Question

For the function f(x) = x^3 + x + 1. We do not have any Rolles point. The coordinate axes are transformed by rotating them by 60 degrees in anti-clockwise sense. The new Rolles point is?

(a) \(\frac{\sqrt{3}}{2}\)

(b) The function can never have a Rolles point

(c) \(3^{\frac{1}{2}}\)

(d) \(\sqrt{\frac{\sqrt{3}-1}{3}}\)

This question was addressed to me in final exam.

This interesting question is from Lagrange’s Mean Value Theorem in chapter Differential Calculus of Engineering Mathematics

Answer 1

Right option is (d) \(\sqrt{\frac{\sqrt{3}-1}{3}}\)

The explanation is: The question is simply asking us to find if there is some open interval in the original function f(x)

where we have f'(x) = tan(60)

We have

f'(x) = 3x^2 + 1 = tan(60)

3x^2 = √3 – 1

x=\(\sqrt{\frac{\sqrt{3}-1}{3}}\)