Question

Answer of this question​

Answer 1

Answer:

The correct answer is option (A) .

Step-by-step explanation:

Given

        $a+\dfrac{1}{a}=12$  

    $\Rightarrow a^2+\dfrac{1}{a^2}+2(a)(\frac{1}{a})=12^2\\\Rightarrow a^2+\dfrac{1}{a^2}+1=144\\\Rightarrow a^2+\dfrac{1}{a^2}=144-2=142$

Now  

         $(a-\dfrac{1}{a})^2=a^2+\dfrac{1}{a^2}-2(a)(\dfrac{1}{a})$

                        $=142-2=140$

Therefore  

             $a-\dfrac{1}{a}=\sqrt{140}$