AP EAMCET · Maths · Application of Derivatives
If \(\left(a^2-1\right) x+a y+(3-a)=0\) is a normal to the curve \(x y=1\), then the interval in which ' \(a\) ' lies is
- A \([-1,1] \cup[2, \infty)\)
- B \((-\infty,-1] \cup(0,1]\)
- C \([-1,1) \cup(1, \infty)\)
- D \((1, \infty)\)
Answer & Solution
Correct Answer
(B) \((-\infty,-1] \cup(0,1]\)
Step-by-step Solution
Detailed explanation
\(x y=1 \Rightarrow \frac{d y}{d x}=-\frac{1}{x^2}\) Slope of normal of \(x y=1\) is \(\Rightarrow m=x^2 \geq 0...(i)\) \(\because\) Equation of normal is…
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