Let \( f(x)=x^3-3ax \), where \( a\gt 0 \), then the values for which \( f(x) \) is decreasing, are
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -\sqrt{a}\lt x\lt\sqrt{a} \)
A function is decreasing where its derivative is negative. Since a > 0,
\( \displaystyle f'(x)=3x^2-3a \)
\( \displaystyle =3(x-\sqrt{a})(x+\sqrt{a}) \)
This is negative between the roots, so f is decreasing when \( -\sqrt{a}\lt x\lt\sqrt{a} \).