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Question

If 498[16cosx+12sinx]=2k+60, then the maximum value of k is

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Solution

498[16cosx+12sinx]=2k+60
1992(4cosx+3sinx)=2k+60
Consider, 4cosx+3sinx
Put 4=rcosα,3=rsinα
r=5,tanα=34
So, 4cosx+3sinx=5cos(xα)
Since, |cos(xα)|1
5cos(xα)5
4cosx+3sinx5
1992(4cosx+3sinx)9960
2k+609960
k4950
Hence, maximum value of k is 4950.

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