MHT CET · Maths · Functions
If \(D_{30}\) is the set of all divisors of \(30, x, y \in D_{30}\), we define \(x+y=\operatorname{LCM}(x, y), x \cdot y=\operatorname{GCD}(x, y)\),
\(x^{\prime}=\frac{30}{x}\) and \(f(x, y, z)=(x+y) \cdot\left(y^{\prime}+z\right)\), then
\(f(2,5,15)\) is equal to
- A 2
- B 5
- C 10
- D 15
Answer & Solution
Correct Answer
(C) 10
Step-by-step Solution
Detailed explanation
\(D_{30}=\{1,2,3,5,6,10,15,30\}\)
\(\begin{aligned} f(2,5,15) &=(2+5) \cdot\left(5^{\prime}+15\right) \\ &=10 \cdot\left(\frac{30}{5}+15\right) \\ &=10(6+15) \\ &=10 \cdot 30 \\ &=10 \end{aligned}\)
\(\begin{aligned} f(2,5,15) &=(2+5) \cdot\left(5^{\prime}+15\right) \\ &=10 \cdot\left(\frac{30}{5}+15\right) \\ &=10(6+15) \\ &=10 \cdot 30 \\ &=10 \end{aligned}\)
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