Applications of Derivatives
Differential Calculus-2
star_batch_jee_advanced_2025
Grade 12
Question:
A polynomial function $P(x)$ of degree 5 with leading coefficient one, increases in the interval $(-\infty, 1)$ and $(3, \infty)$ and decreases in the interval $(1, 3)$. Given that $P(0) = 4$ and $P'(2) = 0$. Find the value $P'(6)$.
Step-by-Step Solution
Key Concept: The derivative reduces the degree by 1 and multiplies the leading coefficient by the original degree.
Since $P(x)$ has degree 5 with leading coefficient 1, $P'(x)$ has degree 4 with leading coefficient 5. Given that $P'(x) = 5(x-1)(x-3)(x-2)^2$, we have $P(x) = 5\int(x-1)(x-3)(x-2)^2\,dx$. Then $P'(6) = 5 \times 5 \times 3 \times 4^2 = 1200$.
Correct Answer: 1200