Question

Let $p(x)=a_0+a_{1}x+\cdots +a_n{x}^n$ 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,
$p(x)=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n$
$p(2)=-15,p(-1)=1,p(0)=7p(1)=9$,
$p(2)=13,p(3)=25$
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 $p(3)$
$\therefore n=3$