Limits, Continuity and Differentiability

Given function,
$f\left( x \right) = \left\{ {\matrix{ {a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr {c{x^2},} & {1 \le x \le 3} \cr {a{x^2} + 2cx,} & {3 < x \le 4} \cr } } \right.$

For continuity at x = 1

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

$ \Rightarrow $ $ae + b{e^{ - 1}} = c$

$ \Rightarrow $ b = ce - $a$e² .....(1)

For continuity at x = 3

$\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right)$

$ \Rightarrow $ 9c = 9a + 6c

$ \Rightarrow $ c = 3a .......(2)

Also given, f'(0) + f'(2) = e

$ \Rightarrow $ (ae^x – be^x )_x=0 + (2c^x)_x=2 = e

$ \Rightarrow $ a – b + 4c = e ........(3)

From (1), (2) & (3)

a – 3ae + ae² + 12a = e

$ \Rightarrow $ a(e² + 13 – 3e) = e

$ \Rightarrow $ a = ${e \over {{e^2} - 3e + 13}}$

A = $\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right]$

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

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

= 4

Now, when x = $\sqrt {A + 1} $ = $\sqrt 5 $, f(x) = [x²]sin($\pi $x) is discontinuous at this non integer point.

But at x = 2, 3 and 5, f(x) is continuous.

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

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

= $\left( {a + 2} \right)$ + 1

= $\left( {a + 3} \right)$

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

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

= $\mathop {\lim }\limits_{x \to {0^ + }} {{1 + x - 1} \over x}$

= 1

And f(0) = b

As f(x) is continuous at x = 0, then

f(0^-) = f(0) = f(0⁺)

$ \Rightarrow $ $a + 3$ = b = 1

$ \therefore $ $a$ = -2 and b = 1

$ \therefore $ $a$ + 2b = -2 + 2 = 0

It is clear from graph that, slope of AC $>$ slope of CB

$ \Rightarrow $ ${{f\left( c \right) - f\left( a \right)} \over {c - a}}$ $>$ ${{f\left( b \right) - f\left( c \right)} \over {b - c}}$

$ \Rightarrow $ ${{f\left( c \right) - f\left( a \right)} \over {f\left( b \right) - f\left( c \right)}}$ $ > {{c - a} \over {b - c}}$

If we consider the case where f(x) is a constant function, then its derivative f'(x) is equal to 0 for all x in the interval (0,1).

Therefore, if we substitute this into the expressions provided in Options A, B and C, we would have :

Option A : |f(c) - f(1)| < |f'(c)| would become |constant - constant| < |0|, which is 0 < 0. This is not true.

Option B : |f(c) + f(1)| < (1 + c)|f'(c)| would become |constant + constant| < (1 + c)$ \times $0, which is a positive number < 0. This is not true.

Option C : |f(c) - f(1)| < (1 - c)|f'(c)| would become |constant - constant| < (1 - c)$ \times $0, which is 0 < 0. This is not true.

Hence, for the case where f(x) is a constant function, none of the options A, B and C are correct.

So, the correct answer would be Option D : None.

Given $\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$

Putting x = 0 we get 1^{$\infty $} form.

$ \therefore $ ${e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{3{x^2} + 2} \over {7{x^2} + 2}} - 1} \right]}}$

= ${e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{ - 4{x^2}} \over {7{x^2} + 2}}} \right]}}$

= e^-4/2

= e^-2 = ${1 \over {{e^2}}}$

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.

$ \begin{array}{r} \text {Given that, } \mathop {\lim }\limits_{x \to 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x} \\ =\mathop {\lim }\limits_{x \to 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \times \frac{(1-\cos x)^2}{x^4} \times \frac{x^4}{\sin ^4 x} \\ =\mathop {\lim }\limits_{x \to 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \times \mathop {\lim }\limits_{x \to 0} \left(\frac{1-\cos x}{x^2}\right)^2 \\ \times \mathop {\lim }\limits_{x \to 0} \left(\frac{x}{\sin x}\right)^4 \end{array} $

$ \begin{aligned} & \text { As } x \rightarrow 0 \\ & (1-\cos x) \rightarrow 0 \\ & =\lim _{(1-\cos x) \rightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \\ & \quad \times \lim _{x \rightarrow 0} \frac{(1-\cos x)^2}{x^2} \times \lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^4 \\ & =\frac{1}{2} \times\left(\frac{1}{2}\right)^2 \times(1)^2 \\ & \left\{\because \lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2} \text { and } \lim _{x \rightarrow 0} \frac{x}{\sin x}=1\right\} \\ & =\frac{1}{8} \end{aligned} $

$ \begin{aligned} &\text { Given that, }\\ &\begin{aligned} & f(x)=\left\{\begin{array}{cc} \frac{x}{|x|+2 x^2}, & x \neq 0 \\ k, & x=0 \end{array}\right. \\ & \mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h) \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{-h}{|-h|+2(-h)^2}=\frac{-h}{h+2 h^2}=\frac{-h}{h(1+2 h)} \\ & =\lim _{h \rightarrow 0} \frac{-1}{1+2 h}=\frac{-1}{1+2 \times 0}=-1 \\ & \text { RHL }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\ & \quad=\lim _{h \rightarrow 0} \frac{h}{|h|+2 h^2}=\frac{h}{h(1+2 h)} \\ & \quad=\lim _{h \rightarrow 0} \frac{1}{1+2 h}=\frac{1}{1+2 \times 0}=1 \end{aligned} $

Also, $f(0)=k$

∵ LHL $\neq$ RHL, therefore given function is discontinuous for all values of $k$.

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \lim _{x \rightarrow k} \frac{\sin \left(2 \pi\left(([x])-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k} \\ & =\lim _{x \rightarrow k} \frac{\sin (2 \pi m-x)+\sin k}{x-k} \\ & {\left[\because[x]-\left[\frac{x}{k}\right]=m \text { is an integer }\right]} \\ & =\lim _{x \rightarrow k} \frac{-\sin x+\sin k}{x-k} \\ & =\lim _{x \rightarrow k} \frac{-\cos x}{1}\left[\text { By } L^{\prime} \text { Hospital's rule }\right] \\ & =-\cos k \end{aligned} \end{aligned} $

We have,

$ \begin{aligned} f(x) & = \begin{cases}1+x, & 0 \leq x \leq 2 \\ 3-x, & 2< x \leq 3\end{cases} \\ f(f(x)) & = \begin{cases}1+f(x), & 0 \leq f(x) \leq 2 \\ 3-f(x), & 2< f(x) \leq 3\end{cases} \\ f f(x) & = \begin{cases}1+(1+x), & 0 \leq x \leq 1 \\ 3-(1+x), & 1< x \leq 2 \\ 1+(3-x), & 2< x \leq 3\end{cases} \\ f \circ f(x) & = \begin{cases}2+x, & 0 \leq x \leq 1 \\ 2-x, & 1< x \leq 2 \\ 4-x, & 2< x \leq 3\end{cases} \end{aligned} $

$\therefore f \circ f(x)$ is discontinuous at $x=1$, and

$ \begin{aligned} & x=2 \\ & \therefore \quad a=1, b=2 \\ & \quad 2 a+3 b=2(1)+3(2)=2+6=8 \end{aligned} $

We have,

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi(x+2)\right.}{x^2} \\ & \lim _{x \rightarrow 0} \frac{1-\cos \left(2 \pi+\pi x+x^2\right)}{x^2} \\ & \lim _{x \rightarrow 0} \frac{1-\cos \left(\pi x+x^2\right)}{x^2} \\ & {[\because \cos (2 \pi+\theta)=\cos \theta]} \end{aligned} $

Applying 'L' Hospital rule

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

Again apply 'L' Hospital rule

$ \begin{aligned} & =\frac{\pi}{2} \lim _{x \rightarrow 0} \frac{\cos \left(\pi x+x^2\right)(2 x+\pi)}{1} \\ & =\frac{\pi}{2} \times \pi=\frac{\pi^2}{2} \end{aligned} $

Given,

$ f(x)=\left\{\begin{array}{ccc} \frac{1-\cos 4 x}{x^2} & , & \text { when } x<0 \\ a & , & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}-4}} & , & x>0 \end{array}\right. $

Since, $f(x)$ is continuous at $x=0$

$ \therefore \quad(\mathrm{LHL})_{X=0}=(\mathrm{RHL})_{X=0}=f(0) $

Now, (LHL)

$ \begin{aligned} x=0 & =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{1-\cos 4 x}{x^2} \\ & =\lim _{h \rightarrow 0} \frac{1-\cos 4(0-h)}{(0-h)^2} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\lim _{h \rightarrow 0} \frac{1-\cos 4 h}{h^2}=\lim _{h \rightarrow 0} \frac{2 \sin ^2 2 h}{h^2} \\ & =2 \lim _{h \rightarrow 0}\left(\frac{\sin 2 h}{2 h}\right)^2 \times 4 \\ & =2 \times(1)^2 \times 4=8 \end{aligned}\\ &\text { From Eq. (i), we have }\\ &\begin{aligned} & (\mathrm{LHL})_x=0=f(0) \Rightarrow 8=a \\ \Rightarrow \quad & a=8 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k \end{aligned} $

$ \begin{array}{lc} \Rightarrow & \mathop {\lim }\limits_{x \to 0} \frac{(1+x) \log (1+x)-x}{x^2}=k \\ \Rightarrow & \mathop {\lim }\limits_{x \to 0} \frac{\log (1+x)+1-1}{2 x}=k \\ \Rightarrow & \quad \text { [using L' Hospital Rule] } \\ \Rightarrow & \frac{1}{2} \mathop {\lim }\limits_{x \to 0} \frac{\log (1+x)}{x}=k \\ \Rightarrow & \frac{1}{2} \times 1=k \quad\left[\because \mathop {\lim }\limits_{x \to 0} \frac{\log (1+x)}{x}=1\right] \\ \Rightarrow & k=\frac{1}{2} \quad \therefore \quad 12 k=12 \times \frac{1}{2}=6 \end{array} $

$ \begin{aligned} & \text { We have, } \\ & f(x) \text { is continuous at } x=1 \\ & \therefore \quad k=\lim _{x \rightarrow 1} \frac{(9 x-1)(\sqrt{x}-1)}{3 x^2+2 x-5} \\ & \Rightarrow \quad k=\lim _{x \rightarrow 1} \frac{9 x^{3 / 2}-9 x-\sqrt{x}+1}{3 x^2+2 x-5} \\ & \Rightarrow \quad k=\lim _{x \rightarrow 1} \frac{9\left(\frac{3}{2} x^{1 / 2}\right)-9-\frac{1}{2 \sqrt{x}}}{6 x+2} \\ & \Rightarrow \quad k=\frac{\frac{27}{2}-9-\frac{1}{2}}{6+2}=\frac{13-9}{8}=\frac{4}{8}=\frac{1}{2} \end{aligned} $

Let $f(x)=\frac{2 x-1}{3 x-4},[1,2]$

Since, $f(x)$ is not defined at

$ x=\frac{4}{3} \in[1,2] $

So, Lagrange's theorem is not applicable on $f(x)$ on $[1,2]$

From f(x) = 5 - | x - 2 |
maximum value of f(x) is at x = 2

From g(x) = | x + 1 |
minimum value of g(x) is at x = -1

$ \therefore $ $\alpha \beta $ = - 2

$ \Rightarrow $ $\mathop {\lim }\limits_{x \to 2} {{(x - 1)(x - 2)(x - 3)} \over {(x - 2)(x - 4)}}$

$ \Rightarrow $ ${{(2 - 1)(2 - 3)} \over {(2 - 4)}} = {1 \over 2}$

$\alpha $ and $\beta $ are the two root of 375x² - 25x - 2 = 0

Both of the roots are lie in (-1, 1) hence sum of given series is finite

$\mathop {\lim }\limits_{n \to \infty } \left( {{\alpha \over {1 - \alpha }} + {\beta \over {1 - \beta }}} \right) = {{\alpha (1 - \beta ) + \beta (1 - \alpha )} \over {(1 - \alpha )(1 - \beta )}}$

$ \Rightarrow {{\left( {\alpha + \beta } \right) - 2\alpha \beta } \over {1 - (\alpha + \beta ) + \alpha \beta }} = {{25 - 2( - 2)} \over {375 - 25 - 2}} = {{29} \over {348}}$

$\mathop {\lim }\limits_{x \to 1} {{{x^2} - ax + b} \over {x - 1}} = 5$

$ \Rightarrow $ ${(1)^2} - a(1) + b = 0$

$ \Rightarrow $$1 - a + b = 0$

$ \Rightarrow $$a - b = 1\,\,......(1)$

Now 'L' hospital rule

2x - a = 5

$ \Rightarrow $2 - a = 5 ($ \because $ x = 1)

$ \Rightarrow $ a = - 3

By putting a = -3 in (1)

$ \Rightarrow $ b = -4

$ \therefore $ a + b = -7

$f(x) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {x < 0} \cr q & {x = 0} \cr {{{\sqrt {{x^2} + x} - \sqrt x } \over {{x^{{3 \over 2}}}}}} & {x > 0} \cr } } \right.$

is continuous at x = 0

So f(0^–) = f(0) = f (0⁺) ... (1)

$f({0^ - }) = \mathop {lt}\limits_{h \to 0} f(0 - h)$

$ \Rightarrow \mathop {lt}\limits_{h \to 0} {{\sin (p + 1)( - h) + \sin ( - h)} \over { - h}}$

$ \Rightarrow \mathop {lt}\limits_{h \to 0} \left[ {{{ - \sin (p + 1)h} \over { - h}} + {{\sinh } \over h}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sin (p + 1)h} \over {h(p + 1)}} \times (p + 1) + \mathop {\lim }\limits_{h \to 0} {{\sinh } \over h}$

= (p + 1) + 1 = p + 2 ...... (2)

Now $f({0^ + }) = \mathop {\lim }\limits_{h \to 0} (0 + h) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {{h^2} + h} - \sqrt h } \over {{h^{3/2}}}}$

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

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sqrt {h + 1} - 1} \over h} \times {{\sqrt {h + 1} + 1} \over {\sqrt {h + 1} + 1}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{h + 1 - 1} \over {h\left( {\sqrt {h + 1} + 1} \right)}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {1 \over {\sqrt {h + 1} + 1}} = {1 \over {1 + 1}} = {1 \over 2}$ ..... (3)

Now, from equation (1)
f(0^–) = f(0) = f(0⁺)

p + 2 = q = 1/2

So, $q = {1 \over 2}$ and $p = {1 \over 2} - 2 = {{ - 3} \over 2}$

$(p,q) \equiv \left( { - {3 \over 2},{1 \over 2}} \right)$

$g'(c) = \mathop {\lim }\limits_{x \to c} {{g(x) - g(c)} \over {x - c}}$

$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right| - \left| {f(c)} \right|} \over {x - c}}$

$ \therefore $ f(c) = 0

$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right|} \over {x - c}}$

$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{f(x)} \over {x - c}}$ if f(x) > 0

and $g'(c) = \mathop {\lim }\limits_{x \to c} {{ - f(x)} \over {x - c}}$ if f(x) < 0

$ \Rightarrow g'(c) = f'(c) = - f'(c)$

$ \Rightarrow $ 2f'(c) = 0

$ \Rightarrow $ f'(c) = 0

If $\mathop {\lim }\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)$

L·H·S·

$\mathop {Lt}\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \left( {{0 \over 0}form} \right)$

$ \Rightarrow \mathop {Lt}\limits_{x \to 1} {{4{x^3}} \over 1} = 4$

Now, $\mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)$ = 4

$ \Rightarrow \mathop {\lim }\limits_{x \to K} {{3{x^2}} \over {2x}} = 4$

$ \Rightarrow {3 \over 2}k = 4 \Rightarrow k = {8 \over 3}$

$f(x) = [x] - \left[ {{x \over 4}} \right]$

Here check continuty at x = 4

LHL = $\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right)$

= $\mathop {\lim }\limits_{x \to {4^ - }} \left[ x \right] - \left[ {{x \over 4}} \right]$

= $\mathop {\lim }\limits_{h \to 0} \left[ {4 - h} \right] - \left[ {{{4 - h} \over 4}} \right]$

= 3 - 0 = 3

h $ \to $ 0 means h > 0 (very small positive value)

So 4 - h = 3.something (means less than 4)

$ \therefore $ [4 - h] = [3.something] = 3

and $\left[ {{{4 - h} \over 4}} \right]$ = [0.something] = 0

RHL = $\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right)$

= $\mathop {\lim }\limits_{x \to {4^ + }} \left[ x \right] - \left[ {{4 \over x}} \right]$

= $\mathop {\lim }\limits_{h \to 0} \left[ {4 + h} \right] - \left[ {{{4 + h} \over 4}} \right]$

= 4 - 1 = 3

Here [4 + h] = [4.something] = 4

and $\left[ {{{4 + h} \over 4}} \right]$ = [1.something] = 1

And f(4) = $\left[ 4 \right] - \left[ {{4 \over 4}} \right]$ = 4 - 1 = 3

As LHL = RHL = f(4)

$ \therefore $ f is continuous at x = 4.

As f(x) is continuous at x = 5 then

$\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = f\left( 5 \right) = \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right)$

$ \Rightarrow $ $\mathop {\lim }\limits_{h \to 0} f\left( {5 - h} \right) = f\left( 5 \right) = \mathop {\lim }\limits_{h \to 0} f\left( {5 + h} \right)$

$ \Rightarrow $ $\mathop {\lim }\limits_{h \to 0} \left( {a\left| {\pi - \left( {5 - h} \right)} \right| + 1} \right) = a\left| {\pi - 5} \right| + 1$

= $\mathop {\lim }\limits_{h \to 0} \left( {b\left| {5 + h - \pi } \right| + 3} \right)$

$ \Rightarrow $ ${a\left| {\pi - 5} \right| + 1}$ = $a\left| {\pi - 5} \right| + 1$

= ${b\left| {5 - \pi } \right| + 3}$

Note : As $\pi $ = 3.14, then $\pi $ - 5 = 3.14 - 5 = -1.84 < 0

$ \therefore $ ${\left| {\pi - 5} \right|}$ = - ($\pi $ - 5) and ${\left| {5 - \pi } \right|}$ = (5 - $\pi $)

$ \Rightarrow $ $ - a\left( {\pi - 5} \right) + 1$ = $ - a\left( {\pi - 5} \right) + 1$ = ${b\left( {5 - \pi } \right) + 3}$

$ \Rightarrow $ $ - a\left( {\pi - 5} \right) + 1$ = ${b\left( {5 - \pi } \right) + 3}$

$ \Rightarrow $ $\left( {a - b} \right)\left( {5 - \pi } \right) = 2$

$ \Rightarrow $ $\left( {a - b} \right) = {2 \over {\left( {5 - \pi } \right)}}$

ƒ(x) = 15 – |x – 10|

g(x) = ƒ(ƒ(x)) = 15 – |ƒ(x) – 10|

= 15 – |15 – |x – 10| – 10|

= 15 – |5 – |x – 10||

As this is a linear expression so it is non differentiable when value inside the modulus is zero.

So non differentiable when

x – 10 = 0 $ \Rightarrow $ x = 10

and 5 – |x – 10| = 0

$ \Rightarrow $ |x – 10| = 5

$ \Rightarrow $ x - 10 = $ \pm $ 5

$ \Rightarrow $ x = 5, 15

$ \therefore $ g(x) is not differentiable at x = 5, 10, 15.

$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$ = f(${{\pi \over 4}}$) = k

$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$ (${0 \over 0}$ form) = k

$ \Rightarrow $ $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - \sqrt 2 \sin x} \over {-\cos e{c^2}x}}$ (Using L Hospital Rule)

$ \Rightarrow $ $\mathop {\lim }\limits_{x \to {\pi \over 4}} \sqrt 2 {\sin ^3}x$ = k

$ \Rightarrow $ k = $\sqrt 2 {\left( {{1 \over {\sqrt 2 }}} \right)^3}$ = ${1 \over 2}$

The general formula for indeterminate form 1^{$\infty $} is

$\mathop {\lim }\limits_{x \to a} {\left( {f\left( x \right)} \right)^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left( {f\left( x \right) - 1} \right)}}$

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

Here I is in 1^{$\infty $} form.

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

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

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

Here ${{{f\left( {3 + x} \right) - f\left( 3 \right) - f\left( {2 - x} \right) + f\left( 2 \right)} \over x}}$ is in ${0 \over 0}$ form.

So using L'Hopital rule we get

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

= ${e^{\left( {f'\left( 3 \right) + f'\left( 2 \right)} \right).1}}$

= e⁰ [ as given ƒ'(3) + ƒ'(2) = 0 ]

= 1

$f(x) = \left\{ {\matrix{ {\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr {x + \left| x \right|} & , & {1 \le x < 2} \cr {x + \left[ x \right]} & , & {2 \le x \le 3} \cr } } \right.$

= $ = \left\{ {\matrix{ { - x - 1,} & { - 1 \le x < 0} \cr {x,} & {0 \le x < 1} \cr {2x,} & {1 \le x < 2} \cr {x + 2,} & {2 \le x < 3} \cr {6,} & {x = 3} \cr } } \right.$

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

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

$ \therefore $ f(x) is continuous at x = - 1

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

= -(0) - 1 = - 1

f(0) = $\mathop {\lim }\limits_{x \to 0} \left( x \right)$ = 0

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

$ \therefore $ f(x) is discontinuous at x = 0

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

f(1) = $\mathop {\lim }\limits_{x \to 1} \left( {2x} \right)$ = 2

f(1⁺) = $\mathop {\lim }\limits_{x \to {1^ + }} \left( {2x} \right)$ = 2

$ \therefore $ f(x) is discontinuous at x = 1

f(3^-) = $\mathop {\lim }\limits_{x \to {3^ - }} \left( {x + 2} \right)$ = 3 + 2 = 5

f(3) = 6

$ \therefore $ f(x) is discontinuous at x = 3

So, f(x) is discontinuous at x = {0, 1, 3}.

$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$

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

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

= $\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {2{{\sin }^2}\left( {{x \over 2}} \right)}}} \right)\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$

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

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

= $4\sqrt 2 $

${{f'(x)} \over {f(x)}} = 1\forall x \in R$

Intergrate & use f(1) = 2

f(x) = 2e^x-1 $ \Rightarrow $ f '(x) = 2e^x$-$1

h(x) = f(f(x)) $ \Rightarrow $ h'(x) = f '(f(x)) f'(x)

h'(1) = f '(f(1)) f'(1)

= f '(2) f '(1)

= 2e . 2 = 4e

$\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }} \times {{\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}} } \over {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} }}$

$\mathop {\lim }\limits_{x \to {1^ - }} {{2\left( {{\pi \over 2} - {{\sin }^{ - 1}}x} \right)} \over {\sqrt {1 - x} \left( {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}$

$\mathop {\lim }\limits_{x \to {1^ - }} {{2{{\cos }^{ - 1}}x} \over {\sqrt {1 - x} }}.{1 \over {2\sqrt \pi }}$

Put  $x = \cos \theta $

$\mathop {\lim }\limits_{x \to {0^ + }} {{2\theta } \over {\sqrt 2 \sin \left( {{\theta \over 2}} \right)}}.{1 \over {2\sqrt \pi }}$

$ = \sqrt {{2 \over \pi }} $

$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$

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

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

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

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

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + 1} \right)} \over {1.\cos \left( {x + {\pi \over 4}} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\left( {1 - {{\tan }^2}x} \right)} \over {\left( {{{\cos x - \sin x} \over {\sqrt 2 }}} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {1 - {{{{\sin }^2}x} \over {{{\cos }^2}x}}} \right)} \over {\left( {\cos x - \sin x} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {{{\cos }^2}x - {{\sin }^2}x} \right)} \over {{{\cos }^2}x\left( {\cos x - \sin x} \right)}}$

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

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos x + \sin x} \right)} \over {{{\cos }^2}x}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos {\pi \over 4} + \sin {\pi \over 4}} \right)} \over {{{\cos }^2}{\pi \over 4}}}$

= ${{2\sqrt 2 \left( {{1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }}} \right)} \over {{1 \over 2}}}$

= ${{2\sqrt 2 \left( {{2 \over {\sqrt 2 }}} \right)} \over {{1 \over 2}}}$

= 8

Other Method :

$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$

This is in ${0 \over 0}$ form so L' Hospital rule is applicable.

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\cot }^3}x\left( { - \cos e{c^2}x} \right) - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}$

= ${{3 \times 1\left( { - {{\left( {\sqrt 2 } \right)}^2}} \right) - {{\left( {\sqrt 2 } \right)}^2}} \over { - 1}}$

= ${{ - 6 - 2} \over { - 1}}$ = 8

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

f(x) = sin$\left| x \right| - \left| x \right|$ + 2(x $-$ $\pi $) cosx

$ \because $  sin$\left| x \right|$ $-$ $\left| x \right|$ is differentiable function at c = 0

$ \therefore $  k = $\phi $

R.H.L. $=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$

(as x $ \to $ 0⁺ $ \Rightarrow $  [x] $=$ 0)

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

$=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\left( {\pi {{\sin }^2}x} \right)}} + 1 = \pi + 1$

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

(as x $ \to $ 0^$-$ $ \Rightarrow $  [x] $=$ $-$1)

$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi $

R.H.L.  $ \ne $  L.H.L.

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

and  $f\left( {\left| x \right|} \right) = {x^2} - 1,x \in \left[ { - 2,2} \right]$

Hence  $g(x) = \left\{ {\matrix{ {{x^2}} & , & {x \in \left[ { - 2,0} \right]} \cr 0 & , & {x \in \left[ {0,1} \right)} \cr {2\left( {{x^2} - 1} \right)} & , & {x \in \left[ {1,2} \right]} \cr } } \right.$

It is not differentiable at x = 1