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
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
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