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
Tushar Watts Grade: 12
        

Let f(x) = ax2+bx+c, where a,b,c belongs to R and a is not equals to zero. If f(x) = x has non real roots, then show that the eqn f(f(x)) = x has all non real roots.

8 years ago

Answers : (3)

Jitender Singh
IIT Delhi
askIITians Faculty
158 Points
										Ans:
Hello Student,
Please find answer to your question below

f(x) = ax^2 + bx + c
f(x) =x
ax^2 + bx + c = x
ax^2 + (b-1)x + c = 0
This equation have non real roots.
(b – 1)2< 4ac
f(f(x)) = x
a(f(x))^2 + 2bf(x) + c = x
a(ax^2 + bx + c)^2 + 2b(ax^2 + bx + c) + c = x
a(ax^4 + b^2x^2 + c^2 + 2abx^3 + 2bcx + 2acx^2) + 2b(ax^2 + bx + c) + c = xa^2x^4 + (b^2 + 2ac + 2ab)x^2 + (c^2 + 2bc + c) + 2abx^3 + (2bc-2b ^2-1)x = 0

a^2x^4 + 2abx^3 + (b^2 + 2ac + 2ab)x^2 + (2bc-2b^2-1)x + (c^2 + 2bc + c) = 0
So this equation also have non real roots.
3 years ago
mycroft holmes
271 Points
										
Without Loss of Generality we can assume that a>0.
 
Then f(x) = x has no real roots implies that f(x)> x for all real x.
 
This also implies that f(f(x))>x for all real x.
 
This means f(f(x)) – x = 0 has non-real roots.
3 years ago
mycroft holmes
271 Points
										
I need to add an explanation:
 
f(x)> x for all x implies f(f(x))>f(x)>x. That’s how we get the inequality f(f(x))>x for all x
3 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

Ask Experts

Have any Question? Ask Experts

Post Question

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