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Questions  

 The maximum value of 3cosx+4sinx+5 is 

a
1
b
5
c
10
d
7

detailed solution

Correct option is C

We have c-a2+b2≤acosθ+bsinθ+c≤ c+a2+b2∴ Max. value =c+a2+b2                         =5+9+16                         =10

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