First slide

Introduction to real valued functions

Question

Complete set of values of x satisfying inequality $| | x - 1 | - 5 | < 2 x - 5$ is

Moderate

A

$(\frac{5}{2}, \infty)$
B

$(\frac{11}{3}, \infty)$
C

$(- 1, \infty)$
D

$(- \infty, \frac{1}{3})$

Solution

We have, $| | x - 1 | - 5 | < 2 x - 5$
$\begin{array}{ll} ∴ 2 x - 5 > 0 \\ \Rightarrow x > \frac{5}{2} (1) \end{array}$
Now from (1),
$\begin{array}{l} - 2 x + 5 < | x - 1 | - 5 < 2 x - 5 \\ \Rightarrow - 2 x + 10 < | x - 1 | < 2 x \end{array}$
Case I : $x \geq 1$
$\begin{array}{ll} ∴ - 2 x + 10 < x - 1 < 2 x \\ \Rightarrow 3 x > 11 and x > - 1 \\ \Rightarrow x > 11 / 3 (2) \end{array}$
Case II : x < 1
$\begin{array}{ll} ∴ - 2 x + 10 < 1 - x < 2 x \\ \Rightarrow x > 9 and x > 1 / 3 \end{array}$
$\Rightarrow x > 9$ , not possible as x < 1
Therefore, $x \in (11 / 3, \infty)$

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