Definite Integration

Let

$ I=\int_{0}^{2}\frac{1}{3^x+3}\,dx $

We will simplify the integrand first.

Since

$ 3^x+3=3(3^{x-1}+1), $

this form is not immediately very helpful. So let us use the substitution

$ 3^x=t $

Then,

$ t=3^x \quad \Rightarrow \quad \frac{dt}{dx}=3^x\ln 3=t\ln 3 $

So,

$ dx=\frac{dt}{t\ln 3} $

Now change the limits:

• When $x=0$, $t=3^0=1$

• When $x=2$, $t=3^2=9$

Thus,

$ I=\int_{1}^{9}\frac{1}{t+3}\cdot \frac{1}{t\ln 3}\,dt $

$ I=\frac{1}{\ln 3}\int_{1}^{9}\frac{1}{t(t+3)}\,dt $

Now use partial fractions:

$ \frac{1}{t(t+3)}=\frac{A}{t}+\frac{B}{t+3} $

So,

$ 1=A(t+3)+Bt $

$ 1=(A+B)t+3A $

Comparing coefficients:

$ A+B=0,\qquad 3A=1 $

Hence,

$ A=\frac{1}{3},\qquad B=-\frac{1}{3} $

Therefore,

$ \frac{1}{t(t+3)}=\frac{1}{3}\left(\frac{1}{t}-\frac{1}{t+3}\right) $

So,

$ I=\frac{1}{3\ln 3}\int_{1}^{9}\left(\frac{1}{t}-\frac{1}{t+3}\right)\,dt $

Now integrate:

$ I=\frac{1}{3\ln 3}\left[\ln t-\ln(t+3)\right]_{1}^{9} $

$ I=\frac{1}{3\ln 3}\left[\ln\frac{t}{t+3}\right]_{1}^{9} $

Substitute the limits:

$ I=\frac{1}{3\ln 3}\left(\ln\frac{9}{12}-\ln\frac{1}{4}\right) $

$ I=\frac{1}{3\ln 3}\left(\ln\frac{3}{4}-\ln\frac{1}{4}\right) $

$ I=\frac{1}{3\ln 3}\ln 3 $

$ I=\frac{1}{3} $

So, the value of the definite integral is

$ \boxed{\frac{1}{3}} $

Hence, the correct option is:

$ \boxed{\text{Option B}} $

Given,

$\int_1^e {{{{{\left( {\log _e^x} \right)}^{1/2}}} \over {x{{\left( {a - {{\left( {\log _e^x} \right)}^{3/2}}} \right)}^2}}} = 1} $

Let $\log _e^x = t$

$ \Rightarrow x = {e^t}$

$ \Rightarrow dx = {e^t}dt$

When $x = 1$ then $t = \log _e^1 = 0$

and when $x = e$ then $t = \log _e^e = 1$

$ \Rightarrow \int_0^1 {{{\sqrt t \,.\,{e^t}dt} \over {{e^t}{{(a - t\sqrt t )}^2}}} = 1} $

$ \Rightarrow \int_0^1 {{{\sqrt t \,dt} \over {{{(a - t\sqrt t )}^2}}} = 1} $

Let ${t^{1/2}} = z$

$ \Rightarrow {1 \over {2\sqrt t }}dt = dz$

$ \Rightarrow dt = 2\sqrt t \,dz \Rightarrow dt = 2z\,dz$

When $t = 0$, then $z = 0$

When $t = 1$, then $z = {1^{1/2}} = 1$

$\therefore$ $\int_0^1 {{{z \times 2z\,dz} \over {{{(a - {z^3})}^2}}} = 1} $

$ \Rightarrow \int_0^1 {{{2{z^2}\,dz} \over {{{(a - {z^3})}^2}}} = 1} $

Now let $a - {z^3} = p$

$ \Rightarrow - 3{z^2}\,dz = dp$

When $z = 0$ then $p = a - 0 = a$

When $z = 1$ then $p = a - {1^3} = a - 1$

$\therefore$ $\int_a^{a - 1} {{{2\,.\,\left( { - {{dp} \over 3}} \right)} \over {{p^2}}} = 1} $

$ \Rightarrow - {2 \over 3}\int_a^{a - 1} {{{dp} \over {{p^2}}} = 1} $

$ \Rightarrow - {2 \over 3}\left[ { - {1 \over p}} \right]_a^{a - 1} = 1$

$ \Rightarrow - {2 \over 3}\left[ { - {1 \over {a - 1}} + {1 \over a}} \right] = 1$

$ \Rightarrow - {2 \over 3}\left[ {{{ - a + a - 1} \over {a(a - 1)}}} \right] = 1$

$ \Rightarrow {2 \over 3}\left( {{1 \over {a(a - 1)}}} \right) = 1$

$ \Rightarrow 3{a^2} - 3a = 2$

$ \Rightarrow 3{a^2} - 3a - 2 = 0$

$ \Rightarrow a = {{3\, \pm \,\sqrt {9 - 4(3)( - 2)} } \over {2\,.\,3}}$

$\therefore$ $a = {{3\, \pm \,\sqrt {33} } \over 6},\,{{3 - \sqrt {33} } \over 6}$

Let $I = \int_{ - \pi /2}^{\pi /2} {{{{x^2}\cos x} \over {1 + {e^x}}}dx} $ ...... (i) [$\because$ $\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx} } $]

$ \Rightarrow I = \int_{ - \pi /2}^{\pi /2} {{{{x^2}\cos ( - x)} \over {1 + {e^{ - x}}}}dx} $ ...... (ii)

On adding Eqs. (i) and (ii), we get

$2I = \int_{ - \pi /2}^{\pi /2} {{x^2}\cos x\left[ {{1 \over {1 + {e^x}}} + {1 \over {1 + {e^{ - x}}}}} \right]dx} $

$ = \int_{ - \pi /2}^{\pi /2} {{x^2}\cos x\,.\,(1)dx} $ [$\because$ $\int_{ - a}^a {f(x)dx} $ $ = 2\int_0^a {f(x)dx} $, when $f( - x) = f(x)$]

$ \Rightarrow 2I = 2\int_0^{\pi /2} {{x^2}\cos x\,dx} $

Using integration by parts, we get

$2I = 2[{x^2}(\sin x) - (2x)( - \cos x) + (2)( - \sin x)]_0^{\pi /2}$

$ \Rightarrow 2I = 2\left[ {{{{\pi ^2}} \over 4} - 2} \right]$

$\therefore$ $I = {{{\pi ^2}} \over 4} - 2$

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

By taking logarithm we have

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

By definite integration, we can write it as

$\ln f(x) = x\int\limits_0^1 {\ln (1 + xy)dy - x\int\limits_0^1 {\ln (1 + {x^2}{y^2})dy} } $

Let xy = u

Then, $\ln f(x) = \int\limits_0^x {\ln (1 + u)du - \int\limits_0^x {\ln (1 + {u^2})du} } $

Using Newton-Leibnitz formula, we get

${1 \over {f(x)}}\,.\,f'(x) = log\left( {{{1 + x} \over {1 + {x^2}}}} \right)$ ...... (i)

Here, at x = 1,

${{f'(1)} \over {f(1)}} = \log (1) = 0$

$\therefore$ $f'(1) = 0$

Now, sign scheme of f'(x) is shown below

$\therefore$ At x = 1, function attains maximum. Since, f(x) increases on (0, 1).

$\therefore$ f(1) > f(1/2)

$\therefore$ Option (a) is incorrect.

f(1/3) < f(2/3)

$\therefore$ Option (b) is correct.

Also, f'(x) < 0, when x > 1

$\Rightarrow$ f'(2) < 0

We have, $f'(x) = {{192{x^3}} \over {2 + {{\sin }^4}\pi x}}$

Given, $f\left( {{1 \over 2}} \right) = 0$ and $m \le \int\limits_{1/2}^1 {f(x)dx \le m} $

$\therefore$ f(x) is increasing in $\left( {{1 \over 2},1} \right)$.

$\therefore$ $f'{(x)_{\max }} = {{192} \over {24}} = 96$

$ \Rightarrow 96 = {{f(1) - f(1/2)} \over {1/2}} \Rightarrow f(1) = 96 \times {1 \over 2} = 48$

$M = {1 \over 2} \times {1 \over 2} \times 48 = 12$

$f'{(x)_{\min .}} = {{\left( {{{192} \over 8}} \right)} \over 3} = {{192} \over {24}}$

$ \Rightarrow {{192} \over {24}} = {{f(1) - 0} \over {(1/2)}} \Rightarrow f(1) = {{192} \over {24}} \times {1 \over 2}$

$\therefore$ $m = {1 \over 2} \times {1 \over 2} \times {{192} \over {24}} \times {1 \over 2} = 1$

$f(x) = 7{\tan ^8}x + 7{\tan ^6}x - 3{\tan ^4}x - 3{\tan ^2}x\,\forall x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right)$

$ = 7{\tan ^6}x\,.\,{\sec ^2}x - 3{\tan ^2}x\,.\,{\sec ^2}x$

$ = (7{\tan ^6}x - 3{\tan ^2}x)\,.\,{\sec ^2}x$

$ \Rightarrow \int\limits_0^{\pi /4} {f(x)dx = \int\limits_0^{\pi /4} {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}dx = \int\limits_0^1 {(7 + 603{t^2})dt = [{t^7} - {t^3}]_0^1 = 0} } } $

Also, $I = \int\limits_0^{\pi /4} {xf(x)dx} $

$ = \left| {x\,.\,\int {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}xdx} } \right|_0^{\pi /4} - \int\limits_0^{\pi /4} {1\,.\,\int {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}x\,dx} } $

$ = \left| {x\,.\,({{\tan }^7}x - {{\tan }^3}x} \right|_0^{\pi /4} - \int\limits_0^{\pi /4} {({{\tan }^7}x - {{\tan }^3}x)dx} $

$ = 0 - \int\limits_0^{\pi /4} {{{\tan }^3}x({{\tan }^4}x - 1)dx} $

$ = - \int\limits_0^{\pi /4} {{{\tan }^3}x({{\tan }^2}x - 1)({{\sec }^2}x)dx} $

$ = - \int\limits_0^1 {({t^5} - {t^3})dt} $

$ = - \left[ {{{{t^6}} \over 6} - {{{t^4}} \over 4}} \right]_0^1 = \left[ {{1 \over 4} - {1 \over 6}} \right] = {1 \over {12}}$

Let ${I_1} = \int_0^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $

$ = \int_0^\pi {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_\pi ^{2\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_{2\pi }^{3\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_{3\pi }^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} } } } $

$\therefore$ ${I_1} = {I_2} + {I_3} + {I_4} + {I_5}$ ...... (i)

Now, ${I_3} = \int_\pi ^{2\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $

Put $t = \pi + t \Rightarrow dt = dt$

$\therefore$ ${I_3} = \int_0^\pi {{e^{\pi + t}}.\,({{\sin }^6}at + {{\cos }^6}at)dt} $

$ = {e^t}\,.\,{I_2}$ ..... (ii)

Now, ${I_4} = \int_{2\pi }^{3\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $

Put $t = 2\pi + t \Rightarrow dt = dt$

$\therefore$ ${I_4} = \int_0^\pi {{e^{t + 2\pi }}({{\sin }^6}at + {{\cos }^6}at)dt} $

$ = {e^{2\pi }}.{I_2}$ ...... (iii)

and ${I_5} = \int_{3\pi }^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $

Put $t = 3\pi + t$

$\therefore$ ${I_5} = \int_0^\pi {{e^{3\pi + t}}({{\sin }^6}at + {{\cos }^6}at)dt} $

$ = {e^{3\pi }}.\,{I_2}$ ...... (iv)

From Eqs. (i), (ii), (iii) and (iv), we get

${I_1} = {I_2} + {e^\pi }.\,{I_2} + {e^{2\pi }}.\,{I_2} + {e^{3\pi }}.\,{I_2}$

$ = (1 + {e^\pi } + {e^{2\pi }} + {e^{3\pi }}){I_2}$

$\therefore$ $L = {{\int_0^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} } \over {\int_0^\pi {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} }}$

$ = (1 + {e^\pi } + {e^{2\pi }} + {e^{3\pi }})$

$ = {{1.\,({e^{4\pi }} - 1)} \over {{e^\pi } - 1}}$ for $a \in R$

(P) Let $p(x)=a x^2+b x+c$

Given, $p(0)=0$

$\Rightarrow \quad c=0$

Also given $\int_0^1 f(x) d x=1$

$\begin{array}{lr} \Rightarrow & \int_0^1\left(a x^2+b x\right) d x=1 \\ \Rightarrow & {\left[\frac{a x^3}{3}+\frac{b x^2}{2}\right]_0^1=1} \\ \Rightarrow & \frac{a}{3}+\frac{b}{2}=1 \\ \Rightarrow & 2 a+3 b=6 \end{array}$

Here, $a$ and $b$ are non negative integer

$\therefore(a, b)=(3,0),(0,2)$

Hence, possible equation of $p(x)$ are $3 x^2$ and $2 x$.

So, P Match with 2.

(Q) Given $f(x)=\sin x^2+\cos x^2$

Differentiate $f(x)$ w.r.t. $x$

$\begin{aligned} & \Rightarrow f^{\prime}(x)=2 x \cdot \cos x^2-2 x \cdot \sin x^2 \\ & \Rightarrow f^{\prime}(x)=2 x\left(\cos x^2-\sin x^2\right) \end{aligned}$

Apply $f^{\prime}(x)=0$

$\begin{aligned} & \Rightarrow x=0, \tan x^2=1 \\ & \Rightarrow x=0, \pm \sqrt{\frac{\pi}{4}}, \pm \sqrt{\frac{5 \pi}{4}}, \pm \sqrt{\frac{9 \pi}{4}}, \pm \sqrt{\frac{13 \pi}{4}} \end{aligned}$

$\begin{aligned} & \text { Differentiate } f^{\prime}(x) \text { w.r.t. } x \\ & \Rightarrow f^{\prime \prime}(x)=2\left(\cos x^2-\sin x^2\right) \\ & +2 x\left(-2 x \sin x^2-2 x \cos x^2\right) \\ & \Rightarrow f^{\prime \prime}(x)=2\left(\cos x^2-\sin x^2\right) \\ & -4 x^2\left(\sin x^2+\cos x^2\right) \\ & \Rightarrow f^{\prime \prime}(x)<0 \text { at } x= \pm \sqrt{\frac{\pi}{4}}, \pm \sqrt{\frac{9 \pi}{4}} \\ \end{aligned}$

So, $f(x)$ is maximum at four points. Hence, Q match with 3. (R) Let $\mathrm{I}=\int_{-2}^2 \frac{3 x^2}{1+e^x} d x\quad \text{... (i)}$

Apply the property $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

$\begin{aligned} & \Rightarrow \quad \mathrm{I}=\int_{-2}^2 \frac{3(-x)^2}{1+e^{-x}} d x \\ & \Rightarrow \quad \mathrm{I}=\int_{-2}^2 \frac{3 x^2 e^x}{e^x+1} d x \quad \text{.... (ii)} \end{aligned}$

Add equation (i) and (ii)

$\begin{aligned} & \Rightarrow \quad 2 \mathrm{I}=\int_{-2}^2 \frac{3 x^2}{e^x+1}\left[1+e^x\right] d x \\ & \Rightarrow \quad \mathrm{I}=\frac{3}{2} \int_{-2}^2 x^2 \cdot d x \\ & \Rightarrow \quad \mathrm{I}=\frac{3}{2}\left[\frac{x^3}{3}\right]_{-2}^2 \\ & \Rightarrow \quad \mathrm{I}=16 \end{aligned}$

(S) We know the property

$\int_{-a}^a f(x) d x=\left\{\begin{array}{l} 2 \int_0^a f(x) d x, \text { when } f(x) \text { is } \\ \text { an even function } \\ 0, \text { when } f(x) \text { is an odd function } \end{array}\right.$

$\begin{aligned} & \text { Let } g(x)=\cos x \cdot \log \left(\frac{1+x}{1-x}\right) \\ & \Rightarrow g(-x)=\cos (-x) \log \left(\frac{1-x}{1+x}\right) \\ & \Rightarrow g(-x)=\cos x \cdot \log \left(\frac{1+x}{1-x}\right)^{-1} \\ & \Rightarrow g(-x)=-\cos x \cdot \log \left(\frac{1+x}{1-x}\right) \end{aligned}$

$\Rightarrow g(-x)=-g(x)$

Hence, $g(x)$ is an odd function

So, $\int_{\frac{-1}{2}}^{\frac{1}{2}} \cos x \cdot \log \left(\frac{1+x}{1-x}\right) d x=0$

$\Rightarrow \frac{\frac{\int_{-1}^2}{\frac{1}{2}} \cos x \cdot \log \left(\frac{1+x}{1-x}\right) d x}{\int_0^{\frac{1}{2}} \cos x \cdot \log \left(\frac{1+x}{1-x}\right) d x}=0$

Hence, $S$ match with 4.

Hint:

(i) Recall the property

$\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

(ii) Recall the property

$ \int_{-a}^a f(x) d x=\left\{\begin{array}{l} 2 \int_0^a f(x) d x, \text { when } f(x) \text { is an even function}\\ 0, \text { when } f(x) \text { is an oden function } \end{array}\right.$

(iii) At the point of local maxima, first derivative of the function is zero and second derivative of the function is negative.

$\text { Let } \mathrm{I}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(2 \operatorname{cosec} x)^{17} \cdot d x \quad \text{... (i)}$

Let, $\operatorname{cosec} x+\cot x=e^u\quad \text{... (ii)}$

We know that $\quad \operatorname{cosec}^2 x-\cot ^2 x=1$

$\begin{aligned} \Rightarrow \quad& (\operatorname{cosec} x+\cot x)(\operatorname{cosec} x-\cot x) =1 \\ \Rightarrow \quad& e^u(\operatorname{cosec} x-\cot x) =1 \\ \Rightarrow \quad& \operatorname{cosec} x-\cot x =e^{-u} \quad \text{... (iii)} \end{aligned}$

From equation (ii) and (iii)

$\begin{array}{rlrl} \Rightarrow & 2 \cot x =e^{+u}-e^{-u} \quad \text{... (iv)}\\ \text { And } & 2 \operatorname{cosec} x =e^u+e^{-u} \quad \text{... (v)} \end{array}$

Differentiate the equation $(v)$ w.r.t. $x$

$\begin{array}{llrl} \Rightarrow & -2 \operatorname{cosec} x \cdot \cot x =\left(e^u-e^{-u}\right) \frac{d u}{d x} \\ \Rightarrow & -\left(e^u+e^{-u}\right) \cdot\left(\frac{e^u-e^{-u}}{2}\right) =\left(e^u-e^{-u}\right) \frac{d u}{d x} \\ \Rightarrow & d x =-\frac{2 d u}{e^u+e^{-u}} \end{array}$

Now, the equation (i) can be written as

$\begin{aligned} & \Rightarrow \mathrm{I}=\int_{\ln (\sqrt{2}+1)}^{\ln (1+0)}\left(e^u+e^{-u}\right)^{17}\left(-\frac{2 d u}{e^u+e^{-u}}\right) \\ & \Rightarrow \mathrm{I}=-\int_{\ln (\sqrt{2}+1)}^0 2\left(e^u+e^{-u}\right)^{16} d u \\ & \Rightarrow \mathrm{I}=\int_0^{\ln (\sqrt{2}+1)} 2\left(e^u+e^{-u}\right)^{16} d u \end{aligned}$

Hint:

(i) Recall $\operatorname{cosec}^2 x-\cot ^2 x=1$

(ii) If $\operatorname{cosec} x+\cot x=e^u$, then $\operatorname{cosec} x-\cot x=e^{-u}, 2 \operatorname{cosec} x=e^u+e^{-u}, 2 \cot x=e^u-e^{-u}$.

Given, $g(a)=\lim _\limits{h \rightarrow 0^{+}} \int_h^{1-h} t^{-a}(1-t)^{a-1} d t$ is differentiable for all $a \in(0,1)$.

$\therefore \quad g^{\prime}\left(\frac{1}{2}\right)=\lim _\limits{\beta \rightarrow 0} \frac{g\left(\frac{1}{2}+\beta\right)-g\left(\frac{1}{2}\right)}{\beta}$

$\begin{aligned} & \therefore g^{\prime}\left(\frac{1}{2}\right)=\lim _{\beta \rightarrow 0} \frac{g\left(\frac{1}{2}+\beta\right)-g\left(\frac{1}{2}\right)}{\beta} \\ & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=\lim _{\beta \rightarrow 0} \frac{1}{\beta} \\ & {\left[\lim _{h \rightarrow 0^{+}} \int_h^{1-h}\left(t^{-\left(\frac{1}{2}+\beta\right)}(1-t)^{\frac{1}{2}+\beta-1}-t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}}\right) d t\right]} \end{aligned}$

$\begin{aligned} \Rightarrow g^{\prime} & \left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}} \\ & {\left[\int_h^{1-h} t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}} \cdot\left(\lim _{\beta \rightarrow 0} \frac{t^{-\beta} \cdot(1-t)^\beta-1}{\beta}\right)\right] } \end{aligned}$

$\begin{aligned} \Rightarrow g^{\prime}\left(\frac{1}{2}\right) & =\lim _{h \rightarrow 0^{+}} \\ & {\left[\int_h^{1-h} t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}} d t\left(\lim _{\beta \rightarrow 0} \frac{\left(\frac{1-t}{t}\right)^\beta-1}{\beta}\right)\right] } \end{aligned}$

$\Rightarrow g^{\prime}\left(\frac{1}{2}\right)=\lim _\limits{h \rightarrow 0} \int_h^{1-h} t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}} d t \log _e\left(\frac{1-t}{t}\right) \quad \text{... (i)}$

Apply the property $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

$\begin{aligned} & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \int_h^{1-h}(1-t)^{\frac{-1}{2}} t^{\frac{-1}{2}} d t \cdot \log _e\left(\frac{t}{1-t}\right) \\ & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \int_h^{1-h} t^{\frac{-1}{2}} \cdot(1-t)^{\frac{-1}{2}} d t \cdot \log _e\left(\frac{1-t}{t}\right)^{-1} \\ & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=-\lim _{h \rightarrow 0} \int_h^{1-h} t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}} d t \cdot \log _e\left(\frac{1-t}{t}\right) \\ & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=-g^{\prime}\left(\frac{1}{2}\right)\\ \begin{aligned} & \Rightarrow 2 g^{\prime}\left(\frac{1}{2}\right)=0 \\ & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=0 \end{aligned} \end{aligned}$

Hint:

(i) If $f(x)$ is differentiable at $x=a$, then

$f^{\prime}(a)=\lim _{\beta \rightarrow 0} \frac{f(a+\beta)-f(a)}{\beta}$

(ii) $\lim _\limits{x \rightarrow 0} \frac{a^x-1}{x}=\ln a$

(iii) Recall the property

$\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

Given, $g(a)=\lim _\limits{h \rightarrow 0^{+}} \int_h^{1-h} t^{-a}(1-t)^{a-1} d t$ is differentiable for $a \in(0,1)$.

$\begin{aligned} & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}} \int_h^{1-h} t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}} d t \\ & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}} \int_h^{1-h} \frac{1}{\sqrt{t-t^2}} d t \\ & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}} \int_h^{1-h} \frac{d t}{\sqrt{\left(\frac{1}{2}\right)^2-\left[t^2-2 \cdot \frac{1}{2} t+\left(\frac{1}{2}\right)^2\right]}} \\ & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}}^{1-h} \frac{d t}{\sqrt{\left(\frac{1}{2}\right)^2-\left(t-\frac{1}{2}\right)^2}} \\ & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}}\left[\sin ^{-1}\left(\frac{\left(t-\frac{1}{2}\right)}{\frac{1}{2}}\right)\right]_h^{1-h} \\ & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}}[\sin (1-2 h)-\sin (2 h-1)] \\ & \Rightarrow g\left(\frac{1}{2}\right)=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi \end{aligned}$

Hint:

(i) For the integral of $\int \frac{d x}{\sqrt{a x^2+b x+c}}$, Convert $a x^2+b x+c$ into perfect square like $a x^2+b x+c=a\left[\left(x+\frac{b}{2 a}\right)^2 \pm \lambda^2\right]$.

(ii) $\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C$

Given, $f(x)$ be a positive, non-constant and differentiable function and $f\left(\frac{1}{2}\right)=1$.

Also given $f^{\prime}(x)<2 f(x)$

$\Rightarrow f^{\prime}(x)-2 f(x)<0$

Multiply both side by $e^{-2 x}$

$\begin{aligned} & \Rightarrow e^{-2 x} f^{\prime}(x)-2 e^{-2 x} f(x)<0 \\ & \Rightarrow d\left(e^{-2 x} f(x)\right)<0 \end{aligned}$

Let $g(x)=e^{-2 x} f(x) \forall x \in\left[\frac{1}{2}, 1\right]$

$\Rightarrow d(g(x))<0$

Hence, $g(x)$ is a decreasing function in the Given, domain $x \in\left[\frac{1}{2}, 1\right]$

$\begin{aligned} & \because \quad g(x)< g\left(\frac{1}{2}\right) \\ & \Rightarrow e^{-2 x} f(x) < e^{-1} \cdot f\left(\frac{1}{2}\right) \\ & \Rightarrow f(x) < e^{2 x-1} \cdot 1 \\ & \therefore \quad f(x) \text { is always positive } \\ & \therefore \quad 0 < f(x) < e^{2 x-1} \\ & \Rightarrow 0 < \int_{\frac{1}{2}}^1 f(x) d x < \int_{\frac{1}{2}}^1 e^{2 x-1} d x \\ & \Rightarrow 0 < \int_{\frac{1}{2}}^1 f(x) d x < \left[\frac{e^{2 x-1}}{2}\right]_{\frac{1}{2}}^1 \\ & \Rightarrow 0 < \int_{\frac{1}{2}}^1 f(x) d x < \frac{e-1}{2} \\ & \Rightarrow \int_{\frac{1}{2}}^1 f(x) d x \in\left(0, \frac{e-1}{2}\right) \end{aligned}$

Hints:

$\Rightarrow$ (i) if $f(x)$ is a decreasing function for $x \in[a, b]$, then $f(b) < f(x) < f(a)$

(ii) Recall that $e^{-2 x} f^{\prime}(x)-2 e^{-2 x} f(x)=d\left(e^{-2 x} f(x)\right)$

We know the property

$\begin{aligned} & \int_{-a}^a f(x) d x \\ & =\left\{\begin{array}{cc} 2 \int_0^a f(x) d x, & \text { when } f(x) \text { is an even function } \\ 0, & \text { when } f(x) \text { is an odd function } \end{array}\right. \end{aligned}$

Let, $\mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^2+\ln \left(\frac{\pi+x}{\pi-x}\right)\right) \cos x \cdot d x$

$\Rightarrow \mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot \cos x d x+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x \cdot \ln \left(\frac{\pi+x}{\pi-x}\right) d x$

Here, $x^2 \cos x$ is an even function and $\cos x \cdot \ln \left(\frac{\pi+x}{\pi-x}\right)$ is an odd function.

$\begin{aligned} & \therefore \mathrm{I}=2 \int_0^{\frac{\pi}{2}} x^2 \cos x d x+0 \\ & \Rightarrow \mathrm{I}=2\left[\left(\sin x \cdot x^2\right)_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} 2 x \sin x \cdot d x\right] \\ & \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4 \int_0^{\frac{\pi}{2}} x \sin x \cdot d x \\ & \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4\left[(x(-\cos x))_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} 1 \cdot(-\cos x) d x\right] \\ & \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4 \int_0^{\frac{\pi}{2}} \cos x d x \\ & \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4[\sin x]_0^{\frac{\pi}{2}} \\ & \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4 \end{aligned}$

${x^2} = t \Rightarrow 2x\,dx = dt$

$I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin t} \over {\sin t + \sin (\ln 6 - t)}}dt} $ and $I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin (\ln 6 - t)} \over {\sin (\ln 6 - t) + \sin t}}dt} $

$2I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {1dt \Rightarrow I = {1 \over 4}\ln {3 \over 2}} $.

$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^3}}}\int_0^x {{{t\log (1 + t)} \over {4 + {t^4}}}dt} $

Using L' Hospital's rule,

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

$ = \mathop {\lim }\limits_{x \to 0} {{\log (1 + x)} \over {3x}}\,.\,{1 \over {4 + {x^4}}}$

$ = {1 \over 3}\,.\,{1 \over 4} = {1 \over {12}}$ [using, $\mathop {\lim }\limits_{x \to 0} {{\log (1 + x)} \over x} = 1$]

$\int\limits_0^1 {{{{x^4}{{(1 - x)}^4}} \over {1 + {x^2}}}dx = \int\limits_0^1 {{{{x^4}{{\{ (1 + {x^2}) - 2x\} }^2}} \over {1 + {x^2}}}dx} } $

$ = \int\limits_0^1 {{x^4}{{{{(1 + {x^2})}^2} - 4x(1 + {x^2}) + 4{x^2}} \over {1 + {x^2}}}dx} $

$ = \int\limits_0^1 {{x^4}\left[ {1 + {x^2} - 4x + {{4\{ 1 + {x^2} - 1\} } \over {1 + {x^2}}}} \right]dx} $

$ = \int\limits_0^1 {{x^4}\left[ {1 + {x^2} - 4x + 4 - {4 \over {1 + {x^2}}}} \right]dx} $

$ = \int\limits_0^1 {\left[ {{x^6} - 4{x^5} + 5{x^4} - 4{{{x^4} - 1 + 1} \over {1 + {x^2}}}} \right]dx} $

$ = \int\limits_0^1 {({x^6} - 4{x^5} + 5{x^4} - 4{x^2} + 4)dx - 4\int\limits_0^1 {{{dx} \over {1 + {x^2}}}} } $

$ = \left[ {{{{x^7}} \over 7} - {{2{x^6}} \over 3} + {x^5} - {{4{x^3}} \over 3} + 4x} \right]_0^1 - 4[{\tan ^{ - 1}}x]_0^1$

$ = \left[ {{1 \over 7} - {2 \over 3} + 1 - {4 \over 3} + 4} \right] - \pi = {{22} \over 7} - \pi $

We have,

${e^{ - x}}f(x) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,dt\,x \in ( - 1,1)} $

On differentiating w.r.t. x, we get

${e^{ - x}}(f'(x) - f(x)) = \sqrt {{x^4} + 1} $

$ \Rightarrow f'(x) = f(x) + \sqrt {{x^4} + 1} \,{e^x}$

$\because$ ${f^{ - 1}}$ is the inverse of f

$\therefore$ ${f^{ - 1}}(f(x)) = x$

$ \Rightarrow {f^{ - 1'}}(f(x))f'(x) = 1$

$ \Rightarrow {f^{ - 1'}}(f(x)) = {1 \over {f'(x)}}$

$ \Rightarrow {f^{ - 1'}}(f(x)) = {1 \over {f(x) + \sqrt {{x^4} + 1} \,{e^x}}}$

At $x = 0$, $f(x) = 2$

${f^{ - 1'}}(2) = {1 \over {2 + 1}} = {1 \over 3}$

(A) We have the integral

${I_n} = \int\limits_{ - \pi }^\pi {{{\sin nx} \over {(1 + {\pi ^x})\sin x}}dx} $

$ = \int\limits_0^\pi {\left( {{{\sin nx} \over {(1 + {\pi ^x})\sin x}} + {{{\pi ^x}\sin nx} \over {(1 + {\pi ^x})\sin x}}} \right)dx = \int\limits_0^\pi {{{\sin nx} \over {\sin x}}} } $

Now, ${I_{n + 2}} - {I_n} = \int\limits_0^\pi {{{\sin (n + 2)x - \sin nx} \over {\sin x}}dx} $

(B) Since ${I_3} = {I_5} = \,...\, = {I_{21}}$, we have

$\sum\limits_{m = 1}^{10} {{I_{2m + 1}} = 10{I_3} = 10\int\limits_0^\pi {{{\sin 3x} \over {\sin x}}dx = 10\int\limits_0^\pi {(3 - 4{{\sin }^2}x)dx} } } $

$ = 10[3x - 2x + \sin 2x]_0^\pi = 2\pi $

(C) Since ${I_2} = {I_4} = \,...\, = {I_{20}}$, we have

$\sum\limits_{m = 1}^{10} {{I_{2m}} = 10\int\limits_0^\pi {{{\sin 2x} \over {\sin x}}dx = 20[\sin x]_0^\pi = 0} } $

Given $g'(x) = \int\limits_0^{{e^x}} {{{f'(t)dt} \over {1 + {t^2}}}} $

$g''(x) = {{dg(x)} \over {d({e^x})}}.\,{{d({e^x})} \over {dx}}$

$ = {{f'({e^x})} \over {1 + {e^{2x}}}}\,.\,{e^x}$

$ = {{{e^x}} \over {1 + {e^{2x}}}}\,.\,{{2a({e^{2x}} - 1)} \over {{{({e^{2x}} + a{e^x} + 1)}^2}}} > 0$

when $x \in (0,\infty )$

$g'(x)$ < 0 when $x \in ( - \infty ,0)$

$y' = {1 \over {3[1 - f{{(x)}^2}]}}$

Clearly $f(x)$ is an odd function then $g'(x)$ is an even function

So,

$\int\limits_{ - 1}^1 {g'(x) = 2\int\limits_0^1 {g'(x)dx} } $

$ = 2[g(x)]_0^1 = 2[g(1) - g(0)]$

$ = 2g(1)$

$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\int\limits_2^{{{\sec }^2}x} {f(t)\,dt} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$ (${0 \over 0}$ form)

$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{f({{\sec }^2}x)2\sec x\sec x\tan x} \over {2x}}$

(Applying L' Hospital Rule)

$ = {{2f(2)} \over {{\pi \over 4}}} = {{8f(2)} \over \pi }$

(A) $\int\limits_{ - 1}^1 {{{dx} \over {1 + {x^2}}} = {{\tan }^{ - 1}}x]_{ - 1}^1} $

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

$ = {\pi \over 4} - \left( { - {\pi \over 4}} \right)$

$ = {{2\pi } \over 4} = {\pi \over 2}$

(B) $\int\limits_0^1 {{{dx} \over {\sqrt {1 - {x^2}} }} = {{\sin }^{ - 1}}x]_0^1} $

$ = {\sin ^{ - 1}}(1) - {\sin ^{ - 1}}(0)$

$ = {\pi \over 2} - 0$

$ = {\pi \over 2}$

(C) $\int\limits_2^3 {{{dx} \over {1 - {x^2}}} = {1 \over 2}\log \left| {{{1 + x} \over {1 - x}}} \right|} $

$ = \left[ {\log \left| {{{1 + 3} \over {1 - 3}}} \right| - \log {{1 + 2} \over {1 - 2}}} \right]$

$ = {1 \over 2}\log \left( {{2 \over 3}} \right)$

(D) $\int\limits_1^2 {{{dx} \over {x\sqrt {{x^2} - 1} }}[{{\sec }^{ - 1}}x]_1^2} $

$ = {\sec ^{ - 1}}2 - {\sec ^{ - 1}}1$

$ = {\pi \over 3} - 0$

$ = {\pi \over 3}$