Exams Joint Entrance Examination Mathematics Functions Question 36 Question 36 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)}(\frac{2{x}^{2}+3x-2}{{x}^{2}-3x-4}) 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 + 770 , 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{gathered} {f}_{1}(x)={\log }_{5}(18x-{x}^{2}-77) \\ \therefore 18x-{x}^{2}-77>0 \\ \, {x}^{2}-18x+770,x-1\ne 1,\frac{2{x}^{2}+3x-2}{{x}^{2}-3x-4}>0 \\ \, x>1,x\ne 2,\frac{(2x-1)(x+2)}{(x-4)(x+1)}>0 \\ \, x>1,x\ne 2, \end{gathered} f 1 ( x ) = log 5 ( 18 x − x 2 − 77 ) ∴ 18 x − x 2 − 77 > 0 x 2 − 18 x + 770 , 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{gathered} \therefore \, x\in (4,\infty) \\ \therefore \gamma =4 \\ \therefore {\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}=49+121+16 \\ =186 \end{gathered} ∴ x ∈ ( 4 , ∞ ) ∴ γ = 4 ∴ α 2 + β 2 + γ 2 = 49 + 121 + 16 = 186 Answer the question to unlock