 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping • Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details

```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 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.
```  ### Course Features

• 728 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution 