Let $p(x)$ be a function defined on $R$ such that $p'(x)=p'(1-x)$ for all $x \in[0,1], p(0)=1$ and $p(1)=41$. Then, $\int\limits_0^1 p(x) d x$ equals
Answer & explanation
Correct answer: option 4
We have,
$p'(x)=p'(1-x)$
i.e., $\frac{d}{d x} p(x)=\frac{d}{d(1-x)}\{p(1-x)\}$
$\Rightarrow p(x)=-p(1-x)+C$ .....(i)
Putting $x=0$ in (i), we get
$p(0)=-p(1)+C \Rightarrow C=42$ ......(ii)
Putting $C=42$ in (i), we get
$p(x)=-p(1-x)+42$
Now, $\int\limits_0^1 p(x) d x=\int\limits_0^1 p(1-x) d x$
$\Rightarrow \int\limits_0^1 p(x)=\int\limits_0^1\{42-p(x)\} d x$
$\Rightarrow 2 \int\limits_0^1 p(x) d x=42$
$\Rightarrow \int\limits_0^1 p(x) d x=21$