If fx=Maxsinx,cosx and gx=Minsinx,cosx, then ∫0πfxdx+∫0πgxdx=

Correct answer: (C) 2. Given fx=Maxsinx,cosx and gx=Minsinx,cosx ∫0πfxdx+∫0πgxdx =∫0(π)/4sinxdx+∫(π)/4πcosxdx+∫0(π)/4cosxdx+∫(π)/4πsinxdx =-cosx0(π)/4+sinx(π)/4π+sinx0(π)/4+-cosx(π)/4π =-1√2+1+0-1√2+1√2-0+1+1√2 =2

Q117 / 160 · AP EAMCET 2022 (04 Jul Shift 2) · Mathematics

If $f (x) = Max \{\sin x, \cos x\}$ and $g (x) = Min \{\sin x, \cos x\}$ , then $\int_{0}^{π} f (x) d x + \int_{0}^{π} g (x) d x =$

A
$2 \sqrt{2} + 2$
B
$2 \sqrt{2} - 2$
Correct answer
C
$2$
D
$2 \sqrt{2}$

Solution

Given $f (x) = Max \{\sin x, \cos x\}$ and $g (x) = Min \{\sin x, \cos x\}$
$\int_{0}^{π} f (x) d x + \int_{0}^{π} g (x) d x$
$= \int_{0}^{\frac{π}{4}} \sin x d x + \int_{\frac{π}{4}}^{π} \cos x d x + \int_{0}^{\frac{π}{4}} \cos x d x + \int_{\frac{π}{4}}^{π} \sin x d x$
$= {[- \cos x]}_{0}^{\frac{π}{4}} + {[\sin x]}_{\frac{π}{4}}^{π} + {[\sin x]}_{0}^{\frac{π}{4}} + {[- \cos x]}_{\frac{π}{4}}^{π}$
$= (- \frac{1}{\sqrt{2}} + 1) + (0 - \frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} - 0) + (1 + \frac{1}{\sqrt{2}})$
$= 2$

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

If fx=Maxsinx,cosx and gx=Minsinx,cosx, then ∫0πfxdx+∫0πgxdx=

Solution

Relevant Questions

If $f (x) = Max \{\sin x, \cos x\}$ and $g (x) = Min \{\sin x, \cos x\}$ , then $\int_{0}^{π} f (x) d x + \int_{0}^{π} g (x) d x =$