Question

f:R→R is a function such that f(x+y)=f(xy)for all x,y∈R and f(34)=34then f(916)=

Answer 1

The function $f:\mathbb{R}\to\mathbb{R}$ is defined such that,

$\longrightarrow f(x+y)=f(xy)\quad\quad\dots(1)$

Put $y=1$ in (1), then,

$\longrightarrow f(x+1)=f(x\cdot1)$

$\longrightarrow f(x+1)=f(x)\quad\quad\dots(2)$

Put $y=0$ in (1), then,

$\longrightarrow f(x+0)=f(x\cdot0)$

$\longrightarrow f(x)=f(0)$

And from (2) we obtain,

$\longrightarrow f(0)=f(1)=f(2)=f(3)=\dots$

Therefore, for some constant k,

$\large\longrightarrow f(x)=k\quad\forall x\in\mathbb{W}$

Here, since $f(34)=34,$

$\longrightarrow f(x)=34\quad\forall x\in\mathbb{W}$

Therefore,

$\longrightarrow\underline{\underline{f(916)=34}}$

Answer 2

Answer:

⭐ Question ⭐

✏ f:R→R is a function such that f(x+y)=f(xy)for all x,y∈R and f(34)=34then f(916)

⭐ Solution ⭐

➡f:R→R is defined such that,

=> f(x+y)=f(xy)…............(1)

➡Put y=1y=1 in (1), then,

=> f(x+1)=f(x⋅1)

=> f(x+1)=f(x)............…(2)

➡Put y=0y=0 in (1), then,

=> f(x+0)=f(x⋅0)

=> f(x)=f(0)⟶f(x)=f(0)

➡And from (2) we obtain,

=> f(0)=f(1)=f(2)=f(3)=…....

Therefore, for some constant k,

=> f(x)=k∀x∈W

Here, since f(34)=34,f(34)=34,

=> f(x)=34∀x∈W

Therefore,

$\bold{\huge{\fbox{\color{blue} {</em></strong><strong><em> </em></strong><strong><em>f</em></strong><strong><em> </em></strong><strong><em>\</em></strong><strong><em>:</em></strong><strong><em> </em></strong><strong><em>(916)</em></strong><strong><em> </em></strong><strong><em>\</em></strong><strong><em>:</em></strong><strong><em> </em></strong><strong><em>=</em></strong><strong><em> </em></strong><strong><em>\</em></strong><strong><em>:</em></strong><strong><em> </em></strong><strong><em>34</em></strong><strong><em> </em></strong><strong><em>}}}}$

Step-by-step explanation:

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