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MODIFIED AND CORRECT QUESTION Find doamin of f(x) x IS IN (0,pi) g(x)=|sinx|+sinx h(x)=sinx+cosx , x is in b/w 0 to pi f(x)=(log h(x) g(x))1/2 log is in square root and h(x) is in base answer is pi/6,pi/2 thanks

MODIFIED AND CORRECT QUESTION 
Find doamin of f(x)
x IS IN (0,pi)

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 and h(x) is in base
answer is pi/6,pi/2 

thanks

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1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
g(x) = |sinx| + sinx
g(x) = 2sinx, 0\leq x\leq \pi
h(x) = sinx+cosx = \sqrt{2}sin(x+\frac{\pi }{4})
f(x) = \sqrt{log_{h(x)}(g(x))}
f(x) = \sqrt{log_{2sinx}(\sqrt{2}sin(x+\frac{\pi }{4}))}
1st Part:
\sqrt{2}sin(x+\frac{\pi }{4}) > 0
\Rightarrow 0< x< \frac{3\pi }{4}
2nd Part:
2sinx > 0
\Rightarrow 0<x<\pi
2sinx \neq 1
\Rightarrow x\neq \frac{\pi }{6}
3rd Part:
log_{2sinx}(\sqrt{2}sin(x+\frac{\pi }{4}))\geq 0
Lets assume
0<2sinx<1
\Rightarrow 0<x<\frac{\pi }{6}, \frac{5\pi }{6}<x<\pi
\sqrt{2}sin(x+\frac{\pi }{4})\leq 1
sin(x+\frac{\pi }{4})\leq \frac{1}{\sqrt{2}}
\Rightarrow \frac{\pi }{2}< x\leq \pi
Final sol. for this part:
\frac{5\pi }{6}<x< \pi
Lets assume
1<2sinx<2
\Rightarrow \frac{\pi }{6}< x < \frac{5\pi}{6}
\sqrt{2}sin(x+\frac{\pi }{4})\geq 1
sin(x+\frac{\pi }{4})\geq \frac{1}{\sqrt{2}}
0\leq x\leq \frac{\pi }{2}
Final sol. of this part:
\frac{\pi }{6}<x\leq \frac{\pi }{2}
Final Solution:
x\in (\frac{\pi }{6}, \frac{\pi }{2}]
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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