Let [x] denote the greatest integer function | vedclass

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(C) For $x \in [-\frac{1}{2}, 0)$,$2x \in [-1, 0)$,so $[2x] = -1$. Thus,$f(x) = 4x^2 - x$.For $x \in [0, \frac{1}{2})$,$f(x) = ax^2 - bx$.For continui

Vedclass · Accepted Answer

$f(x)$ is continuous and differentiable in $(-\frac{1}{2}, \frac{1}{2}) \forall a \in R \& b = 1$.

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(C) For $x \in [-\frac{1}{2}, 0)$,$2x \in [-1, 0)$,so $[2x] = -1$. Thus,$f(x) = 4x^2 - x$.For $x \in [0, \frac{1}{2})$,$f(x) = ax^2 - bx$.For continuity at $x = 0$,$\lim_{x 	o 0^-} f(x) = f(0) = \lim_{x 	o 0^+} f(x)$.$f(0^-) = 4(0)^2 - 0 = 0$ and $f(0^+) = a(0)^2 - b(0) = 0$. Since $0 = 0$,$f(x)$ is continuous at $x = 0$ for all $a, b$.For differentiability at $x = 0$,we check $f'(0^-) = f'(0^+)$.$f'(x) = 8x - 1$ for $x $f'(x) = 2ax - b$ for $x > 0$,so $f'(0^+) = -b$.Equating them,$-1 = -b \Rightarrow b = 1$.Thus,$f(x)$ is differentiable at $x = 0$ if $b = 1$,independent of $a$.

Let $[x]$ denote the greatest integer function and $f(x) = \begin{cases} 4x^2 + [2x]x, & \text{if } x \in [-\frac{1}{2}, 0) \\ ax^2 - bx, & \text{if } x \in [0, \frac{1}{2}) \end{cases}$. Then:

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