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 xRx\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
C20-20correct
D40-40
Solution
Step 1: Integrate the condition f=gf''=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=42=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=394=10C_2 = 3-9-4 = -10. Step 3: Evaluate at x=25x=25
f(25)g(25)=2(25)+(10)=5010=40f(25)-g(25) = 2(25)+(-10) = 50-10 = 40
Answer: (2)
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