Mathematics M.C.Q (1 Marks)

$ = \int_0^{\pi /4} { - (\sin x - \cos x} )dx + \int_{\pi /4}^{\pi /2} {\,(\sin x - \cos x)dx} $

$= 2(\sqrt 2 - 1)$.

$\therefore $ $dx = dt$. 

When $x = - 3 \to 5$, then $t = - 2 \to 6$

Therefore, $\int_{ - 3}^5 {f(x)dx} $

$ = \int_{ - 2}^6 {f(t - 1)dt = } \int_{ - 2}^6 {f(x - 1)dx} $.

$ = \frac{\pi }{4}$,

(Put $x = \sin \theta ,\,dx = \cos \theta \,d\theta $).

$\therefore \int_0^n {\left\{ {x - [x]} \right\}\,dx = n\int_0^1 {(x - [x])\,\,dx} } $

$ = n\left[ {\int_0^1 {x\,\,dx - \int_0^1 {[x]\,dx} } } \right]$

$ = n\left[ {\left( {\frac{{{x^2}}}{2}} \right)_0^1 - 0} \right] = \frac{n}{2}$.

==> $2I = \frac{\pi }{2}\int_0^\pi {\frac{{\tan x}}{{\sec x + \tan x}}dx = \frac{\pi }{2}\int_0^\pi {\frac{{\sin x}}{{1 + \sin x}}dx} } $

$=\frac{\pi }{2}\left[ {\int_0^\pi {1dx - \int_0^\pi {\frac{{dx}}{{1 + \sin x}}} } } \right]$

On solving, we get $I = \frac{{{\pi ^2}}}{2} - \pi = \pi \left( {\frac{\pi }{2} - 1} \right)$.

It gives $I = \frac{\pi }{2}\int_0^\pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}} dx$

Now put $\cos x = t$ and solve, we get 

$I = \frac{\pi }{2} \times \frac{\pi }{2} = \frac{{{\pi ^2}}}{4}$.

So comparing it with the given value, we get $k = a$.

Then  $I=2\int_{0}^{\pi }{{{\cos }^{99}}x\,dx,\,\,\,\{\because {{\cos }^{99}}(2\pi -x)={{\cos }^{99}}x\}}$

Now, $\int_{0}^{\pi }{{{\cos }^{99}}x\,dx\,=0,\,\,\{\because {{\cos }^{99}}(\pi -x)=-{{\cos }^{99}}x\}}$

$\therefore \,\,I = 2 \times 0 = 0$.

$\{\because \,\,\int_{0}^{2a}{f(x)=2\int_{0}^{a}{f(a-x)dx}}$, if $f(2a - x) = f(x) \}$

$I = 2 \times \frac{1}{2} \times \frac{\pi }{2} = \frac{\pi }{2}$.

$\left[ \because \int_{a}^{b}{f(x)dx=\int_{a}^{b}{f(a+b-x)dx}} \right]$ 

$ = \int_{0.5}^{1.5} {(2 - x)f(x)\,dx} = 2\int_{0.5}^{1.5} {f(x)\,dx - I} $

==>$I = \int_{0.5}^{1.5} {f(x)\,dx} $.

We know that, $\int_{ - a}^a {f(x)dx = 0 = \int_{ - a}^0 {f(x)dx + } \int_0^a {f(x)dx} } $

$ \Rightarrow \int_{ - 1}^0 {f(x)\,dx + } \int_0^1 {f(x)\,dx = 0} $

$\Rightarrow \int_{ - 1}^0 {f(x)dx = - 5} $

$ \Rightarrow \int_{ - 1}^0 {f(t)\,dt = - 5} $.

then, $\phi ( - x) = [f( - x) + f(x)]\,[g( - x) - g(x)]$

$\therefore \int_{ - \pi }^\pi {\phi (x)dx = 0} $

==>$\int_{ - \pi }^\pi {[f(x) + f( - x)][g(x) - g( - x)]dx = 0} $.

$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}\,} dx$ .....(ii)

Adding (i) and (ii), 

$2I = \int_{\,\pi /6}^{\,\pi /3} {dx} $; 

$I = \frac{1}{2}\left( {\frac{\pi }{3} - \frac{\pi }{6}} \right) = \frac{\pi }{{12}}$.

Now,$f( - x) = \log \left( {\sqrt {1 + {x^2}} - x} \right) = \log (\sqrt {1 + {x^2}} - x).\frac{{(\sqrt {1 + {x^2}} + x)}}{{(\sqrt {1 + {x^2}} + x)}}$

$ = \log \frac{{[(1 + {x^2}) - {x^2}]}}{{(\sqrt {1 + {x^2}} + x)}}$

$ = \log 1 - \log (\sqrt {1 + {x^2}} + x)$

$ = - \log (\sqrt {1 + {x^2}} + x)$

$ = - f(x)$ 

Hence, $\int_{\, - 1}^{\,1} {\log \,(x + \sqrt {1 + {x^2}} ) = 0} $ 

$\left[ \because \int_{\,-a}^{\,a}{f(x)=0,\,}\text{if }f(-x)=-f(x) \right]$.

$ = \int_{\frac{1}{n}}^{a - \frac{1}{n}} {\frac{{\sqrt {\frac{1}{n} + a - \frac{1}{n} - x} \,\,\,\,\,\,\,\,\,\,\,dx}}{{\sqrt {a - \left( {\frac{1}{n} + a - \frac{1}{n} - x} \right) + } \sqrt {\frac{1}{n} + a - \frac{1}{n} - x} }}} $

$\left[ \because \int_{a}^{b}{f(x)dx=\int_{a}^{b}{f(a+b-x)\,dx}} \right]$

$I = \int_{\frac{1}{n}}^{a - \frac{1}{n}} {\frac{{\sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}dx} $.....$(ii)$

Adding $(i)$ and $(ii),$ we get

$2I = \int_{\,1/n}^{\,a - (1/n)} {1\,dx = \left[ {\,x} \right]_{\,1/n}^{\,a - \frac{1}{n}}} $

$ \Rightarrow 2I = a - \frac{1}{n} - \frac{1}{n} = \frac{{na - 2}}{n}$

$ \Rightarrow I = \frac{{na - 2}}{{2n}}$.

$ = 2\,[ - \cos x]_0^\pi + 0$

$ = - 2\,(\cos \pi - \cos 0)$

$ = - 2\,( - 1 - 1) = 4$.

$I = \int_0^\pi {\frac{{(\pi - x)dx}}{{1 + \sin (\pi - x)}}} $

$I = \int_0^\pi {\frac{{(\pi - x)dx}}{{1 + \sin x}}} $ ... $(ii),$

$\left\{ \because \,\int_{0}^{a}{f(x)\,dx}=\int_{0}^{a}{f(a-x)\,dx} \right\}\,$

Adding $(i)$ and $(ii),$ we get 

$2I = \int_0^\pi {\frac{{\pi \,dx}}{{1 + \sin x}}} $

$2I = \pi \int_0^\pi {\frac{{1 - \sin x}}{{(1 + \sin x)(1 - \sin x)}}dx} $

$2I = \pi \int_0^\pi {\frac{{1 - \sin x}}{{{{\cos }^2}x}}} dx = \pi \int_0^\pi {({{\sec }^2}x - \sec x\tan x)dx} $

$2I = \pi [\tan x - \sec x]_0^\pi = \pi [0 - ( - 1) - (0 - 1)]$,  $2I = 2\pi $

$\therefore$  $I = \pi $.

$= [ - {\theta ^2}\cot \theta ]_0^{\pi /2} + \int_0^{\pi /2} {\,\,\,\,2\theta .\cot \theta .\,d\theta } $

$ = 2[\theta .\log \sin \theta ]_0^{\pi /2} - 2\int_0^{\pi /2} {\log \sin \theta \,d\theta } $

$ \Rightarrow \frac{I}{2} = 0 - \mathop {\lim }\limits_{\theta \to 0} \theta \log .\sin \theta $

$ - \int_0^{\pi /2} {\log \sin \theta \,d\theta } $

==> $\frac{\pi }{2}\log 2$.

Hence $I =\pi \log 2$.

Integrating by parts, we get 

$[x(\log \sin x)]_0^{\pi /2} - \int_0^{\pi /2} {\log \sin x\,dx} $

$I = - \left( { - \frac{\pi }{2}\log 2} \right) = \frac{\pi }{2}\log 2$.

$\therefore $$I = 2 \times 2\int_0^{\pi /2} {\frac{{{{\sin }^4}x}}{{{{\sin }^4}x + {{\cos }^4}x}}\;dx} $.....$(i)$

$I = 4\int_0^{\pi /2} {\frac{{{{\sin }^4}\left( {\frac{\pi }{2} - x} \right)}}{{{{\sin }^4}\left( {\frac{\pi }{2} - x} \right) + {{\cos }^4}\left( {\frac{\pi }{2} - x} \right)}}\;dx} $

$I = 4\int_0^{\pi /2} {\frac{{{{\cos }^4}x}}{{{{\cos }^4}x + {{\sin }^4}x}}\;dx} $.....$(ii)$

Adding $(i)$ and $(ii)$ we get,

$2I = 4\int_0^{\pi /2} {dx = 4 \times \frac{\pi }{2} = 2\pi } $

==> $I = \pi $.

therefore $\int_a^{a + (\pi /2)} {({{\sin }^4}x + {{\cos }^4}x){\rm{ }}dx} $

$ = \int_0^{\pi /2} {({{\sin }^4}x + {{\cos }^4}x)dx} $

$ = 2\int_0^{\pi /2} {{{\sin }^4}x\,dx = \frac{{3\Gamma (5/2)\Gamma (1/2)}}{{2\Gamma \left( {\frac{{4 + 0 + 2}}{2}} \right)}} = \frac{{3\pi }}{8}} $.

we get , $\int_0^\infty {{e^{ - \lambda x}}{x^{n - 1}}dx} $

$ = \frac{1}{{{\lambda ^n}}}\int_0^\infty {{e^{ - t}}{t^{n - 1}}} dt$

$ = \frac{1}{{{\lambda ^n}}}\int_0^\infty {{e^{ - x}}{x^{n - 1}}dx = \frac{{{I_n}}}{{{\lambda ^n}}}} $.

$f(-x)=\sin ^{5}(-x) \cos ^{4}(-x)=-\sin ^{5} x \cos ^{4} x=-f(x),$ i.e., $f$ is an odd function.

Therefore, by $P_{7}(\text { ii }), I=0$

$ \Rightarrow I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{x^3}} dx + \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\cos } xdx + \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\tan }^5}} xdx + \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} 1 .dx$

It is known that if $f ( x )$ is an even function, then $\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$

if $f ( x )$ is an odd function, then $\int_{-a}^{a} f(x) d x=0$

and $I=0+0+0+2 \int_{0}^{\frac{\pi}{2}} 1 d x$

$=2[x]_{0}^{\frac{\pi}{2}}$

$=\frac{2 \pi}{2}$

$=\pi$

Hence, the correct Answer is $A$.

$I=\int_{a}^{b}(a+b-x) f(a+b-x) d x \quad\left(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right)$

$\Rightarrow I=\int_{a}^{b}(a+b-x) f(x) d x$

$\Rightarrow I=(a+b) \int_{a}^{b} f(x) d x-I \quad[\text { using }(1)]$

$\Rightarrow I+I=(a+b) \int_{a}^{b} f(x) d x$

$\Rightarrow 2 I=(a+b) \int_{a}^{b} f(x) d x$

$\Rightarrow I=\left(\frac{a+b}{2}\right) \int_{a}^{b} f(x) d x$

Hence, the correct Answer is $C$.

$ \Rightarrow \frac{{df(x)}}{{dx}} = \frac{d}{{dx}}\left( {\int_a^0 {{t^3}.{e^t}dt} } \right) + \frac{d}{{dx}}\left( {\int_0^x {{t^3}.{e^t}\,dt} } \right) = {x^3}{e^x}$.

Applying gamma function,

$2\int_0^{\pi /2} {{{\sin }^4}x\,dx} = 2\frac{{\Gamma (5/2).\Gamma (1/2)}}{{2.\Gamma (6/2)}} = \frac{{3\pi }}{8}$.

$ = [{\tan ^{ - 1}}x]_{ - t}^t$

$ = 2{\tan ^{ - 1}}t$

Differentiating with respect to $t$,

$f'(t) = \frac{2}{{1 + {t^2}}}$

==> $f'(1) = \frac{2}{2} = 1$.

Applying Leibnitzaes theorem,

$F\,'(x) = \log {x^3}.\frac{d}{{dx}}{x^3} - \log {x^2}.\frac{d}{{dx}}{x^2}$

$ = 3\log x \cdot 3{x^2} - 2\log x \cdot 2x$

$ = (9{x^2} - 4x)\log x$.

Integrating by parts, we obtain

$f(x)=t \int_{0}^{x} \sin t d t-\int_{0}^{x}\left\{\left(\frac{d}{d t} t\right) \int \sin t \, d t\right\} d t$

$=[t(-\cos t)]_{0}^{x}-\int_{0}^{x}(-\cot t) d t$

$=[-t \cos t+\sin t]_{0}^{x}$

$=-x \cos x+\sin x$

$\Rightarrow f^{\prime}(x)=-[\{x(-\sin x)\}+\cos x]+\cos x$

$=x \sin x-\cos x+\cos x$

$=x \sin x$

Hence, the correct Answer is $B$

On differentiating both sides , we get $f(x) = 0 + f'(x)$

We know $\frac{d}{{dx}}({e^x}) = {e^x},\,\,$

$\therefore \,\,f(x) = c{e^x}$.

Put $\cos ^2 x = t$

$\Rightarrow 2 \cos x . \sin x . d x=-d t$

when $x =0, t =1$

$x =\frac{\pi}{2}, t =0$

$=\frac{1}{2} \int \limits_1^0 \frac{- dt }{1+ t ^2}=\frac{1}{2} \int \limits_0^1 \frac{ dt }{1+ t ^2}$

$=\left.\frac{1}{2} \tan ^{-1}( t )\right|_0 ^1=\frac{\pi}{8}$

Put $x =\tan \theta$

$dx =\sec ^2 \theta d \theta$

$I=\int \limits_0^{\pi / 2} \frac{\sec ^2 \theta}{\left(1+\tan ^2 \theta\right)(1+\tan \theta)^2} d \theta$

$=\int \limits_0^{\pi / 2} \frac{\cos ^2 \theta}{(\sin \theta+\cos \theta)^2} d \theta$

$I+I=\int \limits_0^{\pi / 2} \frac{d \theta}{(\sin \theta+\cos \theta)^2}$

$=\int \limits_0^{\pi / 2} \frac{\sec ^2 \theta}{(\tan \theta+1)^2} d \theta$

$=-\left.\frac{1}{(1+\tan \theta)}\right|_0 ^{\frac{\pi}{2}}=1$

$\Rightarrow I =\frac{1}{2}$

$\int \limits_1^M(x-[x])^{[x]} d x$

for $M=2$

$\int \limits_1^2(x-1)^1 d x=\left.\frac{(x-1)^2}{2}\right|_1 ^2=\frac{1}{2}$

for $M=3$

$\int \limits_1^2(x-1)^1 d x+\int_2^3(x-2)^2 d x=\frac{1}{2}+\left.\frac{(x-2)^3}{3}\right|_2 ^3=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$

for $M =4$

$\int \limits_1^2(x-1)^1 d x+\int \limits_2^3(x-2)^2 d x+\int \limits_3^4(x-3)^3 d x$

$=\frac{5}{6}+\left.\frac{(x-3)^4}{4}\right|_3 ^4=\frac{5}{6}+\frac{1}{4}=\frac{26}{24} > 1$

$\int \limits_0^1(f(x))^2 d x-2 \int \limits_0^1 f(x) d x=0$

$\Rightarrow \int \limits_0^1\left(( f ( x ))^2-2 f ( x )+1-1\right) dx =0$

$\Rightarrow \int \limits_0^1( f ( x )-1)^2 dx =\int \limits_0^1 dx =1$

$f ( x )-1=1,-1, \sqrt{2 x },-\sqrt{2 x }, \sqrt{3 x }$

many more such function can be defined

Let $f(x)+x^2=g^2(x) \quad(g(x) > 0)$

$\Rightarrow \int_0^{3 / 2}\left(4\left( g ^2( x )- x ^2\right)+\frac{125}{ g ( x )}\right) dx =108$

$\Rightarrow \int_0^{3 / 2}\left(4 g ^2( x )+\frac{125}{ g ( x )}\right) dx =108+\left.\frac{4 x ^3}{3}\right|_0 ^{3 / 2}$

$=108+\frac{9}{2}=112.5$

Also $4 g^2(x)+2 \cdot \frac{125}{2 g(x)} \geq 3(125-125)^{1 / 3}$

(From A.M. $\geq$ G.M.)

$\Rightarrow 4 g^2(x)+\frac{125}{g(x)} \geq 3 \times 25$

$\Rightarrow \int \limits_0^{3 / 2}\left(4 g ^2( x )+\frac{125}{ g ( x )}\right) dx \geq \int \limits_0^{3 / 2} 75 dx =112.5$

$\Rightarrow$ only equality holds

$\Rightarrow 4 g ^2( x )=\frac{125}{ g ( x )}$

$g ( x )=\frac{5}{2}$ or $f ( x )=\frac{25}{4}- x ^2$

Let $I=\int \limits_1^3\left((x-2)^4 \sin ^3(x-2)\right.$

$\left.+(x-2)^{2019}+1\right) d x$

Since, $\int \limits_a^b f(x) d x=\int_a^b f(a+b-x) d x$

$\therefore \quad I=\int \limits_1^3\left\{(4-x-2)^4 \sin ^3(4-x-2)\right.$

$\left.\quad+(4-x-2)^{2019}+1\right\} d x$

$\therefore \quad I=\int \limits_1^3\left\{-(x-2)^4 \sin ^3(x-2)\right.$

$\left.\quad-(x-2)^{2019}+1\right\} d x$

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

$2 I=\int \limits_1^3 2 d x \Rightarrow I=\int \limits_1^3 d x=2$

$I_n=\int \limits_0^\pi \frac{x \cdot \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$I_n=\int \limits_0^\pi \frac{(\pi-x) \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

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

$2 I_n =\pi \int \limits_0^\pi \frac{\sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$\quad 2 I_n =2 \pi \int \limits_0^{\pi / 2} \frac{\sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$

$\quad I_n =\pi \int_0^{\pi / 2} \frac{\sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x \ldots \text { (iii) }$

$\Rightarrow \quad I_n =\pi \int_0^{\pi / 2} \frac{\cos ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x \ldots \text { (iv) }$

On adding Eqs. $(iii)$ and $(iv)$, we get

$2 I_n=\pi \int_0^{\pi / 2} 1 \cdot d x=\frac{\pi^2}{2} \Rightarrow I_n=\frac{\pi^2}{4}$

$I_n$ is constant for any $n \in N$.

We have, $\int \limits_1^{\sqrt{2}+1} \frac{\left(1+\frac{1}{x^2}\right) d x}{\left(x+\frac{1}{x}\right) \sqrt{x^2+\frac{1}{x^2}}}$

Let $x+\frac{1}{x}=t \Rightarrow\left(1-\frac{1}{x^2}\right) d x=d t$

$\int \limits_2^{2 \sqrt{2}} \frac{d t}{t \sqrt{t^2-2}}$

Let $t=\sqrt{2} z \Rightarrow d t=\sqrt{2} d z$

$=\frac{1}{\sqrt{2}} \int \limits_{\sqrt{2}}^2 \frac{d z}{z \sqrt{z^2-1}}$

$=\left.\frac{1}{\sqrt{2}} \sec ^{-1} z\right|_{\sqrt{2}} ^2=\frac{1}{\sqrt{2}}\left[\frac{\pi}{3}-\frac{\pi}{4}\right]$

$=\frac{\pi}{12 \sqrt{2}}$

We have, $\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+e^x} d x$

$+\int_0^{\pi / 2}\left(\frac{\sin ^2 x}{1+e^x}+\frac{\sin ^2(-x)}{1+e^{-x}}\right) d x$

$=\int_0^{\pi / 2}\left(\frac{\sin ^2 x}{1+e^x}+\frac{e^x \sin ^2 x}{1+e^x}\right) d x$

$=\int_0^{\pi / 2} \sin ^2 x d x=\int_0^{\pi / 2}\left(\frac{1-\frac{\cos 2 x}{2}}{2}\right) d x$

$=\frac{1}{2}\left(x-\frac{\sin 2 x}{2}\right)_0^{\pi / 2}=\frac{\pi}{4}$

Let $n=\frac{1}{m}$

So, $\lim _{m \rightarrow 0} I_n=\lim _{m \rightarrow 0} \frac{\int_0^m x \cdot e^{-x / m}}{\sin x} d x$

Let $x=m t \Rightarrow d x=m d t$

$\lim _{m \rightarrow 0} I_n=\lim _{m \rightarrow 0} \frac{\int_0^1 \frac{m^2 \cdot e^{-t} \cdot t}{\sin (m t)} d t}{\sqrt{m}}$

$=\lim _{m \rightarrow 0} \sqrt{m} \int_0^1\left(\frac{m t}{\sin (m t)}\right) \cdot e^{-t} d t$

$\lim _{m \rightarrow 0} \sqrt{m} \int_0^1 e^{-t} d t=\lim _{m \rightarrow 0} \sqrt{m}\left(1-e^{-1}\right)=0$