Let x&2&x ^2&x&6 &x&6 =ax^4+bx^3+cx^2+dx+e. Then, the value of 5a… - Vidyadip

MCQ

Let $\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}=\text{ax}^4+\text{bx}^3+\text{cx}^2+\text{dx}+\text{e}.$ Then, the value of $5a + 4b + 3c + 2d + e$ is equal to:

A
$0$
B
$-16$
C
$16$
✓
None of these.

✓

Answer

Correct option: D.

None of these.

$\triangle=\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}$
$=\text{x}\begin{vmatrix}\text{x}&6\\\text{x}&6\end{vmatrix}-\text{x}^2\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}+\text{x}\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}$
$= 0 - x^2(12 - x^2) + x(12 - x^2)$
$= x^4 - 12x^2 + 12x - x^3$
$= ax^4 + bx^3 + cx^2 + dx + e$
$\Rightarrow x^4 - 12x^2 + 12x - x^3 = ax^4 + bx^3 + cx^2 + dx + e$
$\Rightarrow a = 1, b = -1, c = -12, d = 12, e = 0$
Thus,
$5a + 4b + 3c + 2d + e $
$= 5 - 4 - 36 + 24 + 0$
$ = -11$

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