Question : If $x-y=1$ and $x^2+y^2=41$ where $x, y \geq 0$, then the value of $x+y$ will be:
Option 1: 9
Option 2: 8
Option 3: 6
Option 4: 7
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Correct Answer: 9
Solution : Given: $x-y=1$ and $x^2+y^2=41$ Consider $x-y=1$ Squaring both sides, we get: ⇒ $(x-y)^2 =1$ ⇒ $x^2 + y^2 - 2xy=1$ ⇒ $41-2xy=1$ ⇒ $2xy = 40$ So, $ (x+y)^2 = x^2+y^2+2xy= 41+40$ $\therefore(x+y) = 9$ Hence, the correct answer is 9.
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Question : If $x^2+y^2=29$ and $xy=10$, where $x>0,y>0$ and $x>y$. Then the value of $\frac{x+y}{x-y}$ is:
Option 1: $- \frac{7}{3}$
Option 2: $\frac{7}{3}$
Option 3: $\frac{3}{7}$
Option 4: $-\frac{3}{7}$
Question : If $x+y+z = 9$, then the value of $(x−4)^3+(y−2)^3+(z−3)^3−3(x−4)(y−2)(z−3)$ is:
Option 1: 6
Option 2: 9
Option 3: 0
Option 4: 1
Question : $\text { If } x^2+y^2+z^2=x y+y z+z x \text { and } x=1 \text {, then find the value of } \frac{10 x^4+5 y^4+7 z^4}{13 x^2 y^2+6 y^2 z^2+3 z^2 x^2}$.
Option 1: 2
Option 2: 0
Option 3: –1
Question : If $x=8(\sin \theta+\cos \theta)$ and $y=9(\sin \theta-\cos \theta)$, then the value of $\frac{x^2}{8^2}+\frac{y^2}{9^2}$ is:
Option 1: 4
Option 2: 6
Option 3: 8
Option 4: 2
Question : Direction: After interchanging + and -, 8 and 7, which one of the following becomes correct?
Option 1: 8 - 7 + 3 x 5 = 35
Option 2: 7 x 8 + 6 - 9 = 25
Option 3: 6 + 8 x 2 - 7 = 0
Option 4: 8 x 2 + 7 - 6 = 9
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