Q44 / 90 · JEE Main 2019 (10 Jan Shift 2) · Mathematics

The value of $\int_{- π / 2}^{π / 2} \frac{d x}{[x] + [\sin x] + 4},$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is

Correct answer
A
$\frac{3}{20} (4 π - 3)$
B
$\frac{3}{10} (4 π - 3)$
C
$\frac{1}{12} (7 π - 5)$
D
$\frac{1}{12} (7 π + 5)$

Solution

We have $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{d x}{[x] + [s i n x] + 4}$

$= \int_{- \frac{π}{2}}^{0} \frac{d x}{[x] + 3} + \int_{0}^{\frac{π}{2}} \frac{d x}{[x] + 4}$

$= \int_{- \frac{π}{2}}^{- 1} \frac{d x}{1} + \int_{- 1}^{0} \frac{d x}{2} + \int_{0}^{1} \frac{d x}{4} + \int_{1}^{\frac{π}{2}} \frac{d x}{5}$

$= {[x]}_{- \frac{π}{2}}^{- 1} + {[\frac{x}{2}]}_{- 1}^{0} + {[\frac{x}{4}]}_{0}^{1} + {[\frac{x}{5}]}_{1}^{\frac{π}{2}} = (- 1 + \frac{π}{2}) + (0 + \frac{1}{2}) + \frac{1}{4} + \frac{π}{10} - \frac{1}{5}$

$= \frac{- 20 + 10 π + 10 + 5 + 2 π - 4}{20} = \frac{12 π - 9}{20} = \frac{3}{20} (4 π - 3)$

Hence, the correct answer is option: 'A'.

The value of ∫-π/2π/2dxx+sin⁡x + 4, where t denotes the greatest integer less than or equal to t, is

Solution

Relevant Questions

The value of $\int_{- π / 2}^{π / 2} \frac{d x}{[x] + [\sin x] + 4},$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is