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
Menu
Chandan Das Grade: Upto college level
        

Let f(x)=(1+b2)x2+2bx+1 n let m(b) be the minimum value of f(x).As b varies,the range of m(b) is 


a.[0,1]


b.[0,1/2]


c.[1/2,1]


d.(0,1].

7 years ago

Answers : (1)

Ramesh V
70 Points
										

f'(x) = (1+b2).2.x + 2b = 0


f'(x) = 0 implies x = -b/(1+b2)


f''(x) = (1+b2).2 > 0 means f(x) has min. at above given x


on solving for min value for f(x) at x we have


m(b) = 1/(1+b2)


its range is [0,1]


--


regards


Ramesh

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details
Get extra Rs. 3,180 off
USE CODE: CART20
Get extra Rs. 339 off
USE CODE: CART20

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details