JEE PYQ: Application Of Derivatives Question 79
Question 79 - 2019 (12 Jan Shift 1)
If the function $f$ given by $f(x) = x^3 - 3(a-2)x^2 + 3ax + 7$, for some $a \in \mathbb{R}$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$, then a root of the equation $\frac{f(x) - 14}{(x-1)^2} = 0$, $(x \neq 1)$ is:
(1) $-7$
(2) $5$
(3) $7$
(4) $6$
Show Answer
Answer: (3)
Solution
$f’(1) = 0$: $3 - 6(a-2) + 3a = 0 \Rightarrow a = 5$. $f(x) = x^3 - 9x^2 + 15x + 7$. $\frac{f(x) - 14}{(x-1)^2} = \frac{(x-1)^2(x-7)}{(x-1)^2} = x - 7 = 0 \Rightarrow x = 7$.
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