Question

If f(x) be an increasing function defined on [a, b] then
max {f(t) such that $a\le t\le x$, $a\le x\le b$}=f(x) & min {f(t), $a\le t\le x$, $a\le x\le b$}=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t), $a\le t\le x$, $a\le x\le b$}=f(a),
min {f(t), $a\le t\le x$, $a\le x\le b$}=f(x).
On the basis of above information answer the following questions.

Let $f\left(x\right)=min\{1,1-\cos x,2\sin x\}$ then ${\int }_0^{\pi }f\left(x\right)dx$ is

• A. $\frac{\pi }{3}+1-\sqrt{3}$
• B. $\frac{2\pi }{3}-1+\sqrt{3}$
• C. $\frac{5\pi }{6}+1-\sqrt{3}$
• D. $\frac{\pi }{6}+1-\sqrt{3}$

Answer 1

The correct option is C $\frac{5\pi }{6}+1-\sqrt{3}$

min $\{1,1-\cos x,2\sin x\forall 0\le x\le \pi \}$
$=\{\begin{array}{cc}1-\cos x& 0\le x\le \frac{\pi }{2}\\ 1& \frac{\pi }{2}\le x\le \frac{5\pi }{6}\\ 2\sin x& \frac{5\pi }{6}\le x\le \pi \end{array}$
$\therefore$ ${\int }_0^{\pi }f\left(x\right)dx={\int }_0^{\pi /2}(1-\cos x)dx+{\int }_{\pi /2}^{5\pi /6}1dx+{\int }_{5\pi /6}^{\pi }2\sin xdx$
$={[x-\sin x]}_0^{\pi /2}+{\left[x\right]}_{\pi /2}^{5\pi /6}-2{\left[\cos x\right]}_{5\pi /6}^{\pi }$

$=\frac{\pi }{2}-1+\frac{5\pi }{6}-\frac{\pi }{2}-2[-1+\frac{\sqrt{3}}{2}]=\frac{5\pi }{6}+1-\sqrt{3}$