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Calculus

GATE CSE 2024 Set 2 · Q16

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Let f(x)f(x) be a continuous function from R\mathbb{R} to R\mathbb{R} such that f(x)=1−f(2−x)f(x) = 1 - f(2 - x) Which one of the following options is the CORRECT value of ∫02f(x)dx\int_0^2 f(x)dx ?

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Official answer: B) 1

Calculus

Let the integral be:

I=∫02f(x) dx— (1)I = \int_0^2 f(x)\,dx \quad \text{--- (1)}

Using the definite integral property ∫abf(x) dx=∫abf(a+b−x) dx\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx, where a=0a = 0 and b=2b = 2:

I=∫02f(2−x) dx— (2)I = \int_0^2 f(2 - x)\,dx \quad \text{--- (2)}

Adding equations (1) and (2):

2I=∫02[f(x)+f(2−x)] dx2I = \int_0^2 \big[f(x) + f(2 - x)\big]\,dx

From the given relation f(x)=1−f(2−x)f(x) = 1 - f(2 - x), we have:

f(x)+f(2−x)=1f(x) + f(2 - x) = 1

Substituting this into the sum of integrals:

2I=∫021 dx=[x]02=2−0=22I = \int_0^2 1\,dx = [x]_0^2 = 2 - 0 = 2
I=1I = 1

Thus, the correct option is (B).

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This Calculus previous year question is Q16 of the GATE CSE 2024 Set 2 paper. Find more GATE CSE PYQs on Calculus in Engineering Mathematics, or browse all GATE CSE PYQs.

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Calculus in GATE CSE

Calculus has 40 GATE CSE PYQs across 25 papers from 2011 to 2026. 14 of them are 1-mark MCQs like this one, and it is the only one in GATE CSE 2024 Set 2.

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GateAI solution cardGATE CSE 2024 Set 2 · Q16 · MCQ

Let f(x) be a continuous function from R to R such that f(x) = 1 - f(2 - x) Which one of the following options is the CORRECT value of 02 f(x)dx ?

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