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  • Complete JEE Main/Advanced Course and Test Series
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Solved Examples on Maxima & Minima

Illustration 1: Let f (x) = (4–x2)2/3, then f has a

 (A) a local maxima at x = 0                   (B) a local maxima at x = 2

 (C) a local maxima at x = –2                 (D) none of these

Solution: The given function is f(x) = (4 – x2)2/3

f'(x) = –4x / 3(4–x2)1/3 = 4x / 3(x2–4)1/3

The numerator gives the points of local maxima while the denominator gievs the points of local minima.

Hence, at x = –2    local minima

                x = 0    local maxima

                x = 2    local minima 

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Illustration 2:The absolute minimum value of x4 – x2 – 2x+ 5

(A) is equal to 5                                    (B) is equal to 3

(C) is equal to 7                                    (D) does not exist

Solution: The given function is x4 – x2 – 2x+ 5.

Hence, f'(x) = 4x3 – 2x – 2

                 = (x – 1) (4x2 + 4x + 2)

Clearly at x = 1, we will set the minimum value which is 3.

Hence, (B) is the correct answer.

Illustration 3:If f(x) = x3 +bx2 + cx+ d and 0 < b2< c, then in (-∞, ∞) (2004)

1. f(x) is strictly increasing function

2. f(x) has local maxima

3. f(x) is strictly decreasing function

4. f(x) is bounded

Solution: Givenf(x) = x3 +bx2 + cx+ d

f’(x) = 3x2 +2bx + c

Now, D = 4b2 -12c

Hence, D < 0

Therefore, f’(x) = 3x2 +2bx + c > 0 for all x in (-∞, ∞)

Hence, f(x) is a strictly increasing function.
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