Question

Let $f\left(x\right),g\left(x\right)$ and $h\left(x\right)$ be continuous function on $[0,a]$ such that $f\left(x\right)=f(a-x),g\left(x\right)=-g(a-x),3h\left(x\right)-4h(a-x)=5$ then ${\int }_0^{a}f\left(x\right)g\left(x\right)h\left(x\right)dx$ is equal to

• A. $1$
• B. $0$
• C. $a$
• D. $-1$

Answer 1

The correct option is B $0$

It is given that
$f(a-x)=f\left(x\right)$
$g(a-x)=-g\left(x\right)$
$h(a-x)=\frac{3h\left(x\right)-5}{4}$
Hence
$I={\int }_0^{a}f\left(x\right).g\left(x\right).h\left(x\right).dx$
$I={\int }_0^{a}f(a-x)g(a-x)h(a-x).dx$
$={\int }_0^a-f\left(x\right)g\left(x\right)\left(\frac{3h\left(x\right)-5}{4}\right).dx$
$={\int }_0^a-\frac{3g\left(x\right).f\left(x\right).h\left(x\right)}{4}+\frac{5f\left(x\right).g\left(x\right)}{4}.dx$
Hence
$2I={\int }_0^{a}f\left(x\right).g\left(x\right).h\left(x\right)-\frac{3g\left(x\right).f\left(x\right).h\left(x\right)}{4}+\frac{5f\left(x\right).g\left(x\right)}{4}.dx$
$={\int }_0^a\frac{f\left(x\right).g\left(x\right).h\left(x\right)+5f\left(x\right).g\left(x\right)}{4}.dx$
$I=\frac{1}{4}{\int }_0^{a}f\left(x\right).g\left(x\right).h\left(x\right).dx+\frac{5}{4}{\int }_0^{a}f\left(x\right).g\left(x\right).dx$
$I=\frac{I}{4}+\frac{5}{4}{\int }_0^{a}f\left(x\right).g\left(x\right).dx$
$\frac{3I}{4}=\frac{5}{4}{\int }_0^{a}f\left(x\right).g\left(x\right).dx$
$I=\frac{5}{3}{\int }_0^{a}f\left(x\right).g\left(x\right).dx$ ...(i)
And
$I=\frac{5}{3}{\int }_0^{a}f(a-x).g(a-x).dx$
$I=\frac{-5}{3}{\int }_0^{a}f\left(x\right).g\left(x\right).dx$ ..(ii)
Adding i and ii, we get
$2I=0$
Or
$I=0$.