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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 .

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  • $\begingroup$ As $$144^{\sin^2x}=(12^2)^{\sin^2x}=(12^{\sin^2x})^2$$ $$144^{\sin^2x}+144^{\cos^2x}=(12^{\sin^2x}+12^{\cos^2x})^2-2\cdot12$$ So, $f(y^2)=f^2(y)-2$(some constant). Can this be reduced recursively ? $\endgroup$ Jun 1 '17 at 6:27
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(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 $max=f(12)=145.$

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Let $$ f(t)=144^t+144^{1-t}. $$ Then $$ f'(t)=\left(144^t-144^{1-t}\right)\ln 144. $$ We see that $f'(t)\le0$ when $t\le1/2$ and $f'(t)\ge0$ when $t\ge1/2$. Therefore $f$ attains its maximum in the interval $t\in[0,1]$ at the end points. The maximum is thus $f(0)=f(1)=145$.

Therefore, by basic properties of continued functions, your function $p=f(\sin^2x)$ takes $145-24+1=122$ distinct integers as its values. Namely those in the range $[24,145]$.

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