If y = \frac{2-x}{1+x} , then what is the value of 1/{y+1}+\frac{2y+1}{y^2-1} ?

Correct Answer is : (1+x)(2−x)1−2x

SSC CGL 5 Aug 2017 Shift 1 Paper

Question : 64 of 100

Marks: +1, -0

y = {2-x}/{1+x}

, then what is the value of

1/{y+1}+{2y+1}/{y^2-1}

Solution:

y = {2-x}/{1+x}

1/{y-1}+{2y+1}/{y^2-1}

=1/{{2-x}/{1+x}+1}+{2[{2-x}/{1+x}]+1}/{[{2-x}/{1+x}]^2-1}

={1+x}/{2-x+1+x}

+(4-2x+1+x)/(4+x^2-4x-1-x^2-2x)

×(1+x)^2/(1+x)

={1+x}/3+{(5-x)(1+x)}/{3-6x}

={1+x}/3[1+(5-x)/{1-2x}]

={1+x}/3[{1-2x+5-x}/{1-2x}]

={1+x}/3(6-3x)/(1-2x)

={(1+x)(2-x) }/(1-2x)

Go to Question:

Prev Question

Next Question