263Another solution can be:
Let a=[x^2-x];b=[x];c=[sinx].And all of these are integers.so
8a+4b-12c=13
or,2a+b-3c=13/4. LHS is an integer while RHS is not.Hence no solution.
Manish Shankar good one :)
Find the numebr of solutions of the equation-
8[x2 - x] + 4[x]=13+12[sinx],where [.]denotes the G.I.F.
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2 Answers
466No solution at all...
- Himanshu Giria Because frankly speakingRHS can have values 1'13 &25 only . LHS On the other hand is always a multiple of 4 or 8 or at least even whereas RHS is always oddUpvote·0· Reply ·2014-06-12 06:11:14
- Himanshu Giria As [] always returns an integer...
- Himanshu Giria […]
- Sushovan Halder yes,your thinking is right.
- Manish Shankar good work Himanshu :)
- Himanshu Giria Thanks Sir...
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