1. Identify the reaction type

Under alkaline conditions, benzaldehyde and acetone undergo cross aldol condensation (Claisen–Schmidt condensation).

• Benzaldehyde: no $\alpha$-hydrogen, so it cannot form enolate.
• Acetone: has $\alpha$-hydrogens, so it forms the enolate.

Since 2 moles of benzaldehyde react with 1 mole of acetone, the condensation occurs on both sides of acetone.

2. Write the overall product

First condensation gives benzylidene acetone: $\mathrm{C_6H_5CHO + CH_3COCH_3 \rightarrow C_6H_5CH=CHCOCH_3 + H_2O}$

Second condensation with another mole of benzaldehyde gives: $\mathrm{C_6H_5CH=CHCOCH_3 + C_6H_5CHO \rightarrow C_6H_5CH=CHCOCH=CHC_6H_5 + H_2O}$

Thus the major product $x$ is dibenzalacetone: $\mathrm{C_6H_5CH=CHCOCH=CHC_6H_5}$

3. Count the $\pi$ bonds in dibenzalacetone

Structure: $\mathrm{Ph-CH=CH-CO-CH=CH-Ph}$

Now count all double bonds:

• Left benzene ring: $3$ $\pi$ bonds
• Right benzene ring: $3$ $\pi$ bonds
• Two alkene bonds: $2$ $\pi$ bonds
• One carbonyl bond: $1$ $\pi$ bond

Total: $3+3+2+1=9$

4. Final answer

The number of $\pi$ bonds in the major product is: $\boxed{9}$