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.

${\int }_0^{\pi /2}min\{\sin x,\cos x\}dx$ equals

• A. $2(\sqrt{3}-1)$
• B. $\sqrt{2}(\sqrt{2}-1)$
• C. $(\sqrt{3}-1)$
• D. $\sqrt{2}(\sqrt{2}+1)$

Answer 1

Answer: (B)

$\sqrt{2}(\sqrt{2}-1)$

min $\{\sin x,\cos x\}$ $\forall 0\le x\le \frac{\pi }{2}$ (from the graph of $y=\sin x$,
$y=\cos x$ $\forall 0\le x\le \frac{\pi }{2}$)
$=\{\begin{array}{cc}\sin x& 0\le x\le \frac{\pi }{4}\\ \cos x& \frac{\pi }{4}\le x\le \frac{\pi }{2}\end{array}$
$\therefore$ Required area $={\int }_0^{\pi /4}\sin xdx+{\int }_{\pi /4}^{\pi /2}\cos xdx$

$={(-\cos x)}_0^{\pi /4}+{\left(\sin x\right)}_{\pi /4}^{\pi /2}=(-\frac{1}{\sqrt{2}}+1)+(1-\frac{1}{\sqrt{2}})$
$=2-\sqrt{2}=\sqrt{2}(\sqrt{2}-1)$