BITSAT-2015 Problem

Calculus Level 4

If $P(x)$ is a quadratic polynomial satisfying

$P(0) = 6$ , $P(1) = 2$ , $P(2) = 0$ .

And $\displaystyle f(x) = \sum_{ n = 0}^{\infty } \dfrac{P(n)}{n!.2^n} x^n$

Then $f(x)$ is

Note:- I have changed the options from the original ones.

(x^2 + 6x)e^{\frac{x}{2}}

(x^2 -2x + 6)e^{\frac{x}{2}}

(x^2 - 4x + 6)e^{\frac{x}{2}}

\dfrac{x^2 - 8x + 24}{4} e^{\frac{x}{2}}

1 solution

Nishu Sharma
May 20, 2015

I guess , you changed option , beacause , in exam option were so , that person can easily check by option's by putting x=0,1 etc , am i right ?

For solution :

$\displaystyle{f(2t)=\sum _{ n=0 }^{ \infty }{ \cfrac { { n }^{ 2 }-5n+6 }{ { n! } } } { \left( t \right) }^{ n }\equiv \sum { \cfrac { { \left( t \right) }^{ n } }{ (n-2)! } } -4\sum { \cfrac { { \left( t \right) }^{ n } }{ (n-1)! } } +6\sum { \cfrac { { \left( t \right) }^{ n } }{ (n)! } } \\ f(2t)={ t }^{ 2 }{ e }^{ t }-4t{ e }^{ t }+6{ e }^{ t }\\ f(x)=\cfrac { ({ x }^{ 2 }-8x+24){ e }^{ t } }{ 4 } }$

Nice solution! The $t$ of the last line should be $\frac{x}{2}$ .

Chew-Seong Cheong - 6 years ago

No, there were $e^x$ in all other options, lol!

Krishna Sharma - 6 years ago

How come u get the question from BITSAT 2015? IS there any website, if any pls mention .Very good one

Ashwin Gopal - 6 years ago

hahaha ! that's funny :D

Keshav Tiwari - 6 years ago

Can we convert the summation to integration pls help!!!!

Ashwin Gopal - 6 years ago

@Nishu sharma , I haven't studied all this stuff yet. If I am not wrong this comes in the chapter of definite integrals.

Aditya Kumar - 6 years ago

No, it actually doesn't. It's just an extension of sequence an series( Exponential ofcourse .)

Keshav Tiwari - 6 years ago

Damn it! How could I miss it!!

Aditya Kumar - 6 years ago

BITSAT-2015 Problem

Note:- I have changed the options from the original ones.

1 solution

0 pending reports