The number of real values of x at which the f | vedclass

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(C) Expanding the determinant $f(x)$ along the first column:$f(x) = 1 \cdot (|x-1|(x-2)^2 - |x-2|(x-1)^2) - |x| \cdot ((x-2)^2 - (x-1)^2) + x^2 \cdot

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$2$

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(C) Expanding the determinant $f(x)$ along the first column:$f(x) = 1 \cdot (|x-1|(x-2)^2 - |x-2|(x-1)^2) - |x| \cdot ((x-2)^2 - (x-1)^2) + x^2 \cdot (|x-2| - |x-1|)$Simplifying the expression:$f(x) = |x-1|(x^2-4x+4) - |x-2|(x^2-2x+1) - |x|(x^2-4x+4 - x^2+2x-1) + x^2(|x-2|-|x-1|)$$f(x) = |x-1|(x^2-4x+4-x^2) + |x-2|(x^2-x^2+2x-1) - |x|(-2x+3)$$f(x) = |x-1|(4-4x) + |x-2|(2x-1) - |x|(3-2x)$We check for non-differentiability at the critical points of the absolute values: $x = 0, 1, 2$.At $x=0$: The term $|x|$ is non-differentiable,but the coefficient $(3-2x)$ is $3 
eq 0$. Thus,$f(x)$ is non-differentiable at $x=0$.At $x=1$: The term $|x-1|$ is non-differentiable,but the coefficient $(4-4x)$ is $0$. Thus,the non-differentiability is removed.At $x=2$: The term $|x-2|$ is non-differentiable,but the coefficient $(2x-1)$ is $3 
eq 0$. Thus,$f(x)$ is non-differentiable at $x=2$.Therefore,the function is non-differentiable at $x=0$ and $x=2$. The number of such values is $2$.

The number of real values of $x$ at which the function $f(x) = \left| \begin{array}{ccc} 1 & |x| & x^2 \\ 1 & |x-1| & (x-1)^2 \\ 1 & |x-2| & (x-2)^2 \end{array} \right|$ is not differentiable is

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