JEE Mains · Maths · STD 12 - 1. relation and function
Let \([t]\) be the greatest integer less than or equal to \(t\). Let \(A\) be the set of al prime factors of \(2310\) and \(f: A \rightarrow \mathbb{Z}\) be the function \(f(x)=\left[\log _2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right]\). The number of one-to-one functions from \(A\) to the range of \(f\) is :
- A \(20\)
- B \(120\)
- C \(25\)
- D \(24\)
Answer & Solution
Correct Answer
(B) \(120\)
Step-by-step Solution
Detailed explanation
\(\mathrm{N}=2310 \)\( =231 \times 10 \) \(= 3 \times 11 \times 7 \times 2 \times 5\) \( A=\{2,3,5,7,11\} \) \( f(x)=\left[\log _2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right]\) \( f(2)=\left[\log _2(5)\right]=2 \) \( f(3)=\left[\log _2(14)\right]=3 \)…
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