Mathematics · Functions

JEE Advanced 2025 — Paper 1 — Question 15

Let $\mathbb{R}$ denote the set of all real numbers. Let $a_{\mathrm{i}}, b_{\mathrm{i}} \in \mathbb{R}$ for $\mathrm{i} \in\{1,2,3\}$ .

Define the functions $f: \mathbb{R} \rightarrow \mathbb{R}, g: \mathbb{R} \rightarrow \mathbb{R}$ , and $h: \mathbb{R} \rightarrow \mathbb{R}$ by

$f(x)=a_{1}+10 x+a_{2} x^{2}+a_{3} x^{3}+x^{4}$ ,

$g(x)=b_{1}+3 x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ ,

$h(x)=f(x+1)-g(x+2)$ .

If $f(x) \neq g(x)$ for every $x \in \mathbb{R}$ , then the coefficient of $x^{3}$ in $h(x)$ is

Option A:
8
Option B:
2
Option C:
-4
Correct
Option D:
-6

Answer: C

Step-by-step solution

Expand $h(x) = f(x+1) - g(x+2)$ . $f(x+1) = a_1 + 10(x+1) + a_2(x+1)^2 + a_3(x+1)^3 + (x+1)^4$ . $g(x+2) = b_1 + 3(x+2) + b_2(x+2)^2 + b_3(x+2)^3 + (x+2)^4$ . The coefficient of $x^3$ in $f(x+1)$ is $a_3 + 4$ (from $a_3(x+1)^3$ and $(x+1)^4$ ). The coefficient of $x^3$ in $g(x+2)$ is $b_3 + 8$ (from $b_3(x+2)^3$ and $(x+2)^4$ ). Thus, coefficient of $x^3$ in $h(x)$ is $(a_3 + 4) - (b_3 + 8) = a_3 - b_3 - 4$ . The condition $f(x) eq g(x)$ for all real $x$ implies $p(x) = f(x) - g(x) = (a_3 - b_3)x^3 + (a_2 - b_2)x^2 + 7x + (a_1 - b_1)$ has no real zero. A cubic polynomial always has at least one real root, so the cubic term must vanish: $a_3 - b_3 = 0$ .

Hence the required coefficient is $-4$ .

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 1
Subject: Mathematics
Chapter: Functions
Topic: Functional Equations

Answer: C

Step-by-step solution

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