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Suppose that `f` is a polynomial of degree `3` and that `f^(x)!=0` at any of the stationary point. Then `f` has exactly one stationary point `f` must have no stationary point `f` must have exactly two stationary points `f` has either zero or two stationary points.

A

f has exactly one stationary point

B

f must have no stationary pint

C

f must have exactly two stationary points

D

f has either zero or two stationary points

Text Solution

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The correct Answer is:
4

The derivative of a degree 3 polynomial is quadratric .Thus must have either 0, 1 or 2 roots .If this has precisely one root then this must be repeated .Hence we have `f(X) =m(x-alpha)^(2)` where `alpha` is the repeated root and m in R . So our origininal funtion f has a critical point at x = `alpha`
Also `f''(x) =2x(x-alpha)` in which case `f''(alpha)` =0 But we aretold that the second derivative is nonzero at critical point hince theremust be either 0 or 2 critical points.
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