Question

If ${\int }_a^b\frac{f\left(x\right)}{f\left(x\right)+f(a+b-x)}\mathrm{dx}=10$, then which values a and b can take?

• A. b = 22, a = 2
• B. b = 15, a = -5
• C. b = 10, a = -10
• D. b = 10, a = -2

Answer 1

Answer: (A), (B), (C)

b = 10, a = -10

${\int }_a^b\frac{f\left(x\right)}{f\left(x\right)+f(a+b-x)}\mathrm{dx}=10...\left[1\right]\Rightarrow {\int }_a^b\frac{f(a+b-x)}{f(a+b-x)+f(a+b-a-b+x)}\mathrm{dx}=10\Rightarrow {\int }_a^b\frac{f(a+b-x)}{f(a+b-x)+f\left(x\right)}\mathrm{dx}=10....\left[2\right]\mathrm{Adding}\left[1\right]\mathrm{and}\left[2\right],\mathrm{we}\mathrm{get}:{\int }_a^{b}1\cdot \mathrm{dx}=20\Rightarrow b-a=20$We choose the options in which b - a =20.