Question

If $P\left(x\right)$ is a polynomial of least degree that has a maximum equal to $6$ at $x=1,$ and a minimum equal to $2$ at $x=3,$ then ${\int }_0^{1}p\left(x\right)dx$ equals:

• A. $17/4$
• B. $13/4$
• C. $19/4$
• D. $5/4$

Answer 1

Answer: (C)

$19/4$

$p\left(x\right)$ has a maxima and a minima so polynomial with least degree would be 3

Suppose $p\left(x\right)=a{x}^3+b{x}^2+cx+d$

$p^{\prime }\left(1\right)=p^{\prime }\left(3\right)=0$

$3a+2b+c=0$.........(1)

$27a+6b+c=\mathrm{0........}\left(2\right)$

From (1) and (2)

$b=-6ac=9aandp\left(1\right)=6a+b+c+d=64a+d=\mathrm{6.......}\left(3\right)p\left(3\right)=227a+9b+3c+d=2d=\mathrm{2.........}\left(4\right)a=1b=-6c=9$

So $p\left(x\right)=x^3-6x^2+9x+2{\int }_0^{1}p\left(x\right)dx={[\frac{x^4}{4}-2x^3+\frac{9x^2}{2}+2x]}_0^1{\int }_0^{1}p\left(x\right)dx=\frac{19}{4}$