Gujarati
Hindi
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

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