20-03-09 2 Past IIT JEE Questions

Question 1)

As the functn is cont. p(a).p(b)<0 implies a root exists b/w (a,b)

as we can c p(46)>0 and and p(45^1/100)<0

Ther4 atleast 1 root exists b/w them

oho .. just when i posted the page refreshed and i saw tapan had already posted the soln.... ... [1]

USING the same techniQ as in Q1

LET p = sinθ to make life better.......

f(1) = 1-sinθ ............ f(1)>0

f(1/2) = -1 - sinθ............ f(1/2) < 0

so der exists atleast 1 root in [.5,1]

NOW TO CHK for its Uniquejess n IDENTITY........

LET THE ROOT BE sinα as it lies b/w [0.5,1]

then we have

4sin³Î± - 3sinα = sinθ

sin(3α) = sinθ

1/3(sin^-1p) = α.............. as p = sinθ

sin(1/3(sin^-1p) ) = ROOT

I m yet to prove that this root is unique in [.5,1] I know, dat I'll do soon

YEAH!!!!!

its uniQue also coz :

[COntradiction methd]

let ther b 2 roots in da range [.5,1]

let them be sinα and sinβ

THEN BOTH SATISFY :

4sin³Î± - 3sinα = sinθ
4sin³Î² - 3sinβ = sinθ

and hence both also satisfy

sin3α = sinθ = sin3β

FOR THIS to happen either α == β
OR α ± 2Π= β

BUT 2Î >6 and hence these values differ by atleast 6units which is >> (1-0.5)

HENCE HO GAYA NAAA......

[1]

ok for question (2)

f'(x) = 12x²-3 > 0 for (1/2 , 12321564564) ;) i meant ∞ [but =0 at 1/2 :)]

f(1) = 1 - p so f(1) >= 0
f(1/2) = -1 - p f(1/2) <=0

we note both can't be =0 at the same time........

hence we have one unique root in [1/2, 1]
( i hope its correct) [23]

as for identifying it [12]
now i noticed that tapan has already done that :)

4 th
g(x) is an increasing function .. given

now i am talking in terms of f(x) .. which we get after integration of g(x) : assumed

f(a) - f(0) + f(b) - f(0)

= f(a) - f(0) + f(4 - a) - f(0)
= f(a) + f(4 - a) + c .. 1

b - a = 4 - 2a .. incresing this means decreasing a

now let a be decreased by h (+ ve)

expr . = f(a - h) + f(4 - a + h) + c ... 2

for this expression to be more than the previous expression
eq. 2 - eq . 1 > 0

f(a - h) + f(4 - a + h) + c - (f(a) + f(4 - a) + c) > 0
=( f(4 - a + h) - f(4 - a) ) - ( f(a) - f(a - h) )> 0

now refer the graph of an increasing function :

on decreasing a .. by increasing the value oh h the difference between area 1 ( f(a) - f(a - h)) and area 2 (f(4 - a + h) - f(4 - a))
increases ... because the area 2 increases at a rate more than area 1 ... reason u can make it out .. integration is dealt in terms of area ..
integration of g from s to d = f(d) - f(s) = area between d and s ...

Ankit : ITs not this lengthyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy sum bro,

HINT :

Remember "Newton dada woh Leibnitz wale"

uske liye condition kya hai : limits of integration must be VARIABLES not constants, try to introduce VARIABLES INSTEAD OF A AND B, UR SUM IS DUN

SOLUTION : (warniong : dont readahead unless u give up )

let A = 2-t; B = 2 + t

ab easily apply leibnitz

SUM HINT PL. for Q5

bhaiya help [94]

that question is coming to kill me!

help!

[3](old smiley :P)

Q5 : HINT PL.

you have written x>=0 twice
i think second time it should be f(x)>=0

then
we get F^'(x)=f(x) and hence from the inequality

F(x)<=F(0)*e^cx

further as F(0)=₀∫⁰f(t)dt=0

hence F(x)<=0
and from the equation f(x)<=cF(x)<=0

hence f(x)<=0 but mentioned in problem f(x)>=0

hence f(x)=0