. Consider the following statements in respect of the determinant 

\( \Delta = \begin{vmatrix} k(k+2) & 2k+1 & 1 \\ 2k+1 & k+2 & 1 \\ 3 & 3 & 1 \end{vmatrix} \)

I. Δ is positive if $k>0$.

II. Δ is negative if $k<0$.

III. Δ is zero if $k=0$.

How many of the statements given above are correct?

Calculation:

Given,

Δ  =  \( \begin{vmatrix} k(k+2) & 2k+1 & 1\\ 2k+1 & k+2 & 1\\ 3 & 3 & 1 \end{vmatrix} \)

Simplify the determinant by the column operation  \(C_1 \rightarrow C_1 - C_2\):

\( \Delta = \begin{vmatrix} k^{2}-1 & 2k+1 & 1\\ k-1 & k+2 & 1\\ 0 & 3 & 1 \end{vmatrix} \)

Expanding along the third row,

\( \Delta = -3 \begin{vmatrix} k^{2}-1 & 1\\ k-1 & 1 \end{vmatrix} \;+\; \begin{vmatrix} k^{2}-1 & 2k+1\\ k-1 & k+2 \end{vmatrix} = (k-1)^{3}. \)

Thus  \( \Delta = (k-1)^{3}\).

Sign analysis

• \(k>0\): if \(0<k<1\), Δ < 0; if \(k>1\), Δ > 0  ⇒ Statement I is false.
• \(k<0\):  \(\Delta<0\)  ⇒ Statement II is true.
• \(k=0\):  \(\Delta=(-1)^{3}=-1\neq0\)  ⇒ Statement III is false.

∴ Only Statement II is correct  ⇒  exactly one statement is true.

Hence, the correct answer is Option 2.