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

Jitender Singh IIT Delhi
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
$2sinx \neq 1$
$\Rightarrow x\neq \frac{\pi }{6}$
3rd Part:
$log_{2sinx}(\sqrt{2}sin(x+\frac{\pi }{4}))\geq 0$
Lets assume
$0<2sinx<1$
$\Rightarrow 0
$\sqrt{2}sin(x+\frac{\pi }{4})\leq 1$
$sin(x+\frac{\pi }{4})\leq \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{\pi }{2}< x\leq \pi$
Final sol. for this part:
$\frac{5\pi }{6}
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}
Final Solution:
$x\in (\frac{\pi }{6}, \frac{\pi }{2}]$
Thanks & Regards
Jitender Singh
IIT Delhi