Question : If $\small a^{4}+1=\left [\frac{a^{2}}{b^{2}}\right ]\left (4b^{2}-b^{4}-1\right),$ then what is the value of $a^{4}+b^{4}?$
Option 1: 2
Option 2: 16
Option 3: 32
Option 4: 64
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Correct Answer: 2
Solution : Given: $a^{4}+1=\frac{a^{2}}{b^{2}}( 4b^{2}-b^{4}-1)$ ⇒ $\frac{(a^{4}+1)}{a^2}=\frac{1}{b^{2}}( 4b^{2}-b^{4}-1)$ ⇒ $a^2+\frac{1}{a^2}=4 -b^2-\frac{1}{b^2}$ ⇒ $a^2+\frac{1}{a^2}+2+b^2+\frac{1}{b^2} +2 =0$ ⇒ $(a-\frac{1}{a})^2+(b-\frac{1}{b})^2=0$ So, $(a-\frac{1}{a})^2=0 \;$and$\; (b-\frac{1}{b})^2=0$ ⇒ $a-\frac{1}{a}=0 \; $ and$\; b-\frac{1}{b}=0$ ⇒ $a=\frac{1}{a} \; $ and$\; b=\frac{1}{b}$ ⇒ $a^2=1 \;$ and $\; b^2 = 1$ $\therefore a^{4}+b^{4}=2$ Hence, the correct answer is 2.
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