For \( \Delta=\begin{vmatrix}1+a & 1 & 1\\1 & 1+b & 1\\1 & 1 & 1+c\end{vmatrix} \) where \( a,b,c \) are roots of equation \( x^3+px+q=0 \), then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( a+b+c \) | I. | \( p-q \) |
| B. | \( \Delta \) | II. | \( -q \) |
| C. | \( ab+bc+ca \) | III. | 0 |
| D. | \( abc \) | IV. | \( p \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-I, C-IV, D-II
For \( x^3+0\cdot x^2+px+q=0 \), the relations between roots and coefficients give
\( a+b+c=0 \)
\( ab+bc+ca=p \)
\( abc=-q \)
Expanding the determinant along the first row:
\( \displaystyle \Delta=(1+a)[(1+b)(1+c)-1]-1\cdot[(1+c)-1]+1\cdot[1-(1+b)] \)
\( \displaystyle =(1+a)(b+c+bc)-c-b \)
\( \displaystyle =abc+ab+bc+ca \)
(A) \( a+b+c \)
This matches III. 0.
(B) \( \Delta \)
This matches I. p − q.
\( \displaystyle \Delta=-q+p=p-q \)
(C) \( ab+bc+ca \)
This matches IV. p.
(D) \( abc \)
This matches II. −q.