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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Question : In $\triangle$ABC, $\angle$C = 90° and CD is perpendicular to AB at D. If $\frac{\text{AD}}{\text{BD}}=\sqrt{k}$, then $\frac{\text{AC}}{\text{BC}}$=?
Option 1: $\sqrt{k}$
Option 2: $\frac{1}{\sqrt{k}}$
Option 3: $\sqrt[4]{k}$
Option 4: $k$
Question : If $a, b, c$ are all non-zero and $a+b+c=0$, then find the value of $\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{ab}$.
Option 1: $3$
Option 2: $4$
Option 4: $\frac{1}{2}$
Question : If $a+b+c=0$, then the value of $\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}$ is:
Option 1: 1
Option 2: 3
Option 3: - 1
Option 4: 0
Question : If $a+b+c=m$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$, then the average of $a^{2}, b^{2},$ and $c^{2}$ is:
Option 1: $m^{2}$
Option 2: $\frac{m^{2}}{3}$
Option 3: $\frac{m^{2}}{9}$
Option 4: $\frac{m^{2}}{27}$
Question : If $A+\frac{1}{1+\frac{1}{2+\frac{1}{3}}}=\frac{9}{10}$, then the value of A is:
Option 1: $\frac{1}{5}$
Option 2: $\frac{3}{10}$
Option 3: $\frac{2}{5}$
Option 4: $\frac{1}{10}$
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