Question

Let minimum value of $f\left(x\right)=2{\tan }^{2}x+8{\cot }^{2}x,x\in (0,\frac{\pi }{2})$ be $m.$ If number of integral values of $N$ for which $m+0.2\le {\log }_{32}N\le m+0.8$ is $(\lambda \cdot 2^{41}+\mu ),$ where $\lambda$ and $\mu$ are co-prime numbers, then the value of $(\lambda +\mu )$ is

Answer 1

$f\left(x\right)=2{\tan }^{2}x+8{\cot }^{2}x$
Applying A.M. $\ge$ G.M.
$\frac{2{\tan }^{2}x+8{\cot }^{2}x}{2}\ge \sqrt{2\left({\tan }^{2}x\right)\left(8{\cot }^{2}x\right)}$

$\Rightarrow f\left(x\right)\ge 8$ for all $x\in (0,\frac{\pi }{2})$
$\therefore m=8$

Now, $m+0.2\le {\log }_{32}N\le m+0.8$
$\Rightarrow 8.2\le {\log }_{32}N\le 8.8$
$\Rightarrow (32{)}^{8.2}\le N\le (32{)}^{8.8}$
$\Rightarrow 2^{41}\le N\le 2^{44}$
$\therefore$ Number of integral values of
$N$ is,
$2^{44}-2^{41}+1$
$=2^{41}(8-1)+1=7\cdot 2^{41}+1$
$\Rightarrow 7\cdot 2^{41}+1\equiv \lambda \cdot 2^{41}+\mu$
$\therefore \lambda +\mu =7+1=8$