The value of parameter a so that the line (3 – a) x + ay + (a 2 – 1) = 0 is normal to the…
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
The value of parameter a so that the line
(3 – a) x + ay + (a2 – 1) = 0 is normal to the curve xy = 1, may lie in the interval
(3 – a) x + ay + (a2 – 1) = 0 is normal to the curve xy = 1, may lie in the interval
- (−∞, 0)
- (1, 3)
- (0, 3)
- (3, ∞)
Answer
(A) (−∞, 0), (D) (3, ∞)
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