Question

If $ p\left( x \right)={{x}^{2}}-2\sqrt{2}x+1 $ then $ p\left( 2\sqrt{2} \right) $ is equal to
\[\begin{align}
  & A.0 \\
 & B.1 \\
 & C.4\sqrt{2} \\
 & D.8\sqrt{2}+1 \\
\end{align}\]

Answer 1

Hint: In this question, we are given a function p(x) and we need to find the value of $ p\left( 2\sqrt{2} \right) $ . For this, we will put the values of x as $ 2\sqrt{2} $ in the function and evaluate the value. We will use the property that $ \sqrt{x}\sqrt{x}=x $ .

Complete step by step solution:
Here we are given the function p(x) as: $ p\left( x \right)={{x}^{2}}-2\sqrt{2}x+1 $ .
We need to find the value of $ p\left( 2\sqrt{2} \right) $ . This means that we need to find value of the function when values of x is equal to $ 2\sqrt{2} $ .
Hence we get $ p\left( 2\sqrt{2} \right)={{\left( 2\sqrt{2} \right)}^{2}}-2\sqrt{2}\left( 2\sqrt{2} \right)+1 $ .
Now let us solve it to get our answer in simplified form.
We know that $ {{x}^{2}} $ can be written as $ x\times x $ . So our function becomes,
 $ p\left( 2\sqrt{2} \right)=\left( 2\sqrt{2}\times 2\sqrt{2} \right)-2\sqrt{2}\left( 2\sqrt{2} \right)+1 $ .
Now, it can be further simplified as,
 $ p\left( 2\sqrt{2} \right)=\left( 4\times \sqrt{2}\times \sqrt{2} \right)-\left( 4\times \sqrt{2}\times \sqrt{2} \right)+1 $ .
We know that $ \sqrt{x}\sqrt{x} $ can be written as x, so similarly we can write $ \sqrt{2}\times \sqrt{2} $ as 2. Hence we get,
 $ p\left( 2\sqrt{2} \right)=\left( 4\times 2 \right)-\left( 4\times 2 \right)+1 $ .
Multiplying 4 by 2, we get 8. So our function becomes
 $ p\left( 2\sqrt{2} \right)=8-8+1 $ .
Subtracting 8 from 8 gives us 0. Hence our function becomes,
 $ p\left( 2\sqrt{2} \right)=0+1 $ .
Adding zero to any number gives us the same number, so adding 0 to 1 we get: $ p\left( 2\sqrt{2} \right)=1 $ which is our required answer.
So, value of the function $ p\left( 2\sqrt{2} \right) $ becomes equal to 1.
Hence, option B is the correct answer.

Note:
Students should take care of the signs while solving these functions. Students can make mistakes while solving the terms like $ {{\left( 2\sqrt{2} \right)}^{2}} $ . Student should note that we can also write $ -\left( 2\sqrt{2} \right)\left( 2\sqrt{2} \right) $ as $ -{{\left( 2\sqrt{2} \right)}^{2}} $ and then cancel it with positive $ {{\left( 2\sqrt{2} \right)}^{2}} $ i.e.
 $ \begin{align}
  & p\left( 2\sqrt{2} \right)={{\left( 2\sqrt{2} \right)}^{2}}-\left( 2\sqrt{2} \right)\left( 2\sqrt{2} \right)+1 \\
 & \Rightarrow p\left( 2\sqrt{2} \right)={{\left( 2\sqrt{2} \right)}^{2}}-{{\left( 2\sqrt{2} \right)}^{2}}+1 \\
 & \Rightarrow p\left( 2\sqrt{2} \right)=0+1 \\
 & \Rightarrow p\left( 2\sqrt{2} \right)=1 \\
\end{align} $