(23)

Given, ${\int }_0^3\left(\right[x^2]+[\frac{x^2}{2}\left]\right)dx=a+b\sqrt{2}-\sqrt{3}-\sqrt{5}+c\sqrt{6}-\sqrt{7}$

Now, ${\int }_0^3\left(\right[x^2]+[\frac{x^2}{2}\left]\right)dx={\int }_0^{1}0 dx+{\int }_1^{\sqrt{2}}1 dx+{\int }_{\sqrt{2}}^{\sqrt{3}}(2+1) dx$

$+{\int }_{\sqrt{3}}^2(3+1) dx+{\int }_2^{\sqrt{5}}6 dx+{\int }_{\sqrt{5}}^{\sqrt{6}}7 dx+{\int }_{\sqrt{6}}^{\sqrt{7}}9 dx+{\int }_{\sqrt{7}}^{\sqrt{8}}10 dx+{\int }_{\sqrt{8}}^{3}12 dx$

$=(\sqrt{2}-1)+(3\sqrt{3}-3\sqrt{2})+(8-4\sqrt{3})+(6\sqrt{5}-12)+(7\sqrt{6}-7\sqrt{5})+(9\sqrt{7}-9\sqrt{6})+(10\sqrt{8}-10\sqrt{7})+(36-12\sqrt{8})$

$=31-6\sqrt{2}-\sqrt{3}-\sqrt{5}-2\sqrt{6}-\sqrt{7}$

$\Rightarrow a=31, b=-6, c=-2$

So, $a+b+c=31-8=23$

(32)

$\underset{n\to \infty }{\lim }(\frac{n}{\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}+\dots \frac{n}{\sqrt{n^4+n^4}})-\underset{n\to \infty }{\lim }(\frac{2n}{(n^2+1)\left(\sqrt{n^4+1}\right)}+\frac{8n}{(n^2+4)\sqrt{n^4+16}}+\dots \frac{2n·n^2}{(n^2+n^2)\sqrt{n^4+n^4}})$

$=\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{1}{n\sqrt{1+\frac{r^4}{n^4}}}-\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{1}{n}\frac{2·{\left(\frac{r}{n}\right)}^2}{(1+{\left(\frac{r}{n}\right)}^2)}\frac{1}{\sqrt{1+\frac{r^4}{n^4}}}$

$={\int }_0^1\frac{dx}{\sqrt{1+x^4}}-2{\int }_0^1\frac{x^2}{(1+x^2)\sqrt{1+x^4}} dx={\int }_0^1\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}} dx$

$={\int }_0^1\frac{(\frac{1}{x^2}-1)dx}{(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}}={\int }_0^1\frac{(\frac{1}{x^2}-1)dx}{(x+\frac{1}{x})\sqrt{{(x+\frac{1}{x})}^2-2}}$

$\text{Put }x+\frac{1}{x}=t,\text{ then (}1-\frac{1}{x^2}\text{)}dx=dt$

$=-{\int }_{\infty }^2\frac{dt}{t\sqrt{t^2-2}}={\int }_2^{\infty }\frac{dt}{t\sqrt{t^2-2}}$

$\text{Put }{t}^2-2=u^2\Rightarrow 2t dt=2u du\Rightarrow dt=\frac{u}{t} du$

$={\int }_{\sqrt{2}}^{\infty }\frac{du}{(u^2+2)}=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{{\displaystyle u}}{{\displaystyle \sqrt{2}}}$${|}_{\sqrt{2}}^{\infty }$

$=\frac{1}{\sqrt{2}}(\frac{{\displaystyle \pi }}{{\displaystyle 2}}-\frac{{\displaystyle \pi }}{{\displaystyle 4}})=\frac{\pi }{4\sqrt{2}}=\frac{\pi }{k}$

$\Rightarrow k=2^{5/2}\Rightarrow k^2=2^5=32$

(450)

$AX=B$ has a negative solution

So, $x,y<0$

Now, $2x-5y=20$                                                 ...(i)

$3x+my=m$                                                          ...(ii)

On solving (i) and (ii), we get

$y=\frac{2m-60}{2m+15}, \text{Since }y<0$

$\Rightarrow \frac{2m-60}{2m+15}<0 \Rightarrow 2m-60<0 \text{and} 2m+15>0$

$\Rightarrow m<30 \text{and} m>\frac{-15}{2}$

$\Rightarrow m\in (\frac{{\displaystyle -15}}{{\displaystyle 2}},30)$

$\text{Also, }x=\frac{25m}{2m+15}<0$

$\Rightarrow m\in (\frac{{\displaystyle -15}}{{\displaystyle 2}},0)$

$\therefore m\in (\frac{{\displaystyle -15}}{{\displaystyle 2}},0)$

$\text{Now, }8{\int }_a^b\left|A\right| dm=8{\int }_{-15/2}^0(2m+15) dm$

$=8{[m^2+15m]}_{-15/2}^0=450$

(8)

Let $I={\int }_{-\pi /2}^{\pi /2}\frac{8\sqrt{2}\cos x dx}{(1+e^{\sin x})(1+{\sin }^{4}x)}$

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

$\Rightarrow 2I={\int }_{-\pi /2}^{\pi /2}(\frac{{\displaystyle 8\sqrt{2}\cos x}}{{\displaystyle (1+e^{\sin x})(1+{\sin }^{4}x)}}+\frac{{\displaystyle e^{\sin x}8\sqrt{2}\cos x}}{{\displaystyle (e^{\sin x}+1)(1+{\sin }^{4}x)}})dx$

$={\int }_{-\pi /2}^{\pi /2}\frac{8\sqrt{2}\cos x }{1+{\sin }^{4}x}dx=2{\int }_0^{\pi /2}\frac{8\sqrt{2}\cos x }{1+{\sin }^{4}x}dx$      $[\because {\int }_{-a}^{a}f(x) dx={\int }_0^{a}f(x) dx,\text{ if }f(-x)=f(x\left)\right]$

Put $t=\sin x \Rightarrow dt=\cos x dx$

$\therefore I=8\sqrt{2}{\int }_0^1\frac{dt}{1+t^4}=4\sqrt{2}{\int }_0^1\frac{2dt}{1+t^4}$

$=4\sqrt{2}{\int }_0^1\left(\frac{{\displaystyle 1+1/t^2}}{{\displaystyle t^2+1/t^2}}\right)dt-4\sqrt{2}{\int }_0^1\frac{1-{\displaystyle \frac{1}{t^2}}}{t^2+\frac{1}{t^2}}dt$

$$\begin{gathered} =4\sqrt{2}{\int }_0^1\frac{1+\frac{1}{t^2}}{{{\displaystyle (}{\displaystyle t}{\displaystyle -}{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle t}}}{\displaystyle )}}^2+2}dt-4\sqrt{2} \\ \frac{1}{2\sqrt{2}}{\left[\ln \frac{{\displaystyle (1+\frac{{\displaystyle 1}}{{\displaystyle t}}-\sqrt{2})}}{{\displaystyle t+\frac{{\displaystyle 1}}{{\displaystyle t}}+\sqrt{2}}}\right]}_0^1 \end{gathered}$$

$=4\sqrt{2}·\frac{1}{\sqrt{2}}{\tan }^{-1}{\left(\frac{{\displaystyle t-\frac{{\displaystyle 1}}{{\displaystyle t}}}}{{\displaystyle \sqrt{2}}}\right)}_0^1-\frac{4\sqrt{2}}{2\sqrt{2}}{\left[\ln \frac{{\displaystyle (t+\frac{{\displaystyle 1}}{{\displaystyle t}}-\sqrt{2})}}{{\displaystyle (t+\frac{{\displaystyle 1}}{{\displaystyle t}}+\sqrt{2})}}\right]}_0^1$

$=2\pi +2\ln (3+2\sqrt{2})=\alpha \pi +\beta {\log }_e(3+2\sqrt{2})$

$\Rightarrow \alpha =\beta =2$

$\text{Then }{\alpha }^2+{\beta }^2=4+4=8$

(219)

We have, $F\left(x\right)={\int }_0^{x}t·f\left(t\right) dt$

$\Rightarrow F\text{'}\left(x\right)=x·f\left(x\right)$                                                                 ...(i)

Also, we have $F\left(x^2\right)=x^4+x^5$

Differentiating on both sides, we get

$\Rightarrow 2x·F\text{'}\left(x^2\right)=4x^3+5x^4$

$\Rightarrow F\text{'}\left(x^2\right)=2x^2+\frac{5}{2}{x}^3\Rightarrow F\text{'}\left(x\right)=2x+\frac{5}{2}{x}^{3/2}$         ...(ii)

From (i) & (ii), we get $\Rightarrow xf\left(x\right)=2x+\frac{5}{2}{x}^{3/2}$

$\Rightarrow f\left(x\right)=2+\frac{5}{2}{x}^{1/2}\Rightarrow f\left(x^2\right)=2+\frac{5}{2}x$

$\therefore \underset{r=1}{\overset{12}{ \sum }}f\left(r^2\right)=\sum _{r=1}^{12}(2+\frac{{\displaystyle 5}}{{\displaystyle 2}}r)=24+\frac{5}{2}·\frac{12·13}{2}=219$

(2)

Given, $f\left(x\right)={\int }_0^{x}g\left(t\right){\log }_e\left(\frac{{\displaystyle 1-t}}{{\displaystyle 1+t}}\right)dt,$ where $g$ is continuous odd function.

$f\text{'}\left(x\right)=g\left(x\right){\log }_e\left(\frac{{\displaystyle 1-x}}{{\displaystyle 1+x}}\right)\Rightarrow f\text{'}(-x)=g(-x){\log }_e\left(\frac{{\displaystyle 1+x}}{{\displaystyle 1-x}}\right)$

           $=-g\left(x\right)[-{\log }_e(\frac{{\displaystyle 1-x}}{{\displaystyle 1+x}}\left)\right]=f\left(x\right)$

$\therefore$     $f\text{'}\left(x\right)$ is an even function.

$\Rightarrow f\left(x\right)$ is an odd function.

Let $I={\int }_{-\pi /2}^{\pi /2}\left(f\right(x)+\frac{{\displaystyle x^2\cos x}}{{\displaystyle 1+e^x}})dx$

$={\int }_{-\pi /2}^{\pi /2}f\left(x\right)dx+{\int }_{-\pi /2}^{\pi /2}\frac{x^2\cos x}{1+e^x}dx$

$I={\int }_{-\pi /2}^{\pi /2}\frac{x^2\cos x}{1+e^x}dx$          ...(i)          $[\because {\int }_{-a}^{a}f(x)dx=0,\text{ as }f(x\left)\text{ is an odd function}\right]$

$I={\int }_{-\pi /2}^{\pi /2}\frac{x^2\cos x e^x}{1+e^x}dx$                               ...(ii)

Adding (i) and (ii), we get

$2I={\int }_{-\pi /2}^{\pi /2}{x}^2\cos x dx=2{\int }_0^{\pi /2}{x}^2\cos x dx ; I={\int }_0^{\pi /2}{x}^2\cos x dx$

$I={\left(x^2\sin x\right)}_0^{\pi /2}-{\int }_0^{\pi /2}2x\sin x dx$

$=\frac{{\pi }^2}{4}-2{(-x\cos x)}_0^{\pi /2}+{\int }_0^{\pi /2}\cos x dx=\frac{{\pi }^2}{4}-2{\left(\sin x\right)}_0^{\pi /2}$

$={\left(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)}^2-2(1-0)={\left(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)}^2-2$

$\therefore \alpha =2$

(12)

Slope of line $45x+5y+3=0$ is $\frac{-45}{5}i.e,-9$

Now, $-9=27(-2)+\frac{9\times 10}{2}.$ So, $r_1=-2$ and $r_2=10,$ we have $\underset{x\to 3}{\lim }{\int }_3^x\frac{8t^2}{15x-10x^2+2x^3-3x} dt$

$=\underset{x\to 3}{\lim }\frac{1}{15x-10x^2+2x^3-3x}{\left[\frac{{\displaystyle 8t^3}}{{\displaystyle 3}}\right]}_3^x$

$=\underset{x\to 3}{\lim }\frac{1}{15x-10x^2+2x^3-3x}[\frac{{\displaystyle 8x^3}}{{\displaystyle 3}}-72]$

$=\underset{x\to 3}{\lim }\frac{8x^2}{15-20x+6x^2-3}=\frac{72}{15-60+54-3}=12$

(6)

Let $I={\int }_{\pi /6}^{\pi /3}\sqrt{1-\sin 2x} dx$

$={\int }_{\pi /6}^{\pi /3}\sqrt{{(\sin x-\cos x)}^2} dx\Rightarrow I={\int }_{\pi /6}^{\pi /3}$$|\sin x-\cos x|$$dx$

$={\int }_{\pi /6}^{\pi /4}(\cos x-\sin x) dx+{\int }_{\pi /4}^{\pi /3}(\sin x-\cos x) dx$

$={[\sin x+\cos x]}_{\pi /6}^{\pi /4}+{[-\cos x-\sin x]}_{\pi /4}^{\pi /3}$

$=(\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}-\frac{{\displaystyle 1}}{{\displaystyle 2}}-\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}})+(\frac{{\displaystyle -1}}{{\displaystyle 2}}-\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}})$

$=\frac{4}{\sqrt{2}}-1-\sqrt{3}=-1+2\sqrt{2}-\sqrt{3}$

$\text{So, }\alpha =-1, \beta =2, \gamma =-1$

$\therefore 3\alpha +4\beta -\gamma =-3+8+1=6$

(155)

$I=9{\int }_0^9\left[\sqrt{\frac{{\displaystyle 10x}}{{\displaystyle x+1}}}\right]dx$

Now, at $x=0,$ $\left[\sqrt{\frac{{\displaystyle 10x}}{{\displaystyle x+1}}}\right]=0$ at $x=9,$ $\left[\sqrt{\frac{{\displaystyle 10x}}{{\displaystyle x+1}}}\right]=3$

Integer values between 0 and 3 are 1 and 2

So  $\left[\sqrt{\frac{{\displaystyle 10x}}{{\displaystyle x+1}}}\right]=1 \text{and} \left[\sqrt{\frac{{\displaystyle 10x}}{{\displaystyle x+1}}}\right]=2$

$\Rightarrow \frac{10x}{x+1}=1 \text{and} \frac{10x}{x+1}=4$

$\Rightarrow 10x=x+1 \text{and} 10x=4x+4$

$\Rightarrow x=\frac{1}{9} \text{and} x=\frac{2}{3}$

$I=9{\int }_0^{1/9}\left[\sqrt{\frac{{\displaystyle 10x}}{{\displaystyle x+1}}}\right]dx+{\int }_{1/9}^{2/3}\left[\sqrt{\frac{{\displaystyle 10x}}{{\displaystyle x+1}}}\right]dx+{\int }_{2/3}^9\left[\sqrt{\frac{{\displaystyle 10x}}{{\displaystyle x+1}}}\right]dx$

$=9[{\int }_0^{1/9}0 dx+{\int }_{1/9}^{2/3}1 dx+{\int }_{2/3}^{9}2 dx]$        [$\because$ $\left[t\right]$ is a greatest integer function]

$=9[0+\frac{{\displaystyle 2}}{{\displaystyle 3}}-\frac{{\displaystyle 1}}{{\displaystyle 9}}+18-\frac{{\displaystyle 4}}{{\displaystyle 3}}]=9[-\frac{{\displaystyle 2}}{{\displaystyle 3}}-\frac{{\displaystyle 1}}{{\displaystyle 9}}+18]$

$=162-6-1=155$

(5)

We have, $f\left(a\right)=\frac{4^a}{4^a+2}$

and $f(1-a)=\frac{4^{(1-a)}}{4^{1-a}+2}=\frac{4^1·4^{-a}}{4^1·4^{-a}+2}=\frac{4^1·4^{-a}}{4^{-a}(4+2·4^a)}=\frac{4}{(4+2·4^a)}=\frac{2}{2+4^a}$

$\Rightarrow f\left(a\right)+f(1-a)=\frac{4^a}{4^a+2}+\frac{2}{2+4^a}=\frac{4^a+2}{4^a+2}=1$

$M={\int }_{f\left(a\right)}^{f(1-a)}x{\sin }^4\left[x\right(1-x\left)\right] dx,$

$N={\int }_{f\left(a\right)}^{f(1-a)}{\sin }^4\left[x\right(1-x\left)\right] dx$

$\Rightarrow M={\int }_{f\left(a\right)}^{f(1-a)}(1-x){\sin }^4\left[\right(1-x\left)\right(1-1+x\left)\right] dx$

$\Rightarrow M={\int }_{f\left(a\right)}^{f(1-a)}(1-x){\sin }^4\left[x\right(1-x\left)\right] dx$

$\Rightarrow {\int }_{f\left(a\right)}^{f(1-a)}{\sin }^4\left[x\right(1-x\left)\right] dx-{\int }_{f\left(a\right)}^{f(1-a)}x{\sin }^4\left[x\right(1-x\left)\right] dx$

$\Rightarrow M=N-M\Rightarrow 2M=N\Rightarrow \frac{M}{N}=\frac{1}{2}$

$\Rightarrow \frac{M}{N}=\frac{\beta }{\alpha }=\frac{1}{2}\Rightarrow \beta =1, \alpha =2$

Then, the least value of ${\alpha }^2+{\beta }^2=4+1=5$