SATHEE: Application of Derivatives 4 Question 55

Application of Derivatives 4 Question 55

####58. Find the coordinates of the point on the curve $y = \frac{x}{1 + x^{2}}$ , where the tangent to the curve has the greatest slope.

(1984, 4M)

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Answer:

Correct Answer: 58. (2)

Solution:

Given, $y = \frac{x}{1 + x^{2}}$

$\begin{aligned} \Rightarrow \frac{d y}{d x} = \frac{(1 + x^{2}) \cdot 1 - x (2 x)}{{(1 + x^{2})}^{2}} = \frac{1 - x^{2}}{{(1 + x^{2})}^{2}} \\ Let \frac{d y}{d x} = g (x) [i.e. slope of tangent] \\ ∴ g (x) = \frac{1 - x^{2}}{{(1 + x^{2})}^{2}} \\ \Rightarrow g^{'} (x) = \frac{{(1 + x^{2})}^{2} \cdot (- 2 x) - (1 - x^{2}) \cdot 2 (1 + x^{2}) \cdot 2 x}{{(1 + x^{2})}^{4}} \\ = \frac{- 2 x (1 + x^{2}) [(1 + x^{2}) + 2 (1 - x^{2})]}{{(1 + x^{2})}^{4}} = \frac{- 2 x (3 - x^{2})}{{(1 + x^{2})}^{3}} \end{aligned}$

For greatest or least values of $m$ , we should have

$g^{'} (x) = 0 \Rightarrow x = 0, x = \pm \sqrt{3}$

Now,

$g^{''} (x) = \frac{{(1 + x^{2})}^{3} (6 x^{2} - 6) - (2 x^{3} - 6 x) \cdot 3 {(1 + x^{2})}^{2} \cdot 2 x}{{(1 + x^{2})}^{6}}$

At $x = 0, g^{''} (x) = - 6 < 0$

$∴ g^{'} (x)$ has a maximum value at $x = 0$ .

$\Rightarrow (x = 0, y = 0)$ is the required point at which tangent to the curve has the greatest slope.

Application of Derivatives 4 Question 55

Answer:

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Application of Derivatives 4 Question 55

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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