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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).
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