Let f: x, y, z → 1, 2, 3 be a one-one mapping such that only one of the following three statements is true and remaining two are false: f (x) ≠ 2, f (y) = 2, f (z) ≠ 1, then

Question

Let f: x, y, z → 1, 2, 3 be a one-one mapping such that only one of the following three statements is true and remaining two are false: f (x) ≠ 2, f (y) = 2, f (z) ≠ 1, then (A) f (x) > f (y) > f (z) (B) f (x) < f (y) < f (z) (C) f (y) < f (x) < f (z) (D) f (y) < f (z) < f (x)

Tardigrade Admin · Accepted Answer

Let f (x) ≠ 2 be true and f (y) = 2, f (z) ≠ 1 are false ⇒ f (x) ≠ 2, f (y) ≠ 2, f (z) = 1 ⇒ f (x) = 3, f (y) = 3, f (z) = 1 but then function is many one, similarly two other cases.

Q. Let $f : x, y, z \to 1, 2, 3$ be a one-one mapping such that only one of the following three statements is true and remaining two are false : $f (x) \neq = 2, f (y) = 2, f (z) \neq = 1$ , then

$f (x) > f (y) > f (z)$

$f (x) < f (y) < f (z)$

$f (y) < f (x) < f (z)$

$f (y) < f (z) < f (x)$

Let $f (x) \neq = 2$ be true and $f (y) = 2, f (z) \neq = 1$ are false
$\Rightarrow f (x) \neq = 2, f (y) \neq = 2, f (z) = 1$
$\Rightarrow f (x) = 3, f (y) = 3, f (z) = 1$ but then function is many one, similarly two other cases.

Q. Let f:x,y,z→1,2,3 be a one-one mapping such that only one of the following three statements is true and remaining two are false : f(x)=2,f(y)=2,f(z)=1, then

f(x)>f(y)>f(z)

f(x)<f(y)<f(z)

f(y)<f(x)<f(z)

f(y)<f(z)<f(x)