Maths

Let

A : Missile hit the target

B : Missile intercepted      

P (B) =    $\frac{1}{3}P\left(A/\overline{B}\right)=\frac{3}{4}$

$P\left(\overline{B}\right)=\frac{2}{3}$  

$\Rightarrow P\left(\overline{B}\cap A\right)=\frac{3}{4}*\frac{2}{3}=\frac{1}{2}$   

Required Probability = $\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}=\frac{1}{8}$  

$\underset{n\to \infty }{lim}{\left(1+\frac{1+\frac{1}{2}+.......+\frac{1}{n}}{n^2}\right)}^n$ limit is in the form of $1^{\infty }$  

$\mathcal{l}=exp\left(\underset{n\to \infty }{lim}\frac{1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}}{n^2}\right)$             

$0\le 1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}\le 1+\frac{1}{\sqrt[]{2}}+\frac{1}{\sqrt[]{3}}+....+\frac{1}{\sqrt[]{n}}\le 2\sqrt[]{n}-1$        

Taking limit   $\left(n\to \infty \right)$

$\mathcal{l}$ = exp (0) (from sandwich)

  $\mathcal{l}=1$          

Second Method :

$1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\ge ln\left(n+1\right).......\left(i\right)$

$1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}\le 1+{\displaystyle {\int }_1^2\frac{1}{2}dx\le 1+lnn..........\left(ii\right)}$

From (i) & (ii)

$ln\left(n+1\right)\le 1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\le 1+Inn,\forall n\in N,n\ge 2$           

As $\underset{n\to \infty }{lim}\frac{ln\left(n+1\right)}{n}=0$  

and $\underset{n\to \infty }{lim}\frac{1+ln\left(n\right)}{n}=0$  

$\therefore$ from sandwich theorem

$\underset{n\to \infty }{lim}\frac{1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}}{n}=0$  

$\therefore e^0=1$   

2x + y = 1

$m_1=\frac{-2}{1}=-2and{m}_1{m}_2=-1$          

$-2.m_2=-1$           

$m_2=\frac{1}{2}$           

y² = 6x

y² = 4 (3/2) x

$y=mx+\frac{a}{m}$         

$y=\frac{1}{2}x+\frac{\frac{3}{2}}{\frac{1}{2}}$        

$y=\frac{x}{2}+3$           

$A\to \left(B\to A\right)$

$\sim A\vee \left(B\to A\right)$

$\sim A\vee \left(\sim B\vee A\right)$

$\left(\sim A\vee B\right)\vee \left(\sim A\vee A\right)=tA\to \left(A\cup B\right)$

$\overrightarrow{O}P=\left(2,-1,1\right)$

Normal vector to the plane

$=\overrightarrow{A}B*A\overrightarrow{C}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left(3,-1,1\right)=\overrightarrow{n}$

Projection of $\overrightarrow{O}Pon\overrightarrow{n}is\left|\frac{\overrightarrow{O}P.\overrightarrow{n}}{\left|\overrightarrow{n}\right|}\right|$

PN = $\frac{6+1+1}{\sqrt[]{11}}=\frac{8}{\text{exception 25:}}$

$\frac{\text{exception 25:}}{}$Projection of OP on plane = ON =  $\sqrt[]{O{P}^2-P{N}^2}=\sqrt[]{6-\frac{64}{11}}=\sqrt[]{\frac{2}{11}}$

  $\frac{x^2}{a}+\frac{y^2}{b}=1,\frac{x^2}{c}-\frac{y^2}{\left(-d\right)}=1$

Ellipse and Hyperbola are orthogonal so these will be confocal.

$\sqrt[]{a-b}=\sqrt[]{c+\left(-d\right)}$            

a – b = c – d

$l={\displaystyle \int \frac{8x^3+20x^2}{x^4+5x^3-7x^2}}dx=4{\displaystyle \int \frac{2x+5}{x^2+5x-7}dx=4ln\left|x^2+5x-7\right|+c}$

 $cosec\left(2co{t}^{-1}\left(5\right)+co{s}^{-1}\left(\frac{4}{5}\right)\right)$

Let cot^-1 (5) = and cos^-1 $\left(\frac{4}{5}\right)=\alpha$

cot = 5 cos = $\frac{4}{5}$

$=cosec\left(2\theta +\alpha \right)=\frac{1}{sin\left(2\theta +\alpha \right)}$

$as\{\begin{array}{l}sin2\theta =\frac{2tan\theta }{1+ta{n}^2\theta }=\frac{2\left(\frac{1}{5}\right)}{1+\frac{1}{25}}=\frac{5}{13}\\ cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }=\frac{1-\frac{1}{25}}{1+\frac{1}{25}}=\frac{12}{13}\end{array}$

Let A = {a, b, c}, B = {1, 2, 3, 4, 5} n (A * B) = 15

x = number of one-one functions from A to B.

${=}^5{C}_3.3!=60$

y = number of one-one functions for A to (A * B)

${=}^{15}{C}_3.3!=15*14*13=2730$

$\frac{Y}{x}=\frac{2730}{60}\Rightarrow 2y=91x$

$f\left(x\right)=x^3-a{x}^2+bx-4$

f (1) = f (2)

->1 – a + b – 4 = 8 – 4a + 2b – 4

->3a – b = 7 . (i)

8a = 16 + 3b . (ii)

(i) and (ii) -> (a, b) = (5, 8)