Maths

$\left|\frac{z+1}{z-1}\right|<1\Rightarrow \left|z+1\right|<\left|z-1\right|\Rightarrow Re\left(z\right)<0$

and $arg\left(\frac{z-1}{z+1}\right)=\frac{2\pi }{3}$  is a part of circle as shown.

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

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

  $x=\pm \omega ,?{\omega }^2$

Now, = ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$  

= ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$  

= 1 + 1 – 1 = 1

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

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

$x=\pm \omega ,?{\omega }^2$

Now, = ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$  

= ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$  

= 1 + 1 – 1 = 1

$f\left(x\right)=\frac{x-1}{x+1}\Rightarrow f\left(f\left(x\right)\right)=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=-\frac{1}{x}$

$f^3\left(x\right)=-\frac{x+1}{x-1}\Rightarrow f^4\left(x\right)=\frac{\frac{x-1}{x+1}+1}{\frac{x-1}{x+1}-1}=x$

$So,f^6\left(6\right)+f^7\left(7\right)=f^2\left(6\right)+f^3\left(7\right)$

= $-\frac{1}{6}-\frac{7+1}{7-1}=-\frac{9}{6}=-\frac{3}{2}$

$9^n-8n-1={\left(1+8\right)}^n-8n-1$  

$=(1+8n{+}^n\text{?}{C}_2\cdot 8^2{+}^n\text{?}{C}_3\cdot 8^3+....]-8n-1$  

So, a = ⁿC₂ + ⁿC₃8 + ⁿC₄8² +….

Similarly, b = ⁿC₂ + ⁿC₃ 5 + ⁿC₄ 5² +….

a - b = ⁿC₃ (8 – 5) + ⁿC₄ (8² – 5²) +….

pva -> $\left(\sim rvp\right)$  

$\sim \left(p\vee q\right)\vee \left(\sim rvp\right)$  

its negation as asked in question

$\left(p\vee q\right)\wedge \left(\sim p\wedge r\right)$  

=  $\left(p\wedge \sim p\wedge r\right)\vee \left(q\wedge r\wedge \sim p\right)$  

=  $\left(p\wedge r\wedge \sim p\right)\left[asp\wedge \sim pisfalse\right]$  

$Mean=\frac{3+12+7+a+\left(43-a\right)}{5}=13$  

Variance =  $\frac{3^2+12^2+7^2+a^2+{\left(43-a\right)}^2}{5}-{\left(13\right)}^2$  

$\Rightarrow \frac{2a^2-a+1}{5}\to Naturalnumber$  

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not  $4\lambda or4\lambda +1from.$  

As each square form is  $4\lambda or4\lambda +1$  

Total number of possible relation = $2^{n^2}=2^4=16$  

Favourable relations = $?,\left\{\left(x,x\right)\right\},\left\{\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right),\left(x,y\right),\left(y,x\right)\right\}$

Probability =  $\frac{5}{16}$  

Mid point of BC is $\frac{1}{2}\left(5\widehat{i}+\left(\alpha -2\right)\widehat{j}+9\widehat{k}\right)$  

$\overline{AB}=\widehat{i}+\left(\alpha -4\right)\widehat{j}+\widehat{k}$  

$\overline{AC}=\widehat{i}+\left(-2-\alpha \right)\widehat{j}+\widehat{k}$

For a = 1,   $\overline{AB}$ and $\overline{AC}$  will be collinear. So for non collinearity

a = 2

$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2\left(1+4+2-16\right)}{1+2^2+2^2}$  

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

$\left|\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z+1\hfill \\ \hfill 3\hfill & \hfill 6\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=0$

->2x – z = 1

option (B) satisfies.