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.

${\int }_0^{\pi }max\{\sin x,\cos x\}dx$ is equal to

• A. $\sqrt{2}-1$
• B. $1-\frac{1}{\sqrt{2}}$
• C. $1+\frac{1}{\sqrt{2}}$
• D. None of these

Answer 1

Answer: (D)

None of these

First plot the graph of $y=\sin x$, $y=cosx$ by a dotted curve & find their point of intersections.
Now we find any two consecutive points of intersections. In between these points either $\sin x>\cos x$ or $\cos x>\sin x$ then in order to get max $(\sin x,\cos x)$ we take those segments for which one function is greater than the other function.
In the figure, A is the point of intersection of $\sin x$ and $\cos x$.
$\therefore$ $A(\frac{\pi }{4},\frac{1}{\sqrt{2}})$
$\therefore$ max $(\sin x,\cos x)$ $\forall 0\le x\le \pi$
$\therefore$ $=\{\begin{array}{cc}\cos x& \forall 0\le x\le \frac{\pi }{4}\\ \sin x& \forall \frac{\pi }{4}\le x\le \pi \end{array}$
$\therefore$ ${\int }_0^{\pi }max\left(\sin x\cos x\right)dx={\int }_0^{\pi /4}\cos xdx+{\int }_{\pi /4}^{\pi }\sin xdx=\sqrt{2}+1$