AP EAMCET · Maths · Functions
The set of all real values of \(x\) such that \(f(x)=\sqrt{\frac{[x]-1}{[x]^2-[x]-6}}\) is a real valued function is
- A \([1, \infty)\)
- B \((-\infty,-2) \cup[4, \infty)\)
- C \([-1,3)\)
- D \([-1,2) \cup[4, \infty)\)
Answer & Solution
Correct Answer
(D) \([-1,2) \cup[4, \infty)\)
Step-by-step Solution
Detailed explanation
Let \(y=[x]\). For \(f(x)\) to be real, \(\frac{y-1}{y^2-y-6} \ge 0\). \(\frac{y-1}{(y-3)(y+2)} \ge 0\). Critical points for \(y\): \(-2, 1, 3\). Denominator must be non-zero. Solution for \(y\): \(y \in (-2, 1] \cup (3, \infty)\). Since \(y=[x]\) must be an integer:…
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