Question

If $-4\le 8x-2\le 4$, then the minimum value of $\frac{1}{x^2}$ is

• A. $16$
• B. $\frac{1}{16}$
• C. $\frac{16}{9}$
• D. $\frac{9}{16}$

Answer 1

The correct option is C $\frac{16}{9}$

Given
$-4\le 8x-2\le 4\Rightarrow -4+2\le 8x-2+2\le 4+2\Rightarrow -2\le 8x\le 6\Rightarrow -\frac{1}{4}\le x\le \frac{3}{4}$
$\Rightarrow 0\le x^2\le \frac{9}{16}$
$\Rightarrow \infty >\frac{1}{x^2}\ge \frac{16}{9}$
Hence, the minimum value of $\frac{1}{x^2}=\frac{16}{9}$

Alternate Solution:
$x\in [-\frac{1}{4},\frac{3}{4}]$
$\therefore$ whenever $0$ lies in the interval while taking the reciprocal, we break it into two parts such that $0$ can be excluded.
$\therefore \frac{1}{x}$ only defined for $[-\frac{1}{4},0)\cup (0,\frac{3}{4}]$
$\to x\in [-\frac{1}{4},0)\Rightarrow \frac{1}{x}\in (-\infty ,-4]\Rightarrow \frac{1}{x^2}\in [16,\infty )\cdots \left(1\right)\to x\in (0,\frac{3}{4}]\Rightarrow \frac{1}{x}\in [\frac{4}{3},\infty )\Rightarrow \frac{1}{x^2}\in [\frac{16}{9},\infty )\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$\therefore \frac{1}{x^2}\in [\frac{16}{9},\infty )$
So, the least value of $\frac{1}{x^2}=\frac{16}{9}$