UPSC NDA I 2019 Mathematics Solved Paper

Concept:The function is exponential: $f(x) = 3^{1+x}$. Multiplying such functions adds exponents because the base is the same.Explanation:Step 1: Write each function value using the given definition:$f(x) = 3^{1+x}$, $f(y) = 3^{1+y}$, $f(z) = 3^{1+z}$.Step 2: Multiply these three expressions. Since they share base $3$, add the exponents:$f(x) f(y) f(z) = 3^{1+x} \times 3^{1+y} \times 3^{1+z} = 3^{(1+x)+(1+y)+(1+z)} = 3^{3 + x + y + z}$.Step 3: Rewrite $3^{3+x+y+z}$ in the form $3^{1+(x+y+z+2)}$ because $3 = 1+2$.Thus, $3^{3+x+y+z} = 3^{1 + (x+y+z+2)} = f(x+y+z+2)$.Step 4: Compare with the options. The expression equals $f(x+y+z+2)$, which is option C.Answer:C. $f(x+y+z+2)$