JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(M\) and \(N\) be the number of points on the curve \(y^{5}-9 x y+2 x=0\), where the tangents to the curve are parallel to \(x\)-axis and \(y\)-axis, respectively. Then the value of \(M + N\) equals \(......\)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(y ^{5}-9 xy +2 x =0\) \(5 y^{4} \frac{d y}{x}-9 x \frac{d y}{d x}-9 y+2=0\) \(\frac{d y}{d x}\left(5 y^{4}-9 x\right)=9 y-2\) \(\frac{d y}{d x}=\frac{9 y-2}{5 y^{4}-9 x}=0\) (for horizontal tangent) \(y=\frac{2}{9} \Rightarrow\) Which does not satisfy the original equation…
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