Question : If $x+y=4, x^{2}+y^{2}=14$ and $x>y$, then the correct value of $x$ and $y$ is:
Option 1: $2+\sqrt{3} \text{ and } 2-\sqrt{3}$
Option 2: $2-\sqrt{2} \text{ and } \sqrt{3}$
Option 3: $3 \text{ and } 1$
Option 4: $2+\sqrt{3} \text{ and } 2\sqrt{2}$
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Correct Answer: $2+\sqrt{3} \text{ and } 2-\sqrt{3}$
Solution : Given: $x^{2}+y^{2}=14$ -----------(i) ⇒ $x+y=4$ ----------------(ii) Squaring on both sides of equation (ii), $(x+y)^{2} = 4^{2}$ ⇒ $x^{2}+y^{2}+2xy = 16$ ⇒ $14+2xy = 16$ ⇒ $2xy = 2$ Subtracting $2xy$ on both sides from equation (i) $x^{2}+y^{2}-2xy = 14-2xy$ ⇒ $(x-y)^{2} = 12$ ⇒ $(x-y) = \sqrt12$ ⇒ $x-y =2\sqrt3$ -----------(iii) From (ii) and (iii) $x = 2+ \sqrt3$ and $y = 2- \sqrt3$ Hence, the correct answer is $2+ \sqrt3 \text{ and } 2- \sqrt3$.
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Question : If $x^3=270+y^3$ and $x=(6+y)$, then what is the value of $(x+y)? $(given that $x>0$ and $y>0$)
Option 1: $2 \sqrt{3}$
Option 2: $\sqrt{3}$
Option 3: $4 \sqrt{3}$
Option 4: $4 \sqrt{2}$
Question : If $x=\sqrt{3}-\frac{1}{\sqrt{3}}, y=\sqrt{3}+\frac{1}{\sqrt{3}}$, then the value of $\frac{x^2}{y}+\frac{y^2}{x}$ is:
Option 1: $\sqrt{3}$
Option 2: $3\sqrt{3}$
Option 3: $16\sqrt{3}$
Option 4: $2\sqrt{3}$
Question : If $x=\frac{1}{x-5}(x>0)$, then the value of $x+\frac{1}{x}$ is:
Option 1: $\sqrt{41}$
Option 2: $\sqrt{29}$
Option 3: $\sqrt{23}$
Option 4: $\sqrt{43}$
Question : If $\left(y^2+\frac{1}{y^2}\right)=74$ and $y>1$, then find the value of $\left(y-\frac{1}{y}\right)$.
Option 1: $6 \sqrt{2}$
Option 2: $-2 \sqrt{19}$
Option 3: $2 \sqrt{19}$
Option 4: $-6 \sqrt{2}$
Question : If $\sin (x - y) = \frac{1}2$ and $\cos (x + y) = \frac{1}2$, then what is the value of $\sin x \cos x + 2\sin^2x + cos^3x \sec x$?
Option 1: $2$
Option 2: $\sqrt{2}+1$
Option 3: $1$
Option 4: $\frac{3}{4}$
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