Sets, Relations & Functions
Functions
star_batch_jee_advanced_2025
Grade 11
Question:
Let $f(x) = (x+1)(x+2)(x+3)(x+4) + 5$ ; where $x \in [-6, 6]$. If the range of the function is $[a, b]$; where $a, b \in \mathbb{N}$, then find the value of $(a + b)$.
Step-by-Step Solution
Key Concept: Substitute $u = x^2+5x+5$ to transform the quartic into a quadratic $u^2+4$, then find the range of $u$ over the given domain.
We rewrite $f(x) = (x+1)(x+2)(x+3)(x+4) + 5$ by grouping: $[(x+1)(x+4)][(x+2)(x+3)] + 5 = [(x^2+5x+4)][(x^2+5x+6)] + 5$. Let $u = x^2+5x+5$, then $(x+1)(x+4) = u-1$ and $(x+2)(x+3) = u+1$, so $f(x) = (u-1)(u+1) + 5 = u^2 - 1 + 5 = u^2 + 4$. For $x \in [-6,6]$, we find the range of $u = x^2+5x+5$. Taking the derivative: $u' = 2x+5 = 0$ gives $x = -2.5$. At $x = -2.5$: $u = 6.25 - 12.5 + 5 = -1.25$ (minimum). At endpoints: $u(-6) = 36-30+5 = 11$ and $u(6) = 36+30+5 = 71$. Thus $u \in [-1.25, 71]$, giving $u^2 \in [0, 5041]$, so $f(x) = u^2+4 \in [4, 5045]$.
Correct Answer: 2049