Question

Let \(k\) be the largest integer such that the equation \((x-1)^2+2kx+11=0\) has no real roots. If \(y\) is a positive real number, then the least possible value of \(\frac{k}{4y}+9y\) is

• A. 4
• B. 6
• C. 8
• D. 10

Answer 1

The Correct Option is B

We are tasked with solving the equation \((x - 1)^2 + 2kx + 11 = 0\) for the condition where the equation has no real roots, and also finding the minimum value of the expression \(\frac{k}{4y} + 9y\) for the value of \(k\).

Step 1: Rewriting the Equation 

The given equation is:

\((x - 1)^2 + 2kx + 11 = 0\)

Expanding the equation, we get:

\((x^2 - 2x + 1) + 2kx + 11 = 0\)

Simplifying it:

\(x^2 + (2k - 2)x + 12 = 0\)

Step 2: Applying the Discriminant Condition

For the quadratic equation to have no real roots, its discriminant \(Δ\) must be less than 0. The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is given by:

\(Δ = b^2 - 4ac\)

In our equation, \(a = 1\), \(b = 2k - 2\), and \(c = 12\). Thus, the discriminant is:

\(Δ = (2k - 2)^2 - 4(1)(12)\)

Simplifying:

\(Δ = (2k - 2)^2 - 48\)

For no real roots, we require \(Δ < 0\), so:

\((2k - 2)^2 < 48\)

Step 3: Solving the Inequality

Solving the inequality:

\(2k - 2 < \sqrt{48}\) and \(2k - 2 > -\sqrt{48}\)

Since \(√48 = 4√3\), we have:

\(2k - 2 < 4\sqrt{3}\) and \(2k - 2 > -4\sqrt{3}\)

Adding 2 to both sides:

\(2k < 4\sqrt{3} + 2\) and \(2k > -4\sqrt{3} + 2\)

Dividing by 2:

\(k < 2\sqrt{3} + 1\) and \(k > 1 - 2\sqrt{3}\)

Since \(k\) is an integer, the largest integer value of \(k\) less than \(2\sqrt{3} + 1\) is 4. Hence:

\(k = 4\)

Step 4: Finding the Minimum Value of \(\frac{k}{4y} + 9y\)

Now, we substitute \(k = 4\) into the expression:

\(f(y) = \frac{4}{4y} + 9y = \frac{1}{y} + 9y\)

To minimize \(f(y)\), we take the derivative and set it equal to 0:

\(f'(y) = -\frac{1}{y^2} + 9 = 0\)

Solving for \(y\):

\(\frac{1}{y^2} = 9\)

\(y^2 = \frac{1}{9}\)

\(y = \frac{1}{3}\)

Substituting \(y = \frac{1}{3}\) into \(f(y)\):

\(f\left(\frac{1}{3}\right) = \frac{1}{\frac{1}{3}} + 9\left(\frac{1}{3}\right) = 3 + 3 = 6\)

Conclusion:

The least possible value of \(\frac{k}{4y} + 9y\)is 6. Hence, the correct option is: (B): 6.