Question

Let $p\left(x\right)=\left|\begin{array}{ccc}{(2^x+2^{-x})}^2& {(3^x+3^{-x})}^2& {(5^x+5^{-x})}^2\\ {(2^x-2^{-x})}^2& {(3^x-3^{-x})}^2& {(5^x-5^{-x})}^2\\ 1& 1& 1\end{array}\right|$ and $q\left(x\right)=\left|\begin{array}{ccc}2x-2& 3x-4& 3x-5\\ x-1& 2x-3& 2x-4\\ x-1& x-1& 2x-4\end{array}\right|$ then

• A. $p(–5)=7$
• B. $p\left(x\right)=q\left(x\right)$ has $3$ solutions
• C. $p(–5)=0$
• D. $p\left(x\right)=q\left(x\right)$ has no solutions

Answer 1

Answer: (B), (C)

$p(–5)=0$

$p\left(x\right)=\left|\begin{array}{ccc}{(2^x+2^{-x})}^2& {(3^x+3^{-x})}^2& {(5^x+5^{-x})}^2\\ {(2^x-2^{-x})}^2& {(3^x-3^{-x})}^2& {(5^x-5^{-x})}^2\\ 1& 1& 1\end{array}\right|$
$R_1\to R_1–R_2$
$p\left(x\right)=\left|\begin{array}{ccc}4& 4& 4\\ {(2^x-2^{-x})}^2& {(3^x-3^{-x})}^2& {(5^x-5^{-x})}^2\\ 1& 1& 1\end{array}\right|$
Hence $p\left(x\right)=0\forall x$
Hence$p(–5)=0$
Now for $q\left(x\right)=p\left(x\right)=0$
$q\left(x\right)=0$
$\begin{array}{cc}& q\left(x\right)=\left|\begin{array}{ccc}2x-2& 3x-4& 3x-5\\ x-1& 2x-3& 2x-4\\ x-1& x-1& 2x-4\end{array}\right|\\ & \\ & R_1\to R_1-R_2-R_3\end{array}$
$q\left(x\right)=\left|\begin{array}{ccc}0& 0& 3-x\\ x-1& 2x-3& 2x-4\\ (x-1)& (x-1)& (2x-4)\end{array}\right|$
$q\left(x\right)=(3-x)\left[{(x-1)}^2\right(x-1\left)\right(2x-3\left)\right]$
$=(3-x)(x-1)[x-1-2x+3]$
$=(x–1)(x–2)(x–3)$
$q\left(x\right)=0$ for $x=1,2,3.$