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__Hint:__When two curves meet each other at a point the x and y intercept at that point is the same. So, we will equate $y$ from both the equations of the line and parabola. When two curves meet each other at a point then the quadratic equation satisfying that condition always has real roots.

__Complete step by step answer:__

Equation of line: \[y=kx................(i)\]

Equation of parabola: $y={{\left( x-1 \right)}^{2}}...........(ii)$

Now, when line & parabola meets, for the given condition we need to equate equation (i) with equation (ii) because the intercepts of $y$ in both the curves at the meeting point will be the same.

$\Rightarrow kx={{(x-1)}^{2}}$

As per identity rule \[{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] ,we can expand ${{(x-1)}^{2}}$

$\Rightarrow kx={{x}^{2}}-2x+1$

Now we will rearrange the equation for getting all the $x$ terms at one side,

$\Rightarrow {{x}^{2}}-2x-kx+1=0.........(iii)$

Now, taking x as common and converting the above equation to quadratic form of $a{{x}^{2}}+bx+c=0$

$\Rightarrow {{x}^{2}}-(2+k)x+1=0$

So here, \[a=1,b=-(k+2),c=1................(iv)\]

As, the line is touching the parabola; the roots of quadratic equation above must be real, which means,

Determinant of the quadratic equation (iii) must be zero, $D=0$

${{b}^{2}}-4ac=0$

Substituting the values from (iv) we get

${{[-(k+2)]}^{2}}-4(1)(1)=0............(v)$

$\Rightarrow {{[-k-2]}^{2}}-4=0$

$\Rightarrow {{k}^{2}}+4k+4-4=0$

$\Rightarrow {{k}^{2}}+4k=0$

$\Rightarrow k(k+4)=0$

$\Rightarrow k=0;k=-4$

Hence, the final answer is option (e).

__Note:__There is a possibility of mistake when substituting the values of b in the determinant. ${{b}^{2}}-4ac=0$ if the sign of the coefficient is not put correctly.

Alternatively, we can rewrite, equation (v) as

${{[-(k+2)]}^{2}}-4=0$

$\Rightarrow {{(k+2)}^{2}}=4$

Taking square roots on both sides

$k+2=\pm 2$

$\Rightarrow k=0;k=-4$

So, the answer in this case is again the same.

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