If |x| < 1,y = x – x2 + x3 – x4 + ……………… ∞,then the

1.	If \|x\| < 1,y = x – x2 + x3 – x4 + ……………… ∞,then the value of x in terms of y isA) 1 - y/yB) y/1 - yC) y/y + 1D) y/y + 1
Answer» Correct option is (B) y/1 - y Since, \(x,-x^2,x^3,-x^4,......\) will form a G.P. whose common difference is -x & first term is x. Now, \(y=x-x^2+x^3-x^4+......\) \(upto\,\infty\,\text{terms (Given)}\) \(=\frac x{1-(-x)}\) \((\because\|x\|<1\Rightarrow S_\infty=\frac a{1-r}\) here r = -x & a = x) \(=\frac x{1+x}\) \(\Rightarrow\frac1y=\frac{1+x}x\) \(\Rightarrow\frac1y=\frac{1}x+1\) \(\Rightarrow\frac1x=\frac1y-1=\frac{1-y}y\) \(\Rightarrow\) \(x=\frac{y}{1-y}\) Correct option is B) y/1 - y

If |x| < 1,y = x – x2 + x3 – x4 + ……………… ∞,then the value of x in terms of y isA) 1 - y/yB) y/1 - yC) y/y + 1D) y/y + 1

Answer»

Correct option is (B) y/1 - y

Since, \(x,-x^2,x^3,-x^4,......\) will form a G.P. whose common difference is -x & first term is x.

Now, \(y=x-x^2+x^3-x^4+......\) \(upto\,\infty\,\text{terms (Given)}\)

\(=\frac x{1-(-x)}\) \((\because|x|<1\Rightarrow S_\infty=\frac a{1-r}\) here r = -x & a = x)

\(=\frac x{1+x}\)

\(\Rightarrow\frac1y=\frac{1+x}x\)

\(\Rightarrow\frac1y=\frac{1}x+1\)

\(\Rightarrow\frac1x=\frac1y-1=\frac{1-y}y\)

\(\Rightarrow\) \(x=\frac{y}{1-y}\)

Correct option is B) y/1 - y