Let $p(x)$ be a function defined on $\mathbb{R}$ such that $p'(x)=p'(1-x)$, for all $x \in [0,1]$, $p(0)=1$ and $p(1)=41$. Then $\int_{0}^{1} p(x) dx$ equals

JEE Main · 2010 · 11 Apr 2010 · Offline · Definite Integrals

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_{0}^{1} p (x) d x

equals

Answer: (A)

The condition

p^{'} (x) = p^{'} (1 - x)

integrates to

p (x) = - p (1 - x) + c

, that is

p (x) + p (1 - x) = c

. Putting

x = 0

gives

p (0) + p (1) = c

, so

1 + 41 = c

and

c = 42

; thus

p (x) + p (1 - x) = 42

for every

x

. Let

I = \int_{0}^{1} p (x) d x

. Replacing

x

1 - x

shows

I = \int_{0}^{1} p (1 - x) d x

as well, so adding the two forms,

2 I = \int_{0}^{1} (p (x) + p (1 - x)) d x = \int_{0}^{1} 42 d x = 42

. Hence

I = 21

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option

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