Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.1k views
in Mathematics by (6.3k points)
Suppose f(x) is a polynomial of degree 5 and with leading co-efficient 2009. Suppose further that f(1)=1 , f(2)=3 , f(3)=5 , f(4)=7 and f(5)=9 then find the value of f(6), f(7) and f(8).

1 Answer

0 votes
by (13.2k points)
selected by
 
Best answer

Let f(x) = g(x) + 2x - 1. Since f(1)=1, f(2)=3, f(3)=5, f(4)=7, f(5)=9 and 1,2,3,4,5 are all the 5 roots of g(x), then follows that g(1) = g(2) = g(3) = g(4) = g(5) = 0.
Since f(x) is 5th degree with leading coefficient 2009, the same is true for g(x).
Therefore, g(x) = 2009(x - 1)(x - 2)(x - 3)(x - 4)(x - 5).

So now, we have
f(x) = 2009(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + 2x - 1
f(6) = 2009(5)(4)(3)(2)(1) + 12 - 1 = 241091

f(7) = 2009(6)(5)(4)(3)(2) + 14 - 1 = 1446493

f(8) = 2009(7)(6)(5)(4)(3) + 14 - 1 = 5062693

No related questions found

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...