SATHEE: Application of Derivatives 1 Question 16

Application of Derivatives 1 Question 16

####16. Let $C$ be the curve $y^{3} - 3 x y + 2 = 0$ . If $H$ is the set of points on the curve $C$ , where the tangent is horizontal and $V$ is the set of points on the curve $C$ , where the tangent is vertical, then $H = \dots$ and $V = \dots$ .

(1994, 2M)

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Answer:

( $H = θ, V = 1, 1$ )

Solution:

Given, $y^{3} - 3 x y + 2 = 0$

On differentiating w.r.t. $x$ , we get

$\begin{aligned} 3 y^{2} \frac{d y}{d x} - 3 x \frac{d y}{d x} - 3 y & = 0 \\ \Rightarrow \frac{d y}{d x} (3 y^{2} - 3 x) & = 3 y \Rightarrow \frac{d y}{d x} = \frac{3 y}{3 y^{2} - 3 x} \end{aligned}$

For the points where tangent is horizontal, the slope of tangent is zero.

i.e.

$\frac{d y}{d x} = 0 \Rightarrow \frac{3 y}{3 y^{2} - 3 x} = 0$

$\Rightarrow y = 0$ but $y = 0$ does not satisfy the given equation of the curve, therefore $y$ cannot lie on the curve.

So, $H = φ$ [null set]

For the points where tangent is vertical, $\frac{d y}{d x} = \infty$

$\begin{aligned} \Rightarrow & \frac{y}{y^{2} - x} & = \infty \\ \Rightarrow & y^{2} - x & = 0 \\ \Rightarrow & y^{2} & = x \end{aligned}$

On putting this value in the given equation of the curve, we get

$\begin{aligned} y^{3} - 3 \cdot y^{2} \cdot y + 2 = 0 \\ \Rightarrow - 2 y^{3} + 2 = 0 \\ \Rightarrow y^{3} - 1 = 0 \Rightarrow y^{3} = 1 \\ \Rightarrow y = 1, x = 1 \\ V = 1, 1 \end{aligned}$

Application of Derivatives 1 Question 16

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Application of Derivatives 1 Question 16

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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