Q167 / 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 $\int_{\frac{1}{3}}^1 f(x) d x$ equal to ?
- Correct answerA$\frac{37}{72}$
- B$\frac{2}{3}$
- C$\frac{17}{72}$
- D$\frac{37}{144}$
Solution
$\begin{aligned}
& \int_{1 / 3}^1 f(x) d x=\int_{1 / 3}^{1 / 2} f(x) d x+\int_{1 / 2}^1 f(x) d x...(1)\\
& \mathrm{f}(\mathrm{x})=\mathrm{xg}(\mathrm{x}) ...(1)
\end{aligned}$
$g\left(\frac{1}{3}\right)=3 \quad g\left(\frac{1}{2}\right)=2 \quad g(1)=1$
The value of $g(x)$ in value $\left(\frac{1}{2}, \frac{1}{3}\right)$ will be 2 and in range $\left(\frac{1}{2}, 1\right)$ it will be 1
from (1)
$\begin{aligned}
& \int_{1 / 3}^1 f(x) d x=\int_{1 / 3}^{1 / 2} x g(x) d x+\int_{1 / 2}^1 x g(x) d x \\
& =\int_{1 / 3}^{1 / 2} x \cdot 2 d x+\int_{1 / 2}^1 x \times 1 d x \\
& =\left[x^2\right]_{1 / 3}^{1 / 2}+\left[\frac{x^2}{2}\right]_{1 / 2}^1 \\
& =\left[\frac{1}{4}-\frac{1}{9}\right]+\frac{1}{2}\left[1-\frac{1}{4}\right]
\end{aligned}$
$=\frac{5}{36}+\frac{3}{8}=\frac{37}{72}$
Hence, the correct answer is option: 'A'.