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← JEE Advanced 2025 · Paper 1
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JEE Advanced 2025 · Paper 1, Mathematics question 1

Let R\mathbb{R} denote the set of all real numbers. Let ai,bi∈Ra_i, b_i \in \mathbb{R} for i∈{1,2,3}i \in \{1, 2, 3\}. Define the functions f:R→Rf : \mathbb{R} \to \mathbb{R}, g:R→Rg : \mathbb{R} \to \mathbb{R}, and h:R→Rh : \mathbb{R} \to \mathbb{R} by

f(x)=a1+10x+a2x2+a3x3+x4,g(x)=b1+3x+b2x2+b3x3+x4,h(x)=f(x+1)−g(x+2).\begin{aligned} f(x) &= a_1 + 10x + a_2x^2 + a_3x^3 + x^4,\\ g(x) &= b_1 + 3x + b_2x^2 + b_3x^3 + x^4,\\ h(x) &= f(x+1) - g(x+2). \end{aligned}

If f(x)≠g(x)f(x) \neq g(x) for every x∈Rx \in \mathbb{R}, then the coefficient of x3x^3 in h(x)h(x) is

Choose one option
Answer and worked solution

Official answer: C

Condition on f − g.
f(x)−g(x)=(a1−b1)+7x+(a2−b2)x2+(a3−b3)x3.f(x)-g(x)=(a_1-b_1)+7x+(a_2-b_2)x^2+(a_3-b_3)x^3.
A real cubic always has a real root. So f≠gf\ne g everywhere forces a3=b3a_3=b_3, and then the quadratic part must have no real root.

Coefficient of x3x^3 in hh. The x4x^4 terms cancel.

  • f(x+1)f(x+1) contributes a3+4a_3+4, since (x+1)4(x+1)^4 has 4x34x^3.
  • g(x+2)g(x+2) contributes b3+8b_3+8, since (x+2)4(x+2)^4 has 8x38x^3.

coefficient=(a3+4)−(b3+8)=a3−b3−4=−4.\text{coefficient}=(a_3+4)-(b_3+8)=a_3-b_3-4=-4.