find the domain of the function
f ( x) = \sqrt{log _{\phi (x) } g(x)}
where
g(x) = \mid sinx\mid + sinx
&
\phi (x) = sinx + cos x
where
0≤ x ≤π
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1 Answers
39R1 : |sinx| + sinx > 0 (Input of log must be strictly positive)
=> |sinx| > -sinx
This happens when x > 0. If x < 0, for eg -pi/2,
|sin(-pi/2)| > -(sin(-pi/2))
=> 1 > 1
which is not true.
So x > 0 is our first restriction(which is anyway there)
R2 : sin(x) + cos(x) ≠1 (base cannot be 1)
x ≠(2n - 1)pi/2 , where n = 1,2,3...
R3 : sin(x) + cos(x) > 0 (base must be positive)
=> -tan(x) > 0
=> tan(x) < 0
x → ( (2n - 1)pi/2, (2n - 1)pi ) U ( (2n + 1)pi/2, 2npi) (II and IV quadrants, hope I got this right)
But as x cannot go further than pi,
x → ( (2n - 1)pi/2, (2n - 1)pi) ONLY.
R4 : logΦ(x)g(x) ≥ 0 (Square root definition)
=> logΦ(x)g(x) ≥ logΦ(x)1
Before removing logarithm, we have to find out whether the logarithm is increasing or decreasing. This we decide by Φ(x)'s values in the domain we have defined for Φ(x). Accordingly we will have to switch the inequalities(if sin(x) + cos(x) < 1) or open it as it is (if sin(x) + cos(x) > 1).
Itna dekh lo agar sahi kiya hai maine...am not sure.
Now...as log needs to be non-negative, Φ(x) cannot be less than 1!
So g(x) ≥ 1 by opening log.
The rest is easy :)
