Definite Integration

$z = {1 \over {1 - \cos \theta + 2i\sin \theta }}$

$ = {{2{{\sin }^2}{\theta \over 2} - 2i\sin \theta } \over {{{(1 - \cos \theta )}^2} + 4{{\sin }^2}\theta }}$

$ = {{\sin {\theta \over 2} - 2i\cos {\theta \over 2}} \over {2\sin {\theta \over 2}\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}}$

${\mathop{\rm Re}\nolimits} (z) = {1 \over {2\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}} = {1 \over 5}$

$\sin {{2\theta } \over 2} + 4{\cos ^2}{\theta \over 2} = {5 \over 2}$$1 - {\cos ^2}{\theta \over 2} + 4\cos {\theta \over 2} = {5 \over 2}$

$3{\cos ^2}{\theta \over 2} = {1 \over 2}$

${\theta \over 2} = n\pi \pm {\pi \over 4}$

$\theta = 2n\pi \pm {\pi \over 2}$

$\theta \in (0,\pi )$

$ \therefore $ $\theta = {\pi \over 2}$

$\int\limits_0^{{\pi \over 2}} {\sin \theta d\theta = {{[ - \cos \theta ]}_0}^{{\pi \over 2}}} $

$ = - (0 - 1)$

$ = 1$

$\because$ $f(x) = \ln \left( {x + \sqrt {{x^2} + 1} } \right)$

$\therefore$ $f(x) + f( - x) = \ln \left( {\sqrt {{x^2} + 1} + x} \right) + \ln \left( {\sqrt {{x^2} + 1} - x} \right)$

$\therefore$ $f(x) + f( - x) = 0$ .... (i)

$\because$ $g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dx$

$ = \int_0^{\pi /2} {\left\{ {\cos \left( {{\pi \over 4}t + f(x)} \right) + \cos \left( {{\pi \over 4}t + f( - x)} \right)} \right\}} dx$

$ = \int_0^{\pi /2} {\left\{ {\cos \left( {{{\pi t} \over 4} + f(x)} \right) + \cos \left( {{{\pi t} \over 4} - f(x)} \right)} \right\}dx} $

$g(t) = 2\int_0^{\pi /2} {\cos {{\pi t} \over 4}.\cos (f(x))dx} $

$\therefore$ $g(1) = \sqrt 2 \int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx$ and

$g(0) = 2\int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx$

$\therefore$ $\sqrt 2 g(1) = g(0)$

a > 0

Let $n \le a < n + 1,n \in W$

$a=\matrix{ {[a]} & + & {\{ a\} } \cr \Downarrow & {} & \Downarrow \cr {G.I.F.} & {} & {Fractional\,part} \cr } $

Here [ a ] = n

Now, $\int_0^a {{e^{x - [x]}}} dx = 10e - 9$

$ \Rightarrow \int\limits_0^n {{e^{\{ x\} }}dx} + \int\limits_n^a {{e^{x - [x]}}dx} = 10e - 9$

$\therefore$ $n\int\limits_0^1 {{e^x}dx} + \int\limits_n^a {{e^{x - n}}dx} = 10e - 9$

$ \Rightarrow n(e - 1) + ({e^{a - n}} - 1) = 10e - 9$

$\therefore$ n = 0 and {a} = log_e 2

So, $a = [a] + \{ a\} = (10 + {\log _e}2)$

$\Rightarrow$ Option (2) is correct.

$\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} )dx} $

We know, $\int\limits_{ - a}^a {f(x)dx = 2\int\limits_0^a {f(x)dx} } $

So, $2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} )dx} $

$l = 2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} ).1dx} $

$ \Rightarrow {l \over 2} = {\log _e}(\sqrt {1 - x} + \sqrt {1 + x} ).\,x]_0^1 - \int\limits_0^1 {{{{1 \over {2\sqrt {1 + x} }} - {1 \over {2\sqrt {1 - x} }}} \over {\sqrt {1 - x} + \sqrt {1 + x} }}.\,x\,dx} $

$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{\sqrt {1 - x} - \sqrt {1 + x} } \over {\sqrt {1 - x} + \sqrt {1 + x} }}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $

$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{(1 - x) + (1 + x) - 2\sqrt {1 - {x^2}} } \over {(1 - x) - (1 + x)}}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $

$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {{{2(1 - \sqrt {1 - {x^2}} )} \over { - 2x}}.{x \over {\sqrt {1 - {x^2}} }}dx} $

$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\int\limits_0^1 {\left( {{1 \over {\sqrt {1 - {x^2}} }} - 1} \right)dx} $

$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}[{\sin ^{ - 1}}x - x]_0^1$

$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\left( {{\pi \over 2} - 1} \right)$

$\therefore$ $l = {\log _e}2 + {\pi \over 2} - 1$

Given, $g(x)=\int_0^x f(t) d t$

$\therefore g(3)=\int_0^3 f(t) d t=\int_0^1 f(t) d t+\int_1^3 f(t) d t$

$\Rightarrow \int_0^1 \frac{1}{3} d t+\int_1^3 0 \cdot d t \leq g(3) \leq \int_0^1 1 d t+\int_1^3 \frac{1}{2} d t$

$\Rightarrow \frac{1}{3} \leq g(3) \leq 1+1$

$\Rightarrow \frac{1}{3} \leq g(3) \leq 2$

F(x) = $\int_0^x {f(t)dt} $

$ \Rightarrow $ F'(x) = f(x) by Leibnitz theorem

I = $\int\limits_0^1 {(F'(x) + f(x)){e^x}dx = \int\limits_0^1 {2f(x){e^x}dx} } $

$I = \int\limits_0^1 {2\sin x\,dx} $

$I = 2(1 - \cos 1)$

$ = 2\left\{ {1 - \left( {1 - {{{1^2}} \over {2!}} + {{{1^4}} \over {4!}} - {1 \over {6!}} + ...} \right)} \right\}$

$ = 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}}} \right)} \right\} < 2(1 - \cos 1) < 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}} - {1 \over {720}}} \right)} \right\}$

${{330} \over {360}} < 2(1 - \cos 1) < {{331} \over {360}}$

${{330} \over {360}} < 1 < {{331} \over {360}}$

Given integral

$\int\limits_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}dx = 10\int\limits_0^1 {{{[\sin 2\pi x]} \over {{e^{\{ x\} }}}}dx} } $ (using property of definite in.)

$ = 10\left[ {\int\limits_0^{1/2} {0.dx} + \int\limits_{1/2}^1 {{{ - 1} \over {{e^x}}}dx} } \right]$

= $ - 10\left[ {{{{e^{ - x}}} \over { - 1}}} \right]_{1/2}^1 = 10\left[ {{e^{ - 1}} - {e^{ - 1/2}}} \right]$

$ = 10{e^{ - 1}} - 10{e^{ - 1/2}}$

comparing with the given relation,

$\alpha$ = 10, $\beta$ = $-$10, $\gamma$ = 0

$\alpha$ + $\beta$ + $\gamma$ = 0

Therefore, the correct answer is (A).

$g(\alpha ) = \int\limits_{\pi /6}^{\pi /3} {{{{{\sin }^\alpha }\left( {{\pi \over 2} - x} \right)} \over {{{\cos }^\alpha }\left( {{\pi \over 2} - x} \right)x + {{\sin }^\alpha }\left( {{\pi \over 2} - x} \right)}}dx} $

$ = \int\limits_{\pi /6}^{\pi /3} {{{{{\cos }^\alpha }x} \over {{{\sin }^\alpha }x + {{\cos }^\alpha }x}}dx} $

$ \therefore $ $2.g(\alpha ) = \int\limits_{\pi /6}^{\pi /3} {{{si{n^\alpha }x + {{\cos }^\alpha }x} \over {{{\sin }^\alpha }x + {{\cos }^\alpha }x}}dx} = \int\limits_{\pi /6}^{\pi /3} {dx} = {\pi \over 3} - {\pi \over 6} = {\pi \over 6}$

$ \Rightarrow $ $g(\alpha ) = {\pi \over {12}}$ i.e. a constant function hence an even function.

$I = \int_0^{10} {[x]\,.\,{e^{[x] + 1 - x}}} dx$

$ = \int_0^1 {0\,dx} + \int_1^2 {{e^{2 - x}}dx + \int_2^3 {2\,.\,{e^{3 - x}}dx} + \int_3^4 {3.{e^{4 - x}}dx} } + ......... + \int_9^{10} {9\,{e^{10 - x}}dx} $

$ = - \{ (1 - e) + 2(1 - e) + 3(1 - e) + ....... + 9(1 - e)\} $

$ = 45(e - 1)$

$(x - 2)Q(x) + 5 = {x^2} + bx + c$

Put x = 2

5 = 2b + c + 4 .... (1)

$\int_0^1 {({x^2} + bx + c)dx} = 1$

$ \Rightarrow {1 \over 3} + {b \over 2} + c = 1$

${b \over 2} + c = {2 \over 3}$ .... (2)

Solve (1) & (2)

$b = {2 \over 9}$

$c = {5 \over 9}$

9(b + c) = 7

$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $ .... (1)

Differentiating both sides w.r.t. x

$f'(x) = {e^x}.f(x) + {e^x}$ (Using Newton L:eibnitz Theorem)

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

Integrating w.r.t. x

$\int {{{f'(x)} \over {f(x) + 1}}dx = \int {{e^x}dx} } $

$ \Rightarrow \ln (f(x) + 1) = {e^x} + c$

Put x = 0

ln 2 = 1 + c ($ \because $ f(0) = 1, from equation (1))

$ \therefore $ $\ln (f(x) + 1) = {e^x} + \ln 2 - 1$

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

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

$f(x) = \int_1^x {{{\ln t} \over {1 + t}}dt} $

then $f\left( {{1 \over x}} \right) = \int_1^{1/x} {{{\ln t} \over {1 + t}}dt} $

Let $t = {1 \over u} \Rightarrow dt = - {1 \over {{u^2}}}du$

$ \Rightarrow f\left( {{1 \over x}} \right) = \int_1^x {{{\ln {1 \over u}} \over {1 + {1 \over u}}}\left( { - {1 \over {{u^2}}}} \right)dx} $

$f\left( {{1 \over x}} \right) = \int_1^x {{{\ln u} \over {u(1 + u)}}} du = \int_1^x {{{\ln t} \over {t(1 + t)}}dt} $

$ \therefore $ $f(x) + f\left( {{1 \over x}} \right) = \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over {t(1 + t)}}} \right)} dt$

$ = \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over t} - {1 \over {t + 1}}} \right)} dt$

$ = \int_1^x {{{\ln t} \over t}dt = {1 \over 2}{{(\ln x)}^2}} $

$ \therefore $ $f(e) + f\left( {{1 \over e}} \right) = {1 \over 2}{(\ln e)^2} = {1 \over 2}$

Let $I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$ .... (1)

Replace x with $-$x,

$ \therefore $ $I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {1 \over {{3^x}}}}}} $ .... (2)

Adding (1) and (2), we get,

$2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x + {3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx$

$ = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x(1 + {3^x})} \over {1 + {3^x}}}dx} $

$ = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}x} \,dx$

[${{{\cos }^2}x}$ is a even function as $f(x) = f( - x)$ for ${\cos ^2}x $]

$= 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,dx$

$ = 2\int\limits_0^{{\pi \over 2}} {\left( {{{1 + \cos 2x} \over 2}} \right)} \,dx$

$ \Rightarrow I = {1 \over 2}\int\limits_0^{{\pi \over 2}} {(1 + \cos 2x)} \,dx$

$ = {1 \over 2}\left[ {x + \sin 2x} \right]_0^{{\pi \over 2}}$

$ = {1 \over 2}\left[ {{\pi \over 2} - 0} \right]$

$ = {\pi \over 4}$

$\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}} dx} $

Here, $n - 1 \le x < n$

$ \therefore $ $[x] = n - 1$

$ \therefore $ $\int\limits_{n - 1}^n {{e^{x - (n - 1)}}} dx$

$ = \left[ {{e^{x - (n - 1)}}} \right]_{n - 1}^n$

$ = {e^1} - {e^0}$

$ = e - 1$

Now, $\sum\limits_{n = 1}^{100} {(e - 1) = 100(e - 1)} $

${I_n} = \int\limits_{\pi /4}^{\pi /2} {{{\cot }^n}xdx} = \int\limits_{\pi /4}^{\pi /2} {{{\cot }^{n - 2}}x(\cos e{c^2}x - 1)dx} $

= $\int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x.co{{\sec }^2}} xdx - \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x} dx$

$ = \left. {{{{{\cot }^{n - 1}}x} \over {n - 1}}} \right]_{\pi /4}^{\pi /2} - {I_{n - 2}}$

$ = {1 \over {n - 1}} - {I_{n - 2}}$

$ \Rightarrow {I_n} + {I_{n - 2}} = {1 \over {n - 1}}$

$ \Rightarrow {I_2} + {I_4} = {1 \over 3}$

${I_3} + {I_5} = {1 \over 4}$

${I_4} + {I_6} = {1 \over 5}$

$ \therefore $ ${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$ are in A.P.

$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ... + {n \over {{{(2n - 1)}^2}}}} \right]$

$\mathop {\lim }\limits_{n \to \infty } \left[ {{n \over {{{\left( {n + 0} \right)}^2}}} + {n \over {{{\left( {n + 1} \right)}^2}}} + {n \over {{{\left( {n + 2} \right)}^2}}} + ... + {n \over {{{\left( {n + \left( {n - 1} \right)} \right)}^2}}}} \right]$

$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{n \over {{{(n + r)}^2}}}} = $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{n \over {{n^2}{{\left( {1 + {r \over n}} \right)}^2}}}} $

$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{1 \over {{{(r/n)}^2} + 2(r/n) + 1}}} $

$ = \int\limits_0^1 {{{dx} \over {{{(x + 1)}^2}}} = \left[ {{{ - 1} \over {(x + 1)}}} \right]_0^1 = {1 \over 2}} $

Note : $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^{pn} {{1 \over n}f\left( {{r \over n}} \right)} = \int_\alpha ^\beta {f(x)} dx$

where, $\alpha = \mathop {\lim }\limits_{n \to \infty } {r \over n} = 0$ (as r = 1)

and $\beta = \mathop {\lim }\limits_{n \to \infty } {r \over n} = p$ (as r = pn)

Here $\alpha = \mathop {\lim }\limits_{n \to \infty } {r \over n} = 0$ (as r = 0)

and $\beta = \mathop {\lim }\limits_{n \to \infty } {r \over n} = \mathop {\lim }\limits_{n \to \infty } {{n - 1} \over n}$ = 1 (as r = n - 1)

$I = \int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}dx} $

Here -1 $ \le $ x $ \le $ 1 then -1 $ \le $ x³ $ \le $ 1

Integer between -1 and 1 is 0. So integration will be divided into two parts, -1 to 0 and 0 to 1.

$ = \int\limits_{ - 1}^0 {{x^2}{e^{[{x^3}]}}dx} + \int\limits_0^1 {{x^2}{e^{[{x^3}]}}dx} $

$ = \int\limits_{ - 1}^0 {{x^2}{e^{ - 1}}dx} + \int\limits_0^1 {{x^2}{e^0}dx} $

= ${1 \over e} \times \left[ {{{{x^3}} \over 3}} \right]_{ - 1}^0 + \left[ {{{{x^3}} \over 3}} \right]_0^1$

$ = {1 \over e} \times \left( {0 - \left( {{{ - 1} \over 3}} \right)} \right) + {1 \over 3}$

$ = {1 \over {3e}} + {1 \over 3} = {{1 + e} \over {3e}}$

$I = \int\limits_1^3 { - 3dx + \int\limits_1^3 {\left[ {{{(x - 1)}^2}} \right]dx} } $

Put x $-$ 1 = t ; dx = dt

$I = ( - 6) + \int\limits_0^2 {\left[ {{t^2}} \right]} dt$

$I = - 6 + \int\limits_0^1 {0dt} + \int\limits_1^{\sqrt 2 } {1dt} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dt} + \int\limits_{\sqrt 3 }^2 {3dt} $

$I = - 6 + \left( {\sqrt 2 - 1} \right) + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 $

$I = - 1 - \sqrt 2 - \sqrt 3 $

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

On integrating both side f(x) = $-$f(2 $-$ x) + c

put x = 0

f(0) + f(2) = c $ \Rightarrow $ c = 1 + e²

$ \Rightarrow $ f(x) + f(2 $-$ x) = 1 + e² ..... (i)

$I = \int\limits_0^2 {f(x)dx} = \int\limits_0^1 {\{ f(x) + f(2 - x)\} dx = (1 + {e^2})} $

$\left| {\matrix{ {f(x)} & {f'(x)} \cr {f'(x)} & {f''(x)} \cr } } \right| = 0$

$ \Rightarrow f(x).f''(x) - {\left( {f'(x)} \right)^2} = 0$

Dividing by ${\left( {f(x)} \right)^2}$, we get

$ \Rightarrow {{f(x).f''(x) - {{\left( {f'(x)} \right)}^2}} \over {{{\left( {f(x)} \right)}^2}}} = 0$

$ \Rightarrow {d \over {dx}}\left( {{{f'(x)} \over {f(x)}}} \right) = 0$

Integrating both side,

${{f'(x)} \over {f(x)}} = c$ (constant)

At, $x = 0$, ${{f'(0)} \over {f(0)}} = c$

$ \Rightarrow {2 \over 1} = c$

$ \Rightarrow c = 2$

$ \therefore $ ${{f'(x)} \over {f(x)}} = 2$

$ \Rightarrow \int {{{f'(x)} \over {f(x)}}} dx = 2\int {dx} $

$ \Rightarrow \ln |f(x)|\, = 2x + c'$

at x = 0,

$ln|f(0)|\, = 0 + c'$

$ \Rightarrow 0 = 0 + c'$

$ \Rightarrow c' = 0$

$ \therefore $ $n|f(x)| = 2x$

$ \Rightarrow f(x) = {e^{2x}}$

$f(1) = {e^2} = {(2.71)^2} = 7.34$

So it lie between (6, 9).

$\mathop {\lim }\limits_{x \to {0^ + }} {{\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \over {{x^3}}}$

This is in ${0 \over 0}$ form, so use L' Hospital rule

$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{d \over {dx}}\left( {\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \right)} \over {{d \over {dx}}\left( {{x^3}} \right)}}$

$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\sin x.2x - 0} \over {3{x^2}}}$

(applying Leibnitz rule)

$ = {2 \over 3}\mathop {\lim }\limits_{x \to {0^ + }} {{\sin x} \over x}$

$ = {2 \over 3}$

${\psi _1}(x) = {e^{ - x}} + x,x \ge 0$

${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2,x \ge 0$

$f(x) = \int_{ - x}^x {(\left| t \right| - {t^2}){e^{ - {t^2}}}dt,\,x > 0} $

$ = 2\int_0^x {(t - {t^2}){e^{ - {t^2}}}dt} $ .... (i)

$g(x) = \int_0^{{x^2}} {\sqrt t \,.\,{e^{ - t}}dt,x > 0} $

Put $t = {z^2} \Rightarrow dt = 2zdz$

$\therefore$ $g(x) = \int_0^x {z.{e^{ - {z^2}}}.\,2zdz} $

$ = 2\int_0^x {{z^2}.{e^{ - {z^2}}}dz = 2\int_0^x {{t^2}.{e^{ - {t^2}}}dt} } $

$f'(x) = 2(x - {x^2}){e^{ - {x^2}}} = 2x(1 - x){e^{ - {x^2}}}$

$\therefore$ f is increasing for x$\in$(0, 1) and f is decreasing for x$\in$(1, $\infty$). Hence, option (d) is incorrect.

Now, $f(x) + g(x) = 2\int_0^x {t.{e^{ - {t^2}}}dt} $

$ = [ - {e^{ - {t^2}}}]_0^x = ( - {e^{ - {x^2}}}) - ( - 1) = 1 - {e^{ - {x^2}}}$

$ \Rightarrow f(x) + g(x) = 1 - {e^{ - {x^2}}}$

$\therefore$ $f(\sqrt {\ln 3} ) + g(\sqrt {\ln 3} ) = 1 - {e^{ - {{(\sqrt {\ln 3} )}^2}}} = 1 - {e^{ - \ln 3}} = 1 - {1 \over 3} = {2 \over 3}$

Hence, option (a) is incorrect.

$\because$ ${\psi _1}(x) = {e^{ - x}} + x$ .... (iii)

$\because$ $ \Rightarrow \psi {'_1}(x) = 1 - {e^{ - x}} < 1$

Then, for $\alpha \in (1,x),{\psi _1}(x) = 1 + \alpha x$ does not true for $\alpha$ > 1. So, option (b) is incorrect.

Now, ${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2$

$ \Rightarrow \psi {'_2}(x) = 2x - 2 + 2{e^{ - x}} = 2(x + {e^{ - x}}) - 2$

$ = \psi {'_2}(x) = 2{\psi _1}(x) - 2$ [from Eq. (iii)]

By LMVT,

$\psi {'_2}(\beta ) = {{{\psi _2}(x) - {\psi _2}(0)} \over {x - 0}}$, for $\beta \in (0,x)$

$ \Rightarrow {{{\psi _2}(x) - 0} \over {x - 0}} = \psi {'_2}(\beta ) \Rightarrow {\psi _2}(x) = x\,.\,\psi {'_2}(\beta ) = 2x({\psi _1}(\beta ) - 1)$

Hence, option (c) is correct.

For option (a)

${\psi _1}(x) = {e^{ - x}} + x$

$\therefore$ $\psi {'_1}(x) = 1 - {e^{ - x}} \Rightarrow \psi {'_1}(x) = 0$

Here, ${\psi _1}(x)$ is always increasing.

Now, ${\psi _1}(x) > 1$

($\because$ ${\psi _1}(0) = 1$)

Thus, option (a) is incorrect.

For option (b)

${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2$

$ \Rightarrow \psi {'_2}(x) = 2x - 2 + 2{e^{ - x}}$

$ \Rightarrow \psi '{'_2}(x) = 2 - 2{e^{ - x}} \ge 0,\forall x > 0$

$ \Rightarrow \psi {'_2}(x)$ is increasing $ \Rightarrow \psi {'_2}(x) > \psi {'_2}(0)$

$ \Rightarrow \psi {'_2}(x) > 0,\forall x > 0$

Thus, option (b) is incorrect.

For option (c)

$f(x) = 2\int_0^x {(t - {t^2}){e^{ - {t^2}}}dt = ( - {e^{ - {t^2}}})_0^x - 2\int_0^x {{t^2}.{e^{ - {t^2}}}} dt} $

$ = 1 - {e^{ - {x^2}}} - 2\int_0^x {{t^2}\left( {1 - {t^2} + {{{t^4}} \over {2!}} + ...} \right)dt} $

$ = 1 - {e^{ - {x^2}}} - {{2{x^3}} \over 3} + {{2{x^5}} \over 5} - {{2{x^7}} \over {2.7}} + {{2{x^9}} \over {6.9}}...$

$ \Rightarrow f(x) - 1 + {e^{ - {x^2}}} + {{2{x^3}} \over 3} - {{2{x^5}} \over 5} < 0$ in $\left( {0,{1 \over 2}} \right)$

Hence, option (c) is also incorrect.

For option (d)

$g(x) = \int_0^{{x^2}} {\sqrt t .{e^{ - t}}dt} $

Put $t = {u^2} \Rightarrow dt = 2u\,du$

$\therefore$ $g(x) = \int_x^x {2.{u^2}{e^{ - {u^2}}}du = \int_0^x {2{u^2}\left( {1 - {u^2} + {{{u^4}} \over {2!}}....} \right)du} } $

$g(x) = {{2{x^3}} \over 3} - {{2{x^5}} \over 5} + {{2{x^7}} \over {2.7}} - {{2{x^9}} \over {9.6}}...$

$ \Rightarrow g(x) - {{2{x^3}} \over 3} + {{2{x^5}} \over 5} - {{2{x^7}} \over {2.7}} \le 0,\forall x \in \left( {0,{1 \over 2}} \right)$

$ \Rightarrow g(x) \le {{2{x^3}} \over 3} - {{2{x^5}} \over 5} + {1 \over 7}{x^7},\forall x \in \left( {0,{1 \over 2}} \right)$

Hence, option (d) is correct.

Given, f(0) = 1 and $\int_0^{\pi /3} {f(t)\,dt = 0} $

For option (a)

Consider a function

$g(x) = \int_0^\pi {f(t)\,dt - \sin 3x} $

g(x) is continuous and differentiable function and g(0) = g($\pi$/3) = 0

$\therefore$ By Rolle's theorem, g'(x) = 0 has at least one solution in (0, $\pi$/3).

i.e. g'(x) = f(x) $\times$ 1 $-$ 3cos 3x = 0 for some $x \in \left( {0,{\pi \over 3}} \right)$

For option (b)

Consider the function

$\phi (x) = \int_0^x {f(t)dt + \cos 3x + {6 \over \pi }x} $

$\phi$(x) is continuous and differentiable function as well as $\phi$(0) = $\phi$($\pi$/3) = 1

Hence, by Rolle's theorem, $\phi$'(x) = 0 has at least one solution in (0, $\pi$ / 3).

i.e. $\phi$'(x) = f(x) $\times$ 1 $-$ 3sin 3x + ${6 \over \pi } = 0$ for some $x \in \left( {0,{\pi \over 3}} \right)$.

For option (c)

Let $L = \mathop {\lim }\limits_{x \to 0} {{x\int_0^x {f(t)dt} } \over {1 - {e^{{x^2}}}}}$ (form ${0 \over 0}$)

$ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} {{xf(x) + \int_0^x {f(t)dt} } \over { - 2x{e^{{x^2}}}}}$ (Using L-Hospital Rule)

Again, using L'-Hospital Rule ($\therefore$ form ${0 \over 0}$)

$L = \mathop {\lim }\limits_{x \to 0} {{xf'(x) + f(x) + f(x)} \over { - 4{x^2}{e^{{x^2}}} - 2{e^{{x^2}}}}} = {{0 + 2f(0)} \over { - 0 - 2}} = - 1$ ($\because$ f(0) = 1)

For option (d)

Let $P = \mathop {\lim }\limits_{x \to 0} {{\sin x.\int_0^x {f(t)dt} } \over {{x^2}}}$ (form ${0 \over 0}$)

Applying L-Hospital Rule,

$P = \mathop {\lim }\limits_{x \to 0} {{\sin x.f(x) + \cos x.\int_0^x {f(t)dt} } \over {2x}}$ (form ${0 \over 0}$)

Again using L-Hospital Rule,

$P = \mathop {\lim }\limits_{x \to 0} {{[\cos x.f(x) + \sin x.f'(x) + \cos x.f(x) - \sin x\int_0^x {f(t)dt]} } \over 2}$

$ \Rightarrow P = {{1 \times f(0) + 0 \times f'(0) + 1 \times f(0) - 0 \times 0} \over 2}$

$ \Rightarrow P = {{1 + 0 + 1 - 0} \over 2} = 1$

$\int_0^\pi \log (\sin x) d x=8 k$

$\begin{aligned} & \text { Let } I=\int_0^{\pi / 4} \log (1+\tan x) d x=\int_0^{\pi / 4} \log (1+\tan \theta) d \theta \\ & =\int_0^{\pi / 4} \log (1+\tan (\pi / 4-\theta)) d \theta \\ & I=\int_0^{\pi / 4} \log \left(\frac{1+\tan \theta+1-\tan \theta}{1+\tan \theta}\right) d \theta \\ & I=\int_0^{\pi / 4} \log 2 d \theta-\int_0^{\pi / 4} \log (1+\tan \theta) d \theta \\ & I=\int_0^{\pi / 4} \log 2 d \theta-I \Rightarrow 2 I=(\log 2)\left(\frac{\pi}{4}\right) \\ & I=\frac{(\log 2) \pi}{8}=\frac{-1}{4} \times\left(\frac{-\pi}{2} \log 2\right)=-\frac{1}{8}(-\pi \log 2) \\ & =\frac{-1}{8} \times 8 k \quad\left[\begin{array}{l} \because \int_0^{\pi / 2} \log \sin x d x=\frac{-\pi}{2} \log 2 \\ =\int_0^\pi \log \sin x=-\pi \log 2 \end{array}\right] \\ & \therefore \quad I=-k \\ \end{aligned}$

$\begin{aligned} \int_0^1 x^n(1-x)^n d x & =\int_0^1(1-x)^n\left(1-(1-x)^n d x\right. \\ & =\int_0^1(1-x)^n(x)^n d x \Rightarrow k=1\end{aligned}$

$\begin{aligned} & \int_0^a \frac{d x}{4+x^2}=\frac{\pi}{8} \\ & \left.\Rightarrow \frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)\right|_0 ^a=\frac{\pi}{8} \Rightarrow \tan ^{-1}\left(\frac{a}{2}\right)=\frac{\pi}{4} \\ & \Rightarrow \quad a / 2=1 \Rightarrow a=2\end{aligned}$

$\left. {\int_1^2 {\left( {{{{x^3} - 1} \over {{x^2}}}} \right)dx = {{{x^2}} \over 2} + {1 \over x}} } \right|_1^2 = {5 \over 2} - {3 \over 2} = 1$

$\begin{aligned} & \text { } \int_0^{\pi / 2} \tan ^n x d x=k \int_0^{\pi / 2} \cot ^n(x) d x \\ & \begin{aligned} & \Rightarrow \int_0^{\pi / 2} \tan ^n\left(\frac{\pi}{2}-x\right) d x=k \int_0^{\pi / 2} \cot ^n(x) d x \\ &\left\{\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right\} \\ & \Rightarrow \int_0^{\pi / 2} \cot ^n x d x=k \int_0^{\pi / 2} \cot ^n x d x \\ & k=1\end{aligned}\end{aligned}$

$\begin{aligned} \int_0^2 x e^x d x & =\left(x e^x-e^x\right)_0^2 \\ & =\left(2 e^2-e^2\right)-\left(0 . e^0-e^0\right)=e^2+1\end{aligned}$

$\begin{aligned} & \text { Let } I=\int_2^4(|x-2|+|x-3|) d x \\ & \because x-a=\left\{\begin{array}{cc}x-a, \quad & x \geq a \\ -(x-a), \quad x < a\end{array}\right. \\ & \therefore \quad I=\int_2^4|x-2| d x+\int_2^3|x-3| d x+\int_3^4|x-3| d x \\ & =\int_2^4(x-2) d x-\int_2^3(x-3) d x+\int_3^4(x-3) d x \\ & =\left(\frac{x^2}{2}-2 x\right)_2^4-\left(\frac{x^2}{2}-3 x\right)_2^3+\left(\frac{x^2}{2}-3 x\right)_3^4 \\ & =\left[\left(\frac{16}{2}-8\right)-\left(\frac{4}{2}-4\right)\right]-\left[\left(\frac{9}{2}-9\right)-\left(\frac{4}{2}-6\right)\right] \\ & =\left|[0-(-2)]-\left[\left(\frac{-9}{2}\right)-(-4)\right]+\left[(-4)-\left(-\frac{9}{2}\right)\right]\right| \\ & =\left|2+\frac{9}{2}-4-4+\frac{9}{2}\right|=[2+9-8]=3\end{aligned}$