JEE PYQ: Application Of Derivatives Question 68
Question 68 - 2019 (12 Apr Shift 2)
Let $f(x) = 5 - |x - 2|$ and $g(x) = |x + 1|$, $x \in \mathbb{R}$. If $f(x)$ attains maximum value at $\alpha$ and $g(x)$ attains minimum value at $\beta$, then $\lim_{x \to -\alpha\beta} \frac{(x-1)(x^2-5x+6)}{x^2-6x+8}$ is equal to:
(1) $1/2$
(2) $-3/2$
(3) $-1/2$
(4) $3/2$
Show Answer
Answer: (1)
Solution
$f(x)$ max at $\alpha = 2$, $g(x)$ min at $\beta = -1$. $-\alpha\beta = 2$. $\lim_{x \to 2}\frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)} = \frac{1 \cdot (-1)}{-2} = \frac{1}{2}$.
BETA
