KCET · Maths · Permutation Combination
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
- A (a) alone is cortect
- B (a) and (c) are correct.
- C (a) and (b) are correct.
- D None is correct.
Answer & Solution
Correct Answer
(B) (a) and (c) are correct.
Step-by-step Solution
Detailed explanation
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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