Exams Joint Entrance Examination Mathematics Functions Question 156 Question 156 of 162 Medium MCQ
If the domain of the function log 5 ( 18 x − x 2 − 77 ) \log_5(18x - x^2 - 77) log 5 ( 18 x − x 2 − 77 ) is ( α , β ) (\alpha, \beta) ( α , β ) and the domain of the function log ( x − 1 ) ( 2 x 2 + 3 x − 2 x 2 − 3 x − 4 ) \log_{(x-1)} \left( \frac{2x^2 + 3x - 2}{x^2 - 3x - 4} \right) log ( x − 1 ) ( x 2 − 3 x − 4 2 x 2 + 3 x − 2 ) is ( γ , δ ) (\gamma, \delta) ( γ , δ ) , then α 2 + β 2 + γ 2 \alpha^2 + \beta^2 + \gamma^2 α 2 + β 2 + γ 2 is equal to:
Solutions · 1 Post your solution Official solution
\[ f 1 ( x ) = log 5 ( 18 x − x 2 − 77 ) ∴ 18 x − x 2 − 77 > 0 x 2 − 18 x + 77 < 0 x ∈ ( 7 , 11 ) α = 7 , β = 11 f 2 ( x ) = log ( x − 1 ) ( 2 x 2 + 3 x − 2 x 2 − 3 x − 4 ) ∴ x − 1 > 0 , x − 1 ≠ 1 , 2 x 2 + 3 x − 2 x 2 − 3 x − 4 > 0 x > 1 , x ≠ 2 , ( 2 x − 1 ) ( x + 2 ) ( x − 4 ) ( x + 1 ) > 0 x > 1 , x ≠ 2 , \begin{aligned}
& f_1(x)=\log _5\left(18 x-x^2-77\right) \\
& \therefore 18 x-x^2-77>0 \\
& \quad x^2-18 x+77<0 \\
& \quad x \in(7,11) \alpha=7, \beta=11 \\
& f_2(x)=\log _{(x-1)}\left(\frac{2 x^2+3 x-2}{x^2-3 x-4}\right) \\
& \therefore \quad x-1>0, x-1 \neq 1, \frac{2 x^2+3 x-2}{x^2-3 x-4}>0 \\
& \quad x>1, x \neq 2, \frac{(2 x-1)(x+2)}{(x-4)(x+1)}>0 \\
& \quad x>1, x \neq 2,
\end{aligned} f 1 ( x ) = log 5 ( 18 x − x 2 − 77 ) ∴ 18 x − x 2 − 77 > 0 x 2 − 18 x + 77 < 0 x ∈ ( 7 , 11 ) α = 7 , β = 11 f 2 ( x ) = log ( x − 1 ) ( x 2 − 3 x − 4 2 x 2 + 3 x − 2 ) ∴ x − 1 > 0 , x − 1 = 1 , x 2 − 3 x − 4 2 x 2 + 3 x − 2 > 0 x > 1 , x = 2 , ( x − 4 ) ( x + 1 ) ( 2 x − 1 ) ( x + 2 ) > 0 x > 1 , x = 2 , \]
\[ ∴ x ∈ ( 4 , ∞ ) ∴ γ = 4 ∴ α 2 + β 2 + γ 2 = 49 + 121 + 16 = 186 \begin{aligned}
& \therefore \quad x \in(4, \infty) \\
& \therefore \gamma=4 \\
& \therefore \alpha^2+\beta^2+\gamma^2=49+121+16 \\
& =186
\end{aligned} ∴ x ∈ ( 4 , ∞ ) ∴ γ = 4 ∴ α 2 + β 2 + γ 2 = 49 + 121 + 16 = 186 \] Answer the question to unlock