Question

Let $f\left(x\right)=2-|x-3|,1\le x\le 5$ and for rest of the values $f\left(x\right)$ can be obtained by using the relation $f\left(5x\right)=\alpha f\left(x\right)\forall x\in R$.
The maximum value of $f\left(x\right)$ in $[5^4,5^5]$ for $\alpha =2$ is:

• A. 16
• B. 32
• C. 64
• D. 8

Answer 1

Answer: (B)

32

$f\left(x\right)=2-|x-3|,1\le x\le 5$
$f\left(1\right)=0$
$f\left(2\right)=1$
$f\left(3\right)=2$
$f\left(4\right)=1$
$f\left(5\right)=0$
$f\left(5x\right)=\alpha f\left(x\right)$
Value of f(x) in $[5^4,5^5]$ will be maximum when $x=3\cdot 5^4$$\because f\left(x\right)$ is minimum in $[1,5]$
$f(3\cdot 5^4)=\alpha f(3\cdot 5^3)$
$={\alpha }^{2}f(3\cdot 5^2)$
$={\alpha }^{3}f(3\cdot 5)$$={\alpha }^{4}f\left(3\right)$
$=2{\alpha }^4$
$\alpha =2$
$\therefore f(3\cdot 5^4)=2{\left(2\right)}^4$
$=32$