Q.

Let f(x)=sin⁡x+∑r=02 cos⁡x+2 rπ3 g(x)=cos⁡x+∑r=02 sin⁡x+2 rπ3 h(x)=max{f(x),g(x)}∫−ππ g{f(x)} sin 2x dx  is equalent to∫π/4π [ h(x) ] {f(x)}2dx where [.] denotes greatest integer function is∫0π {g(x)}337dx=

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*
An Intiative by Sri Chaitanya

a

2∫0π g {f(x)} sin 2x dx

b

4∫0π/2 g {f(x)} sin 2x dx

c

∫−π/2π/2 cos2⁡x⋅sin5⁡xdx

d

1

e

π

f

3π4

g

0

h

5π4

i

0

j

π

k

337π2

l

1

answer is , , .

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

f(x)=sin⁡x+∑r=02 cos⁡x+2rπ3=sin⁡x+cos⁡x+cos⁡x+2π3+cos⁡x+4π3=sin⁡x+cos⁡x−2⋅12cos⁡x=sin⁡xsimilarly g(x) = cosx∫−ππ g {f(x)} sin2x dx=∫−ππ cos(sinx)⋅sin2x dx=0 since integrand is odd∫π/4π [h(x)]{f(x)}2dx=∫x/4π [sin⁡x]⋅sin2⁡xdx=0∫0π {g(x)}337dx=∫0π cos337⁡xdx=∫0π cos337⁡(π−x)dx=−∫0π cos337⁡xdx ∴∫0π cos337⁡xdx=0
Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon