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Monotonically Increasing Functions

If fx=[x], -...

Question

If $f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} [x], & - 2 \leq x \leq - \frac{1}{2} 2 x^{2} - 1, & - \frac{1}{2} < x \leq 2 \end{matrix};$ where $[.]$ represents the greatest integer function, then the function $f (x - 1)$ is discontinous at the points

0, - \frac{1}{2}

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0, 1

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- 1, - \frac{1}{2}

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0, \frac{1}{2}

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Solution

The correct option is D $0, \frac{1}{2}$
$f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} [x], & - 2 \leq x \leq - \frac{1}{2} 2 x^{2} - 1, & - \frac{1}{2} < x \leq 2 \end{matrix}$

Thus $f (x - 1)$ is discontinuous at $x = 0$ and $\frac{1}{2}$

Suggest Corrections

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