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Method of Differentiation — JEE Maths practice question

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

Question
If f(x)=x3+x2f′(1)+xf′′(2)−f′′′(3)f(x) = x^3 + x^2 f'(1) + x f''(2) - f'''(3) for all x∈Rx \in \mathbb{R}, then ∣f(0)+f(3)+16∣|f(0)+f(3)+16| is.
Solution
Answer: 8
Step 1: Set up the unknowns Let A=f′(1)A = f'(1), B=f′′(2)B = f''(2), C=f′′′(3)C = f'''(3). Then f(x)=x3+Ax2+Bx−Cf(x) = x^3+Ax^2+Bx-C. Since f′′′(x)=6f'''(x) = 6, we get C=6C = 6. Step 2: Determine AA and BB f′(x)=3x2+2Ax+Bf'(x) = 3x^2+2Ax+B. From f′(1)=Af'(1) = A:
3+2A+B=A  ⟹  A+B=−3⋯(1)3+2A+B = A \implies A+B = -3 \quad \cdots (1)
 f′′(x)=6x+2Af''(x) = 6x+2A. From f′′(2)=Bf''(2) = B:
12+2A=B⋯(2)12+2A = B \quad \cdots (2)
 From (1)(1) and (2)(2): A=−5A = -5, B=2B = 2. Step 3: Evaluate the expression
f(x)=x3−5x2+2x−6f(x) = x^3 - 5x^2 + 2x - 6
f(0)=−6,f(3)=27−45+6−6=−18f(0) = -6, \quad f(3) = 27 - 45 + 6 - 6 = -18
∣f(0)+f(3)+16∣=∣−6−18+16∣=∣−8∣=8|f(0)+f(3)+16| = |-6-18+16| = |-8| = 8
 Answer: 8
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