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anurag singhal Grade:
        

the sum of max and min values of function f(x)=sin-12x+cos-12x+sec-12x is


a) π


b) π/2


c) 2π


d) 3π/2

 


8 years ago

Answers : (3)

AskIITian Expert Priyasheel - IITD
8 Points
										

4243-1029_6754_sss.JPG

8 years ago
askiitian.expert- chandra sekhar
10 Points
										

Hi Anurag,


f(x)=sin-12x+cos-12x+sec-12x


sin-12x+cos-12x = π/2 for all x


sec-12x belongs to [0,π ]-{π/2}



therefore min of
sec-12x is 0


                   max of sec-12x is π



therefore min of f(x) is
π/2


                   max of f(x) is 3π/2



sum of min and max  values of f(x) is (
π/2 + 3π/2) = 2π



Ans: (c)


All the best


askiitian.expert- chandra sekhar

8 years ago
askIITianexpert IITDelhi
8 Points
										

As Sin-1x  & Cos-1x is defined for x belongs to[-1,1] . Sec-1x is defined for x≤-1 & x≥1.Also Sin-1x+Cos-1x=¶/2 in the domain specified.


Domain of above function f(x) is x={-1/2 , 1/2}   (By solving inequalities -1≤x≤1,x≤-1 & x≥1)


Hence max{f(x)}=¶/2+Sec-1 (2*1/2)=¶/2+¶/2=¶


while min{f(x)}=0.So their sum=¶       ;   Answer is (a).


 

8 years ago
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