If $16 x^{2} < 15 x + 1$ , then find $x .$

(- \frac{1}{16}, 1)

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(- \infty, - \frac{1}{16}) \cup (1, \infty)

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[- \frac{1}{16}, 1]

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(- \infty, - \frac{1}{16}] \cup [1, \infty)

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Solution

The correct option is C $(- \frac{1}{16}, 1)$
Given,
$16 x^{2} < 15 x + 1$
$\Rightarrow 16 x^{2} - 15 x - 1 < 0$
$\Rightarrow 16 x^{2} - 16 x + x - 1 < 0$
$\Rightarrow 16 x (x - 1) + 1 (x - 1) < 0$
$\Rightarrow (16 x + 1) (x - 1) < 0$
Two cases arises:
Case I:
(i) holds,
if $(16 x + 1) > 0$ and $(x - 1) < 0$
if $x > - \frac{1}{16}$ and $x < 1$
if $- \frac{1}{16} < x < 1$
if $x \in (- \frac{1}{16}, 1)$

Case II:
(i) holds,
if $(16 x + 1) < 0$ and $(x - 1) > 0$
if $x < - \frac{1}{16}$ and $x > 1$
No such x exist.
On combining case I and II, we get
$x \in (- \frac{1}{16}, 1)$
Option A is correct.

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