GATE CSE 2023 | GA | Question: 9

Question

$f(x)$ and $g(y)$ are functions of $x$ and $y$, respectively, and $f(x)=g(y)$ for all values of $x$ and $y$. Which one of the following options is necessarily $\text{TRUE}$ for a $x$ and $y?$ 

• A. $ f(x)=0$ and $g(y)=0$
• B. $ f(x)=g(y)=$ constant
• C. $ f(x) \neq$ constant and $g(y) \neq$ constant
• D. $ f(x)+g(y)=f(x)-g(y)$

Comments

Detailed Video Solution:

GATE 2023 Question: f(x) = g(y) for all real values of x,y

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B

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https://gateoverflow.in/398940/gate-cse-2023-memory-based-question-2

Answer 1

When we define "functions", we should also have to define domain and co-domain but sometimes it is not defined in gate.
Here, domain is defined but co-domain is not defined, so I am taking it as set of real numbers.

So, we have $f:\mathcal{R} \rightarrow \mathcal{R}$ and $g:\mathcal{R} \rightarrow \mathcal{R}$

Now, two functions $f$ and $g$ are equal if they have the same domain and codomain and for every element $x$ in the domain (because dom(f)=dom(g)), $f(x)=g(x).$

for example, $f(x)=\dfrac{x}{2}$ and $g(x)=\dfrac{2x}{4}$

But, here, we have given two equal functions where $x$ and $y$ may or may not be same i.e. $f(x)=g(y)$ where $x$ and $y$ belongs to the set of real numbers.

 So, $\forall x,y \in \mathcal{R}, $ $f(x) = g(y)$ and it is true when all $x \in \mathcal{R}$ maps to some element, say, $c$ where $c$ belongs to set $\mathcal{R}$ which is the codomain of $f$ and similarly, all $y \in \mathcal{R}$ maps to same element $c$ where $c$ belongs to set $\mathcal{R}$ which is the codomain of $g$ and in this way both domain and codomain of $f$ and $g$ are equal and $\forall x,y \in \mathcal{R},$ $f(x)=f(y) = c$ where $c$ is some arbitrary constant.

Hence, $\textbf{(B)}$

Answer 2

given f(x)=g(x)   ......eq 1

f(x) depends only upon x and g(y) depends only upon y, for f(x)=g(x) regardless of what specific values x and y take, the only possibility is that both must be equal to a constant value.

when we vary x by keeping y fixed say y',then g(y')= constant C so f(x)=C .....eq2

when we vary y by keeping x fixed say x' then f(x')= constant K so g(x)=K........eq3

from eq 1,2,3

f(x)=g(x)=some constant

opt B)

 

 

Answer 3

Let's check all options one by one.

Given:

f(x)=g(y)∀x,y

Since x and y are independent, the only way this can happen for every x and y is if both functions are the same constant.

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Option A: f(x)=0 and g(y)=0

 Not necessarily true.

Counterexample:

f(x)=5, g(y)=5

Condition is satisfied, but neither function is 0.

So A is not always true.

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Option B: f(x)=g(y)=constant

 Always true.

Proof:

Choose y=0.

f(x)=g(0)

for all x.

Since g(0) is a fixed number, f(x) must be constant.

 

Similarly, choose x=0:

g(y)=f(0)

for all y.

Hence both are constants and equal.

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Option C: f(x)≠constant and g(y)≠constant

 Impossible.

We just proved both must be constant.

Example:

f(x)=x,g(y)=y

Then f(1)=1 and g(2)=2, so

f(1)≠g(2)

Condition fails.

So C cannot be true.

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Option D: f(x)+g(y)=f(x)−g(y) 

Move terms:

               2g(y)=0 

                 g(y)=0

and therefore f(x)=0

Thus D is equivalent to Option A.

It is true only when the constant is 0.

Counterexample:

f(x)=5, g(y)=5

Then

f(x)+g(y)=10, f(x)−g(y)=0

Not equal.

So D is not necessarily true.