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Question

Let $f\left(x\right)={x}^{2}\mathrm{and}g\left(x\right)={2}^{x}$. Then, the solution set of the equation $fog\left(x\right)=gof\left(x\right)$ is (a) R (b) {0} (c) {0, 2} (d) none of these

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Solution

$\text{Since}\left(fog\right)\left(x\right)=\left(gof\right)\left(x\right),\phantom{\rule{0ex}{0ex}}f\left(g\left(x\right)\right)=g\left(f\left(x\right)\right)\phantom{\rule{0ex}{0ex}}⇒f\left({2}^{x}\right)=g\left({x}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒{\left({2}^{x}\right)}^{2}={2}^{{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒{2}^{2x}={2}^{{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=2x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2x=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-2\right)=0\phantom{\rule{0ex}{0ex}}⇒x=0,2\phantom{\rule{0ex}{0ex}}⇒x\in \left\{0,2\right\}\phantom{\rule{0ex}{0ex}}$ So, the answer is (c).

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