If f(x)=(x)/(2)-1 , then on the interval [0, π] which one of the following is correct?

Correct answer: C. [f(x)] , where [·] is the greatest integer function, and 1 f(x) are both discontinuous. aligned & f(x)= x 2 -1, [0, π] & · f(x)= ( x 2 -1) aligned (1)/(f(x))=(1)/((x)/(2)-1) is discontinuous at x=2 . f(x) is discontinuous for x=2 in [0, π]

NDA 2017

Exam: NDA
Year: 2017
Subject: Mathematics
Chapter: Continuity and Differentiability

NDA 2017 · Mathematics · Continuity and Differentiability

If f(x)=(x)/(2)-1 , then on the interval [0, π] which one of the...

f(x)=\frac{x}{2}-1

, then on the interval

[0, \pi]

which one of the following is correct?

Options

A $\tan [f(x)]$ , where $[\cdot]$ is the greatest integer function, and $\frac{1}{f(\mathrm{x})}$ are both continuous.
B $\tan [f(x)]$ , where $[\cdot]$ is the greatest integer functin, and $f^{-1}(\mathrm{x})$ are both continuous.
C $\tan [f(x)]$ , where $[\cdot]$ is the greatest integer function, and $\frac{1}{f(\mathrm{x})}$ are both discontinuous.
D $\tan [f(x)]$ , where $[\cdot]$ is the greatest integer function, is discontinuous but $\frac{1}{f(\mathrm{x})}$ is continuous.

Correct answer

C. $\tan [f(x)]$ , where $[\cdot]$ is the greatest integer function, and $\frac{1}{f(\mathrm{x})}$ are both discontinuous.

Step-by-step solution

\begin{aligned} & \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}-1, \quad[0, \pi] \\ & \tan \cdot \mathrm{f}(\mathrm{x})=\tan \left(\frac{\mathrm{x}}{2}-1\right)\end{aligned}

\frac{1}{f(x)}=\frac{1}{\frac{x}{2}-1}

is discontinuous at

x=2\tan . \mathrm{f}(\mathrm{x}) \text{ is discontinuous\text{ for } } \mathrm{x}=2 \text{ in }[0, \pi]$

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