Suppose that f is a polynomial of degree 3 and that f′′(x)≠0 at any of the stationary point. Then
A
f has exactly one stationary point
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B
f must have no stationary point
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C
f must have exactly 2 stationary point
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D
f has exactly 0 or 2 stationary point.
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Solution
The correct option is Df has exactly 0 or 2 stationary point. The given information is a polynomial of degree 3.
Let the function be f(x)=ax3+bx2+cx+d.
To find the stationary points we differentiate and equate it to zero.
f′(x)=3ax2+2bx+x
The differentiated function is a quadratic equation which will give two real or two complex roots. If we get two real roots then we will have exactly 2 stationary points. Now if the roots are complex as they always occur in pair then we have no stationary points. Thus f has exactly 0 or 2 stationary points. .....Answer