Method of DifferentiationmediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Polynomial f = f′f″: Combined Value = 56 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let ff be a real polynomial of degree nn such that f(x)=f′(x) f′′(x)f(x)=f'(x)\,f''(x) for all x∈Rx\in\mathbb{R}. If f(0)=0f(0)=0, then 36(f′(2)+f′′(2)+∫02f(x) dx)36\left(f'(2)+f''(2)+\displaystyle\int_0^{2}f(x)\,dx\right) is equal to
A4242
B4646
C5656correct
D6666
Solution
Step 1: Degree balance: n=(n−1)+(n−2)⇒n=3n=(n-1)+(n-2)\Rightarrow n=3. With f(0)=0f(0)=0, write f(x)=ax3+bx2+cxf(x)=ax^3+bx^2+cx. Step 2: Then f′(x)=3ax2+2bx+cf'(x)=3ax^2+2bx+c, f′′(x)=6ax+2bf''(x)=6ax+2b, and f=f′f′′f=f'f''. Matching the leading x3x^3 term: 18a2=a⇒a=11818a^2=a\Rightarrow a=\dfrac{1}{18}. Matching the remaining coefficients forces b=0b=0 and c=0c=0, so f(x)=x318f(x)=\dfrac{x^3}{18}. Step 3: Compute each piece: f′(2)=3a(4)=12af'(2)=3a(4)=12a, f′′(2)=6a(2)=12af''(2)=6a(2)=12a, and ∫02ax3 dx=a⋅164=4a\displaystyle\int_0^2 ax^3\,dx=a\cdot\dfrac{16}{4}=4a. Their sum =28a=2818=149=28a=\dfrac{28}{18}=\dfrac{14}{9}. Step 4:
36(149)=56.36\left(\frac{14}{9}\right)=56.
 Correct answer: (3)
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