JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If domain of the function $lo g_{e} (\frac{6 x ^{2} + 5 x + 1}{2 x - 1}) + cos^{- 1} (\frac{2 x ^{2} - 3 x + 4}{3 x - 5})$ is $(α,\ β)∪(γ,\ δ]$ , then, $18 (α^{2} + β^{2} + γ^{2} + δ^{2})$ is equal to

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Explanation

Domain of $lo g_{x} (\frac{6 x ^{2} + 5 x + 1}{2 x - 1})$

So, $\frac{6 x^{2}+5 x+1}{2 x-1}>0$

$\Rightarrow \frac{(3x+1)(2x+1)}{2x-1} > 0$

$\Rightarrow x \in (\frac{- 1}{2}, \frac{- 1}{3}) \cup (\frac{1}{2}, \infty)$

$For domain of cos^{- 1} (\frac{2 x ^{2} - 3 x + 4}{3 x - 5}) domain of cos^{- 1} x \to [- 1, 1]$

$- 1 \leq \frac{2 x ^{2} - 3 x + 4}{3 x - 5} \leq 1$

$\frac{2 x ^{2} - 1}{3 x - 5} \geq 0 and \frac{2 x ^{2} - 6 x + 9}{3 x - 5} \leq 0$

$\Rightarrow x \in [\frac{- 1}{2}, \frac{1}{2}] \cup (\frac{5}{3}, \infty)$

$So, common domain is (\frac{- 1}{2}, \frac{- 1}{3}) \cup [\frac{1}{2}, \frac{1}{2}]$

$∴ 18 (α^{2} + β^{2} + γ^{2} + δ^{2}) = 18 (\frac{1}{4} + \frac{1}{9} + \frac{1}{4} + \frac{1}{2})$

$= 18 (\frac{9 + 4 + 9 + 18}{36}) = \frac{1}{2} (40) = 20$

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