If f (x) = x|x|/1+x² , then f^{-1} (x) equals

Correct Answer is : None of these

Correct Answer is : None of these

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UPSEE Solved Model Paper 7

Question : 124 of 150

Marks: +1, -0

If f (x) =

\frac{x|x|}{1+x^2}

f^{-1}

(x) equals

Solution:

Given

f (x) =

\frac{x|x|}{1+x^2}

We have to find the inverse function

f^{-1}

(x)

Let

y = f (x) =

\begin{cases} \frac{x^2}{1+x^2}, & x \ge 0 \\ \frac{-x^2}{1+x^2}, & x < 0 \end{cases}

If y =

\frac{x^2}{1+x^2}

⇒

x^2 - x^2 y - y

= 0

⇒

x^2

\frac{y}{1-y}

⇒ x =

\sqrt{\frac{y}{1-y}}

for 0 ≤ y < 1

Similarly, if y =

\frac{-x^2}{1+x^2}

⇒ x =

-\sqrt{\frac{-y}{1+y}}

for - 1 < y < 0

Since

f^{-1}(y)

\begin{cases} -\sqrt{\frac{-y}{1+y}}, & -1 < y < 0 \\ \sqrt{\frac{y}{1-y}}, & 0 \le y < 1 \end{cases}

Replacing y by x, we get

f^{-1}(x)

\begin{cases} -\sqrt{\frac{-x}{1+x}}, & -1 < x < 0 \\ \sqrt{\frac{x}{1-x}}, & 0 \le x < 1 \end{cases}

Hence, option 'D' is correct.

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