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How Many Roots f''g + f'g' = 0 Cannot Have in (-1,5) | JEE Advanced

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Question
Let ff, gg be two continuous and twice derivable functions such that f(0)=f(3)=0f(0) = f(3) = 0; f(1)⋅f(2)<0f(1)\cdot f(2) < 0; g(0)=g(3)=0g(0) = g(3) = 0. The number of roots of the equation
f′′(x) g(x)+f′(x) g′(x)=0f''(x)\,g(x) + f'(x)\,g'(x) = 0
 in (−1,5)(-1, 5) cannot be
A22correct
B33
C44
D55
Solution
Step 1: Recognise the left-hand side as a derivative. By the product rule,
ddx[f′(x) g(x)]=f′′(x) g(x)+f′(x) g′(x),\frac{d}{dx}\left[f'(x)\,g(x)\right] = f''(x)\,g(x) + f'(x)\,g'(x) ,
 so the equation is simply
[f′(x) g(x)]′=0.\left[f'(x)\,g(x)\right]' = 0 .
 Step 2: Count the roots of ff. We are given f(0)=f(3)=0f(0) = f(3) = 0, and f(1)f(2)<0f(1)f(2) < 0 means f(1)f(1) and f(2)f(2) have opposite signs, so by the intermediate value theorem ff vanishes somewhere in (1,2)(1,2). That is at least 33 roots of ff in [0,3][0,3]:
0,α∈(1,2),3.0, \quad \alpha \in (1,2), \quad 3 .
 Step 3: Apply Rolle's theorem between consecutive roots of ff. Between 00 and α\alpha, and again between α\alpha and 33, the derivative must vanish:
f′(x)=0 has at least 2 roots in (0,3).f'(x) = 0 \ \text{has at least } 2 \ \text{roots in } (0,3) .
 Step 4: Count the roots of gg. Directly from the data,
g(x)=0 has at least 2 roots, namely 0 and 3.g(x) = 0 \ \text{has at least } 2 \ \text{roots, namely } 0 \ \text{and } 3 .
 Step 5: Combine. The product f′(x)g(x)f'(x)g(x) vanishes wherever either factor does, so
f′(x) g(x)=0 has at least 2+2=4 roots in [0,3].f'(x)\,g(x) = 0 \ \text{has at least } 2+2 = 4 \ \text{roots in } [0,3] .
 Step 6: Apply Rolle's theorem once more, now to the function f′gf'g, between each pair of consecutive roots. Four roots give three gaps, so
[f′(x)g(x)]′=0 has at least 3 roots in (0,3)⊂(−1,5).\left[f'(x)g(x)\right]' = 0 \ \text{has at least } 3 \ \text{roots in } (0,3) \subset (-1,5) .
 Step 7: According to the answer. The number of roots is at least 33, so it can be 33, 44 or 55, but it cannot be 22. Answer: (1).
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