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If \( f(x)=x^{3} \) and \( g(x)=x^{3}-4 x \) in \( -2 \leq x \leq 2 \), then consider the statements:
(a) \( f(x) \) and \( g(x) \) satisfy mean value theorem.
(b) \( f(x) \) and \( g(x) \) both satisfy Rolle's theorem.
(c) Only \( g(x) \) satisfies Rolle's theorem.
Of these statements
Options:
(a) \( f(x) \) and \( g(x) \) satisfy mean value theorem.
(b) \( f(x) \) and \( g(x) \) both satisfy Rolle's theorem.
(c) Only \( g(x) \) satisfies Rolle's theorem.
Of these statements
Solution:
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Verified Answer
The correct answer is:
(a) and (c) are correct.
Given that, $f(x)=x^{3}$ and $g(x)=x^{3}-4 x$
Since, $f(x)$ and $g(x)$ are both continuous at $[-2,2]$ and differentiable at $[-2,2]$
So, $f(x)$ and $g(x)$ satisfy mean value theorem.
Now, $f(-2)=-8, f(2)=8$
So, $f(-2) \neq f(2)$
$g(2)=(2)^{3}-4(2)=0=g(-2)=(-2)^{3}-4(-2)=0$
Therefore, $f(x)$ does not satisfy Rolle's theorem but $g(x)$ satisfy Rolle's theorem.
Since, $f(x)$ and $g(x)$ are both continuous at $[-2,2]$ and differentiable at $[-2,2]$
So, $f(x)$ and $g(x)$ satisfy mean value theorem.
Now, $f(-2)=-8, f(2)=8$
So, $f(-2) \neq f(2)$
$g(2)=(2)^{3}-4(2)=0=g(-2)=(-2)^{3}-4(-2)=0$
Therefore, $f(x)$ does not satisfy Rolle's theorem but $g(x)$ satisfy Rolle's theorem.
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