Question

Let $p(x)=a_{0}+a_{1}x+\dots+a_{n}x^{n}$ If $p(-2)=-15$, $p(-1)=1, p(0)=7, p(1)=9, p(2)=13$ and $p(3)=25$, then the smallest possible value of $n$ is

• A. 5
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (C)

We have,
$p(x)=a_{0}+a_{1}x+a_{2}x^{2}+\dots+a_{n}x^{n}$
$p(2)=-15, p(-1)=1, p(0) =7\,p(1)=9$,
$p(2)=13, p(3)=25$
Clearly, $p(x)$ is an increasing function
$\therefore $ n is an odd number
Here, $n=3$ or $5$
put $n = 3$,
$p(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}$
$p(0)=a_{0}=7$
$p(-1)=1=a_{0}-a_{1}+a_{2}-a_{3}\, \dots(i)$
$p(1)=9=a_{0}+a_{1}+a_{2}+a_{3}\, \dots(ii)$
On adding Eqs. (i) and (ii), we get
$a_{2}=-2$
$\therefore 1=7-a_{1}-2-a_{3}$
$\Rightarrow a_{1}+a_{3}=7-2-1=4\, \dots(iii)$
$\Rightarrow p(2)=13=a_{0}+2a_{1}+4a_{2}+8a_{3}$
$\Rightarrow 13=7+2a_{1}-8+8a_{3}$
$\Rightarrow a_{1}+4a_{3}=7\, \dots(iv)$
On solving Eqs. (iii) and (iv), we get
$a_{1}=3, a_{3}=1$
$\therefore p(x)=7+3x-2x^{2}+x^{3}$
which is satisfies $p(-2)$ and $p(3)$
$\therefore n=3$