Let f(x) be non-constant thrice differentiable function defined on such that f(x) = f(6 – x) and f'(0) = 0 = f'(2) = f'(5). If 'n' is the minimum number of roots of (f”(x))2 + f(x)f”(x) = 0 in the interval x [0, 6] then sum of digits of n equals
Let f(x) be non-constant thrice differentiable function defined on such that f(x) = f(6 – x) and f'(0) = 0 = f'(2) = f'(5). If 'n' is the minimum number of roots of (f”(x))2 + f(x)f”(x) = 0 in the interval x [0, 6] then sum of digits of n equals
f (x) = f (6 – x) Þ f' (x) = -f' (6 – x) …. (1)
put x = 0, 2, 5
f' (0) = f' (6) = f' (2) = f' (4) = f' (5) = f' (1) = 0
and from equation (1) we get f' (3) = -f' (3)
So f' (x) = 0 has minimum 7 roots in
h (x) = f' (x) . f' (x)
h' (x) = (f' (x)2 + f' (x) f' (x)
h (x) = 0 ha
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Maths Continuity and Differentiability 2025
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