Question

Let $A\left(x\right)=\left|\begin{array}{ccc}3& 3x& 3x^2+2a^2\\ 3x& 3x^2+2a^2& 3x^3+6a^{2}x\\ 3x^2+2a^2& 3x^3+6a^{2}x& 3x^4+12a^2{x}^2+2a^4\end{array}\right|,$
where $a\in R$ is a non-zero constant. Then which of the following option(s) is/are INCORRECT ?

• A. There exists $x\in R$ such that $A\left(x\right)=0$
• B. $A\left(x\right)$ is independent of $x$
• C. $\underset{-1}{\overset{1}{\int }}A\left(x\right)dx=16a^6$
• D. Area bounded by the curve $y=A\left(x\right),x=-2,x=3$ and the $x$-axis is $20a^6$
  

Answer 1

Answer: (A), (C)

$\underset{-1}{\overset{1}{\int }}A\left(x\right)dx=16a^6$

$A\left(x\right)=\left|\begin{array}{ccc}3& 3x& 3x^2+2a^2\\ 3x& 3x^2+2a^2& 3x^3+6a^{2}x\\ 3x^2+2a^2& 3x^3+6a^{2}x& 3x^4+12a^2{x}^2+2a^4\end{array}\right|$

$R_2\to R_2-x{R}_1,R_3\to R_3-x^2{R}_1$
$A\left(x\right)=2a^2\cdot 2a^2\left|\begin{array}{ccc}3& 3x& 3x^2+2a^2\\ 0& 1& 2x\\ 1& 3x& 5x^2+a^2\end{array}\right|$

$A\left(x\right)=4a^4\left|\begin{array}{ccc}3& 3x& 3x^2+2a^2\\ 0& 1& 2x\\ 1& 3x& 5x^2+a^2\end{array}\right|$

Expanding along $C_1$, we get
$A\left(x\right)=4a^4\left[3\right(a^2-x^2)-0+(3x^2-2a^2\left)\right]$
$\Rightarrow A\left(x\right)=4a^4\times a^2=4a^6$

$A\left(x\right)\ne 0$ as $a\in R-\left\{0\right\}$

$\underset{-1}{\overset{1}{\int }}A\left(x\right)dx=8a^6$

$y=A\left(x\right)=4a^6$ is a straight line parallel to $x$-axis with $y$-intercept $4a^6$
So, area bounded is a rectangle with area equal to $4a^6\times (3-(-2\left)\right)=20a^6$