If f'(x) = left| begin{array}{ccc} mx & mx - | vedclass

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(A) Given the determinant $f'(x) = \left| \begin{array}{ccc} mx & mx - p & mx + p \ n & n + p & n - p \ mx + 2n & mx + 2n + p & mx + 2n - p \end{arr

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a straight line parallel to $x$-axis

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(A) Given the determinant $f'(x) = \left| \begin{array}{ccc} mx & mx - p & mx + p \ n & n + p & n - p \ mx + 2n & mx + 2n + p & mx + 2n - p \end{array} ight|$.Applying the row operation $R_3 ightarrow R_3 - R_1 - 2R_2$:The first element of $R_3$ becomes $(mx + 2n) - (mx) - 2(n) = 0$.The second element of $R_3$ becomes $(mx + 2n + p) - (mx - p) - 2(n + p) = mx + 2n + p - mx + p - 2n - 2p = 0$.The third element of $R_3$ becomes $(mx + 2n - p) - (mx + p) - 2(n - p) = mx + 2n - p - mx - p - 2n + 2p = 0$.Since all elements of the third row are $0$,the value of the determinant is $f'(x) = 0$.Integrating $f'(x) = 0$ with respect to $x$,we get $f(x) = C$,where $C$ is a constant.The equation $y = C$ represents a straight line parallel to the $x$-axis.

If $f'(x) = \left| \begin{array}{ccc} mx & mx - p & mx + p \\ n & n + p & n - p \\ mx + 2n & mx + 2n + p & mx + 2n - p \end{array} \right|$,then $y = f(x)$ represents

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