IIT JEE , AIEEE Forum - Mathematics , Algebra

elastiboysai

Hi guyz,

here are 3 simple questions:

download img n magnify 4 better view

sure rates to correct answers.

raulrag009

is the summation from r=1 to r=infinity or n

raulrag009

hey man for Q2

when u substitute x=0

u get
lim -1+0/-1+0=1
x-0

elastiboysai

summ frm r= 1 to n

elastiboysai

hey raul rag,

i know it looks like dat exprn. is not indeterminant but if

u simplify the numerator u get it an indeterminant form.

n the ans is not 1

raulrag009

in first q is it 2r^4 or 2r^2

raulrag009

Let T=

tanx-sinx + T

T²-T+sinx-tanx=0

T=[1+-

1+4tanx-4sinx]/2

Let

V=

x³+V

V²-V-x³=0

V= 1+-

1+4x³/2

lim -1 + T/-1+V

x-0

lim [-1 +

1+4tanx-4sinx]/[-1+

1+4x³]

x-0

differentiate

lim [(sec²x-cosx)

1+4x³ ]/12x²root(1+4tanx-4sinx)

^x-0

again differentiate and solve

i think the answer is 1/8

hsbhatt

I couldn't get farther than this:

a_n = a_n-1+

(1+a_n-1²) >2 a_n-1 > 2²a_n-2>...>2^n-1a₁

Also, a_n <2a_n-1+1<2²a_n-2+2+1<...<2^n-1a₁+2^n-2+..+1

Hence a₁<a_n/2^n-1<a₁+1/2+1/2²+...+1/2^n-1.

as n

inf, by Sandwich Theorem, the required limit lies between a₁ and a₁+1 i.e. between 1 and 2.

elastiboysai

these took me some time, but are simple if u hav d idea.

i will post the soln in 2 days.

and raul rag watz the value of the limit, ur soln isnt complete!!

raulrag009

man it's takin a lot of time to solve by L-hospital rule.

my answer is 1/8

elastiboysai

wat abt the denom

raulrag009

man ii hav not solved it further here on the net.

that term again needs to be differentiated twice to get the answer

anywayz what's the answer

hsbhatt

This prob is a beaut! I really enjoyed solving it.

We have a_n = a_n-1+

(1+a_n-1²).

We can see that if a_n-1= cot

then a_n = cot

+cosec

= 1+cos

/sin

= cot

/2

So, let a₁ = cot

, then a₂ = cot

/2, a₃ = cot

/4 ...a_n = cot(

/2^n-1)

Hence a_n/2^n-1 = cot(

/2^n-1)/2^n-1 = 1/2^n-1 cos(

/2^n-1)/sin(

/2^n-1).

This limit is easily evaluated as it reduces to the standard limit sinx/x. The limit in this case = 1/

.

= cot^-1(a₁) = cot^-1(1) =

/4. Hence, required limit = 4/

.

(which as I pointed out in my previous post lies between 1 and 2

).

Thanx again for putting up this problem

hsbhatt

Qn 1: ₁

ⁿ tan^-1(2r/2r⁴+1)

The trick in such problems is to reduce it to a telescopic sum as

f(r+1) - f(r).

And in case of tan^-1 we look for tan^-1a - tan^-1b = tan^-1(a-b/1+ab)

Now (2r/2r⁴+1) = (4r/4r⁴+2)

Dr. = 4r⁴+2 = 4r⁴+1+1 = 4r⁴+4r²+1 - 4r² = (2r²+1)² - (2r)² = (2r²+2r+1)*(2r²-2r+1)

Now we can write Nr as (2r²+2r+1)-(2r²-2r+1).

So if we let f(r) = 2r²-2r+1, we can easily see that f(r+1) = 2r²+2r+1

Hence, the summation reduces to

tan^-1(f(r+1)) - tan^-1(f(r)) =

F(r+1) - F(r) = F(n+1)-F(1) = tan^-1(2n²-2n+1) - tan^-1(1) = tan^-1(2n²-2n+1) -

/4

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