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Functions question

2009 · Shift 0 · Q28
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Functions question

2009 · Shift 0 · Q28

JEE MainMathematicsFunctionsMCQ+4 / −1
For real x, let f(x) = x3 + 5x + 1, then
  1. A
    f is one-one but not onto R
  2. B
    f is onto R but not one-one
  3. C
    f is one-one and onto R
  4. D
    f is neither one-one nor onto R
View written solutionFree

Correct answer: C

  1. Given function

    f(x)=x3+5x+1,x∈Rf(x)=x^3+5x+1, \qquad x\in \mathbb{R}f(x)=x3+5x+1,x∈R

    We must check whether fff is:

    • one-one (injective), and
    • onto R\mathbb{R}R (surjective).
  2. Check if fff is one-one

    A standard way is to study monotonicity using the derivative.

    f′(x)=3x2+5f'(x)=3x^2+5f′(x)=3x2+5

    Since

    3x2+5>0for every x∈R,3x^2+5>0 \quad \text{for every } x\in \mathbb{R},3x2+5>0for every x∈R,

    the function f(x)f(x)f(x) is strictly increasing on all of R\mathbb{R}R.

    A strictly increasing function cannot take the same value at two different points. Hence fff is one-one.

  3. Check if fff is onto R\mathbb{R}R

    Since fff is a cubic polynomial with positive leading coefficient, let us examine its end behavior:

    • As x→+∞x\to +\inftyx→+∞, f(x)=x3+5x+1→+∞f(x)=x^3+5x+1\to +\inftyf(x)=x3+5x+1→+∞
    • As x→−∞x\to -\inftyx→−∞, f(x)=x3+5x+1→−∞f(x)=x^3+5x+1\to -\inftyf(x)=x3+5x+1→−∞

    Also, polynomial functions are continuous on R\mathbb{R}R.

    Therefore, by continuity, f(x)f(x)f(x) takes every real value between −∞-\infty−∞ and +∞+\infty+∞. Hence fff is onto R\mathbb{R}R.

  4. Conclusion

    The function is both:

    • one-one, and
    • onto R\mathbb{R}R.

    So the correct option is:

    C: f is one-one and onto R\boxed{\text{C: } f \text{ is one-one and onto } \mathbb{R}}C: f is one-one and onto R​

  5. Comparison with stored correct answer

    Stored correct answer: C

    Our derived answer: C

    So they agree.

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