If m - n and m - n-1 m-1 - n-1 - mxny- then x - y - 2 -

nbsp; nbsp; (m + n)^ 1 (m^ 1 + n^ 1) = m^xn^y = gt;1/m+n(1/m+1/n)= m^xn^y = gt;1/m+n(m+n/mn)= m^xn^y = gt;1/mn= m^xn^y = g ...

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Byju's Answer

Standard XII

Mathematics

Functions

If m ≠ n an...

Question

If $m \neq n$ and $(m + n)^{- 1} (m^{- 1} + n^{- 1}) = m^{x} n^{y}$ ; then $x + y + 2 =$ .

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Solution

$(m + n)^{- 1} (m^{- 1} + n^{- 1}) = m^{x} n^{y}$
$=> \frac{1}{m + n} (\frac{1}{m} + \frac{1}{n}) = m^{x} n^{y}$
$=> \frac{1}{m + n} (\frac{m + n}{m n}) = m^{x} n^{y}$
$=> \frac{1}{m n} = m^{x} n^{y}$
$=> m^{x} n^{y} \times m n = 1$
$=> m^{x + 1} n^{y + 1} = 1$
$=> m^{x + 1} n^{y + 1} = m^{0} n^{0}$
$=> x + 1 = 0$ .....................(1)
$=> y + 1 = 0$ .......................(2)
On adding 1 and 2, we get;
$=> x + 1 + y + 1 = 0$
$=> x + y + 2 = 0$
Hence proved

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If m≠n and (m+n)−1(m−1+n−1)=mxny; then x+y+2=.

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If $m \neq n$ and $(m + n)^{- 1} (m^{- 1} + n^{- 1}) = m^{x} n^{y}$ ; then $x + y + 2 =$ .