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Method of Differentiation: Let Twice Differentiable Functions Satisfying Value Equal

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

Question
Let f:R→Rf:\mathbb{R}\to\mathbb{R} and g:R→Rg:\mathbb{R}\to\mathbb{R} be twice differentiable functions satisfying f′′(x)=g′′(x)f''(x) = g''(x), 2f′(1)=g′(1)=42f'(1) = g'(1) = 4, and 3f(2)=g(2)=93f(2) = g(2) = 9. Then the value of 15+f(4)−g(4)15+f(4)-g(4) is equal to.
Solution
Answer: 5
Step 1: Integrate the condition f′′=g′′f'' = g''
f′(x)=g′(x)+c1  ⟹  f(x)=g(x)+c1x+c2f'(x) = g'(x)+c_1 \implies f(x) = g(x)+c_1 x+c_2
 Step 2: Determine c1c_1 From g′(1)=4g'(1) = 4 and f′(1)=2f'(1) = 2: c1=f′(1)−g′(1)=−2c_1 = f'(1)-g'(1) = -2. Step 3: Determine c2c_2 From f(2)=3f(2) = 3 and g(2)=9g(2) = 9:
c2=f(2)−g(2)−2c1=3−9−2(−2)=−2c_2 = f(2) - g(2) - 2c_1 = 3 - 9 - 2(-2) = -2
 Step 4: Evaluate the expression
f(4)−g(4)=c1⋅4+c2=−8−2=−10f(4)-g(4) = c_1\cdot 4+c_2 = -8-2 = -10
15+f(4)−g(4)=15−10=515+f(4)-g(4) = 15-10 = 5
 Answer: 5
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