# 2sat approximation related problem.

Let ${ L }_{ \varepsilon }$ to be the language of all 2CNF formulas $\varphi$, such that at least $(\frac { 1 }{ 2 } + \varepsilon )$ of $\varphi$ 's clauses can be satisfied. Prove that there exists ${ \varepsilon }' > 0$ s.t. ${ L }_{ \varepsilon }$ is NP-hard for any $\varepsilon <\varepsilon '$.

I know that $gap-2SAT\left[ \frac { 3 }{ 4 } ,1 \right]$ is in P thus $\frac { 3 }{ 4 }$ -approximation of max2SAT is in P, so if i chose very small constant $\varepsilon$ such that $\frac { 1 }{ 2 } <\frac { 1 }{ 2 } +\varepsilon <\frac { 3 }{ 4 }$ then the statement above is wrong(there is no such ${ \varepsilon }'$).

Any help would be appreciated.

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