a
(a) \({1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x\)

\( \Rightarrow \) \({\log _\pi }3 + {\log _\pi }4 > x\) \( \Rightarrow \) \({\log _\pi }12 > x\)

\({\pi ^2} < 12 < {\pi ^3}\)

\(\therefore 12 > {\pi ^2}\);

\(\therefore {\log _\pi }12 > {\log _\pi }{\pi ^2}\)

i.e., \({\log _\pi }12 > 2\);

\(\therefore x\) will be \(2\).