AP EAMCET · Maths · Functions
Let \(\mathrm{A} \subseteq \mathbb{R}, \mathrm{B} \subseteq \mathbb{R}\) and \(\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}\) be defined by \(\mathrm{f}(\mathrm{x})=\mathrm{x}^2\) \(-3 \mathrm{x}+2\). If \(\mathrm{f}\) is a bijection, then
- A \(A=(-\infty, 0], B=\left(-\infty, \frac{-1}{4}\right]\)
- B \(\mathrm{A}=\left(-\infty, \frac{3}{2}\right], \mathrm{B}=\left[\frac{-1}{4}, \infty\right)\)
- C \(\mathrm{A}=\left[\frac{3}{2}, \infty\right), \mathrm{B}=\left(-\infty, \frac{-1}{4}\right]\)
- D \(A=(-\infty, \infty), B=\left[\frac{-1}{4}, \infty\right)\)
Answer & Solution
Correct Answer
(B) \(\mathrm{A}=\left(-\infty, \frac{3}{2}\right], \mathrm{B}=\left[\frac{-1}{4}, \infty\right)\)
Step-by-step Solution
Detailed explanation
The function \(f\) defined \(f: A \rightarrow B\) by \(f(x)=x^2-3 x+2\) Now \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-3\) For stationary point, \(\mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \mathrm{x}=\frac{3}{2}\)…
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