Question 40: Given the function y = f(x )= ax3+ bx2+ cx+ d with the following variation table: Then |f(x)| = m has 4 distinct solutions \({x_1} < {x_2} < {x_3} < \frac{1}{2} < {x_4}\) if and only if

We have

The given function is \(y = 2{x^3} – 3{x^2} + 1.\)

We see: f(x) = 0

= 0 or \(x= \frac{1}{2} \)Variation table of function y = |f(x)| as follows:

Based on the variation table, deduce the equation |f(x)| = m has four distinct solutions \({x_1} < {x_2} < {x_3} < \frac{1}{2} < {x_4}\) if and only if \(\frac{1}{2} < m < 1\).

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