IIT JEE , AIEEE Forum - Mathematics , Algebra

Sushmi

Q: f(x) = x sin x (x

Z) , then _{[r=0 ]}

^[infinity^] 2/(4r + 1)! is

a.) f ( i ²) + f ( i ) b.) f ( 1 ) + f ( i )

c.) f ( 1 ) - f( i ) d.)none of these

Q: _{[r = -1]}

^[n] (r²+ 3r + 3) . (r+1)! = 998 . 998!

then n is equal to:

a.)1000 b.) 998

c.)996 d.) 994

I hope the problem is properly visible.

computer001

2nd sum
write the 1st thing as (r+2)^2-(r+1)
so v have the term as (r+2)(r+2)!-(r+1)(r+1)!
now it follows telescopic cancellation

Sushmi

Oh thanks..

somebody 1st one.

computer001

so ans wud be 996 btw..

Sushmi

That's obvious but excellent thinking....

I think you made a little mistake in d limit problem

computer001

duh..well if it was SO obvious i dun think ne1 will be posting the ques.. btw 1st q i is iota i suppose??
ok it has 2 b i suppose

computer001

@ sushmi:

yeah i noticed... ive edited it.... i made mistake wid e..e^x=1+x/1! +x^2/2!...
i took it as e^x=x/1! +x^2/2!...

Sushmi

I think you got me the wrong way i praised your thinking abilities, sorry if you got me the wrong way.

and yeah i saw it just now.

computer001

nah...cummon i jus said it in a lighter vein...
but if u wanna say gnly u shud nudge u no...tht wud be better instead of havin conv in a academic thread

EDIT: tht shud be 'an academic thread'
1st q i is iota i suppose??

Sushmi

1st answer is C

Yes you are correct.Its alrite.

elastiboysai

f(1)-f(i) is the answer
just use series expansion

elastiboysai

xsinx=x^2-x^4/4!+x^6/5!-x^8/7!+x^10/9!+...

plug in a,b,c and simplify,
only c gives you
_{[r=0 ]}

^[infinity^] 2/(4r + 1)!

Sushmi

Thank u ........u ppl r really very helpful

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