Skip to main contentSkip to solution

For Δ=∣1+a1111+b1111+c∣\Delta = \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} where a,b,ca, b, c are roots of equation x3+px+q=0x^3 + px + q = 0, then

Match the LIST-I with LIST-II

LIST-ILIST-II
A.a+b+ca + b + cI.p−qp - q
B.Δ\DeltaII.−q-q
C.ab+bc+caab + bc + caIII.00
D.abcabcIV.pp

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

For x3+0⋅x2+px+q=0x^3+0 \cdot x^2+px+q=0, Vieta gives a+b+c=0a+b+c=0 (III), ab+bc+ca=pab+bc+ca=p (IV) and abc=−qabc=-q (II). Row-reducing the determinant gives Δ=abc+ab+bc+ca\Delta=abc+ab+bc+ca, which equals −q+p=p−q-q+p=p-q (I).

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question