If x+1=\sqrt{y}+3, y>0, then the value of \frac{1}{2} ( {x^{3}-6 x^{2}+12 x-8}/\sqrt{y}-y ) is

SSC CPO SI and ASI 2016 Paper

Question : 130 of 200

Marks: +1, -0

x+1=√{y}+3, y>0,

then the value of

{1}/{2} ( {x^{3}-6 x^{2}+12 x-8}/{√{y}}-y )

Solution:

(2)since

, x+1=√{y}+3 ⇒√{y}=x-2

{1}/{2} ( {x^{3}-6 x^{2}+12 x-8}/{√{y}}-y )= {1}/{2} ( {(x-2)^{3}}/{√{y}}-y )

{1}/{2} ( {(x-2)^{3}}/{√{y}}-y )= {1}/{2} ( {\√{y}^{3}}/{√{y}}-y )

= {1}/{2}(y-y)=0

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