Question

The maximum value $M$ of $3^x + 5^x - 9^x + 15^x - 25^x$, as $x$ varies over reals, satisfies

• A. $ 3 < M < 5 $
• B. $0 < M < 2$
• C. $9 < M < 25$
• D. $5 < M < 9$

Answer 1

Answer: (B)

Let
$f\left(x\right) = 3^{x} +5^{x} - 9^{x} + 15^{x}-25^{x} $
$ f\left(x\right) = 3^{x} + 5^{x} -3^{2x} + 3^{x}\cdot5^{x} - 5^{2x} $
Maximum value of $ f\left(x\right)$ is $M$.
$ \therefore M = 3^{x} + 5^{x} - 3^{2x} - 3^{x} \cdot 5^{x} - 5^{2x}$
$ M = a + b -a^{2} + ab - b^{2} $
$ \left[\because 3^{x} = a, 5^{x} = b\right] $
$\Rightarrow a + b + ab - \left( a^{2}+b^{2}\right)$
$ \Rightarrow M \le a +b + ab - 2ab$
$ \left[\because -\left(a^{2}+b^{2}\right) \le - 2ab\right] $
$ \Rightarrow M \le a+b - ab $
$\Rightarrow M \le 1 - \left(a - 1\right)\left(b-1\right) $
So, maximum value of $M$ is $1$ at $x = 0$
$\therefore 0 < M < 2$.