341LHS is ≥2, whereas RHS≤√2
Number of solutions of the equation \mid x-1\mid + \mid x\mid + \mid x+1\mid = \mid sinx\mid + \mid cosx\mid is,
A) 1 B) 0 C) 2 D) 4
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UP 0 DOWN 0 1 6
6 Answers
62This one is not very difficult..
Just plot the graph on both sides..
THe lSH will have a graph with variable slopes.. but the interesting thing to not e is that it will have crossed 3 at x=1 and x=-1
The RHS will never cross 2.. (infact root 2)
Also, LHS will be a straight line from 2 (at x=0, it is 2)
And that also happens to be its minima..
so there will be no solution...
(I hope i have not made anyh clculation mistake..)
341For LHS, Algebraically, you can just observe that |x-1| + |x+1| = |1-x| + |1+x| \ge |1-x+1+x| = 2
So |x-1| + |x+1|+|x| \ge 2
For RHS Cauchy Schwarz Inequality gives (1+1)(\sin^2 x + \cos^2x) \ge (|\sin x| + |\cos x|)^2 which yields that RHS≤√2
1can u post the full soln. please i'm little weak in this subject so can't get hold of ur ans. nishant