Let $f(x) = (x-4)(x-5)(x-6)(x-7)$ then -
Let $f(x) = (x-4)(x-5)(x-6)(x-7)$ then -
- A
$f'(x) = 0$ has four roots
- B
Three roots of $f'(x) = 0$ lie in $(4, 5) \cup (5, 6) \cup (6, 7)$
- C
The equation $f'(x) = 0$ has only one root
- D
Three roots of $f'(x) = 0$ lie in $(3, 4) \cup (4, 5) \cup (5, 6)$
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Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq$ 0 , and $f(x)=g(x) \sin x$.
$STATEMENT$ $-1: \lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.and
$STATEMENT$ $-2: f^{\prime}(0)=g(0)$.
Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq$ 0 , and $f(x)=g(x) \sin x$.
$STATEMENT$ $-1: \lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.and
$STATEMENT$ $-2: f^{\prime}(0)=g(0)$.
- [IIT 2008]