(A) Given that $(x - 1)$ is a factor of $f(x) = x^{3} - kx^{2} + 11x - 6$.According to the factor theorem,if $(x - a)$ is a factor of $f(x)$,then $f(a

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(A) Given that $(x - 1)$ is a factor of $f(x) = x^{3} - kx^{2} + 11x - 6$.According to the factor theorem,if $(x - a)$ is a factor of $f(x)$,then $f(a) = 0$.Here,$a = 1$,so $f(1) = 0$.Substituting $x = 1$ into the polynomial:$(1)^{3} - k(1)^{2} + 11(1) - 6 = 0$$1 - k + 11 - 6 = 0$$6 - k = 0$$k = 6$.