1
chinmaya
·2011-06-01 21:25:02
the RHS can take a minimum value of 3/4
62
Lokesh Verma
·2011-06-01 22:01:13
that much is correct.. but how does it lead to a conclusion!
1
Ricky
·2011-06-01 22:07:17
Biggie hints -
1 . " x " cannot be positive .
2 . " Let graphs tell the story " .
21
Arnab Kundu
·2011-06-01 23:39:33
i think it has no sol.
because sinx=x(x+1/x+1)
for sinx to be zero either x is zero or (x+1/x+1) is zero...........
if x is zero sinx=x2+x+1
= 1
so x is zero is not the case..............
and for (x+1/x +1) to be zero x+1/x has to be =-1
but x+1/x never =-1
so it has no sol.
62
Lokesh Verma
·2011-06-01 23:41:40
arnab.. ur method is wrong because
u are trying to solve for sin x = x^2+x+1 =0
but that is not the question!
106
Asish Mahapatra
·2011-06-01 23:59:39
Draw graphs in the region [-1,1]... Think about why u dont need to draw beyond this region
1
johncenaiit
·2011-06-02 01:11:47
Since sin x < 1,
1 + x + x2 < 1
=> x belongs to (-1,0)
But for -1<x<0 , sin x is negative ..
But minimum value of 1 + x + x2 is 3/4
Hence no solution???
30
Ashish Kothari
·2011-06-02 03:54:37
I think graph tells the whole story. [3] Should be no solutions. [1]
Don't mind the rather hideous looking parabola!
49
Subhomoy Bakshi
·2011-06-02 04:16:43
Welcome back..! [4]
Not u Nishant Bhiaya..u wer always here.. but the QODs! :D