Question : If $\frac{x-a^{2}}{b+c}+\frac{x-b^{2}}{c+a}+\frac{x-c^{2}}{a+b} = 4(a+b+c)$, then $x$ is equal to:
Option 1: $(a+b+c)^{2}$
Option 2: $a^{2}+b^{2}+c^{2}$
Option 3: $ab+bc+ca$
Option 4: $a^{2}+b^{2}+c^{2}-ab-bc-ca$
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Correct Answer: $(a+b+c)^{2}$
Solution : Let values of $a = 1, b = 1, c = 0$ $\frac{x-a^{2}}{b+c}+\frac{x-b^{2}}{c+a}+\frac{x-c^{2}}{a+b} = 4(a+b+c)$ Now, let us find the value of $x$ by substituting the values of $a$, $b$ and $c$. $\frac{x-1}{1}+\frac{x-1}{1}+\frac{x}{2}=4(2)$ ⇒ $(x-1)+(x-1)+\frac{x}{2}=8$ ⇒ $(2x-2)+\frac{x}{2}=8$ ⇒ $2(2x-2)+x=8\times 2$ ⇒ $4x-4+x=16$ ⇒ $5x=20$ ⇒ $x=4$ Now, we will see which option gives answer 4. The answer comes out to be $(a+b+c)^{2}$. Since, $(a+b+c)^{2}=4$ Hence, the correct answer is $(a+b+c)^{2}$.
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Question : If $\frac{a^{2} - bc}{a^{2}+bc}+\frac{b^{2}-ca}{b^{2}+ca}+\frac{c^{2}-ab}{c^{2}+ab}=1$, then the value of $\frac{a^{2}}{a^{2}+bc}+\frac{b^{2}}{b^{2}+ac}+\frac{c^{2}}{c^{2}+ab}$ is:
Option 1: 0
Option 2: 1
Option 3: –1
Option 4: 2
Question : If $\frac{2+a}{a}+\frac{2+b}{b}+\frac{2+c}{c}=4$, then the value of $\frac{ab+bc+ca}{abc}$ is:
Option 1: $2$
Option 2: $1$
Option 3: $0$
Option 4: $\frac{1}{2}$
Question : If $2\cot x=5$, then what is $\frac{2 \cos x-\sin x}{2 \cos x+\sin x}$ equal to?
Option 1: $\frac{3}{4}$
Option 2: $\frac{1}{3}$
Option 3: $\frac{5}{6}$
Option 4: $\frac{2}{3}$
Question : If $x+\frac{1}{x}=c+\frac{1}{c}$, then the value of $x$ is:
Option 1: $c,\frac{1}{c}$
Option 2: $c,c^{2}$
Option 3: $c,2c$
Option 4: $0, 1$
Question : If $\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$, then $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 4
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