6cud nyone post there working.
m getting trouble with mod sinx part.....
Find domain of the following:
log\ [log_{|sin x|}(x^2-8x+23)-\frac{3}{log_{2}|sin x|}]
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7 Answers
6
49log\,[log_{|sin\, x|}(\frac{x^2-8x+23}{8})]
|six x| ≠1
or |sin x|≠0
49let the given thing be log{log|sin x|t}
now -1≤sin x≤1,
so 0≤|sin x|≤1
but as it is base of logarithm, |sin x|≠0 and ≠1
so 0<|sin x|<1
thus sin x = fraction less than 1
let |sin x|=1y where y is a positive real number greater than 1
so, the expression is
log{log(1y)t}
so , log(1y)t≥0
so, log t log 1y≥0
so, log t-log y≥0
so, - log tlog y≥0
so, log tlog y≤0
now, log y cannot be less than 0 since t>1
so, log t≤0
so, t ≤ a0 wer a is any arbitrary positive number not equal to 1
so, t≤1
also sin x≠0,1
x≠0,π,π/2,3π/2
49manmay's question.... divided by eight kaha se aaya..
answer:
log\,[log_{|sin\, x|}({x^2-8x+23})-3log_{|sin\,x|}2 ]=log\,[log_{|sin\, x|}({x^2-8x+23})-log_{|sin\,x|}8 ]=log\,[log_{|sin\, x|}(\frac{{x^2-8x+23}}{8})
1log _{|sinx| }( x^2 - 8x + 23) - \frac{3}{log_{2}|sinx|}>0
log _{|sinx| }( x^2 - 8x + 23) - \3log_{|sinx|}2>0
log _{|sinx| }\left(\frac{x^2- 8x + 23}{2^3} \right)>0
from definition of logarithm funcn
we have
base ≠0 ,1
therefore
\left(\frac{x^2- 8x + 23}{2^3} \right)< |sinx |^0 \left( \texttt{bcoz |sinx| lies between 0 and 1})
hence inequality reverses
so we have
{x^2- 8x + 23} < 8
and
| sinx | ≠0,1
ie
{x^2- 8x + 15} < 0
( x- 5 ) ( x-3 ) < 0
usin no.line rule
x\varepsilon (3,5 )
since
|sinx | ≠0 ,1
sinx ≠0 and sinx ≠±1
ie
x ≠n\pi, \frac{(2n + 1 )\pi }{2}
now we want principal value hence put n= 1
x ≠π , 3π2
so we want domain we shud not include 3, 5 , pi, 3pi/2
DOMAIN IS
\left(3,\pi \right)U (\pi , \frac{3\pi }{2})U ( \frac{3\pi }{2}, 5 )
hope dis is ans !!!!!
6thnks subho nd omi...
a few more doubts may crop up......if they hope u'll help.......:)