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  • . If f(x)=\left|\begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array}\right|, A. f (a) = 0

If f(x)=\left|\begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array}\right|

A. f (a) = 0
B. f (b) = 0
C. f (0) = 0
D. f (1) = 0

Answers (1)

C)

We have:

f(x)=\left|\begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array}\right|

\\ \begin{aligned} &\text { If we put } x=a\\ &f(a)=\left|\begin{array}{ccc} 0 & a-a & a-b \\ a+a & 0 & a-c \\ a+b & a+c & 0 \end{array}\right|\\ &=0\left|\begin{array}{cc} 0 & a-c \\ a+c & 0 \end{array}\right|-0\left|\begin{array}{cc} 2 a & a-c \\ a+b & 0 \end{array}\right|+(a-b)\left|\begin{array}{cc} 2 a & 0 \\ a+b & a+c \end{array}\right| \end{aligned}

\\ \vspace{\baselineskip}= 0 -0 + (a -b) [2a (a + c) -0 (a + b)]\\ \\ \vspace{\baselineskip}= (a -b) [2a\textsuperscript{2} + 2ac -0]\\ \\ \vspace{\baselineskip}= (a -b) (2a\textsuperscript{2} + 2ac) $ \neq $ 0\\ \vspace{\baselineskip} If x = b\\

\\ \begin{array}{l} f(b)=\left|\begin{array}{ccc} 0 & b-a & b-b \\ b+a & 0 & b-c \\ b+b & b+c & 0 \end{array}\right| \\ =0\left|\begin{array}{cc} 0 & b-c \\ b+c & 0 \end{array}\right|-(b-a)\left|\begin{array}{cc} b+a & b-c \\ 2 b & 0 \end{array}\right|+0\left|\begin{array}{cc} b+a & 0 \\ 2 b & b+c \end{array}\right| \end{array}
\\ \vspace{\baselineskip}= 0 -(b - a) [(b + a) (0) -(b -c) (2b)] + 0\\ \\ \vspace{\baselineskip}= - (b -a) [0 - 2b\textsuperscript{2} + 2bc]\\ \\ \vspace{\baselineskip}= (a -b) (2b\textsuperscript{2} -2bc) $ \neq $ 0\\ \vspace{\baselineskip}

If x = 0  according to the given question:

\\ \begin{array}{l} f(0)=\left|\begin{array}{ccc} 0 & 0-a & 0-b \\ 0+a & 0 & 0-c \\ 0+b & 0+c & 0 \end{array}\right| \\ =0\left|\begin{array}{cc} 0 & -c \\ c & 0 \end{array}\right|-(-a)\left|\begin{array}{cc} a & -c \\ b & 0 \end{array}\right|+(-b)\left|\begin{array}{ll} a & 0 \\ b & c \end{array}\right| \end{array}
\\ \vspace{\baselineskip}= 0 + a [a (0) -(-bc)] -b [ac -b (0)]\\ \\ \vspace{\baselineskip}= a [bc] -b [ac]\\ \\ \vspace{\baselineskip}= abc -abc = 0\\ \vspace{\baselineskip}

Then the condition is satisfied.

Posted by

infoexpert22

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