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Method of Differentiation: Let Real Valued Differentiable Functions Value

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

Question
Let ff and gg be two real-valued differentiable functions on R\mathbb{R}. If f′(x)=g(x)f'(x) = g(x) and g′(x)=f(x)g'(x) = f(x) for all x∈Rx \in \mathbb{R}, and f(3)=5f(3) = 5, g(3)=4g(3) = 4, then the value of f2(π)−g2(π)f^2(\pi) - g^2(\pi).
Solution
Answer: 9
Step 1: Show the expression is constant Let ϕ(x)=f2(x)−g2(x)\phi(x) = f^2(x) - g^2(x). Then:
ϕ′(x)=2f(x)f′(x)−2g(x)g′(x)=2f(x)g(x)−2g(x)f(x)=0\phi'(x) = 2f(x)f'(x) - 2g(x)g'(x) = 2f(x)g(x) - 2g(x)f(x) = 0
 So f2(x)−g2(x)=kf^2(x) - g^2(x) = k for all x∈Rx \in \mathbb{R}. Step 2: Determine kk
k=f2(3)−g2(3)=25−16=9k = f^2(3) - g^2(3) = 25 - 16 = 9
 Hence f2(π)−g2(π)=9f^2(\pi) - g^2(\pi) = 9. Answer: 9
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