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Let $f: N \rightarrow N: f(x)=2 x, g: N \rightarrow N: g(y)$ $=3 y+4$ and $h: N \rightarrow N: h(z)$ $=\sin z .$ Show that $h \circ(g \circ f)$ $=(\mathrm{h} \circ \mathrm{g}) \mathrm{o}$ f.


To show: h o (g o f ) = (h o g) o f

Formula used: (i) f o g = f(g(x))

(ii) $g \circ f=g(f(x))$

Given: (i) $f: N \rightarrow N: f(x)=2 x$

(ii) $g: N \rightarrow N: g(y)=3 y+4$

(iii) $h: N \rightarrow N: h(z)=\sin z$

Solution: We have,

$\mathrm{LHS}=\mathrm{h} \mathrm{o}(\mathrm{g} \mathrm{o} \mathrm{f})$

$\Rightarrow \mathrm{h} \circ(\mathrm{g}(\mathrm{f}(\mathrm{x}))$

$\Rightarrow \mathrm{h}(\mathrm{g}(2 \mathrm{x}))$

$\Rightarrow \mathrm{h}(3(2 \mathrm{x})+4)$

$\Rightarrow \mathrm{h}(6 \mathrm{x}+4)$

$\Rightarrow \sin (6 \mathrm{x}+4)$

RHS $=(h \circ g) \circ f$

$\Rightarrow(h(g(x)))$ of

$\Rightarrow(h(3 x+4))$ of

$\Rightarrow \sin (3 x+4)$ of

Now let $\sin (3 x+4)$ be a function $u$

RHS = u o f

$\Rightarrow \mathrm{u}(\mathrm{f}(\mathrm{x}))$

$\Rightarrow \mathrm{u}(2 \mathrm{x})$

$\Rightarrow \sin (3(2 \mathrm{x})+4)$

$\Rightarrow \sin (6 \mathrm{x}+4)=\mathrm{LHS}$

Hence Proved.


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