1thank you .. btw :) :) :)
let f(x) be a polynomial of degree 5 and with leading coefficient 2004 . suppose further , that f(1)= 1,f(2)=3,f(3)=5,f(4)=7 ,f(9)=9. then the value of f(6) is:
a.120351
b.240591
c.100881
d.240491
i know we have 5 unknowns and five equations and we can solve them up to get all the five unknowns and can find out the answer but i feel this is tedious and some kind of conclusions can be drawn out of these hints
.....please help i am not getting them !
i got the question from topic "functions"....may be this can help!!!
in the q it was" leading coefficients" ..i thought it doesn't make any sense and it must be a misprint.... so i wrote "leading coefficient" (i hope i am correct :) )
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6 Answers
62hmm.. yup this one seems to be a bit strange looking
it ideally means
2004x5+ax4+bx3+cx2+dx+e is the polynomial.
After that the question becomes really tough looking..
there can be one way to think in terms of remainders.. but am not able to figure out how this should be approached.
66The options given are not correct. The correct answer is f(6)=-\dfrac{5050007}{7}.
To see how its coming, consider the polynomial
Q(x)=f(x)-(2x-1)
Obviously, Q(x) is of degree 5.
Then, we have
Q(1)=f(1)-1=0
Q(2)=f(2)-3=0
Q(3)=f(3)-5=0
Q(4)=f(4)-7=0
That implies that 1, 2, 3, 4 are the roots of Q(x). Since its a polynomial of degree 5, the remaining root must be real as well. Let it be \alpha. Then, taking into account that Q(x) also has the leading coefficient 2004, we can write Q(x) as
Q(x)=2004(x-1)(x-2)(x-3)(x-4)(x-\alpha)
Hence, f(x)=Q(x)+2x-1
i.e. to say that
f(x)=2004(x-1)(x-2)(x-3)(x-4)(x-\alpha)+2x-1
Finally, since f(9)=9, we get
9=2004\cdot 8 \cdot 7\cdot 6 \cdot 5 \cdot (9-\alpha)+17
which gives us
\alpha = \dfrac{3787561}{420840}
Hence, we finally get
\boxed{f(x)=2004(x-1)(x-2)(x-3)(x-4)\left(x- \dfrac{3787561}{420840}\right)+2x-1}
Hence, we get
\boxed{\boxed{f(6)=-\dfrac{5050007}{7}}}
1yes i followed the same method and got that awkward answer ... was wondering why it isnt in the options ... :(
1i dont know why this one's not in the options but i feel the solution is good and correct .......
i didnt get that mathematica stuff