Question

Which conditional and its converse form a true biconditional?

A) If x>0, then lxl >0
B) If x=3, then x2^2=9
C) If x^2 =4, Then x=2
D) If x=19, then 2x-3=35

Answer 1

Answer: D) If x=19, then 2x-3=35

Explanation:

A) If x > 0, then lxl >0,

L.H.S. x > 0,

R.H.S. lxl >0 ⇒ ± x > 0

⇒ L.H.S. ≠ R.H.S.

B) If x = 3 then
`x^2 = 9`

L.H.S. x = 3,

R.H.S.
`x^2 = 9` ⇒ x = ± 3

⇒ L.H.S. ≠ R.H.S.

C) If
`x^2 =4`, Then x=2

L.H.S.
`x^2 =4` ⇒ x = ± 2,

R.H.S. x = 2,

L.H.S. ≠ R.H.S.

D) If x=19, then 2x-3=35

L.H.S. x = 19,

R.H.S. 2x-3=35 ⇒ 2x = 38 ⇒ x = 19

⇒ L.H.S. = R.H.S.

⇒ Option D is correct.

Answer 2

the answer is
D) If x=19, then 2x-3=35
proof

If x=19, 2(19)-3=38-3=35 so If x=19, then 2x-3=35 is verified
if 2x-3=35, so 2x=35 + 3, 2x=38 implies x = 19, so x =19 is verified


If x=19 if and only if 2x-3=35