Question : If $a^2+b^2=2$ and $c^2+d^2=1$, then the value of $(ad-bc)^2+(ac+bd)^2$ is:
Option 1: $\frac{4}{9}$
Option 2: $\frac{1}{2}$
Option 3: $1$
Option 4: $2$
Correct Answer: $2$
Solution : Given: $a^2+b^2=2$ and $c^2+d^2=1$ $(ad-bc)^2+(ac+bd)^2$ $= (ad)^2+(bc)^2-2abcd+(ac)^2+(bd)^2+2abcd$ $=d^2(a^2+b^2)+c^2(a^2+b^2)$ $=(a^2+b^2)(c^2+d^2)$ $=2 \times 1$ $=2$ Hence, the correct answer is 2.
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