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Question

If f(x)=12tan(πx2)1<x<1 and g(x)=(3+4x4x2). Find domain of (f+g)(x)

A
[12,1)
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B
(12,1]
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C
[12,32]
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D
(1,1)
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Solution

The correct option is A [12,1)
f(x)+g(x)=3+4x4x2+12tan(πx2)
Therefore, for real values of (f+g),
3+4x4x20
4x24x30
4x26x+2x30
2x(2x3)+(2x3)0
(2x+1)(2x3)0
x12 and x3
Therefore
xϵ[12,3] ...(i)
Also consider tan(π.x2). Since tanx is discontinuous for all xϵ(2n+1)π2) hence for tan(π.x2) the domain is R1,3,5...(2n+1) ...(ii)
Hence from i and ii
xϵ[12,1)(1,3)

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