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
$f\left(x\right)=3^x+5^x-9^x+15^x-25^x$
$f\left(x\right)=3^x+5^x-3^{2x}+3^x\cdot 5^x-5^{2x}$
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 $x=0$
$\therefore 0<M<2$.