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 \[ \begin{aligned} & f(x)=a_{1}+10 x+a_{2} x^{2}+a_{3} x^{3}+x^{4}, \\ & g(x)=b_{1}+3 x+b_{2} x^{2}+b_{3} x^{3}+x^{4}, \\ & h(x)=f(x+1)-g(x+2) . \end{aligned} \] If \(f(x) \neq g(x)\) for every \(x \in \mathbb{R}\), then the coefficient of \(x^{3}\) in \(h(x)\) is

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Binomial Theorem

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

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) .

f (x) \neq = g (x)

for every

x \in R

, then the coefficient of

x^{3}

h (x)

Answer: (C)

We compute

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

. The coefficient of

x^{3}

h (x)

a_{3} - b_{3} - 4

.

Since

f (x) \neq = g (x)

for all

x \in R

, we have

f (x) - g (x) = x^{3} (a_{3} - b_{3}) + x^{2} (a_{2} - b_{2}) + 7 x + (a_{1} - b_{1}) \neq = 0

for all

x

. A cubic polynomial cannot be nonzero for all

x

unless it is actually of lower degree. Therefore, the coefficient of

x^{3}

must be zero:

a_{3} - b_{3} = 0

.

Thus, the coefficient of

x^{3}

h (x)

a_{3} - b_{3} - 4 = 0 - 4 = - 4

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Single correct option