Verified Solution • 2025–26 Session moderate

QJEEM25Maba10: JEE Main 2025 Mathematics - Class 11 JEEM25Maba10 Solution: If the domain of the function is and the domain of the function …

If the domain of the function log⁡5(18x−x2−77)\log_5(18x - x^2 - 77) is (α,β)(\alpha, \beta) and the domain of the function log⁡(x−1)(2x2+3x−2x2−3x−4)\log_{(x-1)} \left( \frac{2x^2 + 3x - 2}{x^2 - 3x - 4} \right) is (γ,δ)(\gamma, \delta) , then α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 is equal to:

Multiple Choice Options

  • A 186
  • B 179
  • C 195
  • D 174

Step-by-Step Verified Solution

amp;f1(x)=log⁡5(18x−x2−77)amp;∴18x−x2−77gt;0amp;x2−18x+77lt;0amp;x∈(7,11)α=7,β=11amp;f2(x)=log⁡(x−1)(2x2+3x−2x2−3x−4)amp;∴x−1gt;0,x−1≠1,2x2+3x−2x2−3x−4gt;0amp;xgt;1,x≠2,(2x−1)(x+2)(x−4)(x+1)gt;0amp;xgt;1,x≠2,\begin{aligned}& f_1(x)=\log _5\left(18 x-x^2-77\right) \\& \therefore 18 x-x^2-77>0 \\& \quad x^2-18 x+77<0 \\& \quad x \in(7,11) \alpha=7, \beta=11 \\& f_2(x)=\log _{(x-1)}\left(\frac{2 x^2+3 x-2}{x^2-3 x-4}\right) \\& \therefore \quad x-1>0, x-1 \neq 1, \frac{2 x^2+3 x-2}{x^2-3 x-4}>0 \\& \quad x>1, x \neq 2, \frac{(2 x-1)(x+2)}{(x-4)(x+1)}>0 \\& \quad x>1, x \neq 2,\end{aligned}

JEE Main 2025 (Online) 29th January Evening Shift Mathematics - Functions Question 8 English Explanation

amp;∴x∈(4,∞)amp;∴γ=4amp;∴α2+β2+γ2=49+121+16amp;=186\begin{aligned}& \therefore \quad x \in(4, \infty) \\& \therefore \gamma=4 \\& \therefore \alpha^2+\beta^2+\gamma^2=49+121+16 \\& =186\end{aligned}

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