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Differential Calculus> MODIFIED AND CORRECT QUESTION Find doamin...

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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

vineet nimesh , 10 Years ago

Grade

1 Answers

Jitender Singh

Last Activity: 10 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

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