UPSC NDA I 2021 Mathematics Solved Paper

Concept:
Definite Integrals:

b

∫

a

f(x)dx=

b

∫

a

f(a+b−x)dx
If f(x)=f(2a−x), then

2a

∫

0

f(x)dx=2‌

a

∫

0

f(x)dx
A function f(x) is:
- Even, if f(−x)=f(x). And

∫

−a

af(x)dx=2‌

∫

0

af(x)dx.
- Odd, if f(−x)=−f(x). And

a

∫

−a

f(x)dx=0.
- Periodic, if f(np±x)=f(x), for some number p and n∈Z.
Calculation:
We know that for an even function, f(−x)=f(x) and

a

∫

−a

f(x)dx=2‌

a

∫

0

f(x)dx.
For the given condition

a

∫

0

[f(x)+f(−x)]dx=

a

∫

−a

g(x)dx to be true, f(x) and g(x) both must be even functions, i.e. f(x)=f(−x) and g(x)=f(x)+f(−x).