JEE MAIN - Mathematics (2010 - No. 8)
Let $$p(x)$$ be a function defined on $$R$$ such that $$p'(x)=p'(1-x),$$ for all $$x \in \left[ {0,1} \right],p\left( 0 \right) = 1$$ and $$p(1)=41.$$ Then $$\int\limits_0^1 {p\left( x \right)dx} $$ equals :
$$21$$
$$41$$
$$42$$
$$\sqrt {41} $$
Explanation
$$p'\left( x \right) = p'\left( {1 - x} \right)$$
$$ \Rightarrow p\left( x \right) = - p\left( {1 - x} \right) + c$$
at $$x=0$$
$$p\left( 0 \right) = - p\left( 1 \right) + c \Rightarrow 42 = c$$
Now, $$p\left( x \right) = - p\left( {1 - x} \right) + 42$$
$$ \Rightarrow p\left( x \right) + p\left( {1 - x} \right) = 42$$
$$ \Rightarrow I = \int\limits_0^1 {p\left( x \right)dx\,\,\,\,\,\,\,\,\,\,...\left( i \right)} $$
$$ \Rightarrow I = \int\limits_0^1 {p\left( {1 - x} \right)} dx\,\,\,\,\,\,...\left( {ii} \right)$$
on adding $$(i)$$ and $$(ii),$$
$$\,\,\,\,\,\,\,\,\,\,\,$$$$2I = \int\limits_0^1 {\left( {42} \right)dx \Rightarrow I = 21} $$
$$ \Rightarrow p\left( x \right) = - p\left( {1 - x} \right) + c$$
at $$x=0$$
$$p\left( 0 \right) = - p\left( 1 \right) + c \Rightarrow 42 = c$$
Now, $$p\left( x \right) = - p\left( {1 - x} \right) + 42$$
$$ \Rightarrow p\left( x \right) + p\left( {1 - x} \right) = 42$$
$$ \Rightarrow I = \int\limits_0^1 {p\left( x \right)dx\,\,\,\,\,\,\,\,\,\,...\left( i \right)} $$
$$ \Rightarrow I = \int\limits_0^1 {p\left( {1 - x} \right)} dx\,\,\,\,\,\,...\left( {ii} \right)$$
on adding $$(i)$$ and $$(ii),$$
$$\,\,\,\,\,\,\,\,\,\,\,$$$$2I = \int\limits_0^1 {\left( {42} \right)dx \Rightarrow I = 21} $$
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