One of the possible functions f(x) which satisfies _{-2}² f(x) d x=0 is

Correct Answer is : log⁡(2−x2+x) ≤ft(2-x/2+x)log(2+x2−x)

Correct Answer is : log⁡(2−x2+x) ≤ft(2-x/2+x)log(2+x2−x)

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KCET 2026 Mathematics Solved Paper

Question : 41 of 60

Marks: +1, -0

f(x)

\int_{-2}^{2} f(x) d x=0

Solution:

Concept:

f(x)

is an odd function, then

\int_{-a}^{a} f(x)\,dx = 0

Explanation:

We check which given function is odd.

For Option A,

f(x) = \log\left(\frac{2-x}{2+x}\right)

Now,

f(-x) = \log\left(\frac{2+x}{2-x}\right)

Using

\log\left(\frac{1}{A}\right) = -\log A

f(-x) = -\log\left(\frac{2-x}{2+x}\right) = -f(x)

f(x)

is an odd function.

Since the limits are symmetric from

-2

2

, we get

\int_{-2}^{2} \log\left(\frac{2-x}{2+x}\right) dx = 0

The other options are not odd functions, so they do not satisfy the condition.

Answer:

Option A

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