Maths

$a_{n+1}=\frac{a_{n+2}-a_n}{2}letp={\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}$

$\Rightarrow 64\frac{a_{n+2}}{8^{n+2}}=\frac{16a_{n+1}}{8^{n+1}}+\frac{a_n}{8^n}$

$\Rightarrow 64{\displaystyle \sum _{n=1}^{\infty }\frac{a_{n+2}}{8^{n+2}}=}{\displaystyle \sum _{n=1}^{\infty }\frac{16a_{n+1}}{8^{n+1}}+}{\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}\Rightarrow 64\left(p-\frac{a_1}{8}-\frac{a_2}{8^2}\right)=16\left(p-\frac{a_1}{8}\right)+p$

$\Rightarrow 64\left(p-\frac{1}{8}-\frac{1}{8^2}\right)=16\left(p-\frac{1}{8}\right)+p\Rightarrow 64p-8-1=16p-2+p$

  $P\left(\overline{A}\cap B\right)+P\left(A\cap \overline{B}\right)=1-k,$

$P\left(A\cap B\cap C\right)=k^2$             

$P\left(A\right)+P\left(B\right)-2P\left(A\cap B\right)=1-k$ .(i)

$P\left(B\right)+P\left(C\right)-2P\left(B\cap C\right)=1-k$ .(ii)

$P\left(A\right)+P\left(C\right)-2P\left(A\cap C\right)=1-2k$ .(iii)

Adding (i), (ii) and (iii) we get $P\left(A\cup B\cup C\right)=\frac{-4k+3}{2}+k^2$

$\Rightarrow P\left(A\cup B\cup C\right)=\frac{2k^2-4k+2+1}{2}=\frac{2{\left(k-1\right)}^2+1}{2}\Rightarrow P\left(A\cup B\cup C\right)>\frac{1}{2}$

 Let the smallest angle is C =?

Therefore angle A = 90° -?

And angle B = 90°

i.e. b > a > c

Here, a = 2 R cos? , b = 2R, c = 2R sin?

b² = a² + c²

according to question,

$\begin{array}{l}\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}\\ ?\frac{a^2{b}^2}{c^2}=a^2+b^2\end{array}$     

=> $sin?=?\frac{\sqrt[]{5}?1}{2}$

  $l={\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\left[x\right]+\left[-sinx\right]\right)dx}$ . (i)

$I={\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\left[-x\right]+\left[sinx\right]\right)dx}$ . (ii)

by using property $\left({\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx=}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}\right)$

Adding (i) and (ii) we get 2l = ${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(-2\right)dx=-2\pi \Rightarrow l=-\pi }$

  $lo{g}_{9^{\frac{1}{2}}}x+lo{g}_{9^{\frac{1}{3}}}x+lo{g}_{9^{\frac{1}{4}}}x+.......+lo{g}_{9^{\frac{1}{22}}}x=504$

=> $2lo{g}_{9}x+3lo{g}_{9}x+4lo{g}_{9}x+........+22lo{g}_{9}x=504$

=> (1 + 2 + 3 + .+ 22) log₉ x – log₉ x = 504 Þ x = 81

Truth table

Hence according to option

(4) is most appropriate option

$?f\left(x\right)=ln\left(x+\sqrt[]{1+x^2}\right)\therefore f\left(-x\right)=-f\left(x\right)$ .

Hence f(x) is an odd function.

$Nowg\left(t\right)={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(\frac{\pi }{4}t+f\left(x\right)\right)dx}$

Put t = 0, g(0) = ${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx=2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx}...............\left(i\right)$

where cos(f(x)) is an even function.

Now again put t = 1,

$\therefore g\left(1\right)=\frac{1}{\sqrt[]{2}}*g\left(0\right),from\left(i\right)$

$\therefore g\left(0\right)=\sqrt[]{2}g\left(1\right)$

$f\left(x\right)=\frac{5x+3}{6x-\alpha }=y.....................(i)$

$\Rightarrow x=\frac{\alpha y+3}{6y-5}\Rightarrow f^{-1}\left(x\right)=\frac{\alpha x+3}{6x-5}..............(ii)$

According to question, $f\left(x\right)=f^{-1}\left(x\right)\therefore$  from (i) and (ii) we get $\propto =5$

The projection of $\overrightarrow{BA}on\overrightarrow{BC}=cos\angle ABC=7\left|\frac{7^2+3^2-5^2}{2*7*3}\right|=\frac{11}{2}$