If f r - s - f r - f s - r - s - R- Let m and n be integers- Then f n - m is equal toA- m mB- f m - f n C- m - n f 1D- n-m f1

The correct option is C m/nf(1)Given, f(r + s) = f(r) + f(s) ⇒ f ( 1+1+1+1...x times)=xf(1) Also, we can say that, f(m) =f( m/n+ m/n+ m/n...n times )=n f ( m/ ...

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Standard XII

Mathematics

Ordered Pair

If f r + s = ...

Question

If $f (r + s) = f (r) + f (s) \forall r, s \in R$ . Let $m$ and $n$ be integers. Then $f (\frac{n}{m})$ is equal to

\frac{m}{n}

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\frac{f (m)}{f (n)}

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\frac{m}{n} f (1)

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\frac{n}{m} f (1)

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Solution

The correct option is C $\frac{m}{n} f (1)$
Given, $f (r + s) = f (r) + f (s)$
$\Rightarrow f (1 + 1 + 1 + 1... x times) = x f (1)$
Also, we can say that,
$f (m) = f (\frac{m}{n} + \frac{m}{n} + \frac{m}{n} . . . n times) = n f (\frac{m}{n}) . . . (1)$
$f (m) = f (1 + 1 + . . . m times) = m f (1) . . . (2)$
equating (1) and (2), we get,
$m f (1) = n f (\frac{m}{n})$
Thus, $f (\frac{m}{n}) = \frac{m}{n} f (1)$

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If f(r+s)=f(r)+f(s) ∀ r,s ∈ R. Let m and n be integers. Then f(nm) is equal to

The correct option is C mnf(1)Given, f(r+s)=f(r)+f(s) ⇒f(1+1+1+1...x times)=xf(1) Also, we can say that, f(m)=f(mn+mn+mn...n times)=nf(mn)...(1) f(m)=f(1+1+...m times)=mf(1)...(2) equating (1) and (2), we get, mf(1)=nf(mn) Thus, f(mn)=mnf(1)

If $f (r + s) = f (r) + f (s) \forall r, s \in R$ . Let $m$ and $n$ be integers. Then $f (\frac{n}{m})$ is equal to