Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

the greatest value of the function f(x)=sin2x/sin(2x+pi/4) in the interval [0,pi/4] is aans is root 2 plsssss explain

the greatest value of the function f(x)=sin2x/sin(2x+pi/4)   in the interval [0,pi/4] is 
 
aans is root 2   plsssss  explain

Grade:11

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
f(x) = \frac{sin2x}{sin(2x+\frac{\pi }{4})}
f(x) = \frac{sin2x}{sin(2x).cos(\frac{\pi }{4})+cos(2x).sin(\frac{\pi }{4})}
f(x) = \frac{\sqrt{2}sin2x}{sin(2x)+cos(2x)}
f'(x) = \sqrt{2}\frac{(sin2x+cos2x).(2cos2x)-sin2x(2cos2x-2sin2x)}{(sin(2x)+cos(2x))^{2}}f'(x) = \frac{2\sqrt{2}}{(sin(2x)+cos(2x))^{2}}> 0
=> f(x) is always increasing. So maxima of f(x) will be at x = pi/4.
f(\frac{\pi }{4}) = \frac{sin2.\frac{\pi }{4}}{sin(2.\frac{\pi }{4}+\frac{\pi }{4})}
f(\frac{\pi }{4}) = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty
moidin afsan
20 Points
7 years ago
thank uuuuuuuuuuu

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free