1.

Let p(x) be a function defined on R such that p′(x) = p′(1 – x), for all x ∈ [0, 1], p(0) = 1 and(a) √41(b) 21(c) 41(d) 42

Answer»

(b) : p′(x) = p′(1 – x)
On integration,
p(x) = –p(1 – x) + k,
k being the constant of integration.
Set x = 0 to obtain p(0) = –p(1) + k
⇒ 1 = – 41 + k. ∴ k = 42


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