If φ(x) is differentiable function ∀x∈R and a∈R+ such that φ(0)=φ(2a),φ(a)=φ(3a)andφ(0)=φ(a) then show that there is at least one root of equation φprime(x+a)=φprime(x)∈(0,2a)

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If $φ (x)$ is differentiable function $\forall x \in R$ and $a \in R^{+}$ such that $φ (0) = φ (2 a), φ (a) = φ (3 a) a n d φ (0) \neq = φ (a)$ then show that there is at least one root of equation $φ^{p r i m e} (x + a) = φ^{p r i m e} (x) \in (0, 2 a)$

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φ (x)

is differentiable function

\forall x \in R

and

a \in R^{+}

such that

φ (0) = φ (2 a), φ (a) = φ (3 a) a n d φ (0) \neq = φ (a)

then show that there is at least one root of equation

φ^{p r i m e} (x + a) = φ^{p r i m e} (x) \in (0, 2 a)

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Question Text

φ (x)

is differentiable function

\forall x \in R

and

a \in R^{+}

such that

φ (0) = φ (2 a), φ (a) = φ (3 a) a n d φ (0) \neq = φ (a)

then show that there is at least one root of equation

φ^{p r i m e} (x + a) = φ^{p r i m e} (x) \in (0, 2 a)