Maths

$\underset{x\to 1}{lim}\frac{x^{n}f\left(1\right)-f\left(x\right)}{x-1}$  

$=\underset{x\to 1}{lim}\frac{9x^n-\left(x^6+2x^4+x^3+2x+3\right)}{x-1}$                

= 9n – 19 = 44 -> n = 7

ar (ABC) = 4ar (DEF)

$=4*\frac{1}{2}*\left|2\left(2-5\right)+1\left(5-3\right)+7\left(3-2\right)\right|=2\left|-6+2+7\right|=6$

  $\overline{x}=\sum *p\left(x\right)$  

$\Rightarrow \frac{23}{10}=\frac{-2}{5}-a+1+\frac{4}{5}+6b$            

  $\Rightarrow 6b-a=\frac{9}{10}.........\left(i\right)$             

Also,   $\frac{1}{5}+a+\frac{1}{3}+\frac{1}{5}+b=1$

$\Rightarrow a+b=\frac{4}{15}.......\left(ii\right)$       

(i) & (ii) -> $a=\frac{1}{10},b=\frac{1}{6}$

$\Rightarrow 100{\sigma }^2=781$          

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

$I-A^3-3A+3A^2=I-A^3$

=> 3A² – 3A = 0

=> 3A (A – I) = 0

=>A² = A

$\left[\begin{array}{cc}\hfill a^2\hfill & \hfill ab+bd\hfill \\ \hfill 0\hfill & \hfill d^2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill d\hfill \end{array}\right]$    

Total number of ways = 8

For every a, there  must be a² – 2. So, there will be infinitely many pairs (a, b)

$S=\frac{7}{5}+\frac{9}{5^2}+\frac{13}{5^3}+\frac{19}{5^4}..........\infty$

$\frac{S}{5}=\frac{7}{5^2}+\frac{9}{5^2}+\frac{13}{5^4}+------\infty$

$Let\frac{4S-7}{5}=t$

$\frac{4t}{5}=\frac{2}{25}\left\{\frac{1}{1-\frac{1}{5}}\right\}=\frac{1}{10}$

$t=\frac{1}{8}$

$\frac{4S-7}{5}=\frac{1}{8}$

$S=\frac{61}{32}$

$\therefore 160S=305$

The parabola : ${\left(x-\frac{1}{2}\right)}^2=y-\frac{3}{4}$  

-> y = x² – x + 1 . (i)

$P\left(-\frac{1}{2},\frac{7}{4}\right)$                

$N_P:y=\frac{x}{2}+2.........\left(ii\right)$                

(i) & (ii) -> Q (2, 3)

$P{Q}^2=\frac{125}{16}$                

$a^{10}{b}^{10}{\left(\frac{1}{b}+\frac{2}{a}+4\right)}^{10}$

$a^{10}{b}^{10}\frac{10!{\left(\frac{1}{b}\right)}^{r_1}{\left(\frac{2}{a}\right)}^{r_2}.4^{10-r_1-r_2}}{r_1!r_2!\left(10-r_1-r_2\right)!}$

r₁ = 2, r₂ = 3

$a^7{b}^{8}is\frac{10!2^3.4^{10-2-3}}{2!3!5!}=\frac{2^{13}.10!}{2!3!5!}=2^{16}.315$    

k = 315