For all real values of x, the minimum value of (1 - x + x²)/(1 + x + x²) is
(A) 0   (B) 1   (C) 3   (D) 1/3

Solution:

Maxima and minima are known as the extrema of a function.

Maxima and minima are the maximum or the minimum value of a function within the given set of ranges.

Let f(x) = (1 - x + x²) / (1 + x + x²)

Therefore,

f' (x) = [(1 + x + x²) (- 1 + 2x) - (1 - x + x²)(1 + 2x)] / [(1 + x + x²)²]

= [- 1 + 2x - x + 2x² - x² + 2x³ - 1 - 2x + x + 2x² - x² - 2x³] / [(1 + x + x²)²]

= (2x² - 2)/(1 + x + x²)²

= 2(x² - 1)/(1 + x + x²)²

Now,

f' (x) = 0

⇒ x² = 1

⇒ x = ± 1

f" (x) = [1 + x + x²)2 (2x) - (x² - 1)(2)(1 + x + x²)(1 + 2x)]/[(1 + x + x²)⁴]

= 4 (1 + x + x²) [(1 + x + x²)x - (x² - 1)(1 + 2x)]/[(1 + x + x²)⁴]

= [x + x² + x³ - x² - 2x³ + 1 + 2x]/[(1 + x + x²)³]

= 4 (1 + 3x - x³)/(1 + x + x²)³

Hence,

f" (1) = 4(1 + 3 - 1)/(1 + 1 + 1)³

= 4(3)/(3)³

= 12/27

= 4/9 > 0

Also,

f" (- 1) = 4(1 - 3 - 1)/(1 - 1 + 1)³

= 4 (- 1)

= - 4 < 0

By second derivative test, f is the minimum at x = 1 and the minimum value is given by,

f (1) = (1 - 1 + 1)/(1 + 1 + 1)

= 1/3

Thus, the correct option is D

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 28

For all real values of x, the minimum value of (1 - x + x²)/(1 + x + x²) is (A) 0 (B) 1 (C) 3 (D) 1/3

Summary:

For all the given real values of x, the minimum value of (1 - x + x²)/(1 + x + x²) is 1/3. Maxima and minima are the maximum or the minimum value of a function within the given set of ranges