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
Grade: 12
        

 Let f(x) = x3+x2+100x+7sinx  then show that the eqn  1 /  [y-f(!)]   +  2 / [y-f(2)]   + 3 / [y-f(3)] = 0 has exactly one root lying in (f((1) , f(2)).

10 years ago

Answers : (1)

Ramesh V
70 Points
							

let g(y) = 1 / [y-f(1)] + 2/[y-f(2)] +3/[y-f(3)]

g' (y) = -(1/[y-f(1)]2 + 2/[ y-f(2)]2 + 3/ [y-f(3)]2)

 since g '(y) is negative summ of squares , its always negative

so, g '(y) < 0

g(y) is strictly decreasing function

here f(1) = 102 + 7 sin1

and f(2) = 212 + 7 sin2

clearly f(2) > f(1)

g( f(1) + 0 ) ---> infinity

 g( f(2) - 0 ) ---> - infinity

So, the mean value theorem

between ( -infinity,infinity) as 0 will lie btn y=f(1) & y=f(2)

hence proved

---

Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch

10 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

  • 51 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

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