If \(a+\displaystyle\frac{1}{a}+2=0\), then find the value of \(a^{47}-\displaystyle\frac{1}{a^{98}}\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-2\)
Clearing the fraction turns the condition into a perfect square equal to zero, so \(a\) has only one value, \(a=-1\). An odd power of \(-1\) is \(-1\) and an even power is \(1\).
Multiplying by \(a\): \(a^2+1+2a=0\), so \((a+1)^2=0\) and \(a=-1\)
\(a^{47}=(-1)^{47}=-1\) (odd power)
\(a^{98}=(-1)^{98}=1\) (even power), so \(\displaystyle\frac{1}{a^{98}}=1\)
\(a^{47}-\displaystyle\frac{1}{a^{98}}=-1-1=-2\)