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(x−α) Since, |cos(x−α)|≤1 ⇒5cos(x−α)≤5 ⇒4cosx+3sinx≤5 ⇒1992(4cosx+3sinx)≤9960 ⇒2k+60≤9960 ⇒k≤4950 Hence, maximum value of k is 4950.