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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

moidin afsan , 11 Years ago
Grade 11
anser 2 Answers
Jitender Singh
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
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Last Activity: 11 Years ago
moidin afsan
thank uuuuuuuuuuu
Last Activity: 11 Years ago
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