Question

(i) lf m ≠ n and (m + n)^–1 (m^–1 + n^–1) = m^xn^y then x + y = 2.

(ii) If 2^x = 3^y = 108^z then `2/y + 3/z = 1/x`

विकल्प

• Only (i)

• Only (ii)

• Both (i) and (ii)

• Neither (i) nor (ii)

Answer 1

Neither (i) nor (ii)

Explanation:

Let’s analyze each part of the question:

(i) If m ≠ n and (m + n)^–1 (m^–1 + n^–1) = m^xn^y, then show x + y = –2.

Proof: `(m + n)^-1(m^-1 + n^-1) = 1/(m + n) (1/m + 1/n)`

`(m + n)^-1(m^-1 + n^-1) = 1/(m + n) xx (m + n)/(mn)`

`(m + n)^-1(m^-1 + n^-1) = 1/(mn)`

So, `m^xn^y = 1/(mn)`

`m^xn^y = m^-1n^-1`

By comparing powers,

x = –1 and y = –1,

Hence x + y = –1 + (–1)

x + y = –2

(ii) If \( 2^{x} = 3^{y} = 108^{z} \), then show that \[ \frac{2}{x} + \frac{3}{y} = \frac{1}{z} \]

Proof: Let 2^x = 3^y = 108^z = k.

Then, `2 = k^(1//x), 3 = k^(1//y), 108 = k^(1//z)`

Since 108 = 2² × 3³, 

108^z = (2² × 3³)^z

108^z = 2^2z × 3^3z

108^z = k

Therefore, `k^(1//z) = 2^2 xx 3^3 = k^(2//x) xx k^(3//y)` which implies, `k^(1//z) = k^(2//x + 3//y)`

⇒ `1/z = 2/x + 3/y`

Hence, `2/x + 3/y = 1/z`.