JEE Mains · Maths · STD 12 - 6. Application of derivatives
The normal to the curve \(y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6\) at the point where the curve intersects the \(y - \)axis passes through the point :
- A \(\left( {\frac{1}{2},\frac{1}{3}} \right)\)
- B \(\left( { - \frac{1}{2}, - \frac{1}{2}} \right)\)
- C \(\left( {\frac{1}{2},\frac{1}{2}} \right)\)
- D \(\left( {\frac{1}{2}, - \frac{1}{3}} \right)\)
Answer & Solution
Correct Answer
(C) \(\left( {\frac{1}{2},\frac{1}{2}} \right)\)
Step-by-step Solution
Detailed explanation
We have \(y = \frac{{x + 6}}{{\left( {x - 2} \right)\left( {y - 3} \right)}}\) At \(y-\) axis, \(x = 0 \Rightarrow y = 1\) On differentiating, we get…
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