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Question : If $xy+yz+zx=1$ , then the value of $\frac{1\:+\:y^2}{(x\:+\:y)(y\:+\:z)}$ is:

Option 1: 2

Option 2: 3

Option 3: 4

Option 4: 1


Team Careers360 23rd Jan, 2024
Answer (1)
Team Careers360 24th Jan, 2024

Correct Answer: 1


Solution : Given:
$xy+yz+zx=1$
Now, $\frac{1\:+\:y^2}{(x\:+\:y)(y\:+\:z)}$
= $\frac{1\:+\:y^2}{xy\:+\:yz\:+\:zx\:+\:y^2}$
= $\frac{1\:+\:y^2}{1\:+\:y^2}$
= $1$
Hence, the correct answer is 1.

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