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Question

# Let a function f defined from R→R as f(x)={2m−x,x≤14mx+1,x>1 If function is onto on R, then range of m is

A
[1,)
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B
[1,0)
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C
{-1}
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D
(0,)
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Solution

## The correct option is B [−1,0)Case I: m=−k,where k>0 ∴f(x)={−2k−x,x≤1−4kx+1,x>1 If−∞<x≤1⇒−1≤−x<∞⇒−2k−1≤−2k−x<∞⇒−2k−1≤f(x)<∞ ∀ x≤1 If 1<x<∞⇒−∞≤−x<−1⇒−∞<−4kx<−4k⇒−∞<−4kx+1<−4k+1⇒−∞<f(x)<−4k+1 ∀ x>1 Since, f(x) is an onto function. −4k+1≥−2k−1⇒4m+1≥2m−1⇒2m≥−2⇒m≥−1∴−1≤m<0 [As m is negative] Case II: m=k,where k>0 ∴f(x)={2k−x,x≤14kx+1,x>1 If −∞<x≤1⇒−1≤−x<∞⇒2k−1≤2k−x<∞⇒2k−1≤f(x)<∞ ∀ x≤1 ...[1] If 1<x<∞⇒4k<4kx<∞⇒4k+1<4kx+1<∞⇒4k+1<f(x)<∞ ∀ x>1 ...[2] Since, f(x) is an onto function, the union of [1] and [2] should be R. This is not possible as k is positive. Therefore, m cannot be positive. Case II is not possible. Range of f is [−1,0)

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