A) 0
B) 3
C) \[-1\]
D) none of these
Correct Answer: D
Solution :
| \[fog\left( x \right)=f\left( g\left( x \right) \right)=f\left( 4x\left( 1-x \right) \right)\]\[\Rightarrow \]\[\frac{1-4x\left( 1-x \right)}{1+4x\left( 1-x \right)}\]when \[0\le 4x\left( 1-x \right)\le 1\] |
| and \[0\le x\le 1\] But \[4x-4{{x}^{2}}\ge 0\Rightarrow 0\le x\le 1\] |
| \[4x-4{{x}^{2}}\le 1\Rightarrow {{\left( 2x-1 \right)}^{2}}\ge 0\Rightarrow x\in R\] |
| Hence \[fog\left( x \right)=\frac{1-4x+4{{x}^{2}}}{1+4x-4{{x}^{2}}},0\le x\le 1\] |
| Let \[y=\frac{4{{x}^{2}}-4x+1}{-\left( 4{{x}^{2}}-4x \right)+1},0\le x\le 1\] |
| Put \[4{{x}^{2}}-4x=t\,;\text{ }t\,\in \left[ -1,0 \right]\] |
| \[y=\frac{1+t}{1-t},\frac{dy}{dt}=\frac{1-t+1+t}{{{\left( 1-t \right)}^{2}}}>0\] |
| Range of\[fog\left( x \right)=\left[ 0,1 \right]\]\[\Rightarrow \alpha +\beta =1\] |
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