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Minimum value of 3cosx +4sin + 8 is A.5B.9C.7D.3

Minimum value of 3cosx +4sin + 8 is A.5B.9C.7D.3

Grade:6

2 Answers

Abhiram
18 Points
7 years ago
Minimum value of an equation of the form a cosx+b sinx+c is 
c-\sqrt{a^2+b^2}
Here,c=8 a=3 b=4
=> minimum value= 8-\sqrt{3^2+4^2}
=> minimum value= 8 - 5= 3 
Therefore,the answer is (D)
 
 
Mosaraf Mondal
20 Points
7 years ago
Min value=8- (3^2 + 4^2)^0.5 =8- (9 + 16 )^0.5 =8- 5 =3[ since minumume value of ax+by+c= c- (a^2 + b^2)^0.5]So exact ans is D.3

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