1
well there would be two more cases isn't it??....(HOPEFULLY THEY WON't YIELD any ""NEW"" root)...or there are just two cases??i m getting confused plz confirm!
1For the first question, i think it should be
(b-c)2 , (c-a)2 , (a-b)2 are in AP...
Soln :
(b-c)2 , (c-a)2 , (a-b)2 are in AP...
so,
(c-a)2 - (b-c)2 = (a-b)2 - (c-a)2
or, (b-a)(2c-a-b) = (c-b)(2a-b-c) ...................................(i)
Now,
1b-c , 1c-a , 1a-b are in AP,
iff, 1c-a - 1b-c = 1a-b - 1c-a
iff, a+b-2c(c-a)(c-b) = b+c-2a(a-b)(c-a)
iff, (a-b)(a+b-2c) = (c-b)(b+c-2a)
This is true by (i).
Therefore, 1b-c , 1c-a , 1a-b are in AP
1
kunl
·2011-03-27 04:05:50
hmm...i guess thts y i wasn't getting that one~
ur solution is perfect i guess n question is wrong!
341Another way of doing it is:
(b-c)^2, (c-a)^2, (a-b)^2 are in AP \Rightarrow \frac{(b-c)^2}{(a-b)(b-c)(c-a)}, \frac{(c-a)^2}{(a-b)(b-c)(c-a)}, \frac{(a-b)^2}{(a-b)(b-c)(c-a)} are in AP
\Rightarrow \frac{b-c}{(a-b)(c-a)}, \frac{c-a}{(a-b)(b-c)}, \frac{a-b}{(b-c)(c-a)} are in AP
\Rightarrow -\left[\frac{1}{a-b} + \frac{1}{c-a} \right], etc. are in AP
and for the final step, adding \left[\frac{1}{a-b} + \frac{1}{b-c} + \frac{1}{c-a} \right] to the three terms, we have
\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b} are in AP
1
kunl
·2011-03-27 04:38:21
great solution thank you sir!
1
Abhisek
·2011-03-27 05:00:39
@ Prophet sir , need help in the 2nd qusetion!
Can we do as follows ?
log(x+52) ((x-52x-3)2) < 0
or, 2log(x+52) (x-52x-3) < 0
or, x-52x-3 < 1
or, x-5 < 2x-3
or, x > (-2)
Is it correct ? [7]
1
kunl
·2011-03-27 05:13:15
arey abhishek thats an easy problem....infact i just wanted to know how the "author" knew from beginning itself that among FOUR possible cases only these two will yield "all" the roots and roots of remaining cases will be covered up with these two cases only....or he just did not write it after solving...in short i wanted to know if there was any trick to know which cases are COVERING up all the roots?
11
Joydoot ghatak
·2011-04-02 23:55:29
i think this should be the solution but...
please confirm..
log (x+52) [(x-52x-3) 2] < 0
2 log(x+52) [(x-52x-3)] < 0
log(x+52) [(x-52x-3)] < 0
if the base is greater than equal to 1, then the equality sign holds in the next step,
i.e.,
if (x+52) ≥ 1
thus, x ≥ -32
then,
x-52x-3 < 1
or, x-5 < 2x-3
or, x> -2
thus x can lie in the interval [-3/2 , ∞)
but if the base lies between 0 and 1 , the equality sign gets reversed..
if 0< x + 52 <1
or, -52 < x < -32
then, x-52x-3 > 1
or, x< -2
thus the common region for x is (-5/2 , -2)
is the answer (-52 , -2) U [-32 , ∞ )... ??
