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
• B.
  35
• C.
  42
• D.
  24

Answer 1

The correct option is A

53
When f(x) is divided by $(x-\alpha )$
then remainder =$f\left(\alpha \right)$
Dividend= Divisor X Quotient + Remainder
$f\left(x\right)=(x-\alpha )q+f\left(\alpha \right)f\left(x\right)=(x-1)A+1f\left(x\right)=(x-2)B+2f\left(x\right)=(x-3)C+3\Rightarrow f\left(x\right)=k(x-1)(x-2)(x-3)+xf\left(1\right)=0+1f\left(x\right)=k(x-1)(x-2)(x-3)+x\because f\left(4\right)=16\Rightarrow k(4-1)(4-2)(4-3)+4=16\Rightarrow k.3.2.1=16-4\Rightarrow 6k=12\Rightarrow k=\frac{12}{6}\therefore k=2\therefore f\left(x\right)=2(x-1)(x-2)(x-3)+xf\left(x\right)=2(x-1)(x-2)(x-3)+xf\left(5\right)=2(5-1)(5-2)(5-3)+5\Rightarrow f\left(5\right)=\mathrm{2.4.3.2}+5=48+5\Rightarrow f\left(5\right)=53$
So, option a is correct.