Definite Integration

$\left( {b,c} \right)g\left( {x + \pi } \right) = \int\limits_0^{x + \pi } {\cos \,4t\,dt} $

$ = \int\limits_0^\pi {\cos 4tdt + \int\limits_\pi ^{\pi + x} {\cos 4t\,dt} } $

$ = g\left( \pi \right) + \int\limits_0^x {\cos \,4t\,dt} $

Putting $t = \pi + y$ in second integral, we get

$\int\limits_x^{\pi + x} {\cos \,4t\,dt = \int\limits_0^\pi {\cos \,4t\,dt} } $

$ = g\left( \pi \right) + g\left( x \right) = g\left( x \right) - g\left( \pi \right)$

$\therefore$ $\left. {g\left( \pi \right) = 0} \right)$

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}}$

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

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$

Given differential equation

$f'(x) = {{f(x)} \over {{b^2} + {x^2}}}$

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

$ \Rightarrow {\log _e}|f(x)| = {1 \over b}{\tan ^{ - 1}}\left( {{x \over b}} \right) + c$

$ \because $ f(0) = 1, so c = 0

$ \therefore $ $|f(x)| = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$

$ \Rightarrow f(x) = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$ or $ - {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$

$ \because $ $f(0) = 1 \Rightarrow f(x) = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$

and $f'(x) = {1 \over {{b^2}}}{{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {1 + {{\left( {{x \over b}} \right)}^2}}} = {{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {{b^2} + {x^2}}}$

$ \therefore $ $f'(x) > 0\forall x \in R$ and $b \in {R_0}$.

Therefore, f(x) is an increasing function $\forall b \in {R_0}$.

and f(x) f($-$x)

$ = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}\,.\,{e^{ - {1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}} = 1$

$ \because $ $\cos x = 1 - {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} - {{{x^6}} \over {6!}} + ...$

and $\sin x = x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - {{{x^7}} \over {7!}} + ....$

$ \therefore $ $\int_0^1 {x\cos xdx} \ge \int_0^1 {\left( {x - {{{x^3}} \over {2!}}} \right)} \,dx$

$ = \left[ {{{{x^2}} \over 2} - {{{x^4}} \over 8}} \right]_0^1 = {1 \over 2} - {1 \over 8} = {3 \over 8}$

$ \Rightarrow \int_0^1 {x\cos xdx\, \ge \,{3 \over 8}} $

and, $\int_0^1 {x\sin dx \ge \int_0^1 {\left( {{x^2} - {{{x^4}} \over 6}} \right)} \,dx} $

$\left[ {{{{x^3}} \over 3} - {{{x^5}} \over {30}}} \right]_0^1 = {1 \over 3} - {1 \over {30}} = {9 \over {30}} = {3 \over {10}}$

$ \Rightarrow \int_0^1 {x\sin xdx\, \ge \,{3 \over {10}}} $

and, $\int_0^1 {{x^2}\cos xdx\, \ge \,\int_0^1 {\left( {{x^3} - {{{x^5}} \over 2}} \right)\,dx} } $

$ = \left[ {{{{x^4}} \over 4} - {{{x^6}} \over {12}}} \right]_0^1 = {1 \over 4} - {1 \over {12}} = {2 \over {12}} = {1 \over 6}$

$ \therefore $ $\int_0^1 {{x^2}\cos xdx\, \ge \,{1 \over 6}} $

and, $\int_0^1 {{x^2}\sin xdx\, \ge \int_0^1 {\left( {{x^3} - {{{x^5}} \over 6}} \right)\,dx} } $

$ = \left[ {{{{x^4}} \over 4} - {{{x^6}} \over {36}}} \right]_0^1 = {1 \over 4} - {1 \over {36}} = {8 \over {36}} = {2 \over 9}$

$ \therefore $ $\int_0^1 {{x^2}\sin xdx\, \ge \,{2 \over 9}} $

$I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x(x + 1)}}dx} } $

Clearly, $I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x{{(x + 1)}^2}}}dx} } $

$ \Rightarrow I > \sum\limits_{k = 1}^{98} {(k + 1)\int_k^{k + 1} {{1 \over {{{(x + 1)}^2}}}dx} } $

$ \Rightarrow I > \sum\limits_{k = 1}^{98} {( - (k + 1))\left[ {{1 \over {k + 2}} - {1 \over {k + 1}}} \right]} $

$ \Rightarrow I > \sum\limits_{k = 1}^{98} {{1 \over {k + 2}}} $

$ \Rightarrow I > {1 \over 3} + ... + {1 \over {100}} > {{98} \over {100}}$

$ \Rightarrow I > {{49} \over {50}}$

Also, $I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x(x + 1)}}dx} } $

$ = \sum\limits_{k = 1}^{98} {[{{\log }_e}(k + 1) - {{\log }_e}k]} $

$I < {\log _e}99$

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

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

$\therefore$ Option (c) is correct.

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

$\therefore$ ${{f'(3)} \over {f(3)}} - {{f'(2)} \over {f(2)}} = \log \left( {{4 \over {10}}} \right) - \log \left( {{3 \over 5}} \right)$

$ = \log (2/3) < 0$

$ \Rightarrow {{f'(3)} \over {f(3)}} < {{f'(2)} \over {f(2)}}$

$\therefore$ Option (d) is incorrect.

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

$ = 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 $a \in R$

Given, $f\left( x \right) $= $\int\limits_{{1 \over x}}^x {{{{e^{ - \left( {t + {1 \over t}} \right)}}} \over t}} dt$

Therefore, $\frac{d}{d x} f(x)=\frac{e^{-\left(x+\frac{1}{x}\right)}}{x} \frac{d}{d x} (x)-\frac{e^{-\left(\frac{1}{x}+x\right)}}{1 / x} \times \frac{d}{d x}\left(\frac{1}{x}\right)$

$ \begin{aligned} & =\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}+x e^{-\left(x+\frac{1}{x}\right)} \times\left(-\frac{1}{x^2}\right) \\\\ & =\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}+\frac{1}{x} e^{-\left(x+\frac{1}{x}\right)} \\\\ & =\frac{2 e^{-\left(x+\frac{1}{x}\right)}}{x}>0 \end{aligned} $

As, $ f^{\prime}(x)>0, \forall x \in(0, \infty)$

$\therefore f(x)$ is monotonically increasing on $(0, \infty)$.

$\Rightarrow$ options (a) is correct and (b) is wrong.

Now,

$\begin{aligned} f(x)+f\left(\frac{1}{x}\right) & =\int\limits_{1 / x}^x \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t+\int\limits_x^{\frac{1}{x}} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t \\\\ & =\int\limits_{1 / x}^{1 / x} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t=0\end{aligned}$

Now, let

$ \begin{aligned} g(x) & =f\left(2^x\right)=\int\limits_{2^{-x}}^{2^x} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t \\\\ g(-x) & =f\left(2^{-x}\right) \\\\ & =\int\limits_{2^x}^{2^{-x}} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t \\\\ & =-g(x) \end{aligned} $

$\therefore f\left(2^x\right)$ is an odd function.

In the equation $f(x) = {x^5} - 5x + a$, there are different polynomials depending on the parameter a. Now, for the roots of each of these, in general, f(x) = 0.

That is, $a = 5x - {x^5} = x(5 - {x^5})$

Hence, the parameter a is a function of x. That is,

$a(x) = x(5 - {x^5})$

Now, $a'(x) = 5 - 5{x^4}$

Therefore, extrema occurs at a'(x) = 0

That is, when x⁴ = 1 or x = 1 and x = $-$1 (only real roots considered)

Here, $a''(x) = - 20{x^3}$

$a''(1) < 0$ (maximum)

$a''( - 1) > 0$ (minimum)

Hence, the maximum value is

a(1) = 4

The minimum value is

a($-$1) = $-$4

Hence, when $-$4 < a < 4, there are three points that is, x values where f(x) = 0, that is, three roots of f(x) for any value of a lying in ($-$4, 4)}. .... (1)

When | a | > 4, there is only one x for which f(x) = 0 ...... (2)

Hence, from statements (1) and (2), we can conclude that options (B) and (D) are correct.

(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} } $