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Method of Differentiation — JEE Maths practice question

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

Question
Let a differentiable function ff satisfy f(xy)=2f(x)+f(y)+x+2y+1f(xy) = 2f(x)+f(y)+x+2y+1 for all x,y∈Rx, y \in \mathbb{R}. If f′(1)=1f'(1) = 1, then f(e3)f(e^3) is equal to.
Solution
Answer: 7
Step 1: Find f(1)f(1) Setting x=y=1x = y = 1: f(1)=3f(1)+4  ⟹  f(1)=−2f(1) = 3f(1)+4 \implies f(1) = -2. Step 2: Determine f′(x)f'(x) Differentiating with respect to xx (treating yy as constant): yf′(xy)=2f′(x)+1yf'(xy) = 2f'(x)+1. Setting x=1x = 1: yf′(y)=2f′(1)+1=3  ⟹  f′(y)=3yyf'(y) = 2f'(1)+1 = 3 \implies f'(y) = \dfrac{3}{y}. Step 3: Determine ff and evaluate
f(x)=3ln⁡x+c,f(1)=c=−2  ⟹  f(x)=3ln⁡x−2f(x) = 3\ln x+c, \quad f(1) = c = -2 \implies f(x) = 3\ln x - 2
f(e3)=3(3)−2=7f(e^3) = 3(3)-2 = 7
 Answer: 7
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