Limits, Continuity and Differentiability

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

$ \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, $L = {1 \over {64}}$.

As when $x \in ( - 1,1),f'(x) < 0$

So, f(x) is decreasing on ($-1,1$) at x = 1

$f''(1) = {{4a} \over {{{(a + 2)}^2}}} > 0$

So, local minima at x = 1

Given that,

$g(u) = 2{\tan ^{ - 1}}({e^u}) - {\pi \over 2}$ for $u \in ( - \infty ,\infty )$

$g( - u) = 2{\tan ^{ - 1}}{e^{ - u}} - {\pi \over 2} = 2{\cot ^{ - 1}}({e^u}) - {\pi \over 2}$

$ = 2\left( {{\pi \over 2} - {{\tan }^{ - 1}}({e^u}) - {\pi \over 2}} \right)$

$ = - 2{\tan ^{ - 1}}({e^u}) + {\pi \over 2}$

$ = - g(u)$

$\therefore$ $g(u) = - g(u)$

$ \Rightarrow g(u)$ is an odd function

We have, $g(u) = 2{\tan ^{ - 1}}({e^u}) - {\pi \over 2}$

$g'(u) = {{2{e^u}} \over {1 + 2{e^{2u}}}}$

$g'(u) > 0,\forall u \in R$ [$\because$ ${e^u} > 0$]

So, $g'(u)$ is increasing function.

Given, $g(x) = {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}},0 < x < 2,m \ne 0,n > 0,m,n$ are integer.

Left hand derivative (L.H.D.) of $|x - 1|$ at $x = 1$ is P.

As $|x - 1| = \left\{ {\matrix{ {x - 1,} & {x \ge 1} \cr { - (1 - x),} & {x < 1} \cr } } \right.$

$\therefore$ L.H.D. at $x = 1$ will be $-1$.

[$\therefore$ L.H.D. at $x = 1$ is $\mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over { - h}}$]

$p = - 1$

Also, $\mathop {\lim }\limits_{x \to {1^ + }} g(x) = p = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{{(1 + h - 1)}^n}} \over {\log {{\cos }^m}(1 + h - 1)}} = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{h^n}} \over {\log {{\cos }^m}h}} = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{h^n}} \over {m\log \cos \,h}} = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{n\,.\,{h^{n - 1}}} \over {m\log \cos \,h}} = - 1$ [using L. Hospital rule]

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {{{ - n} \over m}} \right)\,.\,{{{h^{n - 2}}} \over {\left( {{{\tan \,h} \over h}} \right)}} = - 1$

$ \Rightarrow \left( {{n \over m}} \right)\mathop {\lim }\limits_{h \to 0} {{{h^{n - 2}}} \over {\left( {{{\tan \,h} \over h}} \right)}} = 1$

$n - 2 = 0$ and ${n \over m} = 1 \Rightarrow m = n = 2$

or

$[f(x) = |x - 1| = - x + 1,x < 1f'(x) = - 1$

So, L.H.D. at $x = - 1$ is $-1$

$\therefore$ $P = - 1$

Now, $\mathop {\lim }\limits_{x \to {1^ + }} {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}} = p$ .... (i)

Let $x - 1 = t$ in L.H.S. of eq. (i)

$\mathop {\lim }\limits_{t \to } {{{t^n}} \over {m\log \cos \,t}},\left( {{0 \over 0}\,\mathrm{form}} \right)$ [using L' Hospital rule]

We get, $\mathop {\lim }\limits_{t \to 0} {{n{t^{n - 1}}} \over {m\tan \,t}} = {{ - n} \over m}\mathop {\lim }\limits_{t \to 0} {{{t^{n - 1}}} \over {\tan \,t}}\left( {{0 \over 0}\,\mathrm{form}} \right)$

Again using L' Hospital rule,

${{ - n} \over m}\mathop {\lim }\limits_{t \to 0} {{(n - 1){t^{n - 2}}} \over {{{\sec }^2}t}} = P$

$ \Rightarrow {{ + n} \over m}\mathop {\lim }\limits_{t \to 0} {{(n - 1){t^{n - 2}}} \over {{{\sec }^2}t}} = - 1$

For non-zero answer

$n - 2 = 0 \Rightarrow n = 2$ Also $m = n = 2$.

$f(x)$ is a non constant twice differential function such that

$f(x) = f(1 - x) \Rightarrow f'(x) = - f'(1 - x)$ .... (i)

For $x = {1 \over 2}$

We get $f'\left( {{1 \over 2}} \right) = - f'\left( {1 - {1 \over 2}} \right)$

$ = f'\left( {{1 \over 2}} \right) + f'\left( {{1 \over 2}} \right) = 0 \Rightarrow f'\left( {{1 \over 2}} \right) = 0$

For $x = {1 \over 4}$, we get $f'\left( {{1 \over 4}} \right) = - f'\left( {{3 \over 4}} \right)$

But given that $f'\left( {{1 \over 4}} \right) = 0:f'\left( {{1 \over 4}} \right) = f'\left( {{3 \over 4}} \right) = 0$

Hence $f'(x)$ satisfies all conditions of roller's theorem for $x \in \left[ {{1 \over 4},{1 \over 2}} \right]$ and $\left[ {{1 \over 2},{3 \over 4}} \right]$, so there exist at least one point ${C_1} \in \left( {{1 \over 4},{1 \over 2}} \right)$ and at least one point ${C_2} \in \left( {{1 \over 2},{3 \over 4}} \right)$ such that

$f''(G) = 0$ and $f''{(C)_2} = 0$

$\therefore$ $f''(x)$ vanishes at least twice on [0, 1]

Also using $f(x) = f(1 - x)$

$f\left( {x + {1 \over 2}} \right) = f\left( {1 - x - {1 \over 2}} \right) = f\left( { - x + {1 \over 2}} \right)$

$f'\left( {x + {1 \over 2}} \right)$ is an odd function

$ \Rightarrow \sin x\,f\left( {x + {1 \over x}} \right)$ is an even function

$\int\limits_{{{ - 1} \over 2}}^{{1 \over 2}} {f\left( {x + {1 \over x}} \right)\sin x\,dx = 0} $

$f(x)=2+\cos x$ is differentiable everywhere and $f'(x)=-\sin x$

$\therefore f'(c)=0 \Rightarrow c=0, \pm \pi, \pm 2 \pi, \pm 3 \pi \ldots \ldots$.

$\therefore$ for each real $t,[t, t+\pi]$ contains a value from $(0, \pm \pi, \pm 2 \pi, \pm 3 \pi \ldots)$

$\therefore$ Statement 1 is true.

$f(t+2 \pi)=2+\cos (t+2 \pi)=2+\cos t=f(t)$

So, statement 2 is true.

Therefore, $f(x)$ is a periodic function of period $2 \pi$.

$f'(x)=k e^{x}-1 < 0$ if $k \leq 0$

$\therefore f(x)$ is decreasing for all $x$.

Now, $\lim_\limits{x \rightarrow+\infty} f(x)=\lim_\limits{x \rightarrow+\infty} k e^{x}-x=-\infty$

Also, $f(k)=k\left(e^{k}-1\right) \geq 0$

$\left[\therefore k \leq 0 \Rightarrow e^{k}-1 \leq 0\right]$

$\therefore f(x)=0$ has exactly one root in $[k, \infty)$.

$\therefore \quad y=x$ meets $y=k e^{x}$ for $k \leq 0$ at one point.

Let’s call the function $ f(x) = k e^{x} - x $.

To find when it has only one root, check where its graph just touches the x-axis. This happens when $ f(x) = 0 $ and the slope is zero at the same point.

Find the derivative: $ f'(x) = k e^{x} - 1 $.

Set $ f'(x) = 0 $ to find the minimum or maximum point: $ k e^{x} - 1 = 0 $. This gives $ x = -\log k $.

Now check the second derivative: $ f''(x) = k e^{x} $.

Plugging in $ x = -\log k $, we get $ f''(-\log k) = k e^{-\log k} $.

The value of $ f''(-\ln k) = 1 $, which is greater than 0, so this is a minimum point.

The function will only touch the x-axis once if the minimum value is zero. So set $ f(-\ln k) = 0 $:

$ f(-\ln k) = k e^{-\ln k} - (-\ln k) = 1 + \ln k $.

Set $ 1 + \log k = 0 $ to get the value of $ k $:

$ k = \frac{1}{e}$.

We want the equation $ k e^{x} - x = 0 $ to have two different (distinct) solutions for $ x $.

Let’s call the function $ f(x) = k e^{x} - x $. For this equation to have two different solutions, the graph of $ f(x) $ must cross the x-axis at two points.

This will happen if the lowest point (minimum) of $ f(x) $ is below zero, or $ f(x)_{\min} < 0 $.

To find the minimum, take the derivative: $ f'(x) = k e^{x} - 1 $. Set this to zero: $ k e^{x} - 1 = 0 \implies e^{x} = \frac{1}{k} \implies x = -\log k $.

Now, plug $ x = -\log k $ back into $ f(x) $:
$ f(-\log k) = k e^{-\log k} - (-\log k) = k \cdot \frac{1}{k} + \log k = 1 + \log k $.

For two distinct roots, this minimum must be less than zero:
$ 1 + \log k < 0 $

This means $ \log k < -1 $, or $ k < \frac{1}{e} $. We also need $ k > 0 $.

So the possible values for $ k $ are $ 0 < k < \frac{1}{e} $.

Therefore, $ k \in \left(0, \frac{1}{e}\right) $.

First, factor the numerator and denominator of $f(x)$:

$f(x)=\frac{x^{2}-6x+5}{x^{2}-5x+6} = \frac{(x-5)(x-1)}{(x-2)(x-3)}$

(A) When $-1 < x < 1$:

Let's check the sign of each factor in this interval:

For $x$ between $-1$ and $1$:

• $x-5 < 0$ (negative)
• $x-1 < 0$ (negative)
• $x-2 < 0$ (negative)
• $x-3 < 0$ (negative)

So, multiplying two negatives in the numerator and two negatives in the denominator: $f(x) = \frac{(negative) \times (negative)}{(negative) \times (negative)} = \frac{positive}{positive} > 0$

Now check if $f(x) < 1$. Let's look at $f(x) - 1$: $f(x)-1 = \frac{-(x+1)}{(x-2)(x-3)}$

For $-1 < x < 1$, $(x+1) > 0$ and $(x-2), (x-3) < 0$. So numerator is negative, denominator is positive (since two negatives). Therefore, $f(x)-1 < 0 \implies f(x) < 1$.

So, for $-1 < x < 1$: $0 < f(x) < 1$. This matches option (P).

(B) When $1 < x < 2$:

• $x-5 < 0$ (negative)
• $x-1 > 0$ (positive)
• $x-2 < 0$ (negative)
• $x-3 < 0$ (negative)

So, numerator: $negative \times positive = negative$. Denominator: negative $\times$ negative = positive.

So $f(x) = \frac{negative}{positive} = negative$, i.e., $f(x) < 0$. This matches (Q).

(C) When $3 < x < 5$:

• $x-5 < 0$ (negative)
• $x-1 > 0$ (positive)
• $x-2 > 0$ (positive)
• $x-3 > 0$ (positive)

Numerator: $negative \times positive = negative$.
Denominator: $positive \times positive = positive$.

So $f(x) = \frac{negative}{positive} = negative$, so $f(x) < 0$. This matches (Q).

(D) When $x > 5$:

• $x-5 > 0$ (positive)
• $x-1 > 0$ (positive)
• $x-2 > 0$ (positive)
• $x-3 > 0$ (positive)

Numerator: $positive \times positive = positive$.
Denominator: $positive \times positive = positive$.

So $f(x) = \frac{positive}{positive} = positive$, i.e., $f(x) > 0$.

Also, let's check $f(x) < 1$ again. Use previous result: $f(x) - 1 = \frac{-(x+1)}{(x-2)(x-3)}$ For $x > 5$, numerator is negative, denominator is positive, so $f(x)-1 < 0$, so $f(x) < 1$.

Therefore, for $x > 5$, $0 < f(x) < 1$, which matches (P).

(A) $y = x|x| = \left\{ {\matrix{ { - {x^2}} & {if\,x < 0} \cr {{x^2}} & {if\,x \ge 0} \cr } } \right.$

From graph $y=x|x|$ is (continuous in ($-1,1$) (p) differentiable in ($-1,1$) (q) Strictly increasing in ($-1,1$) (r)

(B) $y = \sqrt {|x|} $

$ = \left\{ {\matrix{ {\sqrt { - x} } & {if\,x < 0} \cr {\sqrt x } & {if\,x \ge 0} \cr } } \right.$

${y^2} = - x,x < 0$ {where y can take only +ve values}

and ${y^2} = x,x \ge 0$

$\therefore$ Graph is as follows:

From graph $y = \sqrt {|x|} $ is continuous in ($-1,1$) (P) not differentiable at x = 0 (s)

(C) Note this step

$y = x + [x] = \left\{ {\matrix{ {x - 1,} & { - 1 \le x < 0} \cr {x,} & {0 \le x < 1} \cr {x + 1,} & {x + 1} \cr } } \right.$

$\therefore$ Graph of $y = x + [x]$ is

From graph, $y = x + [x]$ is neither continuous, nor differentiable at x = 0 and hence in ($-1,1$) (s).

(D) $ = |x - 1| + |x + 1|$

$ = \left\{ {\matrix{ { - 2x} & {x < - 1} \cr 2 & { - 1 \le x < 1} \cr {2x} & {x \ge 1} \cr } } \right.$

Graph is as follows:

From graph, $y=f(x)$ is continuous (p) differentiable (q) in ($-1,1$) but not strictly increasing in ($-1,1$).

$ \begin{aligned} & \lim _{x \rightarrow 0}\left[(\sin x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right] \\ = & \lim _{x \rightarrow 0}(\sin x)^{\frac{1}{x}}+\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x} \end{aligned} $

Evaluate the given indeterminate form i.e,

Taking $\ln$ on both sides

$ \begin{aligned} \ln y & =\lim _{x \rightarrow 0} \sin x \log \left(\frac{1}{x}\right) \\ & =\lim _{x \rightarrow 0} \sin x(-\log x) \\ & =\lim _{x \rightarrow 0}\left(\frac{-\log x}{\operatorname{cosec} x}\right) \end{aligned} $

Applying L-Hopital's rule.

$ \begin{aligned} \ln y & =\lim _{x \rightarrow 0} \frac{\frac{-1}{x}}{-\operatorname{cosec} x \cot x} \\ \Rightarrow \quad \ln y & =\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \tan x \end{aligned} $

$ \begin{aligned} & =1 \times 0 & \therefore & \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ & =0 . & & \lim _{x \rightarrow 0} \tan x=0 \\ \Rightarrow \quad y & =e^0=1 & & \end{aligned} $

In this graph we can observe

$ f(x)=1 $(minimum of the given curve as $x>1$ )

i.e, $f(x)=\left\{\begin{array}{rr}x^3 & x \leq 1 \\ 1 & x>1\end{array}\right.$

From graph it is clear that

$\Rightarrow f(x)$ is continuous $\forall x \in \mathbf{R}$

$\Rightarrow$ Since, the obtained $f(x)$ has sharp corner in graph at $x=1$ like $f(x)=|x|$ at $x=0$

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

Concept: Function is non differentiable if it has sharp corner because it indicates that function change its definition abruptly results multiple tangent at that pointed part of curve.

Here, $\quad f(x-y)=f(x) \cdot g(y)-f(y) \cdot g(x)$

Put, $\quad x=y$ then

$f(0)=f(x) \cdot g(x)-f(x) \cdot g(x)=0$ ..... (i)

Also, $\quad g(x-y)=g(x) \cdot g(y)+f(x) \cdot f(y)$

Now put $\quad y=0$ in above, we have

$g(x)=g(x) \cdot g(0)+f(x) \cdot f(0)$

$=g(x) \cdot g(0)+0 \text { using }(\mathrm{i})$

$g(x)=g(x) g(0) \quad \therefore g(0)=1$

Again put $\quad x=y$ in $g(x-y)$, we have

$g(0)=g^{2}(y)+f^{2}(x)$

$\Rightarrow \quad g^{2}(y)+f^{2}(x)=1$ ...... (ii)

Since, RHD is exists then

$\begin{aligned} \text { RHD }=f^{\prime}(0 \text { th }) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{f(0) \cdot g(h)-f(-h) \cdot g(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{-f(-h)}{h} \\ & =\frac{f(0)-f(-h)}{-h}=\text { L.H.D } \end{aligned}$

Hence, $f$ is differentiable at $x=0$

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

$\Rightarrow 2 f^{\prime}(x) \cdot f(x)+2 g^{\prime}(x) \cdot g(x)=0$

$\Rightarrow \quad 2 f(0) \cdot f^{\prime}(0)+2 g^{\prime}(0) \cdot g(0)=0$

$\Rightarrow \quad 0+2 g^{\prime}(0) \cdot 1=0 \quad \because f(0)=0$ and $g(0)=1$

$\Rightarrow \quad g^{\prime}(0)=0$

To find the value of $f(x) - g(x)$ at $x = \frac{3}{2}$, we need to use the given conditions and properties of differentiable functions.

First, we are told that:

$f''(x) - g''(x) = 0$

This implies that:

$f''(x) = g''(x)$

Since the second derivatives of both functions are equal, their difference, $f'(x) - g'(x)$, must be a linear function. Let’s denote it as:

$f'(x) - g'(x) = k$

We'll find the constant $k$ using the initial conditions of the derivatives:

$f'(1) = 2$

$g'(1) = 4$

Thus,

$f'(1) - g'(1) = 2 - 4 = -2$

Therefore,

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

Integrating the above result, we get:

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

To determine the constant $C$, we use the values of the functions at $x = 2$:

$f(2) = 3$

$g(2) = 9$

Thus,

$f(2) - g(2) = 3 - 9 = -6$

Therefore,

$-2 \cdot 2 + C = -6$

$-4 + C = -6$

$C = -2$

So the expression for $f(x) - g(x)$ is:

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

We need to find $f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right)$:

$f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right) = -2 \cdot \frac{3}{2} - 2$

$= -3 - 2$

$= -5$

Thus, the value of $f(x) - g(x)$ at $x = \frac{3}{2}$ is -5.