Question

Let $p\left(x\right)$ be a function defined on $R$ such that $p^{\prime }\left(x\right)=p^{\prime }(1-x)$, for all $xϵ[0,1],p\left(0\right)=1$ and $p\left(1\right)=41$. Then ${\int }_0^{1}p\left(x\right)dx$ equals

• A. $\sqrt{41}$
• B. $21$
• C. $41$
• D. $42$

Answer 1

The correct option is B $21$

Given : $p^{\prime }\left(x\right)=p^{\prime }(1-x)x\in [0,1]p\left(0\right)=1p\left(1\right)=41$

$p^{\prime }\left(x\right)=p^{\prime }(1-x)p\left(x\right)=-p(1-x)+catx=0p\left(0\right)=-p\left(1\right)+c1=-41+cc=42I={\int }_0^{1}p\left(x\right)dx..........\left(1\right)I={\int }_0^{1}p(1-x)dx...........\left(2\right)$

Adding (1) and (2)

$2I={\int }_0^{1}p\left(x\right)+p(1-x)dx=42{\int }_0^{1}xdx2I=42I=21$