Question 22 of 33intermediate🔧 ApplyShort Answer3 marks

Let f={(x,x21+x2):x∈R}f = \left\{ \left( x, \frac{x^2}{1 + x^2} \right) : x \in R \right\} be a function from RR into RR. Determine the range of ff.

Correct Answer

Let y=f(x)=x21+x2y = f(x) = \frac{x^2}{1 + x^2}.

Since x2≥0x^2 \geq 0 for all x∈Rx \in R and 1+x2>01 + x^2 > 0 for all x∈Rx \in R, we have y≥0y \geq 0.

Also, y=x21+x2<x2+11+x2=1y = \frac{x^2}{1 + x^2} < \frac{x^2 + 1}{1 + x^2} = 1 for all x∈Rx \in R.

When x=0x = 0, y=01+0=0y = \frac{0}{1 + 0} = 0. So y=0y = 0 is attained.

As ∣x∣→∞|x| \to \infty, y=x21+x2=11x2+1→1y = \frac{x^2}{1 + x^2} = \frac{1}{\frac{1}{x^2} + 1} \to 1. So yy approaches 11 but never equals 11.

Therefore, the range of ff is [0,1)[0, 1).

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 6 | (Chapter: 17)
For More Understanding

Explanation

The question asks to find the range of the function f(x) = x²/(1 + x²). Following the approach shown in Example 13 of the context where domain and range are determined, we analyze the function algebraically. Since x² ≥\geq 0 and denominator is always positive, y ≥\geq 0. The function approaches 1 as x → ±∞ but never reaches it. The minimum value 0 is attained at x = 0. Hence range is [0, 1).

Solution Steps

  1. Step 1: Let y = f(x) = x²/(1 + x²)

  2. Step 2: Note that x² ≥\geq 0 and 1 + x² > 0 for all x ∈ R, so y ≥\geq 0

  3. Step 3: Show that y < 1 since x² < x² + 1

  4. Step 4: Verify y = 0 is attained at x = 0

  5. Step 5: Show y → 1 as |x| → ∞ but y ≠\neq 1

  6. Step 6: Conclude range = [0, 1)

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