Maths

x² + y² = 36         ……(i)

y² = 9x      .(ii)

Solving (i) & (ii)

x² + 9x – 36 = 0

(x + 12) (x – 3) = 0

x = 3

Let $A={\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{36-x^2}-3\sqrt[]{x}\right)dx}$  

$={\left[\frac{x\sqrt[]{36-x^2}}{2}+18si{n}^{-1}\frac{x}{6}\right]}_0^3-3.{\frac{x^{3/2}}{3/2}]}_0^3$   

  $=3\pi -\frac{3\sqrt[]{3}}{2}$           

$\therefore$ Required area = $=\frac{1}{2}\pi {\left(6\right)}^2+2\left(3\pi -\frac{3\sqrt[]{3}}{2}\right)=\left(24\pi -3\sqrt[]{3}\right)sq.unit$

$f\left(x\right)=\left[x-1\right]cos\left(\frac{2x-}{2}\right)\pi$

$Ifx=k,k\in I$

then f(x) = 0 as $cos\left(\frac{2k-1}{2}\right)\pi =0,\forall k\in I$  

LHL = $\underset{h\to 0}{Lt}\left[k-h-1\right]cos\left(\frac{2k-2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-2\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$  

$RHL=\underset{h\to 0}{Lt}\left[k+h-1\right]cos\left(\frac{2k+2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-1\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$           

  $\therefore f\left(x\right)$  is continuous $\forall x\in R$  

$f\left(x+y\right)=2f\left(x\right)f\left(y\right),x,y\in N$

$f\left(1\right)=2$

${\displaystyle \sum _{k=1}^{10}f\left(\alpha +k\right)={\displaystyle \sum _{k=1}^{10}2f\left(\alpha \right)f\left(k\right)}}$

$=2f\left(\alpha \right){\displaystyle \sum _{k=1}^{10}f\left(k\right)=2.2^{2\alpha -1}.\frac{2}{3}\left(4^{10}-1\right)}$

$=\frac{2^{2\alpha +1}}{3}.\left(4^{10}-1\right)$

$\Rightarrow 2\alpha +1=9$

= 4

  $f\left(x\right)=2x-1g:R-\left\{1\right\}\to R$       

  $g\left(x\right)=\frac{x-1/2}{x-1}$         

  $f\left\{g\left(x\right)\right\}=2\left(\frac{x-1/2}{x-1}\right)-1=\frac{2x-1-x+1}{x-1}=\frac{x}{x-1},x\ne 1$          

$y=\frac{x}{x-1}\Rightarrow x=\frac{y}{y-1}\therefore y\ne 1$           

One – one but not onto

$\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=\frac{2\left(a+b+c\right)}{24}=\frac{c}{5}=\frac{a}{4}=\frac{b}{3}$

$r=\frac{\Delta }{S}=\frac{6k^2}{6k}=k$

$R=\frac{5k}{2}\Rightarrow \frac{R}{r}=\frac{5}{2}$

BC = CN = x = 75

$cot\theta =\frac{3h}{75}$            

$tan\theta =\frac{h}{75}$          

$\frac{3h^2}{75*75}=1\Rightarrow h=\frac{75}{\sqrt[]{3}}=25\sqrt[]{3}$

$\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x^2}{\int }}\left(sin\sqrt[]{t}\right)dt}}{x^3}$ $\left(\frac{0}{0}\right)$   aplying L' Hospital rule

$=\underset{x\to 0}{lim}\frac{sin\left(\sqrt[]{x^2}\right).2x}{3x^2}$

$=\frac{2}{3},forx>0$

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

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

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

p + q = 2

p⁴ + q⁴ = 272

  ${\left(p^2+q^2\right)}^2-2p^2{q}^2=272$          

${\left[{\left(p+q\right)}^2-2pq\right]}^2-2p^2{q}^2=272$           

Let pq = t -> (4 – 2t)² – 2t² = 272

2t² – 16 t – 256 = 0

->t = pq = 16

Required equation x² – 2x + 16 = 0

A  (A -> B) -> B is a tautology