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Question

Let a function f be defined from RR by f(x)={x+m, x12mx1, x>1
If the function f is surjective, then m

A
[2,2]
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B
R{2,1,2}
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C
R{1}
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D
(0,2]
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Solution

The correct option is D (0,2]
Given : f:RR is surjective(onto), Range =R
Now,
When x1:
x+mm+1
Range (,m+1](i)

When x>1,m>0:
mx>m2mx1>2m1
Range (2m1,)(ii)

When x>1,m<0:
mx<m2mx1<2m1
Range (,2m1)(iii)

From (i) and (ii):m>0

For Range to be R,2m1m+1
0<m2(A)

and from (i) and (iii):


Range is either (,m+1] or (,2m1)
Range R, for any value of m.
only possible set of m for which f to be surjective is (A):(0,2]

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