GO Classes 2023 | IIITH Mock Test 1 | Question: 51

Answer 1

The correct answer to the question "If f(x) = x^2 and g(x) = ln(x), then f(x) + g(x) will be" is "None of the above".

The sum function of f(x) and g(x) is:

f(x) + g(x) = x^2 + ln(x)

This function is neither even nor neither odd. A function is even if it is symmetric with respect to the y-axis, that is, f(-x) = f(x) for all x. A function is odd if it is symmetric with respect to the origin, that is, f(-x) = -f(x) for all x.

We can verify that the sum function is not even as follows:

f(-x) + g(-x) = (-x)^2 + ln(-x) = x^2 + ln(-x)

And f(x) + g(x) = x^2 + ln(x)

Since ln(-x) ≠ ln(x), f(x) + g(x) is not even.

We can also verify that the sum function is not odd as follows:

f(-x) + g(-x) = (-x)^2 + ln(-x) = x^2 + ln(-x)

And -[f(x) + g(x)] = -[x^2 + ln(x)] = -x^2 - ln(x)

Since x^2 + ln(-x) ≠ -x^2 - ln(x), f(x) + g(x) is not odd.

Therefore, the correct answer is "None of the above".

Comments

ln(-x) how we can write this ? it is undefined bcs logarithm domain does not take negative values .

---

You are correct ln(-x) is undefined and hence g(x) is undefined so f(x) + g(x) is undefined hence option d is the correct answer

---

okay got it . thanks .

Answer 2

since f(x) = $x^{2}$ is an even function and g(x)=$\log_{e}x$ is undefined 

hence f(x) + g(x) would also be undefined,so the correct option would be D

P.S. – 

an even function is a function that satisfies the following property:

f(-x) = f(x) for all values of x in the domain of the function.

and 

an odd function is a function that satisfies the following property:

f(-x) = -f(x) for all values of x in the domain of the function.