JEE Advanced 2025 — Paper 1 — MATH-I solution

JEE Advanced2025Paper 1MATH-I

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

Define the functions $f : R \to R$ , $g : R \to R$ , and $h : R \to R$ 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 $x \in R$ , then the coefficient of $x^{3}$ in $h (x)$ is

A
$8$
B
$2$
C
$- 4$
D
$- 6$

Solution

Consider the polynomial $p (x) = f (x) - g (x)$ . Expanding,

$p (x) = (a_{3} - b_{3}) x^{3} + (a_{2} - b_{2}) x^{2} + 7 x + (a_{1} - b_{1}) .$

The condition $f (x) \neq = g (x)$ for every real $x$ means $p (x)$ has no real root. A cubic with real coefficients always has at least one real root, so the leading $x^{3}$ coefficient must vanish, giving $a_{3} - b_{3} = 0$ .

Now compute the coefficient of $x^{3}$ in $h (x) = f (x + 1) - g (x + 2)$ . From $f (x + 1)$ the $x^{3}$ contribution comes from $a_{3} (x + 1)^{3}$ and $(x + 1)^{4}$ , giving $a_{3} + 4$ . From $g (x + 2)$ it comes from $b_{3} (x + 2)^{3}$ and $(x + 2)^{4}$ , giving $b_{3} + 8$ .

Therefore the coefficient of $x^{3}$ in $h (x)$ is $(a_{3} + 4) - (b_{3} + 8) = (a_{3} - b_{3}) - 4 = - 4.$

$∴$ The correct option is (C).

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