Let f: R → R be a differentiable function such that f(a)=0=f(b) and f^{prime}(a) f^{prime}(b) >

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5. Continuity and Differentiation

normal

Let $f: R \rightarrow R$ be a differentiable function such that $f(a)=0=f(b)$ and $f^{\prime}(a) f^{\prime}(b) > 0$ for some $a < b$. Then, the minimum number of roots of $f^{\prime}(x)=0$ in the interval $(a, b)$ is

$3$

$2$

$1$

$0$

(KVPY-2010)

Solution

(b)

Given,

$f: R \rightarrow R$ be a differentiable function in $(a, b)$

and $\quad f(a)=f(b)=0, a < b$

and $f^{\prime}(a) f^{\prime}(b) > 0$

Since, $f^{\prime}(a) f^{\prime}(b) > 0$.

$\therefore f^{\prime}(a)$ and $f^{\prime}(b)$ both are positive or negative when both are positive then at least one root. $f^{\prime}(x)=0$ lie in interval, also when both are negative then at least one root lie in the interval.

[by Rolle's theorem]

$\therefore$ Minimum number of roots are $2$.

Standard 12

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normal

Let $f: R \rightarrow R$ be a differentiable function such that $f(a)=0=f(b)$ and $f^{\prime}(a) f^{\prime}(b) > 0$ for some $a < b$. Then, the minimum number of roots of $f^{\prime}(x)=0$ in the interval $(a, b)$ is

Solution

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