1THIS IS THE LINK 4 SOLN 2 QUE1..... A WONDERFUL ONE BY ANURAG.....
ANURAG U STILL GOTTA XPLAIN IT 2 ME TOTALLY ;)
http://targetiit.com/iit_jee_forum/posts/rolle_s_theorem_2287.html
THIS WAS A COMP TYPE QUESTN SO OBVIOUSLY HAD 3 QUESTNS.... I HAD POSTD JUS DA FIRST ONE THINKIN MAYB SUM1 WILL GET THE FUNCTN f(x) but Anurag has come up with a real novel solution so i am forced to post da 2nd and da 3rd que
QUE1 
QUE2
QUE3
the answers are 3 ad -0.5 respectively.......
-
UP 0 DOWN 0 0 12
12 Answers
11Sorry i wasn't clear...
The ans is 1/2
Imposing condition of rolle's theorem, we get φ(b)=0,
=>φ(a)=0; -(1)
Also, Solving φ'(c)=0, we get f''(c)=2A;
Finally, φ(a)=0 gives:
f(b)=f(a)+(b-a)f'(a)+h(b-a)2f''(c)
=f(a)+(b-a)f'(a)+h(b-a)2(2A) -(2) [coz f''(c)=2A]
But φ(a)=0 (From 1)
=>f(b) - f(a)-(b-a)f'(a)-(b-a)2A=0
=>f(b)= f(a) + (b-a)f'(a) + (b-a)2A (3)
Comparing (2) & (3),
h=1/2.
I'm solving the rest...
1Tellme sumthin... is the 'h' in Q2 same as d 1 solvd frm 1?
1Here's the soln 4 Q2:
Simplify f(1+h)-f(1)/h2=kc (sorry, can't type lambda)
We get:
h+3=kc -(1)
Also, applying condition of rolle theorem on φ(x),
φ(1)=0, so φ(1+h)=0;
Simplify, we get,
h+3=A -(2)
Also, φ'(c)=0;
=>f''(c)=2A=6c
=>A=3c -(3)
From (1),(2),(3):
k(or lambda)=3.
Solving Q3...
1Q3 Ans -1/2;
Proceed in the exact same way as q2.
I don't think i need 2 post the whole soln...;)
1There's an incredibly short method for Q3!!
Take h->0 and apply l hospital's rule twice...
U'll getans -1/2
;)
21NICE METHOD anurag!!!!
but hope u hav not taken f(1+h)-f(1)/h2= kc asi it reads in ur soln coz the questn says :
[f(1+h)-f(1)]/h2=kc
i wonder how u got 3 + h = kc
coz i'm gettin
1/h2 + 3 + h = kc
21YEAH!!!!!!!! oh noo.... how cud do this..........
nice one.... ;)
1THNX ANURAG......... THNX MAN!!!!
This is actually the soln. given can sum1 xplain me how do they get the first step itself????
