SATHEE: Application of Derivatives 1 Question 21

Application of Derivatives 1 Question 21

####21. Find all the tangents to the curve $y = \cos (x + y)$ , $- 2 π \leq x \leq 2 π$ , that are parallel to the line $x + 2 y = 0$ .

$(1985, 5 M)$

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Answer:

( $x + 2 y = \frac{π}{2} and x + 2 y = \frac{- 3 π}{2}$ )

Solution:

Given, $y = \cos (x + y)$

$\Rightarrow \frac{d y}{d x} = - \sin (x + y) \cdot 1 + \frac{d y}{d x}$

Since, tangent is parallel to $x + 2 y = 0$ ,

then slope $\frac{d y}{d x} = - \frac{1}{2}$

From Eq. (i), $- \frac{1}{2} = - \sin (x + y) \cdot 1 - \frac{1}{2}$

$\Rightarrow \sin (x + y) = 1$ , which shows $\cos (x + y) = 0$ .

$∴ y = 0$

$\begin{aligned} \Rightarrow & x + y & = \frac{π}{2} or & - \frac{3 π}{2} \\ ∴ & x & = \frac{π}{2} or & - \frac{3 π}{2} \end{aligned}$

Thus, required points are $\frac{π}{2}, 0$ and $- \frac{3 π}{2}, 0$

$∴$ Equation of tangents are

$\frac{y - 0}{x - π / 2} = - \frac{1}{2}$

and

$\frac{y - 0}{x + 3 π / 2} = - \frac{1}{2} \Rightarrow 2 y = - x + \frac{π}{2}$

and

$2 y = - x - \frac{3 π}{2}$

$\Rightarrow x + 2 y = \frac{π}{2}$

and

$x + 2 y = - \frac{3 π}{2}$

are the required equations of tangents.

Application of Derivatives 1 Question 21

Answer:

Solution:

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Application of Derivatives 1 Question 21

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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