TITLE: Number of integral values of p
QUESTION [6 upvotes]: Let $p=144^{\sin ^2 x}+144^{\cos ^2x}$, then we have to find the number of integral value of p .
Using the concept of inequlitiy of AM ,GM I got its minimum value as 24 . 
But how could we find the maximum value .

REPLY [3 votes]: (1) use that $2\cos ^{2}(x) = 1+\cos 2x$ and $2\sin ^{2}(x) = 1-\cos 2x$.
(2) then we have $\ 12 ^{1-\cos 2x}+\ 12 ^{1+\cos 2x}$ = $\tfrac{12}{12^{\ cos 2x}}+\ 12\cdot{12^{\ cos 2x}}$
(3) let ${12^{\ cos 2x}}=t$ then $\ f(t)= \tfrac{12}{t}+\ 12\cdot{t}$, where $1\leqslant  t \leqslant12$
(4) first derivative of $\ f\prime(t)=0$ gives us $t=\pm1$
(5) our function continious on $1\leqslant  t \leqslant12$ and strictly increasing on it, so $min=f(1)=24$ and $max=f(12)=145.$