Limits, Continuity and Differentiability

$f(x) = {\mathop{\rm sgn}} (x)$

$h(x) = 2[x] - {\mathop{\rm sgn}} (x)$

If $x \to {1^ + }$ then $h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)$

$ = 0 - 1 = - 1$

& if $x \to {1^ - }$ then $h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)$

$ = - 2 + 1 = - 1$

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

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

$5f(x + y) = f(x).f(y)$

$5f(3) = f(1).f(2)$

$5f(2) = {(f(1))^2}$

$f(10) = 5$

$f(1) = 20$

$ \Rightarrow f(1).{{{{(f(1))}^2}} \over 5} = 1600$

$\sum\limits_{n = 0}^5 {f(n) = f(0) + 20 + 80 + 320 + 1280 + 5120} $

$ = 1750 + 5120 = 6825$

$\mathop {\lim }\limits_{x \to a} \left( {[x - 5] - [2x + 2]} \right) = 0$

$ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {[x] - 5 - [2x] - 2} \right) = 0$

$ \Rightarrow \mathop {\lim }\limits_{x \to a} [x] - [2x] - 7 = 0$

Case 1 :

If $a \in \left[ {n,n + {1 \over 2}} \right)$

then $\mathop {\lim }\limits_{x \to a} \left( {[x] - [2x] - 7} \right) = 0$

$ \Rightarrow n - 2n - 7 = 0$

$ \Rightarrow n = - 7$

$\therefore$ $a \in \left[ { - 7, - 6.5} \right)$

Case 2 :

If $a \in \left[ {n + {1 \over 2},n + 1} \right)$

then $\mathop {\lim }\limits_{x \to a} \left( {[x] - [2x] - 7} \right) = 0$

$ \Rightarrow n - (2n + 1) - 7 = 0$

$ \Rightarrow - n - 8 = 0$

$ \Rightarrow n = - 8$

$\therefore$ $a \in \left[ { - 7.5, - 7} \right)$

$R.H.L = \mathop {\lim }\limits_{x \to - {{7.5}^ + }} \left( {\left[ x \right] - \left[ {2x} \right] - 7} \right)$

$ = - 8 - \left( { - 15} \right) - 7$

$ = - 8 + 15 - 7 = 0$

$\eqalign{ & L.H.L = \mathop {\lim }\limits_{x \to - {{7.5}^ - }} \left( {\left[ x \right] - \left[ {2x} \right] - 7} \right) \cr \\ & = - 8 - \left( { - 16} \right) - 7 \cr \\ & = - 8 + 16 - 7 = 1 \cr} $

$ \therefore $ R.H.L $ \ne $ L.H.L

$ \therefore $ At x = -7.5, limit does not exists.

$\therefore$ $a \in \left( { - 7.5, - 6.5} \right)$

Given,

$f(x) = \left\{ {\matrix{ {{x^2}\sin \left( {{1 \over x}} \right),} & {x \ne 0} \cr {0,} & {x = 0} \cr } } \right.$

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

Now,

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

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

= Does not exist

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

L.H.D. $ = \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 - h) - f(0)} \over { - h}}$

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

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

R.H.D. $ = \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 + h) - f(0)} \over h}$

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

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

$\therefore$ L.H.D. = R.H.D.

$ \Rightarrow f(x)$ is differentiable at $x = 0$. So, f(x) is continuous.

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

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

$ \Rightarrow 10 = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}(1 + 5x) - {{\log }_e}(1 + \alpha x)} \over x}$

$ = \mathop {\lim }\limits_{x \to 0} {{\log (1 + 5x)} \over {5x}} \times 5 - {{{{\log }_e}(1 + \alpha x)} \over {\alpha x}} \times \alpha $

$ = 1 \times 5 - \alpha $

$ \Rightarrow \alpha = 5 - 10 = - 5$

$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} + \gamma \sin x} \over {x{{\sin }^2}x}} = {2 \over 3}$

$ \Rightarrow \alpha + \beta = 0$ (to make indeterminant form) ...... (i)

Now,

$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} - \beta {e^{ - x}} + \gamma \cos x} \over {3{x^2}}} = {2 \over 3}$ (Using L-H Rule)

$ \Rightarrow \alpha - \beta + \gamma = 0$ (to make indeterminant form) ...... (ii)

Now,

$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} - \gamma \sin x} \over {6x}} = {2 \over 3}$ (Using L-H Rule)

$ \Rightarrow {{\alpha - \beta + \gamma } \over 6} = {2 \over 3}$

$ \Rightarrow \alpha - \beta + \gamma = 4$ ...... (iii)

$ \Rightarrow \gamma = - 2$

and (i) + (ii)

$2\alpha = - \gamma $

$ \Rightarrow \alpha = 1$ and $\beta = - 1$

and $\alpha {\beta ^2} + \beta {\gamma ^2} + \gamma {\alpha ^2} + 3 = 1 - 4 - 2 + 3 = - 2$

$f:R \to R$.

$f(x) = |x - 1|\cos |x - 2|\sin |x - 1| + (x - 3)|{x^2} - 5x + 4|$

$ = |x - 1|\cos |x - 2|\sin |x - 1| + (x - 3)|x - 1||x - 4|$

$ = |x - 1|[\cos |x - 2|\sin |x - 1| + (x - 3)|x - 4|]$

Sharp edges at $x = 1$ and $x = 4$

$\therefore$ Non-differentiable at $x = 1$ and $x = 4$

$f(x) = \mathop {\lim }\limits_{n \to \infty } {{\cos (2\pi x) - {x^{2n}}\sin (x - 1)} \over {1 + {x^{2n + 1}} - {x^{2n}}}}$

For $|x| < 1,\,f(x) = \cos 2\pi x$, continuous function

$|x| > 1,\,f(x) = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {{x^{2n}}}}\cos 2\pi x - \sin (x - 1)} \over {{1 \over {{x^{2n}}}} + x - 1}}$

$ = {{ - \sin (x - 1)} \over {x - 1}}$, continuous

For $|x| = 1,\,f(x) = \left\{ {\matrix{ 1 & {\mathrm{if}} & {x = 1} \cr { - (1 + \sin 2)} & {\mathrm{if}} & {x = - 1} \cr } } \right.$

Now,

$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = - 1,\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1$, so discontinuous at $x = 1$

$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 1,\,\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = - {{\sin 2} \over 2}$, so discontinuous at $x = - 1$

$\therefore$ $f(x)$ is continuous for all $x \in R - \{ - 1,1\} $

$f(x) = {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$

for continuity at $x = 0$, $\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$

Now, $\therefore$ $\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$

$ \Rightarrow p = 3$ (To make indeterminant form)

So, $\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{{{\left( {{3^7} + 3x} \right)}^{{1 \over 7}}} - 3} \over {{{\left( {729 + qx} \right)}^{{1 \over 3}}} - 9}}$

$ = \mathop {\lim }\limits_{x \to 0} {{3\left[ {{{\left( {1 + {x \over {{3^6}}}} \right)}^{{1 \over 7}}} - 1} \right]} \over {9\left[ {{{\left( {1 + {q \over {729}}x} \right)}^{{1 \over 3}}} - 1} \right]}} = {1 \over 3}\,.\,{{{1 \over 7}\,.\,{1 \over {{3^6}}}} \over {{1 \over 3}\,.\,{q \over {729}}}}$

$\therefore$ $f(0) = {1 \over {7q}}$

$\therefore$ Option (B) is correct.

$\beta = \mathop {\lim }\limits_{x \to 0} {{\alpha x - ({e^{3x}} - 1)} \over {\alpha x({e^{3x}} - 1)}},\,\alpha \in R$

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

So, $\alpha = 3$ (to make independent form)

$\beta = \mathop {\lim }\limits_{x \to 0} {{1 - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {3x}} = {{1 - {{\left( {3x + {{9{x^2}} \over 2}\, + \,......} \right)} \over {3x}}} \over {3x}}$

$ = {{ - \left( {{9 \over 2}{x^2} + {{{{(3x)}^3}} \over {31}}\, + \,....} \right)} \over {9{x^2}}} = {{ - 1} \over 2}$

$\therefore$ $\alpha + \beta = 3 - {1 \over 2} = {5 \over 2}$

$f(3x) - f(x) = x$ ...... (1)

$x \to {x \over 3}$

$f(x) - f\left( {{x \over 3}} \right) = {x \over 3}$ ....... (2)

Again $x \to {x \over 3}$

$f\left( {{x \over 3}} \right) - f\left( {{x \over 9}} \right) = {x \over {{3^2}}}$ ...... (3)

Similarly

$f\left( {{x \over {{3^{n - 2}}}}} \right) - f\left( {{x \over {{3^{n - 1}}}}} \right) = {x \over {{3^{n - 1}}}}\,.....\,(n)$

Adding all these and applying $n \to \infty $

$\mathop {\lim }\limits_{n \to \infty } \left( {f(3x) - f\left( {{x \over {{3^{n - 1}}}}} \right)} \right) = x\left( {1 + {1 \over 3} + {1 \over {{3^2}}}\, + \,....} \right)$

$f(3x) - f(0) = {{3x} \over 2}$

Putting $x = {8 \over 3}$

$f(8) - f(0) = 4$

$ \Rightarrow f(0) = 3$

Putting $x = {{14} \over 3}$

$f(14) - 3 = 7 \Rightarrow f(14) = 10$

$f(x) = \left\{ {\matrix{ {{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \cr k & , & {x = 0} \cr } } \right.$

for continuity at $x = 0$

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

$\therefore$ $k = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {\sec x - \cos x}}\left( {{0 \over 0}\,\mathrm{form}} \right)$

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

$ = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {{x^2}}}$

$ = \mathop {\lim }\limits_{x \to 0} {{\ln (1 + {x^2} + {x^4})} \over {{x^2} + {x^4}}}\,.\,{{{x^2} + {x^4}} \over {{x^2}}}$

$ = 1$

$f(x) = \left\{ {\matrix{ {x + a} & , & {x \le 0} \cr {|x - 4|} & , & {x > 0} \cr } } \right.$ and $g(x) = \left\{ {\matrix{ {x + 1} & , & {x < 0} \cr {{{(x - 4)}^2} + b} & , & {x \ge 0} \cr } } \right.$

$\because$ $f(x)$ and $g(x)$ are continuous on R

$\therefore$ $a = 4$ and $b = 1 - 16 = - 15$

then $(gof)(2) + (fog)( - 2)$

$ = g(2) + f( - 1)$

$ = - 11 + 3 = - 8$

$f(x) = \left\{ {\matrix{ {{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr { - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr } } \right.$

If $f(x)$ has maximum value at $x = 1$ then $f(1 + ) \le f(1)$

$ - 2 + {\log _2}({b^2} - 4) \le 1 - 1 + 10 - 7$

${\log _2}({b^2} - 4) \le 5$

$0 < {b^2} - 4 \le 32$

(i) ${b^2} - 4 > 0 \Rightarrow b \in ( - \infty , - 2) \cup (2,\infty )$

(ii) ${b^2} - 36 \le 0 \Rightarrow b \in [ - 6,6]$

Intersection of above two sets

$b \in [ - 6, - 2) \cup (2,6]$

$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{8\sqrt 2 - {{(\cos x + \sin x)}^7}} \over {\sqrt 2 - \sqrt 2 \sin 2x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{0 \over 0}\,\mathrm{form}} \right)$

$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - 7{{(\cos x + \sin x)}^6}( - \sin x + \cos x)} \over { - 2\sqrt 2 \cos 2x}}$ using $\mathrm{L-H}$ Rule

$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{56(\cos x - \sin x)} \over {2\sqrt 2 \cos 2x}}\,\,\left( {{0 \over 0}} \right)$

$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - 56(\sin x + \cos x)} \over { - 4\sqrt 2 \sin 2x}}$ using $\mathrm{L-H}$ Rule

$ = 7\sqrt 2 \,.\,\sqrt 2 = 14$

$\mathop {\lim }\limits_{n \to \alpha } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$

[ This limit will be zero when $\alpha$ < 0 as when $\alpha$ > 0 then overall limit will be $\infty $. ]

$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right)\left( {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$

$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {{n^2} - n - 1} \right) - {{\left( {n\alpha + \beta } \right)}^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$

$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2} - n - 1 - {n^2}{\alpha ^2} - 2n\alpha\beta - {\beta ^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$

$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2}\left( {1 - {\alpha ^2}} \right) - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + {\beta ^2}} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}}$

Here power of "n" in the numerator is 2 and power of "n" in the denominator is 1.

To get the value of limit equal to zero power of "n" should be equal in both numerator and denominator, otherwise value of limit will be infinite ($\infty $).

$\therefore$ Coefficient of n² should be 0 in this case.

$\therefore$ $1 - {\alpha ^2} = 0$

$ \Rightarrow \alpha = \, \pm \,1$

But $\alpha$ should be < 0

$\therefore$ $\alpha = \, + \,1$ not possible

$\therefore$ $\alpha = - 1$.

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

Divide numerator and denominator by n then we get,

$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{ - \left( {1 + 2\alpha \beta } \right) - {{(1 + \beta )} \over n}} \over {\sqrt {1 - {1 \over n} - {1 \over {{n^2}}}} - \alpha - {\beta \over n}}} = 0$

$ \Rightarrow {{ - \left( {1 + 2\alpha \beta } \right) - 0} \over {\sqrt {1 - 0 - 0} - \alpha - 0}} = 0$

$ \Rightarrow {{ - \left( {1 + 2\alpha \beta } \right)} \over {1 - \alpha }} = 0$

$ \Rightarrow - \left( {1 + 2\alpha \beta } \right) = 0$

$ \Rightarrow 1 + 2\alpha \beta = 0$

$ \Rightarrow 2\alpha \beta = - 1$

$ \Rightarrow \beta = - {1 \over {2\alpha }} = - {1 \over {2( - 1)}} = {1 \over 2}$

$\therefore$ $8\left( {\alpha + \beta } \right)$

$ = 8\left( { - 1 + {1 \over 2}} \right)$

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

$ = - 4$

$\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^2} + 2x - 1}}$

$ = \mathop {\lim }\limits_{x \to 1} {{({x^2} - 1)si{n^2}(\pi x)} \over {({x^2} - 1){{(x - 1)}^2}}}$

$ = \mathop {\lim }\limits_{x \to 1} {{{{\sin }^2}(\pi x)} \over {{{(x - 1)}^2}}}$

Let $x = 1 + h$

$\therefore$ when x $\to$ 1 then h $\to$ 0

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

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

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

$ = {\pi ^2} \times 1$

$ = {\pi ^2}$

$\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}$

$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{{(r + 2) - (r + 1)} \over {1 + (r + 2)(r + 1)}}} \right)} } \right\}$

$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {({{\tan }^{ - 1}}(r + 2) - {{\tan }^{ - 1}}(r + 1))} } \right\}$

$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {{{\tan }^{ - 1}}(n + 2) - {{\tan }^{ - 1}}2} \right\}$

$ = 6\tan \left\{ {{\pi \over 2} - {{\cot }^{ - 1}}\left( {{1 \over 2}} \right)} \right\}$

$ = 6\tan \left( {{{\tan }^{ - 1}}\left( {{1 \over 2}} \right)} \right)$

$ = 3$

$f(x) = \left\{ {\matrix{ 0 & {x < 0} \cr {a{e^x} - 1} & {0 \le x < 1} \cr b & {x = 1} \cr {b - 1} & {1 < x < 2} \cr { - c} & {x \ge 2} \cr } } \right.$

To be continuous at x = 0

a $-$ 1 = 0

to be continuous at x = 1

ae $-$ 1 = b = b $-$ 1 $\Rightarrow$ not possible

to be continuous at x = 2

b $-$ 1 = $-$ c $\Rightarrow$ b + c = 1

If a = 1 and b + c = 1 then f(x) is discontinuous at exactly one point.

$\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$ exist & $a \in I$.

$ = \mathop {\lim }\limits_{x \to 7} {{17 - [ - x]} \over {[x] - 3a}}$ exist

$RHL = \mathop {\lim }\limits_{x \to {7^ + }} {{17 - [ - x]} \over {[x] - 3a}} = {{25} \over {7 - 3a}}$ $\left[ {a \ne {7 \over 3}} \right]$

$LHL = \mathop {\lim }\limits_{x \to {7^ - }} {{17 - [ - x]} \over {[x] - 3a}} = {{24} \over {6 - 3a}}$ $\left[ {a \ne 2} \right]$

For limit to exist

$LHL = RHL$

${{25} \over {7 - 3a}} = {{24} \over {6 - 3a}}$

$ \Rightarrow {{25} \over {7 - 3a}} = {8 \over {2 - a}}$

$\therefore$ $a = - 6$

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

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

$ = \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {{{\left( {x + x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}}...} \right)\left( {x - x + {{{x^3}} \over {3!}}...} \right)} \over {{x^4}}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {2 - {{{x^2}} \over {3!}} + {{{x^4}} \over {5!}}...} \right)\left( {{1 \over {3!}} - {{{x^2}} \over {5!}} - 1} \right)$

$ = {1 \over 6}$

$f(x) = \min \{ 1,\,1 + x\sin x\} $, $0 \le x \le x$

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

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

$\therefore$ f(x) is continuous in [0, 2$\pi$]

Now, at x = $\pi$ $L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi - h) - f(\pi )} \over { - h}} = 0$

$R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi + h) - f(\pi )} \over h} = 1 - {{(\pi + h)\sin \,h - 1} \over h} = - \pi $

$\therefore$ f(x) is not differentiable at x = $\pi$

$\therefore$ (m, n) = (1, 0)

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

Let ${\cos ^{ - 1}}x = t$

$ \Rightarrow x = \cos t$

When $x \to {1 \over {\sqrt 2 }}$, then $t \to {\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right) \to {\pi \over 4}$

$\therefore$ $\mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - \tan (t)}}$

$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - {{\sin t} \over {\cos t}}}}$

$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{(\sin t - \cos t)(\cos t)} \over {(\cos t - \sin t)}}$

$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} - \cos t$

$ = - \mathop {\lim }\limits_{t \to {\pi \over 4}} \cos t$

$ = - {1 \over {\sqrt 2 }}$

$\because$ gof is differentiable at x = 0

So R.H.D = L.H.D

${d \over {dx}}(4{e^x} + {k_2}) = {d \over {dx}}\left( {{{( - |x + 3|)}^2} - {k_1}|x + 3|} \right)$

$ \Rightarrow 4 = 6 - {k_1} \Rightarrow {k_1} = 2$

Also $f(f({0^ + })) = g(f({0^ - }))$

$ \Rightarrow 4 + {k_2} = 9 - 3{k_1} \Rightarrow {k_2} = - 1$

Now $g(f( - 4)) + g(f(4))$

$ = g( - 1) + g({e^4}) = (1 - {k_1}) + (4{e^4} + {k_2})$

$ = 4{e^4} - 2$

$ = 2(2{e^4} - 1)$

$\mathop {\lim }\limits_{x \to {\pi \over 2}} {\tan ^2}x\left\{ {\sqrt {2{{\sin }^2}x + 3\sin x + 4} - \sqrt {{{\sin }^2}x + 6\sin x + 2} } \right\}$

$ = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{{{\tan }^2}x({{\sin }^2}x - 3\sin x + 2)} \over {\sqrt {2{{\sin }^2}x + 3\sin x + 4} + \sqrt {{{\sin }^2}x + 6\sin x + 2} }}$

$ = {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(1 - \sin x)(2 - \sin x)} \over {{{\cos }^2}x}}\,.\,{\sin ^2}x$

$ = {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(2 - \sin x){{\sin }^2}x} \over {1 + \sin x}}$

$ = {1 \over {12}}$

Given, $f(x) + f'(x) + f''(x) = {x^5} + 64$ .........(i)

$\Rightarrow f(x)$ is a polynomial in $x$ whose degree is 5.

Let $f(x) = {x^5} + a{x^4} + b{x^3} + c{x^2} + dx + e$

$f'(x) = 5{x^4} + 4a{x^3} + 3b{x^2} + 2cx + d$

$f''(x) = 20{x^3} + 12a{x^2} + 6bx + 2c$

On substituting the value of $f(x), f^{\prime}(x)$ and $f^{\prime \prime}(x)$ in Eq. (i), we get

${x^5}+(a + 5){x^4} + (b + 4a + 20){x^3} + (c + 3b + 12a){x^2} + (d + 2c + 6b)x + e + d + 2c = {x^5} + 64$

Now, equating the coefficient, we get

$ \Rightarrow a + 5 = 0$

$b + 4a + 20 = 0$

$c + 3b + 12a = 0$

$d + 2c + 6b = 0$

$e + d + 2c = 64$

$\therefore$ $a = - 5,\,b = 0,\,c = 60,\,d = - 120,\,e = 64$

$\therefore$ $f(x) = {x^5} - 5{x^4} + 60{x^2} - 120x + 64$

Now, $\mathop {\lim }\limits_{x \to 1} {{{x^5} - 5{x^4} + 60{x^2} - 120x + 64} \over {x - 1}}$ is (${0 \over 0}$ form)

By L' Hospital rule

$\mathop {\lim }\limits_{x \to 1} {{5{x^4} - 20{x^3} + 120x - 120} \over 1}$

$ = - 15$

$f(x) = \left\{ {\matrix{ {{{\sin (x - [x])} \over {x[x]}}} & , & {x \in ( - 2, - 1)} \cr {\max \{ 2x,3[|x|]\} } & , & {|x| < 1} \cr 1 & , & {otherwise} \cr } } \right.$

$f(x) = \left\{ {\matrix{ {{{\sin (x + 2)} \over {x + 2}}} & , & {x \in ( - 2, - 1)} \cr 0 & , & {x \in ( - 1,0]} \cr {2x} & , & {x \in (0,1)} \cr 1 & , & {otherwise} \cr } } \right.$

It clearly shows that f(x) is discontinuous

At x = $-$1, 1 also non differentiable

and at $x = 0$, $L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 - h) - f(0)} \over { - h}} = 0$

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

$\therefore$ f(x) is not differentiable at x = 0

$\therefore$ m = 2, n = 3

f(0) = 0, f(1) = 1 and f(2) = 2

Let h(x) = f(x) $-$ x

Clearly h(x) is continuous and twice differentiable on (0, 2)

Also, h(0) = h(1) = h(2) = 0

$\therefore$ h(x) satisfies all the condition of Rolle's theorem.

$\therefore$ there exist C₁ $\in$(0, 1) such that h'(c₁) = 0

$\Rightarrow$ f'(₁) $-$ 1 = 0 $\Rightarrow$ f'(c₁) = 1

also there exist c₂ $\in$(1, 2) such that h'(c₂) = 0

$\Rightarrow$ f'(c₂) = 1

Now, using Rolle's theorem on [c₁, c₂] for f'(x)

We have f''(c) = 0, c$\in$(c₁, c₂)

Hence, f''(x) = 0 for some x$\in$(0, 2).