Question 65 of 82intermediate🔧 ApplyMCQ1 mark

For all real values of x, the minimum value of (1 - x + x²)/(1 + x + x²) is

Exercise: EXERCISE 6.3 | Q: 28 | (Page 31)
For More Understanding

Explanation

The correct option is (D). To find the minimum value of the function, we use the method of derivatives as illustrated in the provided context (e.g., Example 27 and the soldier/helicopter example). By differentiating the function and setting the derivative to zero, we find the critical points. Evaluating the function at these critical points allows us to determine the minimum value.

Solution Steps

  1. Step 1: Let y=f(x)=1−x+x21+x+x2y = f(x) = \frac{1 - x + x^2}{1 + x + x^2}. Differentiating with respect to xx, we get f′(x)=(1+x+x2)(2x−1)−(1−x+x2)(2x+1)(1+x+x2)2f'(x) = \frac{(1 + x + x^2)(2x - 1) - (1 - x + x^2)(2x + 1)}{(1 + x + x^2)^2}. Simplifying the numerator gives 2x2−22x^2 - 2, so f′(x)=2(x2−1)(1+x+x2)2f'(x) = \frac{2(x^2 - 1)}{(1 + x + x^2)^2}.

  2. Step 2: For critical points, set f′(x)=0f'(x) = 0. This implies 2(x2−1)=02(x^2 - 1) = 0, which gives x=1x = 1 and x=−1x = -1.

  3. Step 3: Evaluate the function f(x)f(x) at these critical points. At x=1x = 1, f(1)=1−1+11+1+1=13f(1) = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3}. At x=−1x = -1, f(−1)=1−(−1)+11+(−1)+1=31=3f(-1) = \frac{1 - (-1) + 1}{1 + (-1) + 1} = \frac{3}{1} = 3.

  4. Step 4: Comparing the values, f(1)=13f(1) = \frac{1}{3} and f(−1)=3f(-1) = 3. Since 13<3\frac{1}{3} < 3, the minimum value is 13\frac{1}{3}.

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