If a^2 + b^2 + c^2 = - 2 and f(x) = | * 20 c 1 + a^2 x & (1 + b^2… - Vidyadip

MCQ

If ${a^2} + {b^2} + {c^2} = - 2$ and $f(x) = \left| {\begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{1 + {b^2}x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}} \right|$ then $f(x)$ is a polynomial of degree

A
$3$
✓
$2$
C
$1$
D
$0$

✓

Answer

Correct option: B.

$2$

b
(b) Applying ${C_1} \to {C_1} + {C_2} + {C_3}$
$f(x) = \left| {\,\begin{array}{*{20}{c}}1&{(1 + {b^2})x}&{(1 + {c^2})x}\\1&{1 + {b^2}x}&{(1 + {c^2})x}\\1&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}\,} \right|$,

$(\because {a^2} + {b^2} + {c^2} + 2 = 0$)

[Applying ${R_2} \to {R_2} - {R_1}$,${R_3} \to {R_3} - {R_1}$]

$ = \left| {\,\begin{array}{*{20}{c}}1&{(1 + {b^2})x}&{(1 + {c^2})x}\\0&{1 - x}&0\\0&0&{1 - x}\end{array}\,} \right| = {(1 - x)^2}.$

Hence degree of $f(x) = 2.$

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