The solution set of the inequality is

JEE MainMathematicsModerateLogarithm and its applications

The solution set of the inequality
$\log_{sin\left( \frac{\pi}{3} \right)}\left( x^{2} - 3x + 2 \right) \geq 2$
is

Show answer and solution

\log_{\sin(\pi/3)}\left( x^{2} - 3x + 2 \right) \geq 2

\log_{\sin\left( \frac{\pi}{3} \right)}x^{2} - 3x + 2 \geq 2\log_{\sin\left( \frac{\pi}{3} \right)}\sin\left( \frac{\pi}{3} \right)

\because\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} < 1

\Rightarrow x^{2} - 3x + 2 \leq \left( \sin\left( \frac{\pi}{3} \right) \right)^{2}

\Rightarrow x^{2} - 3x + 2 \leq 3/4

\Rightarrow 4x^{2} - 12x + 5 \leq 0

\Rightarrow (2x - 5)(2x - 1) \leq 0

x \in \left\lbrack \frac{1}{2},\frac{5}{2} \right\rbrack\ \ldots.(A)

Also,

x^{2} - 3x + 2 > 0

for log to be defined.

\Rightarrow (x - 1)(x - 2) > 0

\Rightarrow x \in ( - \infty,1) \cup (2,\infty)\ \ \ \ldots.(B)

Taking intersection of (A) and (B)

x \in \left\lbrack \frac{1}{2},1 \right) \cup \left( 2,\frac{5}{2} \right\rbrack

Source: Grass-G / M-05 Mathematics Paper 10 October.docx

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