Let f 1 (x) = x + a, f 2 (x) = x 2 + bx + c and ∆ = | array ccc 1 & 1 & 1 f _ 1 ( x _ 1 )…
Mathematics · JEE Advanced · NTA Exams — Matrices and Determinants
Let f1 (x) = x + a, f2 (x) = x2 + bx + c and
∆ = \[\left|\begin{array}{ccc} 1 & 1 & 1 \\ \mathrm{f}_{1}\left(\mathrm{x}_{1}\right) & \mathrm{f}_{1}\left(\mathrm{x}_{2}\right) & \mathrm{f}_{1}\left(\mathrm{x}_{3}\right) \\ \mathrm{f}_{2}\left(\mathrm{x}_{1}\right) & \mathrm{f}_{2}\left(\mathrm{x}_{2}\right) & \mathrm{f}_{2}\left(\mathrm{x}_{3}\right) \end{array}\right|\], then
∆ = \[\left|\begin{array}{ccc} 1 & 1 & 1 \\ \mathrm{f}_{1}\left(\mathrm{x}_{1}\right) & \mathrm{f}_{1}\left(\mathrm{x}_{2}\right) & \mathrm{f}_{1}\left(\mathrm{x}_{3}\right) \\ \mathrm{f}_{2}\left(\mathrm{x}_{1}\right) & \mathrm{f}_{2}\left(\mathrm{x}_{2}\right) & \mathrm{f}_{2}\left(\mathrm{x}_{3}\right) \end{array}\right|\], then
- ∆ is independent of a
- ∆ is independent of b and c
- ∆ is independent of x1, x2, x3
- none of the above
Answer
(A) ∆ is independent of a, (B) ∆ is independent of b and c
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