If f(x) is twice differentiable function such that f(a)=0, f(b)=2, f(c)=1, f(d)=2 , f(e)=0 , where a

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If f(x) is twice differentiable function such that f(a)=0, f(b)=2, f(c)=1, f(d)=2 , f(e)=0 , where a < b < c < d < e , then the minimum number of zeros of g(x)= \f^′(x) \^2+f^′ ′(x) · f(x) in the interval [a, e] is

MARKS Content Team · Accepted Answer

Correct answer: 6. Let, g(x)=d/(d x) [f(x) · f^′(x) ] to get the zero of g(x) we take function h(x)=f(x) · f^′(x) between any two roots of h(x) there lies atleast one root of h^′(x)=0 ⇒ g(x)=0 ⇒ h(x)=0 ⇒ f(x)=0 or f^′(x)=0 If f(x)=0 has 4 minimum solutions, f^′(x)=0 has 3 minimum solutions, h(x)=0 has 7 minimum solutions, then h^′(x)=g(x)=0 has 6 minimum solutions.

If $f(x)$ is twice differentiable function such that $f(a)=0, f(b)=2, f(c)=1, f(d)=2$, $f(e)=0$, where $a < b < c < d < e$, then the minimum number of zeros of $g(x)=\left\{f^{\prime}(x)\right\}^2+f^{\prime \prime}(x) \cdot f(x)$ in the interval $[a, e]$ is

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