First slide

Introduction to real valued functions

Question

$Number of integers satisfying \log_{x} \frac{4 x + 5}{6 - 5 x} < - 1 is$

Moderate

A

0
B

1
C

2
D

3

Solution

$\begin{array}{l} \log_{x} \frac{4 x + 5}{6 - 5 x} < - 1 \\ We must have \\ \frac{4 x + 5}{6 - 5 x} > 0 \\ \Rightarrow \frac{4 x + 5}{5 x - 6} < 0 \\ \Rightarrow x \in (\frac{- 5}{4}, \frac{6}{5}) \end{array}$
$\begin{array}{l} Also x > 0 and x \neq 1 \\ ∴ x \in (0, \frac{6}{5}) - {1} - - - - - (1) \\ Case I: 0 < x < 1 - - - - (2) \end{array}$
$\begin{array}{l} \log_{x} (\frac{4 x + 5}{6 - 5 x}) < - 1 \\ \Rightarrow \frac{4 x + 5}{6 - 5 x} > \frac{1}{x} \\ \Rightarrow \frac{4 x + 5}{6 - 5 x} - \frac{1}{x} > 0 \\ \Rightarrow \frac{4 x^{2} + 5 x + 5 x - 6}{x (6 - 5 x)} > 0 \\ \Rightarrow \frac{4 x^{2} + 10 x - 6}{x (5 x - 6)} < 0 \end{array}$
$\begin{array}{l} \Rightarrow \frac{2 (x + 3) (2 x - 1)}{x (5 x - 6)} < 0 \\ \Rightarrow x \in (- 3, 0) \cup (\frac{1}{2}, \frac{6}{5}) - - - - - (3) \\ From (i), (ii) and (iii), we get \\ ∴ x \in (\frac{1}{2}, 1) \end{array}$
$\begin{array}{l} Case II: x > 1 - - - - (4) \\ \log_{x} (\frac{4 x + 5}{6 - 5 x}) < - 1 \\ \Rightarrow \frac{4 x + 5}{6 - 5 x} < \frac{1}{x} \\ \Rightarrow x \in (- \infty, - 3) \cup (0, \frac{1}{2}) \cup (\frac{6}{5}, \infty) - - - - - (5) \end{array}$
$\begin{array}{l} From (i), (iv) and (v), we get \\ x \in ϕ \\ Thus, x \in (\frac{1}{2}, 1) \end{array}$

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