sanju09 wrote:A cubic polynomial p (x) satisfies the criteria that p (1) = 1, p (2) = 2, p (3) = 3 and p (4) = 5. Then the value of p (6) is
(A) 6
(B) 10
(C) 15
(D) 16
(E) 60

Let the polynomial be
ax3+bx2+cx+d
--> a+b+c+d = 1 .................................i
--> 8a+ 4b+ 2c+ d =2 ........................ii
--> 27a+ 9b +3c + d =3.......................iii
--> 64a + 16b + 4c+ d = 5......................iv

ii - i gives
7a + 3b + c =1......................v
iv-iii gives
37a + 7b + c =2.....................vi

from iii - ii gives
19a + 5b +c= 1.................vii
vi - v gives
30a + 4b = 1..................viii


vi-vii gives
18a + 2b =1 --> 36a+4b =2.........ix
ix-viii 6a =1--> a= 1/6

b = (1-18a)/2 = (1-3)/2 =-1

c= 1-3b-7a = 1-3(-1)-7/6 = 4-7/6 = 17/6

d=1-a-b-c = 1-1/6+1-17/6 = 2-18/6 = -1

so P(6)= 216a+ 36b+6c+d=216x(1/6) +36 (-1)+ 6x(17/6) -1 =[spoiler] 36 -36+17-1=16 Ans D[/spoiler]

now there is an alternate and easy solution which was given by my friend mausam
since P(1)=1, P(2)=2, P(3) =3
so we can easily write the polynomial as A(x-1)(x-2)(X-3) +X
now P(4) =5
A(3)(2)(1) + 4 = 5
--> A=1/6
so P(6) = 1/6x 5x4x3+ 6= 10+6 = 16 Ans D

shashank.ism wrote:now there is an alternate and easy solution which was given by my friend mausam
since P(1)=1, P(2)=2, P(3) =3
so we can easily write the polynomial as A(x-1)(x-2)(X-3) +X
now P(4) =5
A(3)(2)(1) + 4 = 5
--> A=1/6
so P(6) = 1/6x 5x4x3+ 6= 10+6 = 16 Ans D

I would like to know the final polynomial eq. P(x) which satifies the needed critera i.e. p (1) = 1, p (2) = 2, p (3) = 3 and p (4) = 5.

brijesh

If we go by 2nd solution the final polynomial is
A(x-1)(x-2)(x-3) + x --> 1/6(x-1)(x-2)(x-3)+x

Also if we go by 1st soln. we get final polynomial as follows: Only difference is that it is in expanded form
1/6 X^3 - X^2 + 17/6 X -1

Hey Shashank that seems to be a cool trick

But I am stuck at one place (may be I am doing wrong maths)

In the expression A(x-1)(x-2)(x-3)+X.........how do we get the X

Shouldn't it be A(x-1)(x-2)(x-3) + constant(k)?

anybody please reply soon

shashank.ism wrote:now there is an alternate and easy solution which was given by my friend mausam
since P(1)=1, P(2)=2, P(3) =3
so we can easily write the polynomial as A(x-1)(x-2)(X-3) +X
now P(4) =5
A(3)(2)(1) + 4 = 5
--> A=1/6
so P(6) = 1/6x 5x4x3+ 6= 10+6 = 16 Ans D

This is a old post, but i had not got any replies.

Can anybody explain the method given above?

This is a non-GMAT question (the math is way beyond the scope of the test) - no need for GMAT studiers to worry about it.