Maths

Coefficient of x in ${\left(1+x\right)}^p{\left(1-x\right)}^q={-}^p{C_0}^q{C}_1{+}^p{C_1}^q{C}_0=-3\Rightarrow p-q=3$  

              Coefficient of x² in ${\left(1+x\right)}^p{\left(1-x\right)}^q{=}^p{C_0}^q{C}_2{-}^p{C_1}^q{C}_1{-}^p{C_2}^q{C}_0=-5$  

$\Rightarrow \frac{q\left(q-1\right)}{2}-pq+\frac{p\left(p-1\right)}{2}=-5\Rightarrow \frac{q\left(q-1\right)}{2}-\left(q-3\right)q+\frac{\left(q-3\right)\left(q-4\right)}{2}=-5\Rightarrow q=11,p=8$  

              Coefficient of x³ in ${\left(1+x\right)}^8{\left(1-x\right)}^{11}={-}^{11}{C}_3{+}^8{C_1}^{11}{C}_2{-}^8{C_2}^{11}{C}_1{+}^8{C}_3=23$  

$x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1=0$  

$\Rightarrow x^7\left(x-1\right)-x^5\left(x-1\right)+3x^3\left(x-1\right)-x\left(x^2-1\right)+2x\left(1-x\right)+\left(x-1\right)=0$  

$\Rightarrow \left(x-1\right)\left(x^2-1\right)\left(x^5+3x-1\right)=0\therefore x=\pm 1$  are roots of above equation and x⁵ + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or -1.

 3 real roots.

Given series $\left\{3*1\right\},\left\{3*2,3*3,3*4\right\},\left\{3*5,3*6,3*7,3*8,3*9\right\}.........$  

$\therefore$  11^th set will have 1 + (10)2 = 21 terms

Also up to 10^th set total 3 * k type terms will be 1 + 3 + 5 + ……… +19 = 100 terms

$\therefore Set11=\left\{3*101,3*102,......3*121\right\}$ $\therefore$ Sum of elements = 3 * (101 + 102 + ….+121)

$=\frac{3*222*21}{2}=6993.$   

Factors of 36 = 2².3².1

Five-digit combinations can be 

(1, 2, 3, 3), (1, 4, 3, 1), (1, 9, 2, 1), (1, 4, 9, 11), (1, 2, 3, 6, 1), (1, 6, 1, 1)

i.e., total numbers  $\frac{5!5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{3!}+\frac{5!}{2!}+\frac{5!}{3!2!}=\left(30*3\right)+20+60+10=180.$  

Let a and b are the roots of $\left(p^2+q^2\right)x^2-2q\left(p+r\right)x+q^2+r^2=0$  

  $\therefore \alpha +\beta >0and\alpha \beta >0$ Also, it has a common root with x² + 2x – 8 = 0

$\therefore$  The common root between above two equations is 4.

  $\Rightarrow 16\left(p^2+q^2\right)-8q\left(p+r\right)+q^2+r^2=0\Rightarrow \left(16p^2-8pq+q^2\right)+\left(16q^2-8qr+r^2\right)=0$  

      $\Rightarrow {\left(4p-q\right)}^2+{\left(4q-r\right)}^2=0\Rightarrow q=4pandr=16p\therefore \frac{q^2+r^2}{p^2}=\frac{16p^2+256p^2}{p^2}=272$  

$\sim \left(p\iff \sim q\right)\wedge q=\left(p\iff \sim q\right)\wedge qis:$

$\therefore \left(\sim \left(P\iff \sim Q\right)\right)\wedge$ q is equivalent to $\left(p\Rightarrow q\right)\wedge$

$tan\left(2ta{n}^{-1}\frac{1}{5}+se{c}^{-1}\frac{\sqrt[]{5}}{2}+2ta{n}^{-1}\frac{1}{8}\right)tan\left(2ta{n}^{-1}\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}*\frac{1}{8}}+se{c}^{-1}\frac{\sqrt[]{5}}{2}\right)$

$=tan\left(ta{n}^{-1}\frac{3}{4}+ta{n}^{-1}\frac{1}{2}\right)=tan\left(ta{n}^{-1}\frac{\frac{3}{4}+\frac{1}{2}}{1-\frac{3}{8}}\right)$

$=tan\left(ta{n}^{-1}\frac{\frac{5}{4}}{\frac{5}{8}}\right)=2$

$S=\left\{\theta \in \left[0,2\pi \right]:8^{2si{n}^{2}x}+8^{2co{s}^{2}x}=16\right\}$

Now apply AM $\ge GM$ for $\frac{8^{2si{n}^{2}x}+8^{2co{s}^{2}x}}{2}\le {\left(8^{2si{n}^{2}x+2co{s}^{2}x}\right)}^{\frac{1}{2}}\Rightarrow 8^{2si{n}^{2}x}=8^{2co{s}^{2}x}$  

Þ $si{n}^2\theta =co{s}^2\theta$               $\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

$=4+\left[cosec\left(\frac{\pi }{2}+\pi \right)+cosec\left(\frac{\pi }{2}+3\pi \right)+cosec\left(\frac{\pi }{2}+5\pi \right)+cosec\left(\frac{\pi }{2}+7\pi \right)\right]$

$=4-2\left(4\right)=-4$  

$0\le \frac{2+3p}{6}\le 1\Rightarrow p\in \left[-\frac{2}{3},\frac{4}{3}\right],0\le \frac{2-p}{8}\le 1\Rightarrow p\in \left[-6,2\right]and0\le \frac{1-p}{2}\le 1\Rightarrow p\in \left[-1,1\right]$  

  $0<P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)\le 10\le \frac{13}{12}-\frac{p}{8}\le 1\Rightarrow p\in \left[\frac{2}{3},\frac{26}{3}\right]$ Taking intersection to all $p\in \left[\frac{2}{3},1\right]$  

$\therefore p_1+p_2=\frac{5}{3}$  

Given, mean = np = a. and variance = npq = $\frac{\alpha }{3}\Rightarrow q=\frac{1}{3}andp=\frac{2}{3}$   

  $P\left(X=1\right)=n{p}^1{q}^{n-1}=\frac{4}{243}\Rightarrow n{\left(\frac{2}{3}\right)}^1{\left(\frac{1}{3}\right)}^{n-1}=\frac{4}{243}\Rightarrow n=6$  

  $P\left(X=4or5\right){=}^6{C}_4{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^2{+}^6{C}_5{\left(\frac{2}{3}\right)}^5{\left(\frac{1}{3}\right)}^1=\frac{16}{27}$