Question

Mark (✓) against the correct answer in each of the following:
If $p\left(x\right)=(x^4-x^2+x)$, then $p\left(\frac{1}{2}\right)=?$
(a) $\frac{1}{16}$
(b) $\frac{3}{16}$
(c) $\frac{5}{16}$
(d) $\frac{7}{16}$

Answer 1

(c) $\frac{5}{16}$

$$\begin{gathered} p\left(x\right)=x^4-x^2+x \\ \mathrm{Therefore}, \\ p\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^4-{\left(\frac{1}{2}\right)}^2+\frac{1}{2} \\ \Rightarrow p\left(\frac{1}{2}\right)=\frac{1}{16}-\frac{1}{4}+\frac{1}{2} \\ \Rightarrow p\left(\frac{1}{2}\right)=\frac{1-4+8}{16} \\ \Rightarrow p\left(\frac{1}{2}\right)=\frac{5}{16} \end{gathered}$$