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Functions question

2019 · 11 Jan · Shift 2 · Q27
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Functions question

2019 · 11 Jan · Shift 2 · Q27

JEE MainMathematicsFunctionsMCQ+4 / −1
Let a function f : (0, ∞\infty∞) →\to→ (0, ∞\infty∞) be defined by f(x) = ∣1−1x∣\left| {1 - {1 \over x}} \right|​1−x1​​. Then f is :
  1. A
    not injective but it is surjective
  2. B
    neiter injective nor surjective
  3. C
    injective only
  4. D
    both injective as well as surjective
View written solutionFree

Correct answer: A

  1. Given function

    f:(0,infty) o(0,infty),qquad f(x)=\left|1-\frac1x\right|

    We must check whether fff is injective and/or surjective.

  1. Rewrite the function piecewise

    The expression inside modulus changes sign at 1−1x=0  ⟹  x=1.1-\frac1x=0 \implies x=1.1−x1​=0⟹x=1.

    So:

    • For 0<x<10<x<10<x<1, we have 1−1x<01-\frac1x<01−x1​<0, hence f(x)=1x−1.f(x)=\frac1x-1.f(x)=x1​−1.
    • For x≥1x\ge 1x≥1, we have 1−1x≥01-\frac1x\ge 01−x1​≥0, hence f(x)=1−1x.f(x)=1-\frac1x.f(x)=1−x1​.

    Therefore,

    \begin{cases} \frac1x-1, & 0<x<1,\\[4pt] 1-\frac1x, & x\ge 1. \end{cases}$$

  1. Check injectivity

    To test injectivity, look for two different inputs giving the same output.

    Take x=12andx=2.x=\frac12 \quad \text{and} \quad x=2.x=21​andx=2.

    Then f(12)=∣1−2∣=1,f\left(\frac12\right)=\left|1-2\right|=1,f(21​)=∣1−2∣=1, and f(2)=∣1−12∣=12.f(2)=\left|1-\frac12\right|=\frac12.f(2)=​1−21​​=21​.

    These are not equal, so try a general approach.

    Let f(x)=0.f(x)=0.f(x)=0. Then ∣1−1x∣=0  ⟹  1−1x=0  ⟹  x=1.\left|1-\frac1x\right|=0 \implies 1-\frac1x=0 \implies x=1.​1−x1​​=0⟹1−x1​=0⟹x=1.

    That does not show non-injectivity. So let us solve ∣1−1x∣=12.\left|1-\frac1x\right|=\frac12.​1−x1​​=21​.

    Then 1−1x=±12.1-\frac1x=\pm \frac12.1−x1​=±21​.

    • If 1−1x=121-\frac1x=\frac121−x1​=21​, then 1x=12  ⟹  x=2.\frac1x=\frac12 \implies x=2.x1​=21​⟹x=2.
    • If 1−1x=−121-\frac1x=-\frac121−x1​=−21​, then 1x=32  ⟹  x=23.\frac1x=\frac32 \implies x=\frac23.x1​=23​⟹x=32​.

    Since f(2)=f(23)=12,f(2)=f\left(\frac23\right)=\frac12,f(2)=f(32​)=21​, with 2≠232\ne \frac232=32​, the function is not injective.

  1. Check surjectivity

    Codomain is (0,∞)(0,\infty)(0,∞), so every positive real number must be attained by fff.

    Let us find the range of fff.

    • For 0<x<10<x<10<x<1, f(x)=1x−1.f(x)=\frac1x-1.f(x)=x1​−1. As x→0+x\to 0^+x→0+, 1x→∞\frac1x\to \inftyx1​→∞, so f(x)→∞f(x)\to \inftyf(x)→∞. As x→1−x\to 1^-x→1−, 1x−1→0+\frac1x-1\to 0^+x1​−1→0+. Hence on (0,1)(0,1)(0,1), range is (0,∞).(0,\infty).(0,∞).

    • For x≥1x\ge 1x≥1, f(x)=1−1x.f(x)=1-\frac1x.f(x)=1−x1​. Here f(1)=0f(1)=0f(1)=0, and as x→∞x\to\inftyx→∞, f(x)→1−f(x)\to 1^-f(x)→1−. So this branch gives [0,1).[0,1).[0,1).

    Combining both branches, overall range is [0,∞).[0,\infty).[0,∞).

    But the codomain is (0,∞)(0,\infty)(0,∞).

    For surjectivity onto (0,∞)(0,\infty)(0,∞), every y>0y>0y>0 should have some x>0x>0x>0 such that f(x)=yf(x)=yf(x)=y.

    Since the branch 0<x<10<x<10<x<1 already gives all values in (0,∞)(0,\infty)(0,∞), every positive real number is attained.

    Therefore fff is surjective onto (0,∞)(0,\infty)(0,∞).

  1. Conclusion

    • fff is not injective
    • fff is surjective

    Hence the correct option is: A: not injective but surjective\boxed{\text{A: not injective but surjective}}A: not injective but surjective​

  1. Comparison with stored answer

    Stored correct answer is B: neither injective nor surjective.

    But from the range calculation, fff attains every value in (0,∞)(0,\infty)(0,∞), so it is surjective. Hence the stored answer is incorrect.

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