Maths

$S=\frac{7}{5}+\frac{9}{5^2}+\frac{13}{5^3}+\frac{19}{5^4}..........\infty$

$\frac{S}{5}=\frac{7}{5^2}+\frac{9}{5^2}+\frac{13}{5^4}+------\infty$

$Let\frac{4S-7}{5}=t$

$\frac{4t}{5}=\frac{2}{25}\left\{\frac{1}{1-\frac{1}{5}}\right\}=\frac{1}{10}$

$t=\frac{1}{8}$

$\frac{4S-7}{5}=\frac{1}{8}$

$S=\frac{61}{32}$

$\therefore 160S=305$

The parabola : ${\left(x-\frac{1}{2}\right)}^2=y-\frac{3}{4}$  

-> y = x² – x + 1 . (i)

$P\left(-\frac{1}{2},\frac{7}{4}\right)$                

$N_P:y=\frac{x}{2}+2.........\left(ii\right)$                

(i) & (ii) -> Q (2, 3)

$P{Q}^2=\frac{125}{16}$                

$a^{10}{b}^{10}{\left(\frac{1}{b}+\frac{2}{a}+4\right)}^{10}$

$a^{10}{b}^{10}\frac{10!{\left(\frac{1}{b}\right)}^{r_1}{\left(\frac{2}{a}\right)}^{r_2}.4^{10-r_1-r_2}}{r_1!r_2!\left(10-r_1-r_2\right)!}$

r₁ = 2, r₂ = 3

$a^7{b}^{8}is\frac{10!2^3.4^{10-2-3}}{2!3!5!}=\frac{2^{13}.10!}{2!3!5!}=2^{16}.315$    

k = 315

Total 4 digit number

4 digit divisible by 7

1001, 1008, -9996

9996 = 1001 + (n₁ – 1) 7

n₁ = 1286

4 digit no divisible by 3

1002, 1005, -9999

9999 = 1002 + (n₂ – 1)3

n₂ = 3000

4 digit number visible by 21

1008, 1031, -9996

n₃ = 429

4 digit number divisible by 7 or 3

= 9000 – 1286 – 3000 + 429

= 5143

$AP=\sqrt[]{9+1-6+4+1}$

AP = 3 = AQ

$r=\sqrt[]{1+4-1}=2$        

$tan\theta =\frac{3}{2}$  

$\frac{Areaof\Delta APQ}{Areaof\Delta BPQ}=\frac{AR}{RB}=\frac{3sin\theta }{2cos\theta }=\frac{9}{4}$

$f\left(2\right)=f\left(4\right)=0andf\left(x\right)=x^3-6x^2+ax+b$  

$\Rightarrow f\left(x\right)=\left(x-2\right)\left(x-4\right)x=x^3-6x^2+8x$               

->a = 8, b = 0

$f\text{'}\left(x\right)=0\Rightarrow x=2\pm \frac{2}{\sqrt[]{3}},x_4=2+\frac{2}{\sqrt[]{3}},f\left(x_4\right)=\frac{-16}{3\sqrt[]{3}}$           

$f\text{'}\left(x_3\right)=\frac{\sqrt[]{3}}{2}f\left(x_4\right)=\frac{-8}{3}$               

The tangent to the parabola

$y^2=8xat\left(2,-4\right)is-4y=4\left(x+2\right)$  

x + y + 2 = 0

$OA=\sqrt[]{a}$         

$\Rightarrow \left|\frac{0+0+2}{\sqrt[]{2}}\right|=\sqrt[]{a}$  

$\sqrt[]{2}=\sqrt[]{a}$      

a = 2

$\frac{1}{2}{\displaystyle \underset{0}{\overset{3/2}{\int }}\left(1+4x-x^2\right)dx}$        

$=\frac{1}{2}{\left[x+2x^2-\frac{x^3}{3}\right]}_0^{3/2}=\frac{39}{8}$

${\displaystyle \underset{0}{\overset{3/2}{\int }}mx=\frac{9m}{8}}$              

(as per question)

$\Rightarrow \frac{39}{8}=\frac{9m}{4}$

$m=\frac{39}{18}$

12m = 26

Required number =   $\frac{4*2+24*3+36*4}{2}=112$

${\displaystyle \int \frac{sinx}{si{n}^{3}x+co{s}^{3}x}dx}$

$I={\displaystyle \int \frac{tanxse{c}^{2}x}{ta{n}^{3}x+1}dx}$

Let, tan x = t

sec² xdx = dt

${\displaystyle \int \frac{tdt}{t^3+1}={\displaystyle \int \frac{t}{\left(t+1\right)\left(t^2-t+1\right)}dt}}$    

$\frac{t}{\left(t+1\right)\left(t^2-t+1\right)}=\frac{A}{\left(t+1\right)}+\frac{Bt+c}{t^2-t+1}$

$l=\frac{-1}{3}{\displaystyle \int \frac{1}{t+1}dt+\frac{1}{3}{\displaystyle \int \frac{t+1}{t^2-t+1}dt}}$

The value of 18 $\left(\alpha +\beta +{\gamma }^2\right)$ is equal to 3.