y=3x\(^2\) + 5x - 3

dy/dx = 6x + 5

as dy/dx = 0

6x + 5 = 0

x = \(\frac{-5}{6}\)

Maximum value: 3 \( ^2{\frac{-5}{6}}\)  + 5 \(\frac{-5}{6}\) - 3

3 \(\frac{75}{36}\) - \(\frac{25}{6}\) - 3

Using the L.C.M. 36

= \(\frac{25 - 50 - 36}{36}\)

= \(\frac{-61}{36}\)

No correct option