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Hint- Use the concept that for a quadratic equation or any function the zero of that quadratic equation or the function will always satisfy the equation or the value of that function will be zero.

Complete step-by-step solution -

Given that the function is $f\left( x \right) = k{x^2} - 3kx + 3k - 1$

And 1 is the zero of the equation. So by the help of hint we have

$f\left( 1 \right) = 0$

So proceeding forward with the function, we have

$

f\left( {x = 1} \right) = {\left. {k{x^2} - 3kx + 3k - 1} \right|_{x = 1}} = 0 \\

\Rightarrow f\left( 1 \right) = k{\left( 1 \right)^2} - 3k\left( 1 \right) + 3\left( 1 \right) - 1 = 0 \\

\Rightarrow k - 3k + 3 - 1 = 0 \\

$

Proceeding forward in order to find the value of $k$

\[ \Rightarrow k - 3k + 3 - 1 = 0 \\

\Rightarrow - 2k + 2 = 0 \\

\Rightarrow - 2k = - 2 \\

\Rightarrow k = 1 \\

\]

Hence, the value of k is 1.

Note- A zero of a function is an input value to the function that produces an output of 0. Also remember that roots of an equation are also called its zero. If one of the roots of the function is given, other roots can also be found out easily. Remember these points while solving such problems.

Complete step-by-step solution -

Given that the function is $f\left( x \right) = k{x^2} - 3kx + 3k - 1$

And 1 is the zero of the equation. So by the help of hint we have

$f\left( 1 \right) = 0$

So proceeding forward with the function, we have

$

f\left( {x = 1} \right) = {\left. {k{x^2} - 3kx + 3k - 1} \right|_{x = 1}} = 0 \\

\Rightarrow f\left( 1 \right) = k{\left( 1 \right)^2} - 3k\left( 1 \right) + 3\left( 1 \right) - 1 = 0 \\

\Rightarrow k - 3k + 3 - 1 = 0 \\

$

Proceeding forward in order to find the value of $k$

\[ \Rightarrow k - 3k + 3 - 1 = 0 \\

\Rightarrow - 2k + 2 = 0 \\

\Rightarrow - 2k = - 2 \\

\Rightarrow k = 1 \\

\]

Hence, the value of k is 1.

Note- A zero of a function is an input value to the function that produces an output of 0. Also remember that roots of an equation are also called its zero. If one of the roots of the function is given, other roots can also be found out easily. Remember these points while solving such problems.