For the first one it is a falling line and the second is 0
Answer:
I THINK 20 mm
Step-by-step explanation:
Answer:
c
Step-by-step explanation:
<u>Solution-</u>
The equation given here,
![\frac{1}{x-1} =\frac{x-2}{2x^2-2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx-1%7D%20%3D%5Cfrac%7Bx-2%7D%7B2x%5E2-2%7D)
![\Rightarrow 2x^2-2=(x-1)(x-2)](https://tex.z-dn.net/?f=%5CRightarrow%202x%5E2-2%3D%28x-1%29%28x-2%29)
![\Rightarrow 2x^2-2=x^2-3x+2](https://tex.z-dn.net/?f=%5CRightarrow%202x%5E2-2%3Dx%5E2-3x%2B2)
![\Rightarrow x^2+3x-4=0](https://tex.z-dn.net/?f=%5CRightarrow%20x%5E2%2B3x-4%3D0)
![\Rightarrow x^2 +4x-x-4=0](https://tex.z-dn.net/?f=%5CRightarrow%20x%5E2%20%2B4x-x-4%3D0)
![\Rightarrow (x-1)(x+4)=0](https://tex.z-dn.net/?f=%5CRightarrow%20%28x-1%29%28x%2B4%29%3D0)
![\Rightarrow x=1 \ and -4](https://tex.z-dn.net/?f=%5CRightarrow%20x%3D1%20%5C%20and%20-4)
But, at x=1,
becomes infinity so it can't be a solution to the equation.
∴ x=1 is the extraneous solution of the equation