CAT 1999QA Question 35

Basics (Functions)Easy
Passage / Data

Directions: Answer the questions based on the following information.
Let x and y be real numbers and let
f (x,y) = |x + y| , F(f (x,y)) = −f (x,y)
and G(f (x, y)) = −F(f (x, y))

Which of the following statements is true?

Answer & solution

  • A

    F(f(x, y)) ⋅ G(f(x, y)) = –F(f(x, y)) ⋅ G(f(x, y))

  • B

    F(f(x, y)) ⋅ G(f(x, y)) > –F(f(x, y)) ⋅ G(f(x, y))

  • C

    F(f(x, y)) ⋅ G(f(x, y)) ≠ G(f(x, y)) ⋅ G(f(x, y))

  • F(f(x, y)) + G(f(x, y)) + f(x, y) = f(–x, –y)

Solution

f (x,y) = |x + y| − − − This is always positive
F(f (x,y)) = − f (x,y) = −|x + y| − − − This is always negative
G(f (x,y)) = − F(f (x,y)) = − (−|x + y|) = |x + y| − − − This is always positive

F(f(x, y))G(f(x, y)) = -|x + y)2

and G(f(x, y)) G(f(x, y)) = |x + y|2

From the choices, we observe that:
Option (a): LHS of the expression is -|x + y)2, which is always non positive. RHS of the expression
is |x + y)2, which is always non negative. The only situation when LHS is equal to RHS is when each is equal to zero. Hence, (a) is not necessarily true. Option (b): The given expression can be written as

-|x + y|2 > |x + y|2 or 0 > 2|x + y|2

This implies that 0 > |x + y|, which is not true. Hence, (b) is not true.
Option (c): F(f(x, y)) G(f(x, y)) = - |x + y|2

and G(f(x, y)) G(f(x, y)) = |x + y|2

These two expressions can be equal if |x + y| = 0.
Hence, (c) is not necessarily true.

Option (d): F(f (x,y)) + G(f (x,y)) + f (x,y)

=-|x + y| + |x + y| + |x + y| = |x + y|
f (−x,−y) = |(−x) + (−y)| = |−x − y| = |−(x + y)| = |x + y|

Therefore, the two expressions are equal.

CAT 1999 QA Q35: Which of the following statements is true? — Solution | TheCATExam