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For a third degree monic polynomial, it is seen that the sum of roots are zero. What is the relation between the minimum angle to be rotated to have a Rolles point (α in Radians) and the cyclic sum of the roots taken two at a time c

(a) α = ^π⁄180 * tan^-1(c)

(b) Can never have a Rolles point

(c) α = ^180⁄π tan^-1(c)

(d) α = tan^-1(c)

The question was asked in an online interview.

Query is from Lagrange’s Mean Value Theorem in division Differential Calculus of Engineering Mathematics

1 Answer

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Right choice is (d) α = tan^-1(c)

For explanation I would say: From Vietas formulas we can deduce that the x^2 coefficient of the monic polynomial is zero (Sum of roots = zero). Hence, we can rewrite our third degree polynomial as

y = x^3 + (0) * x^2 + c * x + d

Now the question asks us to relate α and c

Where c is indeed the cyclic sum of two roots taken at a time by Vietas formulae

As usual, Rolles point in the rotated domain equals the Lagrange point in the existing domain. Hence, we must have

y ^‘ = tan(α)

3x^2 + c = tan(α)

To find the minimum angle, we have to find the minimum value of α

 such that the equation formed above has real roots when solved for x.

So, we can write

tan(α) – c > 0

tan(α) > c

α > tan^-1(c)

Thus, the minimum required angle is

α = tan^-1(c).

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