Maths

$=8*7*6*5*4+\frac{9*8}{2}$

= 6756

(y – 2) = m (x – 8)

⇒   x-intercept

⇒     $\left(\frac{-2}{m}+8\right)$

⇒   y-intercept

⇒   (–8m + 2)

⇒   OA + OB = $\frac{-2}{m^2}$  + 8 – 8m + 2

$f\text{'}\left(m\right)=\frac{2}{m^2}-8=0$  

-> $m^2=\frac{1}{4}$

-> $m=\frac{-1}{2}$

-> $f\left(\frac{-1}{2}\right)=18$

->Minimum = 18

Take e^sinx = t (t > 0)

⇒ $t-\frac{2}{t}=2$

⇒   $\frac{t^2-2}{t}=2$

->t² – 2t – 2 = 0

->t² – 2t + 1 = 3

⇒   (t −1)² = 3

⇒   t = 1 ± $\sqrt[]{3}$  

⇒   t = 1 ± 1.73

⇒   t = 2.73 or –0.73 (rejected as t > 0)

⇒   e^{sin x} = 2.73

->log_e e^{sin x} = log_e 2.73

⇒ sin x = log_e 2.73 > 1

So no solution.

z₁ + z₂ = 5

$z_1^3+z_2^3=20+15i$            

  $z_1^3+z_2^3={\left(z_1+z_2\right)}^3-3z_1{z}_2\left(z_1+z_2\right)$           

$z_1^3+z_2^3=125-3z_1\cdot z_2\left(5\right)$            

 ⇒ 20 + 15i = 125 – 15z₁z₂

⇒ 3z₁z₂ = 25 – 4 – 3i

3z₁z₂ = 21– 3i

z₁⋅z₂ = 7 – i

(z₁ + z₂)² = 25

$z_1^2+z_2^2=25-27\left(7-i\right)$     

= 11 + 2i

  ${\left(z_{\text{?}1}^2+z_2^2\right)}^2$         = 121 − 4 + 44i

⇒   $z_1^4+z_2^4+2{\left(7-i\right)}^2=117+44i$

⇒   $z_1^4+z_2^4$ = 117 + 44i − 2(49 −1−14i )

= 21 + 72i

⇒   $\left|Z_1^4+Z_2^4\right|=75$

f is increasing function

x < 5x < 7x

f (x) < f (5x) < f (7x)

  $\frac{f\left(x\right)}{f\left(x\right)}<\frac{f\left(5x\right)}{f\left(x\right)}<\frac{f\left(7x\right)}{f\left(x\right)}$           

$\underset{x\to \infty }{lim}\frac{f\left(x\right)}{f\left(x\right)}<\underset{x\to \infty }{lim}\frac{f\left(5x\right)}{f\left(x\right)}<\underset{x\to \infty }{lim}\frac{f\left(7x\right)}{f\left(x\right)}$             

-> $1<\underset{x\to \infty }{lim}\frac{f\left(5x\right)}{f\left(x\right)}<1$ $\underset{x\to \infty }{lim}\frac{f\left(5x\right)}{f\left(x\right)}=1$

$\underset{x\to \infty }{lim}\left(\frac{f\left(5x\right)}{f\left(x\right)}-1\right)=0$            

Area =   $\left|{\displaystyle \underset{1}{\overset{4}{\int }}\left[{\left(4-x\right)}^2-\frac{{\left(x-4\right)}^2}{3}\right]}\right|dx$

Area =   $=\left|\left(16-\frac{64}{3}\right)-\left(4-\frac{1}{3}+\frac{27}{9}\right)\right|$

$=\left|\left(16-\frac{64}{3}\right)-\left(4-\frac{1}{3}+\frac{27}{9}\right)\right|$

$=\left|16-\frac{64}{3}-4+\frac{1}{3}+3\right|$

$=\left|15-2\right|=6$

a + d, a + 7d and a + 43d are 1^st, 2^nd, 3^rd term of G.P.

$\frac{a+7d}{a+d}=\frac{a+43d}{a+7d}$              

⇒ (a + 7d)² = (a + d) (a + 43d)

⇒ a² + 49d² + 14d = a² + 44ad + 43d³

⇒ 6d² = 30ad

⇒ d² = 5d

⇒ d = 0, 5

a = 1, d = 5

  $S_{20}=\frac{20}{2}\left[2+\left(19\right)5\right]$          

= 10 [95 + 2]

= 970

  $\frac{a+b+68+44+40+60}{6}=55$

212 + a + b = 330

⇒ a + b = 118

$\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(\overline{x}\right)}^2=194$          

$\frac{a^2+b^2+{\left(68\right)}^2+{\left(44\right)}^2+{\left(40\right)}^2+{\left(60\right)}^2}{6}={\left(55\right)}^2=194$

= 3219

11760 + a² + b² = 19314

⇒ a² + b² = 19314 – 11760

= 7554

(a + b)² –2ab = 7554

From here b = 41.795

a + b = 118

⇒ a + b + 2b = 118 + 83.59

= 201.59

Let probability of tail is   $\frac{1}{3}$

⇒ Probability of getting head = $\frac{2}{3}$  

∴ Probability of getting 2 heads and 1 tail

$=\left(\frac{2}{3}*\frac{2}{3}*\frac{1}{3}\right)*3$

$=\frac{4}{27}*3$

$=\frac{4}{9}$