Question 118

If the operation Θ is defined for all real numbers a and b by the relation $$aΘ b =a^{2}\frac{b}{{3}}$$ then $$2Θ {3Θ(­-1)} = ?$$

Solution

It is given that $$aΘ b =a^{2}\frac{b}{{3}}$$

Applying the same rule for $$2Θ {3Θ(-1)}$$

= 2 Θ $${\frac{3^2 \times (-1)}{3}}$$

= 2 Θ -3

= $$\frac{2^2 \times (-3)}{3} = -4$$


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