CAT 2023 Slot 2 QA Question 5

The problem says: Let k be the largest integer such that the equation (x - 1)^2 + 2kx + 11 = 0 has no real roots. Then, if y is a positive real number, we need to find the least possible value of k/(4y) + 9y.

Alright, so first, We find the largest integer k for which the quadratic equation has no real roots. Then, using that k, find the minimum value of the expression involving y.

Starting with the first part: the quadratic equation. Let me write it down clearly.

The equation is (x - 1)^2 + 2kx + 11 = 0.

I should expand this to get it into standard quadratic form. Let's do that.

Expanding (x - 1)^2 gives x^2 - 2x + 1. So, substituting back into the equation:

x^2 - 2x + 1 + 2kx + 11 = 0.

Now, let's combine like terms. The x^2 term is just x^2. The x terms are -2x and 2kx, which combine to (2k - 2)x. The constants are 1 and 11, which add up to 12.

So, the equation becomes:

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

Now, for a quadratic equation ax^2 + bx + c = 0, the discriminant D is given by D = b^2 - 4ac. If D < 0, the equation has no real roots.

So, Computing the discriminant for our equation.

Here, a = 1, b = (2k - 2), and c = 12.

So, D = (2k - 2)^2 - 4 1 12.

Calculating that:

First, expand (2k - 2)^2: that's (2k)^2 - 22k2 + 2^2 = 4k^2 - 8k + 4. Wait, actually, that's incorrect. Let me correct that.

Wait, (a - b)^2 = a^2 - 2ab + b^2. So, (2k - 2)^2 = (2k)^2 - 2(2k)(2) + (2)^2 = 4k^2 - 8k + 4. Yes, that's correct.

So, D = 4k^2 - 8k + 4 - 48.

Simplify that: 4k^2 - 8k + 4 - 48 = 4k^2 - 8k - 44.

We need D < 0 for no real roots, so:

4k^2 - 8k - 44 < 0.

Let me divide both sides by 4 to simplify:

k^2 - 2k - 11 < 0.

Now, we have a quadratic inequality: k^2 - 2k - 11 < 0.

To find the values of k where this inequality holds, we can find the roots of the equation k^2 - 2k - 11 = 0 and then determine the intervals where the quadratic is negative.

Using the quadratic formula: k = [2 ± \] / 2 = [2 ± \] / 2.

Simplify \: that's 4*\$\sqrt{3}$ , so \$\sqrt{48}$ = 4√3 ≈ 6.928.

So, the roots are:

k = [2 + 4√3]/2 = 1 + 2√3 ≈ 1 + 3.464 ≈ 4.464,

and

k = [2 - 4√3]/2 = 1 - 2√3 ≈ 1 - 3.464 ≈ -2.464.

So, the quadratic k^2 - 2k - 11 is a parabola opening upwards (since the coefficient of k^2 is positive). It will be below zero between its two roots.

So, the inequality k^2 - 2k - 11 < 0 holds for k between -2.464 and 4.464.

But k is an integer, and we need the largest integer k such that the inequality holds. So, the upper bound is approximately 4.464, so the largest integer less than that is 4.

Wait, but let me confirm: if k=4, does the discriminant D become negative?

Compute D when k=4.

D = 4(4)^2 - 8(4) - 44 = 4*16 - 32 - 44 = 64 - 32 - 44 = 64 - 76 = -12.

Yes, that's negative, so the equation has no real roots when k=4.

What about k=5? Let's check D when k=5.

D = 4(5)^2 - 8(5) - 44 = 4*25 - 40 - 44 = 100 - 40 - 44 = 100 - 84 = 16.

That's positive, so the equation has real roots when k=5. So, k=5 is too big.

Therefore, the largest integer k is 4.

Now, moving on to the second part: find the least possible value of k/(4y) + 9y, where y is a positive real number.

Since we found k=4, substitute that into the expression:

Expression: (4)/(4y) + 9y = (1/y) + 9y.

We need to find the minimum value of (1/y) + 9y for y > 0.

This is a standard optimization problem. We can use calculus or the AM-GM inequality.

Let me try both methods.

First, using calculus:

Let f(y) = (1/y) + 9y.

Find f'(y) and set it to zero to find critical points.

f'(y) = -1/y^2 + 9.

Set f'(y) = 0:

-1/y^2 + 9 = 0 => 9 = 1/y^2 => y^2 = 1/9 => y = 1/3 (since y > 0).

Now, check the second derivative to confirm it's a minimum.

f''(y) = 2/y^3.

At y = 1/3, f''(