Question : If $x^2 = y+z$, $y^2=z+x$, $z^2=x+y$, then the value of $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$ is:
Option 1: –1
Option 2: 1
Option 3: 2
Option 4: 4
Answer (1)
10th Jan, 2024
Correct Answer: 1
Solution :
Given: $x^2 = y+z,y^2 = z+x$ and $z^2 = x+y$
Now, $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$
= ($\frac{x}{x+x^2}+\frac{y}{y+y^2}+\frac{z}{z+z^2})$
= ($\frac{x}{x+y+z}+\frac{y}{y+x+z}+\frac{z}{z+x+y})$
= ($\frac{x+y+z}{x+y+z}$)
= 1
Hence, the correct answer is 1.
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