Question

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 \(x=\beta\))}

• A. \(28\)
• B. \(23\)
• C. \(5\)
• D. \(1\)

Hint:

For a quadratic \(ax^2+bx+c\), the vertex \(x\)-coordinate is \(-\tfrac{b}{2a}\).
- Always substitute back carefully to find the extremum (maximum or minimum) value.

Answer 1

The Correct Option is B

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 \(\boxed{-\tfrac{7}{34}}\).