Integrals

$r{\cdot }^n\text{?}{C}_r=n{\cdot }^{n-1}\text{?}{C}_{r-1}$

${\displaystyle \sum _{k=1}^{10}{\left(k{\cdot }^{10}\text{?}{C}_k\right)}^2}={\displaystyle \sum _{k=1}^{10}{\left(10{\cdot }^9\text{?}{C}_{K-1}\right)}^2}$

= $100{\cdot }^{18}\text{?}{C}_9$

22000L = $100{\cdot }^{18}\text{?}{C}_9$

L =

$18!=2^{16}\cdot 3^8\cdot 5^3\cdot 7^2\cdot 11^1\cdot 13\cdot 17$

$9+4+2+1$ $9!=2^7\cdot 3^4\cdot 5^1\cdot 7^1$

6 + 24 + 2 + 1 $\frac{18!}{{\left(9!\right)}^2}=2^2\cdot 5\cdot 11\cdot 13\cdot 17$

3 +3 + 1

= 221

$\underset{x\to 10}{lim}{\left(\frac{{\left(x+2cosx\right)}^3+2{\left(x+2cosx\right)}^2+3sin\left(x+2cosx\right)}{{\left(x+2\right)}^3\pm 2{\left(x+2\right)}^2+3sin\left(x+2\right)}\right)}^x$

$e^{\frac{100}{16+3sin2}\left[\frac{12-3\left(4\right)+8*1-8+3cos2-3cos2}{1}\right]}$ , using L.H.L Rule, e0 = 1

$2x^2-12x+7=0or3x^2+5x+12=x^2+3x+10$

$x^2+x+1=0$

Sum of roots = 6, no solution.

  $l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{e^{cosx}sinxdx}{\left(1+co{s}^{2}x\right)\left(e^{cosx}+e^{-cosx}\right)}}$ ${\displaystyle \underset{}{\overset{}{}}\frac{{}^{}}{{}^{}{}^{}{}^{}}}$

$2l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{sindx}{1+co{s}^{2}x}=2{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{sinxdx}{1+co{s}^{2}x}}}$

$l={\displaystyle \underset{1}{\overset{0}{\int }}\frac{-dt}{1+t^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^2}=\frac{\pi }{4}}}$

$l={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{\left(1+e^x\right)\left(si{n}^{6}x+co{s}^{6}x\right)}}$ ……. (i)

$=2{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{dt}{4+t^2}=2{\left(ta{n}^{-1}\left(\frac{t}{2}\right)\right)}_0^{\infty }}=\pi$

$f\left(x\right)=\{\begin{array}{cc}\hfill \left[x\right],x<0\hfill & \hfill \left|1-x\right|,x\ge 0\hfill \end{array}$ $g\left(x\right)=\{\begin{array}{cc}\hfill c^x-x,x<0\hfill & \hfill {\left(x-1\right)}^2-1,x\ge 0\hfill \end{array}$

fog

$fog(\begin{array}{}\end{array}\left[e^x-x\right],\left(e^x-x<0\right)\left(x<0\right)\left[{\left(x-1\right)}^2-1\right]{\left(x-1\right)}^2-1<0x\ge 0\left|1-e^x+x\right|,e^x-x\ge 0x<0\left|1-{\left(x+1\right)}^2+1\right|,{\left(x-1\right)}^2-1\ge 0x\ge 0,x\in \left(2,\infty \right)($

 Not possible as of inequalities give $?.$  

e^x – x < 0, x < 0                 (x 1)² – 1 < 0

Not possible                     (x – 1 + 1)(x – 1 – 1) < 0

(x)(x – 2) < 0

x   $\in \left(0,2\right)$

continuous  $\forall$ x < 0

$fog\{\begin{array}{ccc}\hfill \left|1-e^x+x\right|,x<0\hfill & \hfill \left|1-{\left(x-1\right)}^2+1\right|,x=0\hfill & \hfill \left|1-{\left(x-1\right)}^2+1\right|,x\ge 2\hfill \end{array}$ Discontinuous at 0

continuous   $\forall x$

  $\therefore$  fog is discontinuous at 0

 ${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$

$f\left(x\right)=x+x{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}-{\displaystyle \underset{0}{\overset{1}{\int }}{t}_{0}f\left(t\right)dt}$

Let $1+{\displaystyle \underset{0}{\overset{1}{\int }}f}\left(t\right)dt=\alpha$

${\displaystyle \underset{0}{\overset{1}{\int }}tf\left(t\right)dt=\beta }$

So, f (x) = αx – β

Now, $\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt+1}$

$\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}\left(at-\beta \right)dt+1}$

$\beta ={\displaystyle \underset{0}{\overset{1}{\int }}t\cdot f\left(t\right)dt}$

$\beta =\frac{4}{13},\alpha =\frac{18}{13}$

f (x) = αx – β

$=\frac{18x-4}{13}$

option (D) satisfies

$?s_1+s_2=k$                

76x² + 3πr² = k

$\therefore 152x\frac{dx}{dr}+6\pi r=0$

Now

$V=40x^3+\frac{2}{3}\pi r^3$

$\Rightarrow \left(\frac{x}{r}\right)=\frac{152}{3}\frac{1}{120}=\frac{19}{45}$

 $co{s}^{-1}\frac{1}{\sqrt[]{5}}=co{t}^{-1}\frac{1}{2}$

$tan\left(2\left(ta{n}^{-1}\frac{1}{2}+co{t}^{-1}\frac{1}{2}\right)\right)+ta{n}^{-1}\left(\frac{1}{2}\right)=tan\left(\pi +ta{n}^{-1}\frac{1}{2}\right)=\frac{1}{2}$

$tan2\alpha =2\sqrt[]{2}\Rightarrow \frac{2tan\alpha }{1-ta{n}^2\alpha }=2\sqrt[]{2}$

$\sqrt[]{2}ta{n}^2\alpha +tan\alpha -\sqrt[]{2}=0$

$tan\alpha =\frac{1}{\sqrt[]{2}},\left(-\sqrt[]{2}\right)\to rejected$

 $I={\displaystyle \underset{0}{\overset{5}{\int }}cos}\left(\pi x-\pi \left[\frac{x}{2}\right]\right)dx$

$I={\displaystyle \underset{0}{\overset{2}{\int }}cos\left(\pi x\right)}dx+{\displaystyle \underset{2}{\overset{4}{\int }}cos}\left(\pi x-\pi \right)dx+{\displaystyle \underset{4}{\overset{5}{\int }}cos\left(\pi x-2\pi \right)dx}$

$I={\frac{sin\pi x}{\pi }|}_0^2+{\frac{sin\left(\pi x-\pi \right)}{\pi }|}_2^4+{\frac{sin\left(\pi x-2\pi \right)}{\pi }|}_4^5=0$