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Let f:Rf: \mathbb{R} -> R\mathbb{R} be a function defined as f(x)=x4f(x) = x^4. Which one of the following is true?

Solution

✅ Correct Option: 4

The function f(x)=x4f(x) = x^4 maps real numbers to real numbers. Two properties need to be checked:

  • One-one (Injective): Each output comes from only one input
  • Onto (Surjective): Every real number can be obtained as an output

A function is one-one if f(x1)=f(x2)f(x_1) = f(x_2) implies x1=x2x_1 = x_2.

Consider f(2)f(2) and f(−2)f(-2):

f(2)=24=16f(2) = 2^4 = 16

f(−2)=(−2)4=16f(-2) = (-2)^4 = 16

Since f(2)=f(−2)f(2) = f(-2) but 2≠−22 \neq -2, the function is not one-one.


A function f:R→Rf: \mathbb{R} \to \mathbb{R} is onto if every real number can be obtained as an output.

For x4x^4 with any real number xx:

When x=0x = 0: f(0)=04=0f(0) = 0^4 = 0

When x=2x = 2: f(2)=24=16f(2) = 2^4 = 16

When x=−2x = -2: f(−2)=(−2)4=16f(-2) = (-2)^4 = 16

Since the power is even, x4≥0x^4 \geq 0 for all real xx.

To obtain y=−1y = -1 as output, we would need x4=−1x^4 = -1, which has no real solution.

The range of ff is [0,∞)[0, \infty), not all of R\mathbb{R}.

Since negative numbers cannot be outputs, ff is not onto.


The function ff is neither one-one nor onto.

  • Not one-one because f(2)=f(−2)=16f(2) = f(-2) = 16
  • Not onto because negative numbers cannot be outputs

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