Question

If `|("x"^"k", "x"^("k" + 2), "x"^("k" + 3)),("y"^"k", "y"^("k" + 2), "y"^("k" + 3)),("z"^"k", "z"^("k" + 2), "z"^("k" + 3))|` = (x - y) (y - z) (z - x)`(1/"x"+ 1/"y" + 1/"z") ` then

पर्याय

• k = –3

• k = –1

• k = 1

• k = 3

Answer 1

k = – 1

Explanation:

L.H.S →

`|(x^k, x^(k + 2), x^(k + 3)),(y^k, y^(k + 2), y^(k + 3)),(z^k, z^(k + 2), z^(k + 3))| = x^k y^k z^k |(1, x^2, x^3),(1, y^2, y^3),(1, z^2, z^3)|`

= (xyz)^k(x – y) (y – z) (z – x) (xy + yz + zx)

R.H.S. →

= `(x - y) (y - z) (z - x) (1/x+ 1/y + 1/z)`

= `(x - y) (y - z) (z - x) ((xy + yz + zx)/(xyz))`

L.H.S. = R.H.S.

∴ `(xyz)^k (x - y) (y - z) (z - x) (xy + yz + zx) = (x - y) (y - z) (z - x) ((xy + yz + zx)/(xyz))`

∴ `(xyz)^k cancel((x - y)) cancel((y - z)) cancel((z - x)) (xy + yz + zx) = cancel((x - y)) cancel((y - z)) cancel((z - x)) ((xy + yz + zx)/(xyz))`

∴ `(xyz)^k  cancel((xy + yz + zx)) = cancel((xy + yz + zx))/(xyz)`

∴ (xyz)^k = `1/"xyz"`

∴ (xyz)^k = xyz^{- 1}

∴ k = – 1