Question

If f and g are continuous functions in [0, 1] satisfying f(x) = f(a – x) and g(x) = g(a – x) = a, then $\underset{0}{\overset{a}{\int }}f\left(x\right)g\left(x\right)dx$ is equal to

(a) $\frac{a}{2}$

(b) $\frac{a}{2}\underset{0}{\overset{a}{\int }}f\left(x\right)dx$

(c) $\underset{0}{\overset{a}{\int }}f\left(x\right)dx$

(d) $a\underset{0}{\overset{b}{\int }}f\left(x\right)dx$

Answer 1

Given: f(x) = f(a – x) and g(x) + g(a – x) = a

$$\begin{gathered} \mathrm{Let}I=\underset{0}{\overset{a}{\int }}f\left(x\right)g\left(x\right)dx...\left(1\right) \\ \Rightarrow I=\underset{0}{\overset{a}{\int }}f\left(a-x\right)g\left(a-x\right)dx\left(\because \underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f\left(a-x\right)dx\right) \\ \Rightarrow I=\underset{0}{\overset{a}{\int }}f\left(x\right)\left(a-g\left(x\right)\right)dx\left(\because f\left(x\right)=f\left(a-x\right)\mathrm{and}g\left(a-x\right)=a-g\left(x\right)\right) \\ \Rightarrow I=a\underset{0}{\overset{a}{\int }}f\left(x\right)dx-\underset{0}{\overset{a}{\int }}f\left(x\right)g\left(x\right)dx \\ \Rightarrow I=a\underset{0}{\overset{a}{\int }}f\left(x\right)dx-I\left(\mathrm{From}\left(1\right)\right) \\ \Rightarrow 2I=a\underset{0}{\overset{a}{\int }}f\left(x\right)dx \\ \Rightarrow I=\frac{a}{2}\underset{0}{\overset{a}{\int }}f\left(x\right)dx \end{gathered}$$

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Hence, the correct option is (b).