Sets

$A=\left\{z\in c:1\le \left|z-\left(1+i\right)\right|\le 2\right\}$

$\left|z-\left(1+i\right)\right|\ge 1$

and $$ $\left|z-\left(1+i\right)\right|\le 2$

and also $\left|z-\left(1-i\right)\right|=1$  $$

hence B has infinite set.

Any point on line L is

M (3r + 6, 2r + 1, 3r + 2)                                                          

$\overrightarrow{PM}\perp rtoL$          

$\therefore 3\left(3r+5\right)+2\left(2r-1\right)+3\left(3r-1\right)=0$          

$r=\frac{-5}{11}$

$x^4?3x^3?2x^2+3x+1=0$

$?x=\pm 1$

and let, are roots of x2 – 3x – 1 = 0

$??+?=3??=?1$

$?1^3+{\left(?1\right)}^3+{?}^3+{?}^3$

= 27 + 3 (3) = 36

$f\left(x-y\right)=2^{x}f\left(y\right)+4^{y}f\left(x\right)$ ……. (i)

$$ $f\left(y+x\right)=2^{y}f\left(x\right)+4^{x}f\left(y\right)$  ……. (ii)

(i) – (ii)

$\therefore \frac{f\text{'}\left(4\right)}{f\text{'}\left(2\right)}=\frac{k\left(16\mathcal{l}n2-256\mathcal{l}n4\right)}{k\left(4\mathcal{l}n2-16\mathcal{l}n4\right)}$

$=\frac{\left(16-512\right)\mathcal{l}n2}{\left(4-32\right)\mathcal{l}n2}=\frac{496}{28}$

$14\frac{f\text{'}\left(4\right)}{f\text{'}\left(2\right)}=248$

3cos²2θ + 6cos2θ -   $10\frac{\left(1+cos2\theta \right)}{2}+5=0$

$cos2\theta \left(3cos2\theta +1\right)=0$

 Þ cos2θ = 0,     $\frac{-1}{3}$

Draw y = cos2θ, y = 0 and y =  $\frac{-1}{3},$  find the pt. of intersection.

 $C\subseteq AandC\cap B=\phi$

If C is formed only by {1, 2, 4, 5} total number of subsets of A = 27.

Total number of subsets of {1, 2, 4, 5} = 24

$\therefore$ Number of subsets where $C\cap B\ne \phi$

= 27 – 24 = 112

 $tan?+\sqrt[]{5}tan2?tan?=\sqrt[]{5}?tan2?$

$tan3?=\sqrt[]{5}$

$?=\frac{n?}{3}+\frac{?}{3}tan?=\sqrt[]{5}$

Five solution.

$S\cap T=\left\{-5,-4,3\right\}$

$\mathrm{m}\mathrm{a}\mathrm{x}\left\{n\right(A),n(B\left)\right\}\le n(A\cup B)\le n\left(U\right)$

$\Rightarrow 76\le 76+63-x\le 100$

$\Rightarrow -63\le -x\le -39$

$\Rightarrow 63\ge x\ge 39$

The Commission prepared four sets of question paper having identical questions but at different serial number. The candidates were distributed these question papers as per the seating plan prepared and approved by the local exam designated authority. The Tamil Nadu Public Service Commission will conduct the TNPSC Group 4 exam on December 20, 2026 at the designated exam centres.