Let x^2+3x&x-1&x+ 3 +1&-2x&x-4 -3&x+4&3x =ax^4+bx^3+cx^2+dx+e be … - Vidyadip

MCQ

Let $\begin{vmatrix}\text{x}^2+3\text{x}&\text{x}-1&\text{x}+ 3\\\text{x}+1&-2\text{x}&\text{x}-4\\\text{x}-3&\text{x}+4&3\text{x}\end{vmatrix}=\text{ax}^4+\text{bx}^3+\text{cx}^2+\text{dx}+\text{e}$ be an identity in $x,$ where $a, b, c, d, e$ are independent of $x$. Then the value of $e$ is:

A
$4$
✓
$0$
C
$1$
D
None of these.

✓

Answer

Correct option: B.

$0$

Let $\begin{vmatrix}\text{x}^2+3\text{x}&\text{x}-1&\text{x}+ 3\\\text{x}+1&-2\text{x}&\text{x}-4\\\text{x}-3&\text{x}+4&3\text{x}\end{vmatrix}$
$=(\text{x}^2+3\text{x})\begin{vmatrix}-2\text{x}&\text{x}-4\\\text{x}+4&3\text{x}\end{vmatrix}-(\text{x}-1)\begin{vmatrix}\text{x}+1&\text{x}-4\\\text{x}-3&3\text{x}\end{vmatrix}+(\text{x}+3)\begin{vmatrix}\text{x}+1&-2\text{x}\\\text{x}-3&\text{x}+4\end{vmatrix}$
$= (x^2 + 3x)(-6x - x^2 + 16) - (x - 1)(3x^2 + 3x - x^2 + 7x - 12) + (x + 3)(x^2 + 5x + 4 + 2x^2 - 6x)$
$= -7x^4 + 16x^2 + 48x + 21x^3 + 8x^2 - 22x - 2x^3 - 12 + 8x^2 + x + 3x^3 + 12$
$= -7x^4 + 22x^3 + 32x^2 + 27x + 0$
But $x$ is a root of $ax^4 + bx^3 + cx^2 + dx + e$
$e= 0$

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