Question

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

विकल्प

• 4

• 0

• 1

•  none of these

Answer 1

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