Question

If $p\left(x\right)=x^2-4x+3$, then the value of $p\left(2\right)-p(-1)+p\left(\frac{1}{2}\right)$ is.

• A. $\frac{31}{4}$
• B. $\frac{33}{4}$
• C. $-\frac{31}{4}$
• D. $-\frac{33}{4}$

Answer 1

The correct option is C

$-\frac{31}{4}$

Explanation of the correct answer:

Evaluate the value $p\left(2\right)-p(-1)+p\left(\frac{1}{2}\right)$

We have, $p\left(x\right)=x^2-4x+3$………………..$\left(1\right)$

Putting the value$x=2,x=-1$, $x=\frac{1}{2}$in equation $\left(1\right)$

$$\begin{gathered} p\left(2\right)-p(-1)+p\left(\frac{1}{2}\right) \\ =(2^2-4\times 2+3)-\{{(-1)}^2-4\times (-1)+3\}+\left\{{\left(\frac{1}{2}\right)}^2-4\times \frac{1}{2}+3\right\} \\ =(4-8+3)-(1+4+3)+\left(\frac{1}{4}-2+3\right) \\ =-1-8+\frac{1-8+12}{4} \\ =-9+\frac{5}{4} \\ =\frac{-36+5}{4} \\ =\frac{-31}{4} \end{gathered}$$

Hence, the correct option is ‘C’.