Q101 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
If $\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60}=\frac{11}{9}$, then $\lim _{x \rightarrow a} \frac{x^2+9 x+20}{x^2-x-20}=$
- A-9
- B-4
- C$-\frac{1}{4}$
- Correct answerD$-\frac{1}{9}$
Solution
$\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60}$
[Use L'Hospital]
$\Rightarrow \lim _{x \rightarrow 4} \frac{4 x+2 c t+3}{3 x^2-4 x-23}=\frac{19+2 a}{9}=\frac{11}{9} \Rightarrow a=-4$
Now, $\lim _{x \rightarrow-4} \frac{x^2+9 x+20}{x^2-x-20}$
[Use L'Hospital]
$=\lim _{x \rightarrow-4} \frac{2 x+9}{2 x-1}=\frac{-1}{9}$.
Hence, the correct answer is option: 'D'.