JEE Mathematics 2024 Question with Solution

Given: the solution states that the correct option is B.

Find: The interval to which the roots of $x^2 + nx + (n-3) = 0$ belong.

The solution is internally inconsistent: it first mentions $n = 1.5$, then substitutes $n = 3$, and concludes the correct option is B. Using the working shown in the working after substitution, for $n = 3$ the quadratic becomes

$$x^2 + 3x + (3-3) = 0$$

so

$$x^2 + 3x = 0$$

Factorizing,

$$x(x+3) = 0$$

Hence the roots are

$$x = 0, \,-3$$

The page nevertheless marks Option B as correct. Therefore, based on the solution, the recorded answer is B, while noting that the displayed root $x=0$ lies on the boundary and makes the interval statement in the working inconsistent with the option text.

Therefore, the correct option is B according to the solution.

Given: the solution concludes Option B and also shows substitution with $n=3$.

Find: Whether the shown algebra matches the marked option.

From the shown substitution,

$$x^2 + nx + (n-3) = 0$$

with $n=3$ gives

$$x^2 + 3x + 0 = 0$$

which factors as

$$x(x+3)=0$$

Thus the roots are $0$ and $-3$. These do not both belong strictly to $(-\infty,0)$ because $0$ is not included in that open interval. So the source has a discrepancy between the marked option and the algebra displayed.

Since the