Limits, Continuity and Differentiability

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

$ = \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

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

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

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

$f(x)$ 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} $

In this graph we can observe

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

$f(x)=x^3$ (minimum of the 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.