Question

If ${\int }_a^b\frac{f\left(x\right)}{f\left(a\right)+f(a+b-x)}dx=10$, then

• 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)

$b=15,a=-5$
$b=10,a=-10$
$b=22,a=2$

To find $a$ and $b$

We know that

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dxI={\int }_a^b\frac{f\left(x\right)}{f\left(x\right)+f(a+b-x)}dx={\int }_a^b\frac{f(a+b-x)}{f(a+b-x)+f(a+b-(a+b-x\left)\right)}dx$

$\Rightarrow I={\int }_a^b\frac{f(a+b-x)}{f(a+b-x)+f\left(x\right)}dxI+I={\int }_a^b\frac{f\left(x\right)}{f\left(x\right)+f(a+b-x)}dx+{\int }_a^b\frac{f(a+b-x)}{f(a+b-x)+f\left(x\right)}dx$

$2I={\int }_a^b\frac{f\left(x\right)+f(a+b-x)}{f\left(x\right)+f(a+b-x)}dx\Rightarrow 2I={\int }_a^{b}1.dx={\left[x\right]}_a^b=b-a\Rightarrow I=\frac{1}{2}(b-a)$

Given that, $\frac{1}{2}(b-a)=10$

$b-a=20$

Hence the correct answer are those which $(b-a)=20$

A,B and C are correct.