SATHEE: Application of Derivatives 1 Question 1

Application of Derivatives 1 Question 1

####1. If the tangent to the curve $y = \frac{x}{x^{2} - 3}, x \in R, (x \neq \pm \sqrt{3})$ , at a point $(α, β) \neq (0, 0)$ on it is parallel to the line $2 x + 6 y - 11 = 0$ , then

(2019 Main, 10 April II)

(a) $| 6 α + 2 β | = 19$

(b) $| 6 α + 2 β | = 9$

(d) $| 2 α + 6 β | = 11$

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Answer:

Correct Answer: 1. (a)

Solution:

Equation of given curve is

$y = \frac{x}{x^{2} - 3}, x \in R, (x \neq \pm \sqrt{3})$

On differentiating Eq. (i) w.r.t. $x$ , we get

$\frac{d y}{d x} = \frac{(x^{2} - 3) - x (2 x)}{{(x^{2} - 3)}^{2}} = \frac{(- x^{2} - 3)}{{(x^{2} - 3)}^{2}}$

It is given that tangent at a point $(α, β) \neq (0, 0)$ on it is parallel to the line

$2 x + 6 y - 11 = 0 .$

$∴$ Slope of this line $= - \frac{2}{6} = {\frac{d y}{d x} |}_{(α, β)}$

$\Rightarrow - \frac{α^{2} + 3}{{(α^{2} - 3)}^{2}} = - \frac{1}{3}$

$\Rightarrow 3 α^{2} + 9 = α^{4} - 6 α^{2} + 9$

$\Rightarrow α^{4} - 9 α^{2} = 0$

$\Rightarrow α = 0, - 3, 3$

$\Rightarrow α = 3$ or $- 3, [∵ α \neq 0]$

Now, from Eq. (i),

$β = \frac{α}{α^{2} - 3} \Rightarrow β = \frac{3}{9 - 3} or \frac{- 3}{9 - 3} = \frac{1}{2} or - \frac{1}{2}$

According to the options, $| 6 α + 2 β | = 19$

at $(α, β) = \pm 3, \pm \frac{1}{2}$

Application of Derivatives 1 Question 1

Answer:

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Application of Derivatives 1 Question 1

Answer:

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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