1Is there any shortcut for problem 1 ??
Thnx 4 the answer sir...
And can sum1 try prob 2....
1) A sequence of real number xn is defined as follows...
x0 , x1 are arbitary +ve real numbers and
xn+2 =1+xn+1 xn ,n = 0,1,2,........
Then x2011 is
a) 1 b) x0 c) x1 d) x2
2)The eq. log2x2x(log2x)2 + (log2x)4 = 1 has
a) A root less than 1 b) has only 1 root greater than 1 c)Two irrational roots d)No real roots
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8 Answers
3411) Repeated use of the recursion yields x_{n+5}=x_n
Since 2011 is of the form 5k+1, we have x_{2011}=x_1
1
71Why it seems to me that second one can be solved.. though not in first try!!
1(log2x)4 ≥0 (as it has a power of 4)
or, log2x ≥ 0
or, x ≥ 20
or, x ≥ 1....
from this expression we can deduce that Option.2 MAY BE correct....
By the way, in the LHS, is it
log2x2x . (log2x)2
OR
log2x[2x(log2x)2 ]
I mean, is the 2x(log2x)2 part inside the log2x operator ?
1for 2..if i ve perceivd the operators corrctly then by observation
for x=2 the eqn is satisfied