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Let \( \Large f : \left[ 0,\ 1 \right] \rightarrow \left[ 0,\ 1 \right] \) defined by \( \Large f \left(x\right)=\frac{1-x}{1+x},\ 0 \le x \le 1 \) and let \( \Large g : \left[ 0,\ 1 \right] \rightarrow \left[ 0,\ 1 \right] \) be defined by \( \Large g \left(x\right) = 4x \left(1-x\right),\ 0 \le x \le 1 \), then fog and gof is:


A) \( \Large \frac{ \left(2x-1\right)^{2} }{1+4x+4x^{2}},\ \frac{8x \left(1-x\right) }{ \left(1+x\right)^{2} } \)

B) \( \Large \frac{ \left(2x-1\right) }{1+4x+4x^{2}},\ \frac{8 \left(1-x\right)x }{ \left(1+x\right)^{2} } \)

C) \( \Large \frac{ \left(2x+1\right)^{2} }{1+4x+4x^{2}},\ \frac{8}{ \left(1+x\right)^{2} } \)

D) \( \Large \frac{ \left(2x+1\right)^{2} }{ \left(1+x\right)^{2} },\ \frac{8 \left(1-x\right) }{ \left(1+x\right)^{2} } \)

Correct Answer:
A) \( \Large \frac{ \left(2x-1\right)^{2} }{1+4x+4x^{2}},\ \frac{8x \left(1-x\right) }{ \left(1+x\right)^{2} } \)

Description for Correct answer:
Given that \( \Large f: \left[ 0,\ 1 \right] \rightarrow \left[ 0,\ 1 \right] \) defined on \( \Large f \left(x\right)=\frac{1-x}{1+x} \)

and\( \Large g \left(x\right)=4x \left(1-x\right) \)

Therefore, \( \Large \left(fog\right)x=f\{ g \left(x\right) \} \)

\( \Large \frac{1-4x \left(1-x\right) }{1+4x \left(1-x\right) } = \frac{ \left(2x-1\right)^{2} }{1+4x+4x^{2}} \)

and \( \Large \left(gof\right)x = g\{ f \left(x\right) \} \)

= \( \Large 4 \left(\frac{1-x}{1+x}\right) \left(1-\frac{1-x}{1+x}\right) \)

=\( \Large 4 \left(\frac{1-x}{1+x}\right) \left(\frac{2x}{1+x}\right)=8x \left(\frac{1-x}{(1+x)^{2}}\right) \)

Part of solved Set theory questions and answers : >> Elementary Mathematics >> Set theory




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