Question:

The equation of the line parallel to x-axis and tangent to the curve $ y=\frac{1}{{{x}^{2}}+2x+5} $ is

Updated On: Jun 8, 2024
  • $ y=\frac{1}{4} $
  • $ y=4 $
  • $ y=\frac{1}{2} $
  • $ y=0 $
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The Correct Option is A

Solution and Explanation

Curve, $ y=\frac{1}{{{x}^{2}}+2x+5} $ ..(i) Let the equation of line which is parallel to $ x- $ axis is, $ y=c $ ...(ii) The line (ii) is a tangent to curve (i), then slope of curve = slope of line $ \frac{-(2x+2)}{{{({{x}^{2}}+2x+5)}^{2}}}=0 $ $ \left( \because \frac{dy}{dx}=\frac{-(2x+2)}{{{({{x}^{2}}+2x+5)}^{2}}} \right) $ $ \Rightarrow $ $ x=-1 $ From E (i), $ y=\frac{1}{1-2+5}=\frac{1}{4} $ From E (ii), $ c=\frac{1}{4} $ Hence, the required equation of line is, $ y=\frac{1}{4}. $
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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives