Maths M.C.Q (1 Marks)

As we know, $1+x^8<2, \forall x \in(0,1)$

$\Rightarrow \quad \frac{1}{2}<\frac{1}{1+x^8}, \forall x \in(0,1)$

$\Rightarrow \quad \frac{x}{2}<\frac{x}{1+x^8}, \forall x \in(0,1)$

$\Rightarrow \int_0^1 \frac{x}{2} d x<\int_0^1 \frac{x}{1+x^8} d x$

$\Rightarrow \int_0^1 \frac{x}{1+x^8} d x>\frac{1}{4}$

$\Rightarrow \quad \text { Similarly, as we know, } 1+x^8<1+x^4$

$\forall x \in(0,1)$

$\Rightarrow \int_0^1 \frac{x}{1+x^4} d x<\int_0^1 1+x^8 d x$

$\Rightarrow J>\frac{1}{2}\left[\tan { }^{-1}\left(x^2\right)\right]_0^1$

$\Rightarrow J>\frac{\pi}{8}$

$\therefore$ Statement $J=\int_0^1 \frac{x}{1+x^8} d x>\frac{1}{4}$ is true, while the statement $J=\int_0^1 \frac{x}{1+x^8} d x<\frac{\pi}{8}$ is not true.

It is given that a continuous function

$f:[0, \infty) \rightarrow R$ satisfies

$f(x) =2 \int_0^x t f(t) d t+1, \forall x \geq 0$

$\therefore  f^{\prime}(x) =2 x f(x)$

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

On integrating both sides, we get

$\ln |f(x)|=x^2+c$

$\Rightarrow f(x)=K e^{x^2}, \text { where } K=e^c$

$\because  f(0)=2 \int_0^0 t f(t) d t+1=1$

$\therefore  K=1$

$\therefore  f(1)=e$

Given, for $x \in R$

$f(x)=|\sin x|$

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

$p(x)=g(x)-\frac{2}{\pi} x$

$\because p(x+\pi)=g(x+\pi)-\frac{2}{\pi}(x+\pi)$

$=\int \limits_0^{x+\pi}|\sin (t)| d t-\frac{2}{\pi} x-2$

$=\int \limits_0^\pi|\sin t| d t+\int \limits_\pi^{\pi+x}|\sin t| d t-\frac{2}{\pi} x-2$

$=2 \int \limits_0^{\pi / 2} \sin t d t+\int \limits_0^x|\sin t| d t-\frac{2}{\pi} x-2$

$[\because|\sin t|$ is periodic function having period $\pi$ ]

$2[-\cos t]_0^{\pi / 2}+g(x)-\frac{2}{\pi} x-2$

$2(0-(-1))+g(x)-\frac{2}{\pi} x-2$

$2+g(x)-\frac{2}{\pi} x-2=g(x)-\frac{2}{\pi} x=p(x)$

$\therefore p(x+\pi)=p(x)$ for all $x$.

For a continuous function $f:[0,1] \longrightarrow[0,1]$, it is given that

$x^2+(f(x))^2 \leq 1$

$\Rightarrow(f(x))^2 \leq 1-x^2$

$\left.\Rightarrow \quad f(x) \leq \sqrt{1-x^2} \quad \quad \quad \because f(x) \in[0,1]\right\}$

$\therefore \int_0^1 f(x) d x \leq \int_0^1 \sqrt{1-x^2} d x$

$\Rightarrow \int_0^1 f(x) d x \leq\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right]_0^1$

$\Rightarrow \int_0^1 f(x) d x \leq \frac{\pi}{4}$

So, $f(x)=\sqrt{1-x^2}$, because it is given that

$lint_0^1 f(x)=\frac{\pi}{4}$

Now,

$\int_{1 / 2}^{1 / \sqrt{2}} \frac{f(x)}{1-x^2} d x=\int_{1 / 2}^{1 / \sqrt{2}} \frac{d x}{\sqrt{1-x^2}}=\left[\sin ^{-1} x\right]_{1 / 2}^{1 / \sqrt{2}}$

$=\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12}$

Let $I=\int_0^\pi(1-|\sin 8 x|) d x$

$I=\int_0^\pi d x-\int_0^\pi|\sin 8 x| d x$

$I=\pi-\int_0^{\frac{\pi}{8} \times 8}|\sin 8 x| d x$

$I=\pi-8 \int_0^{\frac{\pi}{8}} \sin 8 x d x[\because \sin 8 x B$ periodic

$I=\pi-8\left[\frac{-\cos 8 x}{8}\right]_0^{\pi / 8}$

$I=\pi+(\cos \pi-\cos 0)=\pi-2$

We have,

$L= \lim _{n \rightarrow \infty} \sqrt{n} \int \limits_0^1 \frac{1}{\left(1+x^2\right)^n} d x$ $\left(1+x^2\right)^n>1+n x^2$

$\left(1+x^2\right)^n > 1+n x^2$

$\frac{1}{\left(1+x^2\right)^n} < \frac{1}{1+n x^2}$

$\int \limits_0^1 \frac{d x}{\left(1+x^2\right)^n} < \int_0^1 \frac{d x}{1+n x^2}$

$\int \limits_0^1 \frac{d x}{\left(1+x^2\right)^n}<\frac{1}{\sqrt{n}}\left[\tan ^{-1} \sqrt{n} x\right]_0^1$

$\int \limits_0^1 \frac{d x}{\left(1+x^2\right)^n}<\frac{1}{\sqrt{n}} \tan ^{-1} \sqrt{n}$

$L=\lim _{n \rightarrow \infty} \sqrt{n} \int_0^1 \frac{1}{\left(1+x^2\right)^n} d x$

$<\lim _{n \rightarrow \infty} \frac{\sqrt{n}}{\sqrt{n}}\left(\tan ^{-1} \sqrt{n}\right)$

$L<\tan ^{-1} \infty=\frac{\pi}{2}$

$\frac{1}{2} < L < 2$

$\int \limits_0^x f(t) d t =p f(x)$

$f(x) =p f^{\prime}(x)$

$\frac{f^{\prime}(x)}{f(x)} =\frac{1}{p}$

$\log f(x) =\frac{x}{p}+C$

$f(x) =A e^{x / p}$

Putting $x=0$

$f(0)=A e^0 \Rightarrow A=0 \quad[\because f(0)=0]$

$f(x) =0$

$p(0) =0$

$f(0) =0\,\,p \neq 0$

Case$I$

There is no non-zero continuous to $f(x)$.

Case$II$

$p=0$

$\int \limits_0^x f(t) d t=0, \forall x \in R$

If is possible when $f(x)=0$.

Hence, $\forall p \in R$. There is no non-zero continous function. Hence, $S \in R$.

We have,

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

$f(x) =\int \ln \left(x^2-1\right) d x$

$f(x) =x \ln \left(x^2-1\right)-\int \frac{2 x^2}{x^2-1} d x$

$f(x)=x \ln \left(x^2-1\right)-2 \int\left(\frac{x^2-1}{x^2-1}+\frac{1}{x^2-1}\right) d x$

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

$f(2)=2 \ln (3)-4-\ln \left(\frac{1}{3}\right)+C=0$

$\Rightarrow C=4-3 \ln 3$

$\therefore f(x)=x \ln \left(x^2-1\right)-2 n \ln \frac{(x-1)}{(x+1)}+4-3 \ln 3$

defincd for Sinfinite $C$ values possible in set $S$ such that $f^{\prime}(x)=\ln \left(x^2-1\right)$

We have, $J_n=\int_0^1 e^{-y} y^n d y$

$I_n =\left[-y^n e^{-y}\right]_0^1+\int_0^1 n y^{n-1} e^{-y} d y$

$I_n =-\frac{1}{e}+n I_{n-1}$

$\Rightarrow \frac{I_n}{n !} =\frac{-1}{n ! e}+\frac{I_{n-1}}{(n-1) !}$

$\Rightarrow \frac{I_n}{n !}-\frac{I_{n-1}}{(n-1) !}=\frac{-1}{n ! e}$

$n =1 \frac{I_1}{1 !}-\frac{I_0}{0 !}=-\frac{1}{e}$

$n =2 \frac{I_2}{2 !}-\frac{I_1}{1 !}=-\frac{1}{2 ! e}$

$n =3 \frac{I_3}{3 !}-\frac{I_1}{2 !}=-\frac{1}{3 ! e}$

$n =n \frac{I_n}{n !}-\frac{I_n}{n-1}=\frac{1}{n ! e}$

$\text { Adding these form, we get }$

$\frac{I_n}{n !}-I_0 =-\frac{1}{e}\left(1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots+\frac{1}{n !}\right)$

$\frac{I_n}{n !} =\frac{I_0}{n !}-\frac{1}{e}\left(1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots+\frac{1}{n !}\right)$

$n ! =\frac{1}{e}+1-\frac{1}{e}(e)\left[\because e=1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots\right]$

$=\frac{1}{e}+1-1$

We have,

$f(x)=\max \left\{3, x^2, \frac{1}{x^2}\right\} \text { for } x \in\left[\frac{1}{2}, 2\right]$

$\int \limits_{1 / 2}^2 f(x) d x=\int \limits_{1 / 2}^{1 / \sqrt{3}} \frac{1}{x^2} d x+\int \limits_{1 / \sqrt{3}}^{\sqrt{3}} 3 d x+\int \limits_{\sqrt{3}}^2 x^2 d x$

$\left[-\frac{1}{x}\right]_{1 / 2}^{1 / \sqrt{3}}+[3 x]_{1 / \sqrt{3}}^{\sqrt{3}}+\left[\frac{x^3}{3}\right]_{\sqrt{3}}^2$

$=-\sqrt{3}+2+3 \sqrt{3}-3 \sqrt{3}+\frac{8}{3}-\frac{3 \sqrt{3}}{3}=\frac{14}{3}$

We have,

$g(x) =\int_0^{|x|^{3 / 4}} t^{23} \sin \frac{1}{t} d t$

$g^{\prime}(x) =|x|^{\frac{1}{2}} \sin \frac{1}{|x|^{3 / 4}}$

$\lim _{x \rightarrow 0} \frac{g(x)}{x} =\lim _{x \rightarrow 0} \frac{g^{\prime}(x)}{1}$

$=\lim _{x \rightarrow 0}||^{1 / 2} \sin \frac{1}{|x|^{3 / 4}}$

$=0$

We have,

$a_n=\int \limits_{-\pi}^\pi|x-1| \cos n x d x$

$a_n=\int \limits_{-\pi}^1-(x-1) \cos n x d x$

$a_n=-\left[\frac{(x-1) \sin n x]^1}{n}\right]_{-\pi}^{-}-\left[\frac{\cos n \pi}{n^2}\right]_{-\pi}^1$

$\quad+\left[\frac{(x-1) \sin n x}{n}\right]_1^\pi-\left[\frac{\cos n \pi}{n^2}\right]_1^\pi$

$a_n=\frac{2 \pi \sin n \pi}{n}+\frac{2}{n^2} \cos n \pi-\frac{2}{n^2} \cos x \pi$

$a_n=\frac{2 \pi \sin n \pi}{n}$

$\lim _{x \rightarrow \infty} a_n=\lim _{n \rightarrow \infty} \frac{2 \pi \sin n \pi}{n}=0$

We have,

$\int_0^1 x f(x) d x=\frac{1}{3}+\frac{1}{4} \int \limits_0^1(f(x))^2 d x$

$\Rightarrow -\frac{1}{3}=\frac{1}{4}\left[\int \limits_0^1\left[(f(x))^2-4 x f(x)\right] d x\right.$

$\Rightarrow-\frac{1}{3}=\frac{1}{4}\left[\int \limits_0^1(f(x))^2-4 x f(x)+(2 x)^2-(2 x)^2\right] d x$

$\Rightarrow-\frac{1}{3}=\frac{1}{4}\left[\int \limits_0^1[f(x)-2 x]^2 d x-\int \limits_0^1 4 x^2 d x\right]$

$\Rightarrow-\frac{4}{3}=\int \limits_0^1[f(x)-2 x]^2 d x-\left[\frac{4 x^3}{3}\right]_0^1$

$\Rightarrow-\frac{4}{3}=\int \limits_0^1[f(x)-2 x]^2 d x-\frac{4}{3}$

$\Rightarrow \quad \int \limits_0^1[f(x)-2 x]^2 d x=0$

$\therefore$ Only one continuous function.

We have,

$\lim _{x \rightarrow \infty} x^2 \int \limits_0^x e^{t^3-x^3} d t =\lim _{x \rightarrow \infty} x^2 e^{-x^3} \int \limits_0^x e^{e^3} d t$

$=\lim _{x \rightarrow \infty} \frac{x^2 \int \limits_0^x e^{t^3} d t}{e^{x^3}}$

Apply $L$ Hospital's rule

$=\lim _{x \rightarrow \infty} \frac{2 x \int \limits_0^x e^{t^3} d t+x^2 e^{x^3}}{3 x^2 e^{x^3}}$ $=\lim _{x \rightarrow \infty} \frac{2 \int \limits_0^x e^{t^3} d t+x^2 e^{x^3}}{3 x^2 e^{x^3}}$

Again Apply $L$ Hospital's rule, we get

$=\lim _{x \rightarrow \infty} \frac{2 e^{x^3}+e^{x^3}+3 x^3 e^{x^3}}{3 e^{x^3}+9 x^3 e^{x^3}}$

$=\lim _{x \rightarrow \infty} \frac{e^{x^3}\left(3+3 x^3\right)}{e^{x^3}\left(3+9 x^3\right)}=\frac{1}{3}$

We have, $f(x)=\min \{x-[x], 1-x+[x]\}$ $g(x)=\max \{x-[x], 1-x+[x]\}$ Graph of $f(x)$ is

$\int \limits_0^n g(x) d x=\int_0^n g(x) d x=\frac{3 n}{4}$

$\int \limits_0^n f(x) d x=\frac{n}{4}$

$\because \quad \int_0^n(g(x)-f(x)) d x=\frac{3 n}{4}-\frac{n}{4}=100$

$\Rightarrow \quad \frac{n}{2}=100 \Rightarrow n=200$

We have,

$\int \limits_0^1 f^2(x) d x=\left(\int  \limits_0^1 f(x) d x\right)^2, x \in[0,1]$

We know Cauchy Schwartz inequality

${\left[\int_a^b f(x) \cdot g(x) d x\right]^2 \leq \int_a^b(f(x))^2 d x }$

$\int_a^b(g(x))^2 d x$

Here, $g(x)=1$ and equality holds only when $\frac{f(x)}{g(x)}=\lambda$

So, $f(x)$ is constant function.

$\therefore f(x)$ is a singleton.

We have,

$f(x)=\max \{|x|,|x-1|, \ldots,|x-2 n|\}$

$\int \limits_0 ^{2 n} f(x) d x=\int \limits_0^n f(x) d x+\int \limits_n^{2 n} f(x) d x$

$=\int \limits_0^n|x-2 n| d x+\int \limits_n^{2 n}|x| d x$

$=\int \limits_0^n(2 n-x) d x+\int \limits_n^{2 n} x d x$

$=\left[2 n x-\frac{x^2}{2}\right]_0^n+\left[\frac{x^2}{2}\right]_n^{2 n}$

$=\left(2 n^2-\frac{n^2}{2}\right)+\left(\frac{4 n^2}{2}-\frac{n^2}{2}\right)$

$=2 n^2-\frac{n^2}{2}+2 n^2-\frac{n^2}{2}=3 n^2$

We have,

$p(x)$ is cubic polynomial.

Let $p(x)=a x^3+b x^2+c x+d$

$p(1)=a+b+c+d=3$

$p(0)=d=2$

$p(-1)=-a+b-c+d=4$

From Eqs.$(i)$ and $(iii)$, we get

$2(b+d)=7 \Rightarrow b+d=\frac{7}{2}$

On putting the value of $d$ in Eq. $(iv)$, we get

$b=\frac{3}{2}$

Now, $\int \limits_{-1}^1 p(x) d x=\int \limits_{-1}^1\left(a x^3+b x^2+c x+d\right) d x$

$=2 \int \limits_{-1}^1\left(b x^2+d\right) d x$

$\because a x^3$ and $c x$ are odd functions]

$=2\left[\frac{b x^3}{3}+d x\right]_0^1=2\left(\frac{b}{3}+d\right)$

$=2\left[\frac{3}{2 \times 3}+2\right]=2\left[\frac{5}{2}\right]=5$

Let $I=\int \limits_0^{\infty} e^{-t}|x-t| d t, x > 0$

$\Rightarrow \quad I=\int \limits_0^x e^{-t}(x-t) d t+\int \limits_x^{\infty} e^{-t}(t-x) d t$

$\Rightarrow \quad I=\left[-(x-t) e^{-t}\right]_0^x+\int \limits_0^x \frac{d(x-t)}{d x} e^{-t} d t$

$+\left[\frac{(t-x) e^{-t}}{-1}\right]_x^{\infty}+\int \limits_x^{\infty} e^{-t} d t$

$\Rightarrow I=x+\left[e^{-t}\right]_0^x+\left[-e^{-t}\right]_x^{\infty}$

$\Rightarrow I=x+e^{-x}-1+e^{-x}$

$\Rightarrow I=x+2 e^{-x}-1$

We have,

$f(x)+\int_0^x t f(t) d t+x^2=0$

On differentiating, we get

$f^{\prime}(x)+x f(x)+2 x =0$

$\Rightarrow f^{\prime}(x) =-x(f(x)+2)$

$\Rightarrow \frac{f^{\prime}(x)}{f(x)+2} =-x$

On integrating, we get

$\quad \log (f(x)+2)=-\frac{x^2}{2}+C$

$\Rightarrow \quad f(x)=A e^{-x^2 / 2}-2$

$\Rightarrow \quad f(0)=0=A-2 \Rightarrow A=2$

$\therefore \quad \lim _{x \rightarrow \infty} f(x)=-2$

$\qquad \quad \lim ^{-x^2 / 2}-2$

$\Rightarrow(x) \text { is an even function. }$

$f(x) \text { intersect } X \text {-axis at one point }(0,0)$

$\therefore \text { Option (b) is correct. }$

Let $I=\int \limits_1^n[x][\sqrt{x}] d x$

$\Rightarrow I=\int \limits_1^2 d x+\int \limits_2^3 d x+\int \limits_3^4 3 d x+\int \limits_4^5 8 d x+\int \limits_5^6 10 d x$

$+\int \limits_6^7 12 d x+\int \limits_7^8 14 d x+\int \limits_8^9 16 d x+\int \limits_9^{10} 27 d x+\ldots$

But $I > 60$ $I=1+2+3+8+10+12+14+16=66$

So, least value of $n=9$

Let $I=\int \limits_0^n \cos (2 \pi[x]\{x\}) d x$

$\begin{aligned} \{x\}=x-[x] \\ \therefore \quad I =\int\limits_0^n \cos [2 \pi[x](x-[x])] d x \end{aligned}$

$\Rightarrow \quad I=\int \limits_0^1 \cos 0 d x+\int \limits_1^2 \cos 2 \pi(x-1) d x$

$+\int \limits_2^3 \cos 2 \pi(x-2) d x$

$+\ldots+\int \limits_{n=1}^n \cos 2 \pi(x-(n-1) d x$

$\Rightarrow \quad I=1+\int \limits_1^2 \cos 2 \pi x d x+\int \limits_2^3 \cos 4 \pi x d x$

$+\ldots+\int \limits_{n-1}^n \cos 2 \pi(n-1) x d x$

$\Rightarrow \quad I=1+\left[\frac{\sin 2 \pi x}{2 \pi}\right]_1^2+\left(\frac{\sin 4 \pi x}{4 \pi}\right)_2^3+\ldots$

$+\left[\frac{\sin 2 \pi(n-1) x}{2 \pi(n-1)}\right]_{n-1}^n$

$\Rightarrow \quad I=1+0+0+\ldots+0 \Rightarrow I=1$

We have, $f(x)=x+\int \limits_0^x f(t) d t$

$f^{\prime}(x)=1+f(x)$

On integrating, we get

$\log (1+f(x))=x+c$

$1+f(x)=A e^x$

$f(x)=A e^x-1$

$f(0)=A-1$

$0=A-1 \quad[\because f(0)=0]$

$\Rightarrow \quad A=1$

$\therefore \quad f(x)=e^x-1$

$f(y)=e^y-1$

$f(x+y)=e^{x+y}-1=e^x \cdot e^y-1$

$f(x)+f(y)+f(x) f(y)$

$=e^x-1+e^y-1+\left(e^x-1\right)\left(e^y-1\right)$

$=e^x-1+e^y-1+e^{x+y}-e^x-e^y+1$

$=e^{x+y}-1$

Hence, $f(x+y)=f(x)+f(y)+f(x) f(y)$

$\therefore F ( x ) \geq 0$

$\int_0^1 f(x)^2 d x \leq 100$

not necessarily true.

because $(f(x))^2$ can take very high values then area bounded by $(f(x))^2, x$-axis $x$ $=0$ to $1$ may cross $100$

We have,

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

$\Rightarrow \quad I_n =\left[x^n \sin x\right]_0^{\pi / 2}-\int \limits_0^{\pi / 2} n x^{n-1} \sin x d x$

$\Rightarrow \quad I_n =\left(\frac{\pi}{2}\right)^n-\left[n x^{n-1}(-\cos x)\right]_0^{\pi / 2}$

$\Rightarrow \quad I_n =\left(\frac{\pi}{2}\right)^n-n(n-1) I_{n-2}^{\pi / 2} n(n-1) x^{n-2} \cos x d x$

$\Rightarrow \quad I_n+n n(n-1) I_{n-2}=\left(\frac{\pi}{2}\right)^n$

$\text { Now, } \sum \limits_{n=2}^{\infty}\left[\frac{I_n}{n !}+\frac{I_{n-2}}{(n-2) !}\right]$

$\Rightarrow \quad \sum \limits_{n=2}^{\infty}\left[\frac{I_n}{n !} n(n-1) I_{n-2}\right]$

$\sum \limits_{n=2}^{\infty} \frac{\left(\frac{\pi}{2}\right)^n}{n !}=e^{\pi / 2}-1-\frac{\pi}{2}$

We have,

$p_t(x)=(\sin t) x^2-(2 \cos t) x+\sin t$

$A(t)=\int_0^1 p_t(x) d x$

$A(t)=\int_0^1\left((\sin t) x^2-(2 \cos t) x+\sin t\right) d x$

$A(t)=\left[\frac{x^3}{3} \sin t-x^2 \cos t+x \sin t\right]_0^1$

$A(t)=\frac{1}{3} \sin t-\cos t+\sin t$

$A(t)=\frac{4}{3} \sin t-\cos t$

$A^{\prime}(t)=\frac{4}{3} \cos t+\sin t$

$A(t) \text { has infinitely critical points. }$

$A(t)=0 \text { for infinitely many value of } t$

$\text { Hence, II and III statements are truc. }$

Let $I=\int \limits_1^n[x]\{x\} d x$

$I=\int \limits_1^2\{x\} d x+\int\limits_2^3 2\{x\} d x+\ldots+\int\limits_{n-1}^n(n-1)\{x\} d x$

$I=(1+2+3+4+\ldots+(n-1)) \int\limits_0^1\{x\} d x$

$L=\frac{n(n-1)}{2} \int\limits_0^1 x d x$

$I=\frac{n(n-1)}{2}\left[\frac{x^2}{2}\right]_0^1=\frac{n(n-1)}{4}$

Given, $I \geq 2013$

$\frac{n(n-1)}{4} \geq 2013$

$\begin{aligned} \Rightarrow & & n^2-n & \geq 4052 \\ \Rightarrow & & n^2-n+\frac{1}{4} & \geq 4052+\frac{1}{4} \end{aligned}$

$\Rightarrow \quad\left(n-\frac{1}{2}\right)^2 \geq \frac{32209}{4}$

$\Rightarrow \quad n \geq \frac{\sqrt{32209}}{2}+\frac{1}{2}$

$\Rightarrow \quad n \geq \frac{17946+1}{2}$

$=\frac{180.46}{2}$

$\Rightarrow \quad n \geq 90.23$

$\therefore$ Least value of $x$ is $91$

Let

$I=\int \limits_1^{n+1} \frac{(\{x\})^{[x]}}{[x]} d x$

$I=\int \limits_1^2 \frac{\{x\}}{1} d x+\int \limits_2^3 \frac{\{x\}^2}{2} d x+\int \limits_3^4 \frac{\{x\}^2}{3} d x+$ $\cdots \int \limits_n^{n+1} \frac{\{x\}^n}{n} d x$

$\Rightarrow \quad I=\int \limits_0^1\left(\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\ldots+\frac{x^n}{n}\right) d x$

$\Rightarrow I=\left[\frac{x^2}{1 \times 2}+\frac{x^3}{2 \times 3}+\frac{x^4}{3 \times 4}+\ldots+\frac{x^{n+1}}{n(n+1)}\right]_0^1$

$\Rightarrow I=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots+\frac{1}{n(n+1)}$

$\Rightarrow I=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots$

$\Rightarrow I=1-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}$

We have,

$f(x)=\int \limits_{-10}^x 2^{[t]} d t, x \in[-10,10]$

Let $r$ be an integer $r \in[-10,10]$

For LHL $\lim \limits_{x \rightarrow r^{-}} \int \limits_{-10}^x 2^{[t]} d t$

$\lim \limits_{x \rightarrow r-h} \int \limits_{-10}^{r-h} 2^{[t]} d t$

$\lim \limits_{h \rightarrow 0}\left[\int \limits_{-1}^{-9} 2^{[t]} d t+\int \limits_{-9}^{-8} 2^{[t]} d t+\ldots+\int \limits_{r-1}^{r-h} 2^{[t]} d t\right]$

$\lim \limits_{h \rightarrow 0}\left[2^{-10}+2^{-9}+2^{-8}+\ldots+2^{r-1}(1-h)\right]$

$2^{-10}+2^{-9}+2^{-8}+\ldots+2^{r-1}$

$RHL \lim \limits_{x \rightarrow r^{+}} \int\limits_{-10}^x 2^{[t]} d t \Rightarrow \lim \limits_{x \rightarrow r+h} \int\limits_{-10}^{r+h} 2^{[t]} d t$

$\lim \limits_{h \rightarrow 0}\left[\int\limits_{-10}^{-9} 2^{[t]} d t+\int\limits_{-9}^{-8} 2^{[t]} d t+\ldots+\int\limits_r^{r+h} 2^{[t]} d t\right]$

$=2^{-10}+2^{-9}+2^{-8}+\ldots+2^{r-1}$

$f(r)=\int\limits_{-10}^r 2^{[t]} d t=2^{-10}+2^{-9}+\ldots+2^{r-1}$

$\lim \limits_{x \rightarrow r^{-}} f(x)=\lim \limits_{x \rightarrow r^{+}} f(x)=f(x)$

Hence, $f(x)$ is continuous for all $r \in(-10,10)$

Let

$I=\frac{\int \limits_0^{\pi / 2}(\sin x)^{\sqrt{2}+1} d x}{\int \limits_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} d x}$

$I=\frac{\int \limits_0^{\pi i 2}(\sin x)^{\sqrt{2}} \sin x d x}{\int \limits_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} d x}$

$\quad\left[-(\sin x)^{\sqrt{2}} \cos x\right]_0^{\pi / 2}+\sqrt{2}$

$I=\frac{\int \limits_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} \cos ^2 x d x}{\int \limits_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} d x}$

$I=\frac{\sqrt{2} \int \limits_0^{\pi / 2}(\sin x)^{\sqrt{2}-1}\left(1-\sin ^2 x\right) d x}{\sqrt{2}\left[\int \limits_0^{\pi / 2}(\sin x)^{\sqrt{2}-1}-\int \limits_0^{\pi / 2}(\sin x)^{\sqrt{2}+1} d x\right]}$

$I=\frac{\pi(2 \sin x)^{\sqrt{2}-1} d x}{\int \limits_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} d x}$

$I=\sqrt{2}\left[1-\frac{\int \limits_0^{\pi / 2}(\sin x)^{\sqrt{2}+1} d x}{\pi / 2} \int \limits_0^{\sqrt{2}}(\sin x)^{\sqrt{2}-1} d x\right]$

$I=\sqrt{2}[1-I]$

$I=\sqrt{2}-\sqrt{2} I \Rightarrow I+\sqrt{2} I=\sqrt{2}$

$I=\frac{\sqrt{2}}{\sqrt{2}+1} \Rightarrow I=\frac{\sqrt{2}(\sqrt{2}-1)}{2-1}$

$I=2-\sqrt{2}$

Let

$I =\int \limits_{-2012}^{2012}\left(\sin \left(x^3\right)+x^5+1\right) d x$

$\Rightarrow I =\int \limits_{-2012}^{2012} \sin x^3 d x+\int \limits_{-2012}^{2012} x^5 d x+\int \limits_{-2012}^{2012} d x$

$\Rightarrow I =0+0+2 \int \limits_0^{2012} d x$

$\Rightarrow I {\left[\because \sin x^3 \text { and } x^5 \text { are odd function }\right] }$

$=2[x]_0^{2012}=2 \times(2012)=4024$

Let $I=\int \limits_0^5[x]\{x\} d x$

$I=\int \limits_0^1 0\{x\} d x+\int \limits_1^2 1 \cdot\{x\} d x+\int \limits_2^3 2\{x\} d x$

$+\int \limits_3^4 3\{x\} d x+\int \limits_4^5 4\{x\} d x$

$I=0+1 \int \limits_0^1 x d x+2 \int \limits_0^1 x d x+3 \int \limits_0^1 x d x$

$I=(1+2+3+4) \int \limits_0^1 x d x \quad+4 \int \limits_0^1 x d x$

$I=10\left[\frac{x^2}{2}\right]_0^1 \Rightarrow I=10\left(\frac{1}{2}-0\right)=5$

Let $I=\int \limits_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x$

$\Rightarrow \quad I=\int \limits_{-\pi}^\pi \frac{\cos ^2(\pi-\pi-x)}{1+a^{\pi-\pi-x}} d x$

$\quad\left[\because \int \limits_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]$

$\Rightarrow \quad I=\int \limits_{-\pi}^\pi \frac{\cos ^2 x}{1+a^{-x}} d x$

$\Rightarrow \quad 2 I=\int \limits_{-\pi}^\pi \cos ^2 x d x$

$\Rightarrow \quad 2 I=2 \int \limits_0^\pi \cos ^2 x d x$

$\Rightarrow \quad I=\int \limits_0^\pi \cos ^2 x d x$

$\Rightarrow \quad I=2 \int \limits_0^{\pi / 2} \cos ^2 x d x$

$\Rightarrow \quad I=2 \int \limits_0^{\pi / 2} \cos ^2\left(\frac{\pi}{2}-x\right) d x$

$I=2 \int \limits_0^{\pi / 2} \sin ^2 x d x$

$2 I=2 \int \limits_0^{\pi / 2}\left(\cos ^2 x+\sin ^2 x\right) d x$

$2 I=2 \int \limits_0^{\pi / 2} x d x \Rightarrow I=\lfloor x]_0^{\pi / 2}=\frac{\pi}{2}$

We have,

$L=(2012)^{1 / 3}+(2013)^{1 / 3}+\ldots+(3011)^{1 / 3}$

$R=(2013)^{1 / 3}+(2014)^{1 / 3}+\ldots+(3012)^{1 / 3}$

and $I=\int \limits_{2012}^{3012}(x)^{1 / 3} d x$

Let $f(x)=x^{1 / 3}$

$n h=b-a$

$n h=3012-2012=1000$

$\Rightarrow \quad n=1000 \quad[\because h=1]$

$I=\frac{b-a}{h}[f(a)+f(a+h)+f(a+2 h)+$

$\quad \ldots+f(a+(n-1) h]$

$I=1000\left[(2012)^{1 / 3}+(2011)^{1 / 3}\right.$

$\left.+\ldots+(3011)^{1 / 3}\right]$

$2 I=1000 \times 2[2012]^{1 / 3}+(2013)^{1 / 3}$

$\left.+\ldots+(3011)^{1 / 3}\right]$

$2 I=1000\left[(2012)^{1 / 3}+L+R-(3012)^{1 / 3}\right]$

$\therefore L+R < 2 I$

Explanation :-

$I =2012 \int \limits_0^1 \frac{e^{\cos \pi x}}{e^{\cos \pi x}+e^{-\cos \pi x}} d x$

using king property

$I =2012 \int \limits_0^1 \frac{e^{-\cos \pi x}}{e^{-\cos \pi x}+e^{\cos \pi x}} d x$

$\Rightarrow 2 I =2012 \Rightarrow I =1006$

We have, $f:(2, \infty) \rightarrow N$

$f(x)=$ The largest prime factor of $[x]$.

Let $I=\int_2^8 f(x) d x$

$\Rightarrow \quad I=\int \limits_2^3 2 d x+\int \limits_3^4 3 d x+\int \limits_4^5 2 d x+\int \limits_5^6 5 d x$

$+\int \limits_6^7 3 d x+\int \limits_7^8 7 d x$

$\Rightarrow \quad I=2+3+2+5+3+7=22$

We have,

$f_n(x) =\operatorname{minimum}\left(\frac{x^n}{n !}, \frac{(1-x)^n}{n !}\right), x \in[0,1]$

$I_n =\int \limits_0^1 f_n(x) d x$

$I_n =\left|\int \limits_0^{1 / 2} \frac{x^n}{n !} d x\right|+\left|\int_{1 / 2}^1 \frac{(1-x)^n}{n !} d x\right|$

$I_n =\left|\left[\frac{x^{n+1}}{(n+1) n !}\right]_0^{1 / 2}\right|+\mid\left[\left.\frac{(1-x)^{n+1}}{(n+1) n !}\right|_{1 / 2} ^1 \mid\right.$

$I_n =\frac{2}{(n+1) !}\left[\left(\frac{1}{2}\right)^{n+1}\right]$

$\sum \limits_{n=1}^{\infty} I_n =2\left[\frac{1}{2 !}\left(\frac{1}{2}\right)^2+\frac{1}{3 !}\left(\frac{1}{2}\right)^3\right]$

$\sum \limits_{n=1}^{\infty} I_n =2\left[e^{1 / 2}-\left(1+\frac{1}{2}\right)\right]$

$=2 \sqrt{e}-3$

Let

$I=\lim _{n \rightarrow \infty}\left(\begin{array}{l}\frac{1}{\sqrt{4 n^2-1}}+\frac{1}{\sqrt{4 n^2-4}}+\ldots \\ +\frac{1}{\sqrt{4 n^2-n^2}}\end{array}\right)$

$=I=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{\sqrt{4-\left(\frac{r}{n}\right)^2}}$

$I=\int \limits_0^1 \frac{1}{\sqrt{4-x^2}} d x$

$I=\left[\sin ^{-1} \frac{x}{2}\right]_0^1=\sin ^{-1} \frac{1}{2}-\sin ^{-1} 0=\frac{\pi}{6}$

We have,

$I_n=\int_1^{\bullet}(\log x)^n d x$

Put $\log x=t \Rightarrow x=e^t$

$\Rightarrow \quad d x=e^t d t$

When $x=1, t=0$

and $\quad x=e \Rightarrow t=1$

$\therefore \quad I_n=\int_0^1 t^n e^t d t$

$\Rightarrow \quad I_n=\left[t^n e^t\right]_0^1-n \int_0^1 t^{n-1} e^t d t$

$\Rightarrow \quad I_n=e-n I_{n-1} \quad[\because$ by parts]

$\Rightarrow \quad I_n+n I_{n-1}=e$

$\Rightarrow \quad I_{2011}+2011 I_{2010}=e$

Similarly, $I_{100}+100 I_{99}=e$

Hence, option $(c)$ is correct.

Let

$I =\int \limits_0^1 \cos (\pi x) \cos [2 x] \pi d x$

$\Rightarrow I =\int \limits_0^{1 / 2} \cos (\pi x) \cos 0 d x+\int \limits_{1 / 2}^1 \cos \pi x \cos \pi d x$

$\Rightarrow I =\int \limits_0^{1 / 2} \cos \pi x d x-\int \limits_{1 / 2}^1 \cos \pi x d x$

$\Rightarrow I =\left[\frac{\sin \pi x]^{1 / 2}}{\pi}\right]_0-\left[\frac{\sin \pi x}{\pi}\right]_{1 / 2}^1$

$\Rightarrow I =\frac{1}{\pi}\left[\sin \frac{\pi}{2}-\sin 0\right]-\frac{1}{\pi}\left[\sin \pi-\sin \frac{\pi}{2}\right]$

$\Rightarrow I =\frac{1}{\pi}[1-0]-\frac{1}{\pi}[0-1]=\frac{2}{\pi}$

Let $I=\int \limits_0^1 x^{10} \sin n x d x$

$\Rightarrow I=\left[\frac{-x^{10} \cos n x}{n}\right]_0^1+\frac{10}{n} \int \limits_0^1 x^9 \cos n x d x$

$\Rightarrow I=-\frac{\cos n}{n}+\frac{10 \sin n}{n^2}-\frac{90}{n} \int \limits_0^1 x^8 \sin n x d x$

$\lim _{n \rightarrow \infty} I=0 \quad n \rightarrow \infty I=0$

$ \int_{1 / 4}^{3 / 4} \cos \left(2\left(\tan ^{-1} \sqrt{\frac{1+\mathrm{x}}{1+\mathrm{x}}}\right)\right) \mathrm{dx} $

$ \left.\int_{1 / 4}^{3 / 4} \frac{1-\tan ^2\left(\tan ^{-1} \sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\right.}{1+\tan ^2\left(\tan ^{-1} \sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\right.}\right) d x $

$ =\int_{1 / 4}^{3 / 4} \frac{1-\left(\frac{1+x}{1-x}\right)}{1+\left(\frac{1+x}{1-x}\right)} d x=\int_{1 / 4}^{3 / 4} \frac{-2 x}{2} d x $

$ =\int_{1 / 4}^{3 / 4}(-x) d x=-\left(\frac{x^2}{2}\right)_{1 / 4}^{3 / 4} $

$ =-\frac{1}{2}\left[\frac{9}{16}-\frac{1}{16}\right] $

$ =-\frac{1}{4} $

$\int_0^{\pi / 4} \frac{\tan ^2 x \sec ^2 x d x}{\left(1+\tan ^3 x\right)^2} d x$

Let $1+\tan ^3 \mathrm{x}=\mathrm{t}$

$\tan ^2 \mathrm{x} \sec ^2 \mathrm{x} d \mathrm{x}=\frac{\mathrm{dt}}{3}$

$\frac{1}{3} \int_1^2 \frac{\mathrm{dt}}{\mathrm{t}^2}=\frac{1}{6}$

$ I=\int_{-\pi / 2}^{\pi / 2}\left(\frac{x^2 \cos x}{1+\pi^{-x}}+\frac{1+\sin ^2 x}{1+e^{\sin (-x)^{2023}}}\right) d x$

On Adding, we get

$2 I=\int_{-\pi / 2}^{\pi / 2}\left(x^2 \cos x+1+\sin ^2 x\right) d x$

On solving

$ I=\frac{\pi^2}{4}+\frac{3 \pi}{4}-2 $

$ a=3$

Using $\int_0^{2 a} f(x) d x=0$ where $f(2 a-x)=-f(x)$

Here $\quad f(1-x)=-f(x)$

$\therefore \mathrm{I}=0$

$ \frac{1}{2}\left[\int_0^1 \sqrt{3+x} d x-\int_0^1(\sqrt{1+x}) d x\right]$

$ \frac{1}{2}\left[2 \frac{(3+x)^{\frac{3}{2}}}{3}-\frac{2(1+x)^{\frac{3}{2}}}{3}\right]_0^1 $

$ \frac{1}{2}\left[\frac{2}{3}(8-3 \sqrt{3})-\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)\right] $

$ \frac{1}{3}[8-3 \sqrt{3}-2 \sqrt{2}+1] $

$ =3-\sqrt{3}-\frac{2}{3} \sqrt{2}=a+b \sqrt{2}+c \sqrt{3} $

$ a=3, b=-\frac{2}{3}, c=-1 $

$ 2 a+3 b-4 c=6-2+4=8$

$ =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}|\sin x-\cos x| d x $

$ =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}(\sin x-\cos x) d x $

$ =-1+2 \sqrt{2}-\sqrt{3} $

$ =\alpha+\beta \sqrt{2}+\gamma \sqrt{3} $

$ \alpha=-1, \beta=2, \gamma=-1 $

$ 3 \alpha+4 \beta-\gamma=6$

$\frac{10 x}{x+1}=4$           $\Rightarrow x=\frac{2}{3}$

$\frac{10 x}{x+1}=9$           $\Rightarrow x=9$

$\mathrm{I}=9\left(\int_0^{1 / 9} 0 \mathrm{dx}+\int_{1 / 9}^{2 / 3} 1 . \mathrm{dx}+\int_{2 / 3}^9 2 \mathrm{dx}\right)$

$=155$

$ =\int_0^{\pi / 3}\left(\frac{1+\cos 2 x}{2}\right)^2 d x $

$ =\frac{1}{4} \int_0^{\pi / 3}\left(1+2 \cos 2 x+\cos ^2 2 x\right) d x $

$ =\frac{1}{4}\left[\int_0^{\pi / 3} d x+2 \int_0^{\pi / 3} \cos 2 x d x+\int_0^{\pi / 3} \frac{1+\cos 4 x}{2} d x\right] $

$ =\frac{1}{4}\left[\frac{\pi}{3}+(\sin 2 x)_0^{\pi / 3}+\frac{1}{2}\left(\frac{\pi}{3}\right)+\frac{1}{8}(\sin 4 x)_0^{\pi / 3}\right] $

$ =\frac{1}{4}\left[\frac{\pi}{3}+(\sin 2 x)_0^{\pi / 3}+\frac{1}{2}\left(\frac{\pi}{3}\right)+\frac{1}{8}(\sin 4 x)_0^{\pi / 3}\right] $

$ =\frac{1}{4}\left[\frac{\pi}{2}+\frac{\sqrt{3}}{2}+\frac{1}{8} \times\left(-\frac{\sqrt{3}}{2}\right)\right] $

$ =\frac{\pi}{8}+\frac{7 \sqrt{3}}{64} $

$ \therefore \mathrm{a}=\frac{1}{8} ; b=\frac{7}{64} $

$ \therefore 9 \mathrm{a}+8 \mathrm{~b}=\frac{9}{8}+\frac{7}{8}=2$

$ =\int_0^1 0 d x+\int_1^{12} 1 d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x$

$ +\int_{\sqrt{3}}^2 3 \mathrm{dx}+\int_2^{\sqrt{5}} 4 \mathrm{dx}+\int_{\sqrt{5}}^{\sqrt{6}} 5 \mathrm{dx} $

$ +\int_{\sqrt{6}}^{\sqrt{7}} 6 \mathrm{dx}+\int_{\sqrt{7}}^{\sqrt{8}} 7 \mathrm{dx}+\int_{\sqrt{8}}^3 8 \mathrm{dx} $

$ +\int_0^{\sqrt{2}} 0 \mathrm{dx}+\int_{\sqrt{2}}^2 1 \mathrm{dx} $

$ +\int_2^{\sqrt{6}} 2 \mathrm{dx}+\int_{\sqrt{6}}^{\sqrt{8}} 3 \mathrm{dx}+\int_{\sqrt{8}}^3 4 \mathrm{dx}=31-6 \sqrt{2}-\sqrt{3}-\sqrt{5} $

$ -2 \sqrt{6}-\sqrt{7} $

$ \mathrm{a}=31 \quad b=-6 \quad c=-2 $

$ \mathrm{a}+\mathrm{b}+\mathrm{c}=31-6-2=23$