(\sin a - \sin b)} \over {{b^3} + 2{b^2} + 3\sin b}}}}$

$\because$ $\mathop {\lim }\limits_{x \to 0} {{a - b} \over x} = \mathop {\lim }\limits_{x \to 0} {{2(\cos x - 1)} \over x} = 0$

$ = {e^0}$

$ = 1$

$f(x)= \begin{cases}\left|4 x^{2}-8 x+5\right|, & \text { if } 8 x^{2}-6 x+1 \geq 0 \\ {\left[4 x^{2}-8 x+5\right],} & \text { if } 8 x^{2}-6 x+1<0\end{cases}$

$ = \begin{cases}4 x^{2}-8 x+5, & \text { if } x \in\left[-\infty, \frac{1}{4}\right] \cup\left[\frac{1}{2}, \infty\right) \\ {\left[4 x^{2}-8 x+5\right]} & \text { if } x \in\left(\frac{1}{4}, \frac{1}{2}\right)\end{cases} $

$f(x)=\left\{\begin{array}{cc} 4 x^2-8 x+5 & \text { if } x \in\left(-\infty, \frac{1}{4}\right] \cup\left[\frac{1}{2}, \infty\right) \\ 3 & x \in\left(\frac{1}{4}, \frac{2-\sqrt{2}}{2}\right) \\ 2 & x \in\left[\frac{2-\sqrt{2}}{2}, \frac{1}{2}\right) \end{array}\right.$


$\therefore \quad$ Non-diff at $x=\frac{1}{4}, \frac{2-\sqrt{2}}{2}, \frac{1}{2}$

Given,

$F(x) = \left| {\matrix{ {a + \sin {x \over 2}} & { - b\cos x} & 0 \cr { - b\cos x} & 0 & {a + \sin {x \over 2}} \cr 0 & {a + \sin {x \over 2}} & { - b\cos x} \cr } } \right|$

$ = \left( {a + \sin {x \over 2}} \right)\left( { - {{\left( {a + \sin {x \over 2}} \right)}^2}} \right) + b\cos x \times {b^2}{\cos ^2}x$

$ = - {\left( {a + \sin {x \over 2}} \right)^3} - {b^3}{\cos ^3}x$

Now,

$\mathop {\lim }\limits_{x \to 0} {{F(x)} \over {{x^3}}}$

$ = \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {a + \sin {x \over 2}} \right)}^3} - {b^3}{{\cos }^3}x} \over {{x^3}}}$

Given limit exists, it only possible when a = 0 and b = 0.

$ = \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {\sin {x \over 2}} \right)}^3}} \over {{x^3}}}$

$ = \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2} \times \left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)} \right)^3}$

$ = \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2}} \right)^3} \times {\left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)^3}$

$ = - {1 \over 8} \times 1 = L$

$\therefore$ $ - 112L = - 112 \times - {1 \over 8} = 14$

$ \begin{aligned} & \lim _{x \rightarrow 1} \frac{\left(\frac{\sin \left(3 x^{2}-4 x+1\right)}{3 x^{2}-4 x+1}\right)\left(3 x^{2}-4 x+1\right)-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\ \Rightarrow & \lim _{x \rightarrow 1} \frac{3 x^{2}-4 x+1-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\ \Rightarrow & \lim _{x \rightarrow 1} \frac{2(x-1)^{2}}{2 x^{3}-7 x^{2}+a x+b}=-2 \end{aligned} $

So $f(x)=2 x^{3}-7 x^{2}+a x+b=0$ has $x=1$ as repeated root, therefore $f(1)=0$ and $f^{\prime}(1)=0$ gives

$ a+b+5 \text { and } a=8 $

So, $a-b=11$

$f(x) = [1 + x] + {{{a^{2[x] + \{ x\} }} + [x] - 1} \over {2[x] + \{ x\} }}$

$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \alpha - {4 \over 3}$

$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} 1 + [x] + {{{\alpha ^{x + [x]}} + [x] - 1} \over {x + [x]}} = \alpha - {4 \over 3}$

$ \Rightarrow \mathop {\lim }\limits_{h \to {0^ - }} 1 - 1 + {{{\alpha ^{ - h - 1}} - 1 - 1} \over { - h - 1}} = \alpha - {4 \over 3}$

$\therefore$ ${{{\alpha ^{ - 1}} - 2} \over { - 1}} = \alpha - {4 \over 3}$

$ \Rightarrow 3{\alpha ^2} - 10\alpha + 3 = 0$

$\therefore$ $\alpha = 3$ or ${1 \over 3}$

$\because$ $\alpha$ in integer, hence $\alpha$ = 3

$ \mathrm{f}(\mathrm{g}(\mathrm{x}))=\left[2 \mathrm{~g}^2(\mathrm{x})\right]+1 $

$ =\left\{\begin{array}{l} {\left[2(2 x-3)^2\right]+1 ; x<0} \\ {\left[2(2 x+3)^2\right]+1 ; x \geq 0} \end{array}\right. $

$\therefore$ fog is discontinuous whenever $2(2 x-3)^2$ or $2(2 x+3)^2$ belongs to integer except $x=0$

$\therefore 62$ points of discontinuity.

$\because f(-1)=2$ and $f(1)=3$

For $x \in(-1,1),\left(4 x^{2}-1\right) \in[-1,3)$

hence $f(x)$ will be discontinuous at $x=1$ and also

whenever $4 x^{2}-1=0,1$ or 2

$ \Rightarrow x=\pm \frac{1}{2}, \pm \frac{1}{\sqrt{2}} \text { and } \pm \frac{\sqrt{3}}{2} $

So there are total 7 points of discontinuity.

$f(x) = {x^6} + 2{x^4} + {x^3} + 2x + 3$

$\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44$

$\mathop {\lim }\limits_{x \to 1} {{9{x^n} - ({x^6} + 2{x^4} + {x^3} + 2x + 3)} \over {x - 1}} = 44$

$\mathop {\lim }\limits_{x \to 1} {{9n{x^{n - 1}} - (6{x^5} + 8{x^3} + 3{x^2} + 2)} \over 1} = 44$

$\Rightarrow$ 9n $-$ (19) = 44

$\Rightarrow$ 9n = 63

$\Rightarrow$ n = 7

$f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1]$

$f\left( x \right) = \left\{ {\matrix{ { - 2\left| {{x^2} - 1} \right| + 1,} & { - 2 < x < - 1} \cr { - \left| {{x^2} - 1} \right| + 1,} & { - 1 \le x < 0} \cr {\sin {\pi \over 3} + 1,} & {0 \le x < 1} \cr {\left| {{x^2} - 1} \right| + {1 \over {\sqrt 2 }} - 2,} & {1 \le x < 2} \cr } } \right.$

$ \therefore $ at x = -1, $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = 1$ and $\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = 1$

Hence continuous at x = –1

Similarly check at x = 0,

$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1$ and $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$

So, f(x) discontinuous and at x = 0

$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$ and $\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = {1 \over {\sqrt 2 }} - 2$

So, f(x) discontinuous and at x = 1

Hence 2 points of discontinuity.

$f(x) = \left\{ {\matrix{ {a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \cr {{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x > 0} \cr } } \right.$

For continuity at '0'

$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$

$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{\tan 2x - \sin 2x} \over {b{x^3}}} = - a$

$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{{{8{x^3}} \over 3} + {{8{x^3}} \over {3!}}} \over {b{x^3}}} = - a$

$ \Rightarrow 8\left( {{1 \over 3} + {1 \over {3!}}} \right) = - ab$

$ \Rightarrow 4 = - ab$

$ \Rightarrow 10 - ab = 14$

1 $-$ {x} = 1 $-$ x; 0 $\le$ x < 1



Non differentiable at

$x = {1 \over 2},1,{3 \over 2},2,{5 \over 2}$

$f(x) = \left\{ {\matrix{ {{{P(x)} \over {\sin (x - 2)}},} & {x \ne 2} \cr {7,} & {x = 2} \cr } } \right.$

P''(x) = const. $\Rightarrow$ P(x) is a 2 degree polynomial

f(x) is cont. at x = 2

f(2⁺) = f(2^$-$)

$\mathop {\lim }\limits_{x \to {2^ + }} {{P(x)} \over {\sin (x - 2)}} = 7$

$\mathop {\lim }\limits_{x \to {2^ + }} {{(x - 2)(ax + b)} \over {\sin (x - 2)}} = 7 \Rightarrow 2a + b = 7$

P(x) = (x $-$ 2)(ax + b)

P(3) = (3 $-$ 2)(3a + b) = 9 $\Rightarrow$ 3a + b = 9

a = 2, b = 3

P(5) = (5 $-$ 2)(2.5 + 3) = 3.13 = 39

$f(x) = \left\{ {\matrix{ {3\left( {{{1 - \left| x \right|} \over 2}} \right)} & {if\,\left| x \right| \le 2} \cr 0 & {if\,\left| x \right| > 2} \cr } } \right.$

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

$f(x) = \left\{ {\matrix{ {0,} & {x < - 2} \cr {{3 \over 2}(1 + x),} & { - 2 \le x < 0} \cr {{3 \over 2}(1 - x),} & {0 \le x < 2} \cr {0,} & {x > 2} \cr } } \right.$

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

$f(x - 2) = \left\{ {\matrix{ {0,} & {x < 0} \cr {{3 \over 2}(x - 1),} & {0 \le x < 2} \cr {{3 \over 2}( - 1 - x),} & {2 \le x < 4} \cr {0,} & {x > 4} \cr } } \right.$

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

$ = \left\{ {\matrix{ {{{3x} \over 2} + 6,} & { - 4 \le x \le 2} \cr { - {{3x} \over 2},} & { - 2 < x < 2} \cr {{{3x} \over 2} - 6,} & {2 \le x \le 4} \cr {0,} & {\left| x \right| > 4} \cr } } \right.$

So, n = 0 and m = 4

$\therefore$ m + n = 4

$f(x) = {x^3} - 6{x^2} + 9x - 3$

$f(x) = 3{x^2} - 12x + 9 = 3(x - 1)(x - 3)$

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

$g(x) = \left[ {\matrix{ {f(9x)} & {0 \le x \le 1} \cr 0 & {1 \le x \le 3} \cr { - 1} & {3 < x \le 4} \cr } } \right.$

g(x) is continuous

$g'(x) = \left[ {\matrix{ {3(x - 1)(x - 3)} & {0 \le x \le 1} \cr 0 & {1 \le x \le 3} \cr { - 1} & {3 < x \le 4} \cr } } \right.$

g(x) is non-differentiable at x = 3

$\mathop {\lim }\limits_{x \to 0} {{\alpha x\left( {1 + x + {{{x^2}} \over x}} \right) - \beta \left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3}} \right) + \gamma {x^2}(1 - x)} \over {{x^3}}}$

$\mathop {\lim }\limits_{x \to 0} {{x(\alpha - \beta ) + {x^2}\left( {\alpha + {\beta \over 2} + \gamma } \right) + {x^3}\left( {{\alpha \over 2} - {\beta \over 3} - \gamma } \right)} \over {{x^3}}} = 10$

For limit to exist

$\alpha - \beta = 0,\alpha + {\beta \over 2} + \gamma = 0$${\alpha \over 2} - {\beta \over 3} - \gamma = 10$ ..... (i)

$\beta = \alpha ,\gamma = - 3{\alpha \over 2}$

Put in (i)

${\alpha \over 2} - {\alpha \over 3} + {{3\alpha } \over 2} = 10$

${\alpha \over 6} + {{3\alpha } \over 2} = 10 \Rightarrow {{\alpha + 9\alpha } \over 6} = 10$

$ \Rightarrow \alpha = 6$

$\alpha$ = 6, $\beta$ = 6, $\gamma$ = $-$9

$\alpha$ + $\beta$ + $\gamma$ = 3

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

form : 1^$\infty$ $ = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right) \times (x + 2)}}$

Now, $\mathop {\lim }\limits_{x \to 0} {{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}} = \mathop {\lim }\limits_{x \to 0} {{\sin x\sqrt {\cos 2x} - \cos x \times {1 \over {2\sqrt {\cos 2x} }} \times ( - 2sin2x)} \over {2x}}$ (by L' Hospital Rule)

$\mathop {\lim }\limits_{x \to 0} {{\sin x\cos 2x + \sin 2x.\cos x} \over {2x}} = {1 \over 2} + 1 = {3 \over 2}$

So, ${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right)(x + 2)}}$

$ = {e^{{3 \over 2} \times 2}} = {e^3}$

$\Rightarrow$ a = 3

Given, $f(x + y) = f(x)\,.\,f(y)\,\forall x,y \in R$

$\therefore$ $f(x) = {a^x} \Rightarrow f'(x) = {a^x}\,.\,\log (a)$

Now, $f'(0) = \log (a) \Rightarrow 3 = \log (a) \Rightarrow a = {e^3}$

$\therefore$ $f(x) = {({e^3})^x} = {e^{3x}}$

$\therefore$ $f(h) = {e^{3h}}$

Now, $\mathop {\lim }\limits_{h \to 0} \left( {{{f(h) - 1} \over h}} \right) = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over h}} \right)$

$ = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over {3h}} \times 3} \right) = 3 \times 1 = 3$

$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} {{\cos \left( {\sin x} \right) - \cos x} \over {{x^4}}}$

$ \Rightarrow $ ${1 \over k} = \mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{\sin x + x} \over 2}} \right)\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$

= $\mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{x + \sin x} \over 2}} \right)} \over {\left( {{{x + \sin x} \over 2}} \right)}} \times {{\sin \left( {{{x - \sin x} \over 2}} \right)} \over {\left( {{{x - \sin x} \over 2}} \right)}} \times {{{x^2} - {{\sin }^2}x} \over {4{x^4}}}$

= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{x + \sin x} \over x}} \right)\left( {{{x - \sin x} \over {{x^3}}}} \right) \times {1 \over 4}$

= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + \cos x} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$

= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$

= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 + \sin x} \over {6x}}} \right) \times {1 \over 4}$

= $2 \times 2 \times {1 \over 6} \times {1 \over 4}$ = ${1 \over 6}$

$g[f(x)] = \left[ {\matrix{ {f(x) + 1} & {f(x) < 0} \cr {{{(f(x) - 1)}^2} + b} & {f(x) \ge 0} \cr } } \right.$

$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x + a < 0\& x < 0} \cr {|x - 1| + 1} & {|x - 1| < 0\& x \ge 0} \cr {{{(x + a - 1)}^2} + b} & {x + a \ge 0\& x < 0} \cr {{{(|x - 1| - 1)}^2} + b} & {|x - 1| \ge 0\& x \ge 0} \cr } } \right.$

$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)\& x \in ( - \infty ,0)} \cr {|x - 1| + 1} & {x \in \phi } \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,\infty )\& x \in [0,\infty )} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in R\& x \in [0,\infty )} \cr } } \right.$

$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)} \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,0)} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in [0,\infty )} \cr } } \right.$

g(f(x)) is continuous.

At x = $-$a

-a + a + 1 = (-a + a - 1)² + b

$ \Rightarrow $ 1 = b + 1

$ \Rightarrow $ b = 0

at x = 0

(a $-$1)² + b = (|0 - 1| - 1)² + b

$ \Rightarrow $ (a $-$1)² + b = b

$ \Rightarrow $ a = 1

$ \Rightarrow $ a + b = 1

$\mathop {\lim }\limits_{x \to 0} {{\left\{ {a\left( {1 + x + {{{x^2}} \over {2!}} + .....} \right) - b\left( {1 - {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}}......} \right) + c\left( {1 - x + {{{x^2}} \over {2!}}......} \right)} \right\}} \over {x\left( {x - {{{x^3}} \over {3!}} + .....} \right)}} = 2$

$ \therefore $ $\mathop {\lim }\limits_{x \to 0} {{(a - b + c) + x(a - c) + {x^2}\left( {{a \over 2} + {b \over 2} + {c \over 2}} \right) + ....} \over {{x^2}\left( {1 - {{{x^2}} \over 6}....} \right)}} = 2$

For this limit to exist

a $-$ b + c = 0 & a $-$ c = 0

& ${a \over 2} + {b \over 2} + {c \over 2} = 2$

$ \Rightarrow $ a + b + c = 4

Points of non-differentiability in ($-$3, 3) are at x = $-$2, $-$1, 0, 1, 2.

i.e. 5 points.

$\mathop {\lim }\limits_{x \to 0} {{ax - \left( {{e^{4x}} - 1} \right)} \over {ax\left( {{e^{4x}} - 1} \right)}}$

Applying L' Hospital Rule

$\mathop {\lim }\limits_{x \to 0} {{a - 4{e^{4x}}} \over {a\left( {{e^{4x}} - 1} \right) + ax\left( {4{e^{4x}}} \right)}}$

This is ${{a - 4} \over 0}$.

limit exist only when $a - 4 = 0$ $ \Rightarrow $ a = 4

Applying L' Hospital Rule

$\mathop {\lim }\limits_{x \to 0} {{ - 16{e^{4x}}} \over {a\left( {4{e^{4x}}} \right) + a\left( {4{e^{4x}}} \right) + ax\left( {16{e^{4x}}} \right)}}$

= ${{ - 16} \over {4a + 4a}} = {{ - 16} \over {32}} = - {1 \over 2} = b$

$a - 2b = 4 - 2\left( {{{ - 1} \over 2}} \right) = 4 + 1 = 5$

$f(x) = |2x + 1| - 3|x + 2| + |{x^2} + x - 2|$

$f(x) = \left\{ {\matrix{ {{x^2} - 7;} & {x > 1} \cr { - {x^2} - 2x - 3;} & { - {1 \over 2} < x < 1} \cr { - {x^2} - 6x - 5;} & { - 2 < x < {{ - 1} \over 2}} \cr {{x^2} + 2x + 3;} & {x < - 2} \cr } } \right.$

$ \therefore $ $f'(x) = \left\{ {\matrix{ {2x;} & {x > 1} \cr {2x - 3;} & { - {1 \over 2} < x < 1} \cr { - 2x - 6;} & { - 2 < x < {{ - 1} \over 2}} \cr {2x + 2;} & {x < - 2} \cr } } \right.$

Check at 1, $-$2 and ${{ - 1} \over 2}$

Non. differentiable at x = 1 and ${{ - 1} \over 2}$

${\tan ^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {{{r + 1 - r} \over {1 + r(r + 1)}}} \right)$

$ = {\tan ^{ - 1}}(r + 1) - {\tan ^{ - 1}}r$

$ \therefore $ $\sum\limits_{r = 1}^n {\left( {{{\tan }^{ - 1}}(r + 1) - {{\tan }^{ - 1}}(r)} \right)} $

$ = {\tan ^{ - 1}}(2) - {\tan ^{ - 1}}(1) + ta{n^{ - 1}}(3) - {\tan ^1}(2) + ta{n^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n)$

$ = {\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(1)$

$ = {\tan ^{ - 1}}\left( {{{n + 1 - 1} \over {1 + (n + 1)1}}} \right)$

$ = {\tan ^{ - 1}}\left( {{n \over {n + 2}}} \right)$

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

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

$ = \mathop {\lim }\limits_{x \to \infty } {n \over {n + 2}}$

$ = 1$