Question : If $\frac{1}{(x-2)}+x=8$, then what is the value of $\frac{1}{(x-2)^2}+(x-2)^2$?
Option 1: 38
Option 2: 36
Option 3: 40
Option 4: 34
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Correct Answer: 34
Solution : Given: $\frac{1}{(x-2)}+x=8$ Subtracting 2 from both sides, we get: ⇒ $\frac{1}{(x-2)}+x-2=6$ Squaring both sides, we get, ⇒ $\frac{1}{(x-2)^2}+(x-2)^2+2\times \frac{1}{(x-2)} \times(x-2) = 36$ ⇒ $\frac{1}{(x-2)^2}+(x-2)^2=36-2$ ⇒ $\frac{1}{(x-2)^2}+(x-2)^2=34$ Hence, the correct answer is 34.
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