CAT 2023 Slot 2 · Quant · Question 7

First expand the quadratic into standard form.

{\left( x-1 \right)}^{2}+2kx+11={x}^{2}-2x+1+2kx+11.

Collecting terms, this is {x}^{2}+\left( 2k-2 \right)x+12.

No real roots means the discriminant is negative: {\left( 2k-2 \right)}^{2}-4\times 12<0.

So {\left( 2k-2 \right)}^{2}<48, giving \left| 2k-2 \right|<\sqrt{48}=4\sqrt{3}\approx 6.93.

That is -6.93<2k-2<6.93, so -4.93<2k<8.93 and -2.46<k<4.46.

The largest integer in that range is k=4.

Substituting, the expression becomes \frac{4}{4y}+9y=\frac{1}{y}+9y.

Both terms are positive for y>0, so apply AM-GM: \frac{1}{y}+9y\ge 2\sqrt{\frac{1}{y}\times 9y}.

The product inside the root is the constant 9, so the bound is 2\times 3=6.

Equality holds when \frac{1}{y}=9y, that is y=\frac{1}{3}, which is a legitimate positive value.

The least possible value is 6.

Shortcut. AM-GM applies precisely because the two terms multiply to a constant — the y cancels. So the minimum is 2\sqrt{\frac{k}{4}\times 9}=3\sqrt{k}, and k=4 gives 6 immediately.