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Let f: [a, b] -> R be a function, conti...

Let `f: [a, b] -> R` be a function, continuous on `[a, b]` and twice differentiable on `(a, b)`. If, `f(a) = f(b) and f'(a) = f'(b)`, then consider the equation `f''(x) - lambda (f'(x))^2 = 0`. For any real `lambda` the equation hasatleast M roots where `3M + 1` is

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The correct Answer is:
4
Let `g(x)=f^(')(x) e^(-lamdaf(x))`
`g(a)=g(b)` so using Rolle's theorem in `Cepsilon(a,b)`
`g^(')(c)=0`
`impliesg^(')(x)=e^(-lamdaf(x))(f^('')(x))-lamda(f^(')(x)^(2))`
`impliesf^('')(c)-lamda(f^(')(c))^(2)=0`
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