If \(4+3x-7x^2\) attains its maximum value \(M\) at \(x=\alpha\) and \(5x^2-2x+1\) attains its minimum value at \(x=\beta\), then
\[
\frac{28\,(M - \alpha)}{5\,(m + \beta)}
=\,?
\]
\textit{(Assume \(m\) is that minimum value of \(5x^2 -2x +1\) at \(x=\beta\))}
Show Hint
- Always substitute back carefully to find the extremum (maximum or minimum) value.
- \(28\)
- \(23\)
- \(5\)
- \(1\)
The Correct Option is B
Solution and Explanation
Step 1: Eliminate the radicals and rearrange.
Given the equation:
\[ \frac{x - 1}{\sqrt{2x^2 - 5x + 2}} = \frac{41}{60}, \]
cross-multiply to obtain:
\[ 60(x - 1) = 41\sqrt{2x^2 - 5x + 2}. \]
Next, square both sides (being careful) and move all terms to one side to form a polynomial equation in \(x\).
Step 2: Solve the resulting equation.
Expanding both sides, we get:
\[ 3600(x - 1)^2 = 1681(2x^2 - 5x + 2). \]
Simplify this expression and solve for \(x\). This process should yield two real solutions, though there may be extraneous solutions to check.
Step 3: Select the root in the interval \(-\tfrac{1}{2} < x < 0\).
Among the real solutions, determine which one falls between \(-\tfrac{1}{2}\) and \(0\). The correct root is \(\boxed{-\tfrac{7}{34}}\).
Top Questions on Parabola
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