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 $\text{a, b, c, d, e}$ 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 $\begin{vmatrix}\text{x}^2+3\text{x}&\text{x}-1&\text{x}+ 3\\\text{x}+1&-2\text{x}&\text{x}-4\\\text{x}-3&\text{x}+4&3\text{x}\end{vmatrix}$
$=(\text{x}^2+3\text{x})\begin{vmatrix}-2\text{x}&\text{x}-4\\\text{x}+4&3\text{x}\end{vmatrix}-(\text{x}-1)\begin{vmatrix}\text{x}+1&\text{x}-4\\\text{x}-3&3\text{x}\end{vmatrix}+(\text{x}+3)\begin{vmatrix}\text{x}+1&-2\text{x}\\\text{x}-3&\text{x}+4\end{vmatrix}$
$=\left(x^2+3 x\right)\left(-6 x-x^2+16\right)-(x-1)\left(3 x^2+3 x-x^2+7 x-12\right)+(x+3)\left(x^2+5 x+4+2 x^2-6 x\right)$
$=-7 x^4+16 x^2+48 x+21 x^3+8 x^2-22 x-2 x^3-12+8 x^2+x+3 x^3+12$
$=-7 x^4+22 x^3+32 x^2+27 x+0$
But $x$ is a root of $a x^4+b x^3+c x^2+d x+e$
$e=0$

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