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

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

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).