Question

If $f\left(x\right)=$max$\{\sin x,\cos x,\frac{1}{2}\}$, then the area of the region bounded by the curves $y=f\left(x\right),x-$axis $y-$axis and $x=2\pi$ is

• A. $(\frac{5\pi }{12}+3)$.sq.unit
• B. $(\frac{5\pi }{12}+\sqrt{2})$.sq.unit
• C. $(\frac{5\pi }{12}+\sqrt{3})$.sq.unit
• D. $(\frac{5\pi }{12}+\sqrt{2}+\sqrt{3})$.sq.unit

Answer 1

The correct option is D $(\frac{5\pi }{12}+\sqrt{2}+\sqrt{3})$.sq.unit

$\because f\left(x\right)=$max$\{\sin x,\cos x,\frac{1}{2}\}$
Interval value of $f\left(x\right)$
For $0\le x<\frac{\pi }{4},\cos x$
For $\frac{\pi }{4}\le x<\frac{5\pi }{6},\sin x$
For $\frac{5\pi }{6}\le x<\frac{5\pi }{3},\frac{1}{2}$
For $\frac{5\pi }{3}\le x<2\pi ,\cos x$
Hence, required area
$={\int }_0^{\frac{\pi }{4}}\cos xdx+{\int }_{\frac{\pi }{4}}^{\frac{5\pi }{6}}\sin xdx+{\int }_{\frac{5\pi }{6}}^{\frac{5\pi }{3}}\frac{1}{2}dx+{\int }_{\frac{5\pi }{3}}^{2\pi }\cos xdx$
$={\left[\sin x\right]}_0^{\frac{\pi }{4}}-{\left[\cos x\right]}_{\frac{\pi }{4}}^{\frac{5\pi }{6}}+\frac{1}{2}{\left[x\right]}_{\frac{5\pi }{6}}^{\frac{5\pi }{3}}+{\left[\sin x\right]}_{\frac{5\pi }{3}}^{2\pi }$
$=(\frac{1}{\sqrt{2}}-0)-(-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}})+\frac{1}{2}(\frac{5\pi }{3}-\frac{5\pi }{6})+(0+\frac{\sqrt{3}}{2})$
On simplification, we get
$=(\frac{5\pi }{12}+\sqrt{2}+\sqrt{3})$.sq.unit.