Definite Integration

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

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$

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

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

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

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

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

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

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

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

$\begin{aligned} & \text { (c) Let } I=\int_{-1 / 2}^{1 / 2}\left\{[x]+\log \left(\frac{1+x}{1-x}\right) d x\right\} \\ & \because \int_{-a}^a f(x) d x=\left\{\begin{array}{cr} 2 \int_0^a f(x) d x, & f(-x)=f(x) \\ 0, & f(-x)=-f(x) \end{array}\right. \\ & =\int_{-1 / 2}^{1 / 2}[x] d x+\int_{-1 / 2}^{1 / 2} \log \left(\frac{1+x}{1-x}\right) d x \end{aligned}$

Let $f(x)=\log \left(\frac{1+x}{1-x}\right)$

$\begin{aligned} & f(-x)=\log \left(\frac{1-x}{1+x}\right)=-f(x) \quad \quad \text { [odd function] } \\ & =\int_{-1 / 2}^{1 / 2}[x] d x+0=\int_{-1 / 2}^0[x] d x+\int_0^{1 / 2}[x] d x \\ & =\int_{-1 / 2}^0(-1) d x+\int_0^{1 / 2} 0 d x \\ & =-(x)_{-1 / 2}^0+0=-\left(0+\frac{1}{2}\right)=-\frac{1}{2} \quad \end{aligned}$

$\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx$

= $\int\limits_1^2 {{e^x}{x^x}\left[ {1 + \left( {1 + {{\log }_e}x} \right)} \right]} dx$

= $\int\limits_1^2 {{e^x}\left[ {{x^x} + {x^x}\left( {1 + {{\log }_e}x} \right)} \right]} dx$

= $\left[ {{e^x}{x^x}} \right]_1^2$

= e² $ \times $ 4 - e $ \times $ 1

= 4e² - e

= e(4e - 1)

Note : $\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} $ = e^xf(x) + c

I₂ = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$

I₂ = $\int\limits_0^1 {\left( {1 - {x^{50}}} \right){{\left( {1 - {x^{50}}} \right)}^{100}}dx} $

I₂ = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int\limits_0^1 {{x^{50}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} $

I₂ = I₁ - $\int\limits_0^1 {x.{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} $

Now apply IBP

I₂ = I₁ -
$\left[ {x\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int {{{d\left( x \right)} \over {dx}}.\int {{{d\left( x \right)} \over {dx}}\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} } } } \right]$

Let (1 – x⁵⁰) = t

$ \Rightarrow $ -50x⁴⁹dx = dt

I₂ = I₁ -
$\left[ {x.\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}} \right]_0^1$ - $ - \int\limits_0^1 {\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}dx} $

= I₁ - 0 - ${ - {1 \over {50}}.{1 \over {101}}{I_2}}$

I₂ = I₁ - ${1 \over {5050}}{I_2}$

$ \Rightarrow $ I₂ + ${1 \over {5050}}{I_2}$ = I₁

$ \Rightarrow $ ${{5051} \over {5050}}{I_2}$ = I₁

$ \Rightarrow $ I₂ = ${{5050} \over {5051}}$I₁

Given I₂ = $\alpha $I₁

$ \therefore $ $\alpha $ = ${{5050} \over {5051}}$

$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$$\left( {{0 \over 0}} \right)$

Apply L Hospital Rule

= $\mathop {\lim }\limits_{x \to 1} {{2\left( {x - 1} \right).{{\left( {x - 1} \right)}^2}\cos {{\left( {x - 1} \right)}^4} - 0} \over {\left( {x - 1} \right).\cos \left( {x - 1} \right) + \sin \left( {x - 1} \right)}}$ $\left( {{0 \over 0}} \right)$

= $\mathop {\lim }\limits_{x \to 1} {{2{{\left( {x - 1} \right)}^3}\cos {{\left( {x - 1} \right)}^4}} \over {\left( {x - 1} \right).\left[ {\cos \left( {x - 1} \right) + {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}} \right]}}$

= $\mathop {\lim }\limits_{x \to 1} {{2{{\left( {x - 1} \right)}^2}\cos {{\left( {x - 1} \right)}^4}} \over {\cos \left( {x - 1} \right) + {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}}}$

on taking limit

= ${0 \over {1 + 1}}$ = 0

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

Replacing x with $\left( {{\pi \over 2} - {\pi \over 2} + x} \right)$, we get

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

= $\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{e^{\sin x}}dx} \over {1 + {e^{\sin x}}}}} $ .....(2)

Adding (1) and (2), we get

2I = $\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {1dx} $ = $\left( {{\pi \over 2} - \left( { - {\pi \over 2}} \right)} \right)$ = $\pi $

$ \Rightarrow $ I = ${{\pi \over 2}}$

Given,

I = $\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} $

$I = \int\limits_{\pi /6}^{\pi /3} ({2.{{\tan }^3}} x{\sec ^2}x{\sin ^4}3x + 3{\tan ^4}x{\sin ^2}3x.\,2\sin 3xcos\,3x\,)dx$

$ = {1 \over 2}\int\limits_{\pi /6}^{\pi /3} ({4{{\tan }^3}} x{\sec ^2}x{\sin ^4}3x + 3.4{\tan ^4}x{\sin ^3}3xcos\,3x\,)dx$

$ = {1 \over 2}\int\limits_{\pi /6}^{\pi /3} {{d \over {dx}}\left( {{{\tan }^4}x{{\sin }^4}3x} \right)} dx$

$ = {1 \over 2}\left[ {{{\tan }^4}x{{\sin }^4}3x} \right]_{\pi /6}^{\pi /3}$

$ = {1 \over 2}\left[ {9.(0) - {1 \over 3}.{1 \over 3}(1)} \right] = - {1 \over {18}}$

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

$ \therefore $ $f(f(x)) = \left| {|x - 2| - 2} \right| = g(x)$

$ \Rightarrow g(x) = \left| {|x - 2| - 2} \right| = \left\{ {\matrix{ {|x - 4|} & {if\,x \ge 2} \cr {| - x|} & {if\,x < 2} \cr } } \right.$

$ \therefore $ $\int\limits_0^3 {(g(x) - f(x))dx} $

$ = \int\limits_0^3 {g(x)dx} - \int\limits_0^3 {f(x)dx} $

$ = \int\limits_0^2 {| - x|dx} + \int\limits_2^3 { - (x - 4)dx} - \int\limits_0^2 { - (x - 2)dx} - \int\limits_2^3 {(x - 2)dx} $

$ = \int\limits_0^2 {x\,dx} - \int\limits_2^3 {(x - 4)dx} + \int\limits_0^2 {(x - 2)dx} - \int\limits_2^3 {(x - 2)dx} $

$ = \left[ {{{{x^2}} \over 2}} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 4x} \right]_2^3 + \left[ {{{{x^2}} \over 2} - 2x} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 2x} \right]_2^3$

$ = 2 - \left\{ {{9 \over 2} - 12 - 2 + 8} \right\} + \{ 2 - 4\} - \left\{ {{9 \over 2} - 6 - 2 + 4} \right\}$

$ = 2 - \left\{ {{9 \over 2} - 6} \right\} - 2 - \left\{ {{9 \over 2} - 4} \right\}$

$ = 10 - {9 \over 2} - {9 \over 2} = {{20 - 9 - 9} \over 2} = {2 \over 2} = 1$

$I = \int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{(1 - {x^2})}^{{3 \over 2}}}}}} dx$

Let $x = \sin \theta $

$ \Rightarrow dx = \cos \theta d\theta $

When x = 0 then sin$\theta $ = 0 $ \Rightarrow $ $\theta $ = 0

When $x = {1 \over 2}$ then $\sin \theta = {1 \over 2}$ $ \Rightarrow $ $\theta = {\pi \over 6}$

$ \therefore $ $I = \int\limits_0^{{\pi \over 6}} {{{{{\sin }^2}\theta } \over {{{({{\cos }^2}\theta )}^{{3 \over 2}}}}}\cos \theta d\theta } $

$ = \int\limits_0^{{\pi \over 6}} {{{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}d\theta } $

$ = \int\limits_0^{{\pi \over 6}} {{{\tan }^2}d\theta } = \int\limits_0^{{\pi \over 6}} {({{\sec }^2}\theta - 1)d\theta } $

$ = \left[ {\tan \theta - \theta } \right]_0^{{\pi \over 6}}$

$ = {1 \over {\sqrt 3 }} - {\pi \over 6}$

Given, ${k \over 6} = {1 \over {\sqrt 3 }} - {\pi \over 6} = {{2\sqrt 3 - \pi } \over 6}$

$ \Rightarrow k = 2\sqrt 3 - \pi $

Critical points = $ - $1, 0, 1.

$ \therefore $ f'(x) = a(x $ - $ 1)(x + 1)x

$ \therefore $ f(x) = a$\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C$

$ \because $ f(x) = f(0)

$ - a\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C = C$

$a{{{x^2}} \over 4}\left( {{x^2} - 2} \right) = 0$

$ \therefore $ x = 0, $\sqrt 2 $, $ - \sqrt 2 $

$ \therefore $ T = $\left\{ {0,\sqrt 2 , - \sqrt 2 } \right\}$

Sum of square of elements of $T = {0^2} + {\left( {\sqrt 2 } \right)^2} + {\left( { - \sqrt 2 } \right)^2} = 4$

$\int\limits_{ - \pi }^\pi {\left| {\pi - \left| x \right|} \right|dx} $

= $2\int\limits_0^\pi {\left| {\pi - \left| x \right|} \right|} dx$ [As it is even function]

= $2\int\limits_0^\pi {\left( {\pi - x} \right)} dx$

= $2\left[ {\pi x - {{{x^2}} \over 2}} \right]_0^\pi $

= ${\pi ^2}$

$F(x) = \int\limits_1^x {{t^2}g(t)dt} $

$ \Rightarrow $ F'(x) = x²g(x) = x²$\int\limits_1^t {f(u)du} $

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

Now, F''(x) = 2xg(x) + x²g'(x)

F''(1) = 2g(1) + g'(1) = 0 + g'(1) = 3

[ As g'(t) = f(t); g'(1) = f'(1) = 3 ]

So, at x = 1, F'(1) = 0 and F"(1) = 3 > 0

$ \therefore $ For the function f(x), x = 1 is a point of local minima.

I = $\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$ .....(1)

I = $\int\limits_0^{2\pi } {{{\left( {2\pi - x} \right){{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$ ......(2)

Adding (1) and (2)

2I = $\int\limits_0^{2\pi } {{{2\pi {{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$

$ \Rightarrow $ I = $2\pi \int\limits_0^\pi {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$

$ \Rightarrow $ I = $2\pi \left[ {\int\limits_0^{\pi /2} {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx + \int\limits_0^{\pi /2} {{{{{\cos }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx} \right]$

= ${2\pi \int\limits_0^{\pi /2} 1 dx}$

= $2\pi .{\pi \over 2}$ = ${\pi ^2}$

ƒ(x) = a + bx + cx²

$\int\limits_0^1 {f\left( x \right)dx} $ = $\left[ {ax + {{b{x^2}} \over 2} + {{c{x^3}} \over 3}} \right]_0^1$

= ${a + {b \over 2} + {c \over 3}}$

= ${1 \over 6}\left[ {6a + 3b + c} \right]$

f(1) = a + b + c

f(0) = a

$f\left( {{1 \over 2}} \right) = a + {b \over 2} + {c \over 4}$

By checking each option you have to find the solution.

${1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}$

= ${1 \over 6}\left[ {a + a + b + c + 4\left( {a + {b \over 2} + {c \over 4}} \right)} \right]$

= ${1 \over 6}\left[ {6a + 3b + c} \right]$

$ \therefore $ Option (A) is correct option.

Let f(x) = ${1 \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}$

f'(x) = $ - {1 \over 2}{{ - 6{x^2} - 18x + 12} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}$

= ${{ - 6\left( {x - 1} \right)\left( {x - 2} \right)} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}$



f_max = f(1)

f_min = f(2)

f(1) = ${1 \over {\sqrt {2 - 9 + 12 + 4} }}$ = ${1 \over {\sqrt 9 }}$ = ${1 \over 3}$

f(2) = ${1 \over {\sqrt {16 - 36 + 24 + 4} }}$ = ${1 \over {\sqrt 8 }}$

$ \therefore $ ${1 \over 3} < {I} <$ ${1 \over {\sqrt 8 }}$

So ${1 \over 9} < {I^2} < {1 \over 8}$

$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$

This is in ${0 \over 0}$ form.

So apply newton leibniz rule

$\mathop {\lim }\limits_{x \to 0} {{x.\sin \left( {10x} \right) - 0} \over 1}$ = 0

$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$

$ \Rightarrow $ $4\alpha \int\limits_{ - 1}^0 {{e^{\alpha x}}dx} + 4\alpha \int\limits_0^2 {{e^{ - \alpha x}}dx} $ = 5

$ \Rightarrow $ $4\alpha \left[ {{{{e^{\alpha x}}} \over \alpha }} \right]_{ - 1}^0 + 4\alpha \left[ {{{{e^{ - \alpha x}}} \over { - \alpha }}} \right]_0^2$ = 5

$ \Rightarrow $ $4{e^{ - 2\alpha }} + 4{e^{ - \alpha }} - 3$ = 0

Let e^{-$\alpha $} = t

$ \therefore $ 4t² + 4t - 3 = 0

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

$ \therefore $ e^{-$\alpha $} = ${1 \over 2}$

$ \Rightarrow $ $\alpha $ = ${\log _e}2$

2cot²$\theta $ - ${5 \over {\sin \theta }}$ + 4 = 0

$ \Rightarrow $ $2{{{{\cos }^2}\theta } \over {{{\sin }^2}\theta }} - {5 \over {\sin \theta }} + 4$ = 0

$ \Rightarrow $ 2sin² $\theta $ – 5sin$\theta $ + 2 = 0

$ \Rightarrow $ (2sin$\theta $ – 1)(sin$\theta $ – 2) = 0

$ \therefore $ sin$\theta $ = ${1 \over 2}$ [ sin$\theta $ = 2 not possible]

$ \therefore $ $\theta $₁ = ${\pi \over 6}$ and $\theta $₂ = ${{5\pi } \over 6}$ as $\theta $₁ $<$ $\theta $₂

$ \therefore $ I = $\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{\cos }^2}3\theta d\theta } $

= $\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{1 + \cos 6\theta } \over 2}d\theta } $

= ${1 \over 2}\left[ {\theta + {{\sin 6\theta } \over 2}} \right]_{{\pi \over 6}}^{{{5\pi } \over 6}}$

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

I = ${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$ ...(1)

x $ \to $ a + b - x

I = ${1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {a + b - x} \right) + f\left( {a + b - x + 1} \right)} \right)dx} $

I = ${1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} $ .....(2)

[As ƒ(x) = ƒ(a + b + 1 - x)

$ \Rightarrow $ ƒ(x + 1) = ƒ(a + b - x)]

Adding (1) and (2) we get

2I = ${{a + b} \over {a + b}}\int\limits_a^b {\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} $

$ \Rightarrow $ I = ${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {x + 1} \right)dx} $

$ \Rightarrow $ I = ${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {a + b - x + 1} \right)dx} $

$ \Rightarrow $ I = ${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( x \right)dx} $

$ \Rightarrow $ I = $\int\limits_a^b {f\left( x \right)dx} $

Let x = t + 1

$ \therefore $ I = $\int\limits_{a - 1}^{b - 1} {f\left( {t + 1} \right)dt} $

= $\int\limits_{a - 1}^{b - 1} {f\left( {x + 1} \right)dx} $