Q166 / 270 · NDA 2016 (Phase 1) · Mathematics
Consider the functions
$f(x)=x g(x) \text { and } g(x)=\left[\frac{1}{x}\right]$
Where $[\cdot]$ is the greatest integer function.
What is $\int_{\frac{1}{3}}^{\frac{1}{2}} g(x) d x$ equal to?
- A$\frac{1}{6}$
- Correct answerB$\frac{1}{3}$
- C$\frac{5}{18}$
- D$\frac{5}{36}$
Solution
$\begin{aligned}
& \int_{1 / 3}^{1 / 2} \mathrm{~g}(x) \mathrm{d} x \\
& \mathrm{~g}(\mathrm{x})=\left\{\begin{array}{lc}
2, & \text { if } \frac{1}{3} \lt x \leq \frac{1}{2} \\
3, & \text { if } x=\frac{1}{3}
\end{array}\right.
\end{aligned}$
As $g(x)$ is a gretest integer function so value of $g(x)$ in integral limit will be
So $\int_{1 / 3}^{1 / 2} g(x) d x=\int_{1 / 3}^{1 / 2} 2 d x$
$\begin{aligned}
& =2[x]_{1 / 3}^{1 / 2} \\
& =2\left[\frac{1}{2}-\frac{1}{3}\right] \\
& =\frac{1}{3}
\end{aligned}$
Hence, the correct answer is option: 'B'.