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 $x\in [0,1],p\left(0\right)=1$ and $p\left(1\right)=41$. Then ${\int }_0^{1}p\left(x\right)$ dx equals

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

Answer 1

The correct option is A 21

$p^{\prime }\left(x\right)=p^{\prime }(1-x)$
$\Rightarrow p\left(x\right)=-p(1-x)+c$
At $x=0$
$p\left(0\right)=-p\left(1\right)+c=42=c$
Now $p\left(x\right)=-p(1-x)+42$
$\Rightarrow p\left(x\right)+p(1-x)=42$ ... (i)
We know that $I={\int }_0^{1}p\left(x\right)$ dx $={\int }_0^{1}p(1-x)dx$
Now, $I={\int }_0^{x}p\left(x\right)dx$
From eq (i), we get
$2I={\int }_0^1\left(42\right)dx$ $\Rightarrow I=21$ .