Method of DifferentiationmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Method of Differentiation: Let Twice Differentiable Equals (JEE Main 2026)

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

Question
Let f(x)f(x) and g(x)g(x) be twice differentiable with f′′(x)=g′′(x)f''(x)=g''(x) for all x∈Rx\in\mathbb{R}, f′(1)=2g′(1)=4f'(1)=2g'(1)=4, and g(2)=3f(2)=9g(2)=3f(2)=9. Then f(25)−g(25)f(25)-g(25) equals:
A2020
B4040
C−20-20correct
D−40-40
Solution
Step 1: Integrate the condition f′′=g′′f''=g''
f′(x)=g′(x)+C1f'(x) = g'(x)+C_1
 From 2g′(1)=42g'(1)=4: g′(1)=2g'(1)=2. From f′(1)=4f'(1)=4: C1=4−2=2C_1 = 4-2 = 2. Step 2: Integrate again
f(x)=g(x)+2x+C2f(x) = g(x)+2x+C_2
 From 3f(2)=93f(2)=9: f(2)=3f(2)=3. From g(2)=9g(2)=9: C2=3−9−4=−10C_2 = 3-9-4 = -10. Step 3: Evaluate at x=25x=25
f(25)−g(25)=2(25)+(−10)=50−10=40f(25)-g(25) = 2(25)+(-10) = 50-10 = 40
 Answer: (2)
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