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Question

# If f:R→R defined by f(x)=⎧⎨⎩x,x<1x2,1≤x≤48√x,x>4 , then f−1(x) is

A
f1(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪x,x<1x,1x16x264,x>16
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B
f1(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪x,x<1x,1x4x24,x>4
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C
f1(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪x,x<1x,1x8x28,x>8
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D
f1(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪x,x<1x,1x4x264,x>4
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Solution

## The correct option is A f−1(x)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩x,x<1√x,1≤x≤16x264,x>16It can be inferred from the options that f is a bijective function. Given f(x)=⎧⎨⎩x,x<1x2,1≤x≤48√x,x>4 Let f(x)=y⇒x=f−1(y) ⋯(1) ∴x=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩y,y<1√y,1≤√y≤4y264,y264>4⇒f−1(y)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩y,y<1√y,1≤y≤16y264,y>16∴f−1(x)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩x,x<1√x,1≤x≤16x264,x>16

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