CAT 2022 Slot 3 · Quant · Question 1

Each condition is a statement about a discriminant, so handle them one at a time.

For 2{x}^{2}+kx+5=0 to have no real roots, its discriminant must be negative: {k}^{2}-40<0.

So {k}^{2}<40, giving -\sqrt{40}<k<\sqrt{40}, that is roughly -6.32<k<6.32.

For integers, that means k runs from -6 to 6.

For {x}^{2}+\left( k-5 \right)x+1=0 to have two distinct real roots, its discriminant must be positive: {\left( k-5 \right)}^{2}-4>0.

So {\left( k-5 \right)}^{2}>4, which means \left| k-5 \right|>2.

That splits into k-5>2 or k-5<-2, giving k>7 or k<3.

Combining with the first range, k must lie between -6 and 6 and also be below 3.

The branch k>7 is impossible, since k cannot exceed 6.

So k runs over -6,-5,-4,-3,-2,-1,0,1,2.

That is 9 values — the third option.

Shortcut. The two conditions pull in opposite directions, so the answer is the overlap of two short integer ranges. Noting that the second condition's upper branch (k>7) is already excluded by the first saves half the work.