Question : If $x^{2}-12x+33=0$, then what is the value of $(x-4)^{2}+\frac{1}{(x-4)^{2}}?$
Option 1: $16$
Option 2: $14$
Option 3: $18$
Option 4: $20$
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Correct Answer: $14$
Solution :
Put $(x-4) = m$
⇒ $x = m+4$
⇒ $(m + 4)^{2} - 12(m + 4) + 33 = 0$
⇒ $m^{2} + 16 + 8m - 12m - 48 + 33 = 0$
⇒ $m^{2}- 4m + 1 = 0$
On dividing the equation by $m$,
⇒ $m + \frac{1}{\text{m}} = 4$
Now putting $(x - 4) = m$ in $(x - 4)^{2} + [\frac{1}{(x - 4)^{2}}]$
⇒ $m^{2} + \frac{1}{m^{2}} = (m +\frac{1}{m})^{2}-2$
⇒ $m^{2} + \frac{1}{m^{2}} = 16 -2 = 14$
So, $(x-4)^{2}+\frac{1}{(x-4)^{2}}=14$
Hence, the correct answer is $14$.
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