wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If f(x) is a cubic polynomial such that f(1) =1, f(2) =2, f(3) =3 and f(4) =16 then, find the value of f(5).

A

53
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

35
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

42
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

24
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A
53
When f(x) is divided by (xα)
then remainder =f(α)
Dividend= Divisor X Quotient + Remainder
f(x)=(xα)q+f(α)f(x)=(x1)A+1f(x)=(x2)B+2f(x)=(x3)C+3f(x)=k(x1)(x2)(x3)+xf(1)=0+1f(x)=k(x1)(x2)(x3)+xf(4)=16k(41)(42)(43)+4=16k.3.2.1=1646k=12k=126k=2f(x)=2(x1)(x2)(x3)+xf(x)=2(x1)(x2)(x3)+xf(5)=2(51)(52)(53)+5f(5)=2.4.3.2+5=48+5f(5)=53
So, option a is correct.

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Division Algorithm
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon