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

Question

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 · Accepted Answer

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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Question : If xy+yz+zx=1 , then the value of 1+y2(x+y)(y+z) is:Option 1: 2Option 2: 3Option 3: 4Option 4: 1

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Question : If $x y + y z + z x = 1$ , 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