- Home
- Standard 11
- Mathematics
4-2.Quadratic Equations and Inequations
normal
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
$5 < M < 9$
(KVPY-2012)
Solution
(b)
Let
$f(x)=3^x+5^x-9^x+15^x-25^{-x}$
$f(x)=3^x+5^x-3^{2 x}+3^x \cdot 5^x-5^{2 x}$
Maximum value of $f(x)$ is $M$.
$\therefore \quad M=3^x+5^x-3^{2 x}-3^x \cdot 5^x-5^{2 x}$
$M=a+b-a^2+a b-b^2$
$\left[\because 3^x=a, 5^x=b\right]$
$\Rightarrow \quad a+b+a b-\left(a^2+b^2\right)$
$\Rightarrow \quad M \leq a+b+a b-2 a b$
$\left.M \leq a+b-a b-\left(a^2+b^2\right) \leq-2 a b\right]$
$\Rightarrow \quad M \leq 1-(a-1)(b-1)$
So, maximum value of $M$ is 1 at $x=0$.
$\therefore \quad 0 < M < 2$
Standard 11
Mathematics