MCQ
If $\text{p}(\text{x})=\text{x}^2-2\sqrt2\text{x}+1$ then $\text{p}(2\sqrt2)=?$
- A0
- B1
- C$4\sqrt2$
- D-1
✓
Answer
- 1
Solution:
$\text{p}(\text{x})=\text{x}^2-2\sqrt2\text{x}+1$
$\text{p}(2\sqrt2)=(2\sqrt2)^2-2\sqrt2(2\sqrt2)+1$
$=8-8+1$
$=1$
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