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Q.
If $f(x) = x, g(x) = 2x^2 + 1$ and $h(x) = x+ 1$ then $(hogof)(x) = $
Relations and Functions - Part 2
Solution:
Given $f(x)=x, g(x)=2 x^{2}+1$ and $h(x)=x+1$
hogof $(x)$
$=h(g(f(x)))$
$=h(g(x)) \ldots \ldots[\because f(x)=x]$
$=h\left(2 x ^{2}+1\right) \ldots\left[\because g ( x )=2 x ^{2}+1\right]$
$=2 x ^{2}+1+1$
$=2 x^{2}+2$
$=2\left(x^{2}+1\right)$
$\therefore \operatorname{hogof}( x )=2\left( x ^{2}+1\right)$