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Mathematics · Application of Derivatives

JEE Advanced 2019 — Paper 2 — Question 25

The function sin⁡(x+a)sin⁡(x+b)\dfrac{\sin (x+a)}{\sin (x+b)} has no maxima or minima if

  1. Option A:

    b−a=nπ,n∈I\quad b-a=n \pi, n \in I

    Correct
  2. Option B:

    b−a=(2n+1)π,n∈I\quad b-a=(2 n+1) \pi, n \in I

    Correct
  3. Option C:

    b−a=2nπ,n∈I\quad b-a=2 n \pi, n \in I

    Correct
  4. Option D:

    none of these

Answer: A, B, C

Step-by-step solution

Given

f(x)=sin⁡(x+a)sin⁡(x+b)f(x)=\frac{\sin(x+a)}{\sin(x+b)}

Differentiate

f′(x)=cos⁡(x+a)sin⁡(x+b)−sin⁡(x+a)cos⁡(x+b)sin⁡2(x+b)f'(x)=\frac{\cos(x+a)\sin(x+b)-\sin(x+a)\cos(x+b)}{\sin^2(x+b)}

Using identity

sin⁡Acos⁡B−cos⁡Asin⁡B=sin⁡(A−B)\sin A\cos B-\cos A\sin B=\sin(A-B) cos⁡(x+a)sin⁡(x+b)−sin⁡(x+a)cos⁡(x+b)=sin⁡((x+b)−(x+a))\cos(x+a)\sin(x+b)-\sin(x+a)\cos(x+b) = \sin((x+b)-(x+a))

Thus

f′(x)=sin⁡(b−a)sin⁡2(x+b)f'(x)=\frac{\sin(b-a)}{\sin^2(x+b)}

For no maxima or minima,

f′(x)=0f'(x)=0 sin⁡(b−a)=0\sin(b-a)=0 b−a=nπ,n∈Zb-a=n\pi,\quad n\in\mathbb Z

This includes both

b−a=(2n+1)πb-a=(2n+1)\pi

and

b−a=2nπb-a=2n\pi

Correct options: A, B, C

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2019
Paper
Paper 2
Subject
Mathematics
Chapter
Application of Derivatives
Topic
Maxima and Minima
The function sin (x+a)/sin (x+b) has no maxima or minima if | JEE Advanced 2019 PYQ with Solution · DhiX AI