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Find doamin of f(x) g(x)=|sinx|+sinx h(x)=sinx+cosx , x is in b/w 0 to pi f(x)=(logh(x) g(x))1/2 log is in square root answer is pi/6,pi/2 pls give two solution graph and analytical. thanks

Find doamin of f(x)


g(x)=|sinx|+sinx

h(x)=sinx+cosx ,   x is in b/w 0 to pi
f(x)=(logh(x) g(x))1/2 
log is in square root
answer is pi/6,pi/2 
pls give two solution graph and analytical.
thanks

Grade:

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
1st Part:
g(x) = |sinx| + sinx
g(x) = 2sinx, 0\leq x\leq \pi
g(x) = 0, \pi \leq x\leq 2\pi
2nd Part:
h(x) = sinx + cosx, 0\leq x\leq \pi
f(x) = \sqrt{log(h(x).g(x))}
1st Condition:
\Rightarrow h(x).g(x)>0
2sinx (sinx + cosx)>0
sinx \geq 0, 0\leq x\leq \pi
sinx + cosx >0
\sqrt{2}sin(x+\frac{\pi }{4})>0
0\leq x<\frac{3\pi }{4}…...........(1)
2nd Condition:
log(2sinx (sinx + cosx))\geq 0
2sinx (sinx + cosx)\geq 1
2sinx.cosx\geq 1-2sin^{2}x
sin2x\geq cos2x
\Rightarrow \frac{\pi }{4}\leq 2x\leq \frac{5\pi }{4}
\Rightarrow \frac{\pi }{8}\leq x\leq \frac{5\pi }{8}…...............(2)
Final Solution:
Intersection of (1) & (2)
x\in [\frac{\pi }{8}, \frac{5\pi }{8}]
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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