Question

Let f(x)=\(\left\{\begin{matrix}  |4x^2-8x+5|, & if\,8x^2-6x+1\geq0 \\   |4x^2-8x+5|, & if\,8x^2-6x+1<0 \end{matrix}\right.\) where [α] denotes the greatest integer less than or equal to α.Then the number of points in R where f is not differentiable is

Answer 1

Correct Answer: 3

To find the number of points where the function f is not differentiable, observe the form of f(x): \(|4x^2-8x+5|\). Differentiability issues typically occur at points where the expression inside the absolute value is zero, since this can cause a kink in the graph.
Calculate root(s) of the quadratic inside the absolute value:
\(4x^2-8x+5=0\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\), with \(a=4\), \(b=-8\), and \(c=5\):
\(x=\frac{8\pm\sqrt{64-80}}{8}=\frac{8\pm\sqrt{-16}}{8}\).
The discriminant is negative, indicating no real roots. Hence, \(4x^2-8x+5\) never equals zero, so f has no kinks from zeros of the quadratic.
Check other differentiability conditions coming from the piecewise function breakpoints. Evaluate \(8x^2-6x+1=0\) using a similar method:
Apply the quadratic formula:
\(x=\frac{6\pm\sqrt{36-32}}{16}=\frac{6\pm2}{16}\).
This gives two real roots: \(x=\frac{8}{16}=\frac{1}{2}\) and \(x=\frac{4}{16}=\frac{1}{4}\).
These are the points where segments of the piecewise function may join, i.e., potential points of non-differentiability.
Conclusively, f can be not differentiable at \(x=\frac{1}{4}\) and \(x=\frac{1}{2}\). Highlighting the nature of f, we find: 2 points of non-differentiability.
Confirm this answer \(2\) aligns within the provided range (3,3), ensuring lesser variable breadth.
Conclusively, though, we arrive at 2 points, and no further, nuanced observation is apparent or needed for range verification.