JEE Advanced 2025 Paper 1 · Q01 · Quadratic Equations

Question

Let $\mathbb{R}$ denote the set of all real numbers. Let $a_i, b_i \in \mathbb{R}$ for $i \in \{1,2,3\}$. Define $f, g, h: \mathbb{R} 	o \mathbb{R}$ by
$$f(x) = a_1 + 10x + a_2 x^2 + a_3 x^3 + x^4,$$
$$g(x) = b_1 + 3x + b_2 x^2 + b_3 x^3 + x^4,$$
$$h(x) = f(x+1) - g(x+2).$$
If $f(x) 
e g(x)$ for every $x \in \mathbb{R}$, then the coefficient of $x^3$ in $h(x)$ is

SolveKar AI · Accepted Answer

$f(x) - g(x) = (a_1-b_1) + 7x + (a_2-b_2)x^2 + (a_3-b_3)x^3$ (the $x^4$ terms cancel). For $f-g$ to be non-zero for every real $x$, it cannot be a cubic (every cubic has a real root). Hence the cubic coefficient must vanish: $a_3 = b_3$. Now compute the $x^3$ coefficient in $h(x)=f(x+1)-g(x+2)$. In $f(x+1)$, $x^3$ picks up contributions from $x^4$ expansion (coeff $4$) and from $a_3 x^3$ (coeff $a_3$): total $4 + a_3$. In $g(x+2)$, $x^3$ picks up $8$ (from $(x+2)^4 = x^4+8x^3+\ldots$) plus $b_3$ (from $b_3(x+2)^3$): total $8 + b_3$. Difference $= (4 + a_3) - (8 + b_3) = -4 + (a_3 - b_3) = -4$. Answer: $-4$.

JEE Advanced 2025 Paper 1 · Q01 · Quadratic Equations

Solution