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Differential Calculus
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f(x) = ln(xx + 1) ; x belongs to (0,1)
f'(x) = xx(1 + lnx) / (xx + 1)
f'(x) = 0
1 + ln x = 0 ; x = 1/e
f'(x) > 0 for x >1/e & f'(x) < 0 for x < 1/e
f(1/e) = ln( 1/e1/e + 1)
We need to know the value of the following for the determination of range :
f(0+) , f(1-) , f(1/e)
f(0+)
f(0+) = limx->0+ f(x) = limh->0 f(0 + h) = limh->0 ln(hh + 1)
Let A = limh->0 ln(hh + 1)
eA = limh->0 hh + 1
eA - 1 = limh->0 hh
ln(eA - 1) = limh->0 h*ln(h)
ln(eA - 1) = limh->0 ln(h)/(1/h)
Apply L'Hospital's Rule
ln(eA -1) = limh->0 (1/h)/ (-1/h2)
ln(eA -1) = limh->0 (-h)
ln(eA - 1) = 0 ==> solving it we get A = ln 2
Hence f(0+) = ln2
f(1-)
f(1-) = limx->1- f(x) = limh->0 f(1-h) = limh->0 ln{(1-h)(1-h) + 1} = ln 2
So ,
f(0+) = f(1-) = ln 2 where as f(1/e) = ln( 1/e1/e + 1)
Hence Range is ( ln( 1/e1/e + 1) , ln 2 ) Answer
Note : It may confuse that out of ln( 1/e1/e + 1) & ln 2 which one is greater .. though it is clear that f(x) is decreasing for x <1/e & increasing for x > 1/e .. You can also use the following analysis;
1/e1/e < 1 because ax < 1 for x > 0 & a<1
1/e1/e + 1 < 2
Since ln x is an increasing function So , ln x > ln y <=> x > y
ln( 1/e1/e + 1) < ln 2
. Note that it is very important to calculate f(0+) , because inspite of the decreasing nature of the curve in x <1/e , it is quite possible that f(0+) > f(1-) -- though it is not in this question but in some questions where unknown variables may be included or nature of function is not as clear as in this question , you should always follow the right approach...even though you see that f(0+ ) = f(1-) and calculation of f(0+) is in itself a good question ..
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f(x) = ln (xx + 1)
f '(x) = xx (1+lnx) / (xx + 1)
Clearly f '(x) = at x = 1/e. f '(x) < 0 when x < 1/e and f '(x) > 0 when x > 1/e
so minima occurs at x = 1/e
now check limiting values of f(x) when x tends to 0 and 1 and select the one with higher value for maxima. Both x -> 0 and x -> 1 gives f(x) = ln2
so the range is [ ln ((1/e)1/e+1) , ln2 )