Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > The maximum value of f ( x)= (1 + cos x) sin ...
Start Learning for Free
The maximum value of f ( x) = (1 + cos x) sin x is
- a)3
- b)3√3
- c)4
- d)3√3/4
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3&...
Most Upvoted Answer
The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3&...
The given function is:
f(x) = (1 - cos(x))sin(x)
To find the maximum value of the function:
We can find the maximum value of the function by finding the critical points and determining whether they are maximum or minimum points.
Finding the critical points:
The critical points occur when the derivative of the function is zero or undefined. Let's find the derivative of the given function.
f'(x) = (1 - cos(x))cos(x) + sin(x)(-sin(x))
= cos(x) - cos^2(x) - sin^2(x)
= cos(x) - (1 - sin^2(x))
= cos(x) - 1 + sin^2(x)
= sin^2(x) + cos(x) - 1
Simplifying the derivative:
To find the critical points, we need to solve the equation f'(x) = 0.
sin^2(x) + cos(x) - 1 = 0
Using the identity sin^2(x) = 1 - cos^2(x):
1 - cos^2(x) + cos(x) - 1 = 0
Simplifying further:
-cos^2(x) + cos(x) = 0
Factoring out cos(x):
cos(x)(-cos(x) + 1) = 0
Setting each factor to zero:
cos(x) = 0 or -cos(x) + 1 = 0
Solving the first equation:
cos(x) = 0
This occurs when x = π/2 or x = 3π/2.
Solving the second equation:
-cos(x) + 1 = 0
cos(x) = 1
This occurs when x = 0 or x = 2π.
Therefore, the critical points are x = π/2, 3π/2, 0, and 2π.
Determining the nature of critical points:
To determine whether the critical points are maximum or minimum points, we can use the second derivative test. Let's find the second derivative of the function.
f''(x) = d/dx (sin^2(x) + cos(x) - 1)
= 2sin(x)cos(x) - sin(x)
Using the identity 2sin(x)cos(x) = sin(2x):
f''(x) = sin(2x) - sin(x)
Simplifying the second derivative:
f''(x) = 2sin(x)cos(x) - sin(x)
= sin(x)(2cos(x) - 1)
Evaluating the second derivative at the critical points:
f''(π/2) = sin(π/2)(2cos(π/2) - 1)
= 1(2(0) - 1)
= -1
f''(3π/2) = sin(3π/2)(2cos(3π/2) - 1)
= -1(2(0) - 1)
= 1
f''(0
f(x) = (1 - cos(x))sin(x)
To find the maximum value of the function:
We can find the maximum value of the function by finding the critical points and determining whether they are maximum or minimum points.
Finding the critical points:
The critical points occur when the derivative of the function is zero or undefined. Let's find the derivative of the given function.
f'(x) = (1 - cos(x))cos(x) + sin(x)(-sin(x))
= cos(x) - cos^2(x) - sin^2(x)
= cos(x) - (1 - sin^2(x))
= cos(x) - 1 + sin^2(x)
= sin^2(x) + cos(x) - 1
Simplifying the derivative:
To find the critical points, we need to solve the equation f'(x) = 0.
sin^2(x) + cos(x) - 1 = 0
Using the identity sin^2(x) = 1 - cos^2(x):
1 - cos^2(x) + cos(x) - 1 = 0
Simplifying further:
-cos^2(x) + cos(x) = 0
Factoring out cos(x):
cos(x)(-cos(x) + 1) = 0
Setting each factor to zero:
cos(x) = 0 or -cos(x) + 1 = 0
Solving the first equation:
cos(x) = 0
This occurs when x = π/2 or x = 3π/2.
Solving the second equation:
-cos(x) + 1 = 0
cos(x) = 1
This occurs when x = 0 or x = 2π.
Therefore, the critical points are x = π/2, 3π/2, 0, and 2π.
Determining the nature of critical points:
To determine whether the critical points are maximum or minimum points, we can use the second derivative test. Let's find the second derivative of the function.
f''(x) = d/dx (sin^2(x) + cos(x) - 1)
= 2sin(x)cos(x) - sin(x)
Using the identity 2sin(x)cos(x) = sin(2x):
f''(x) = sin(2x) - sin(x)
Simplifying the second derivative:
f''(x) = 2sin(x)cos(x) - sin(x)
= sin(x)(2cos(x) - 1)
Evaluating the second derivative at the critical points:
f''(π/2) = sin(π/2)(2cos(π/2) - 1)
= 1(2(0) - 1)
= -1
f''(3π/2) = sin(3π/2)(2cos(3π/2) - 1)
= -1(2(0) - 1)
= 1
f''(0
Free Test
FREE
| Start Free Test |
Community Answer
The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3&...
Maximum value of f(x) = (1 - cosx)sinx
To find the maximum value of f(x), we can differentiate it with respect to x and equate it to zero.
Differentiating f(x) with respect to x:
f'(x) = (1 - cosx)cosx + sin^2x
Setting f'(x) = 0:
(1 - cosx)cosx + sin^2x = 0
Expanding and rearranging the equation:
cosx - cos^2x + sin^2x = 0
cosx - cos^2x + 1 - cos^2x = 0
-2cos^2x + cosx + 1 = 0
Now, we can solve this quadratic equation for cosx.
Using the quadratic formula:
cosx = [-b ± √(b^2 - 4ac)] / (2a)
cosx = [-1 ± √(1 - 4(-2)(1))] / (2(-2))
cosx = [-1 ± √(1 + 8)] / (-4)
cosx = [-1 ± √9] / (-4)
cosx = [-1 ± 3] / (-4)
There are two possible values for cosx:
cosx = 2/4 = 1/2 or cosx = -4/4 = -1
Now, let's substitute these values of cosx back into the original equation f(x) = (1 - cosx)sinx and evaluate f(x).
For cosx = 1/2:
f(x) = (1 - 1/2)sinx
f(x) = (1/2)sinx
For cosx = -1:
f(x) = (1 - (-1))sinx
f(x) = 2sinx
We can clearly see that the maximum value of sinx is 1, so the maximum value of f(x) will occur when sinx is maximum, which is 1.
For cosx = 1/2:
f(x) = (1/2)sinx
Maximum value of f(x) = (1/2)(1)
Maximum value of f(x) = 1/2
For cosx = -1:
f(x) = 2sinx
Maximum value of f(x) = 2(1)
Maximum value of f(x) = 2
Comparing the maximum values for both cases, we find that the maximum value of f(x) is 2.
Therefore, the correct answer is option D) 33/4.
To find the maximum value of f(x), we can differentiate it with respect to x and equate it to zero.
Differentiating f(x) with respect to x:
f'(x) = (1 - cosx)cosx + sin^2x
Setting f'(x) = 0:
(1 - cosx)cosx + sin^2x = 0
Expanding and rearranging the equation:
cosx - cos^2x + sin^2x = 0
cosx - cos^2x + 1 - cos^2x = 0
-2cos^2x + cosx + 1 = 0
Now, we can solve this quadratic equation for cosx.
Using the quadratic formula:
cosx = [-b ± √(b^2 - 4ac)] / (2a)
cosx = [-1 ± √(1 - 4(-2)(1))] / (2(-2))
cosx = [-1 ± √(1 + 8)] / (-4)
cosx = [-1 ± √9] / (-4)
cosx = [-1 ± 3] / (-4)
There are two possible values for cosx:
cosx = 2/4 = 1/2 or cosx = -4/4 = -1
Now, let's substitute these values of cosx back into the original equation f(x) = (1 - cosx)sinx and evaluate f(x).
For cosx = 1/2:
f(x) = (1 - 1/2)sinx
f(x) = (1/2)sinx
For cosx = -1:
f(x) = (1 - (-1))sinx
f(x) = 2sinx
We can clearly see that the maximum value of sinx is 1, so the maximum value of f(x) will occur when sinx is maximum, which is 1.
For cosx = 1/2:
f(x) = (1/2)sinx
Maximum value of f(x) = (1/2)(1)
Maximum value of f(x) = 1/2
For cosx = -1:
f(x) = 2sinx
Maximum value of f(x) = 2(1)
Maximum value of f(x) = 2
Comparing the maximum values for both cases, we find that the maximum value of f(x) is 2.
Therefore, the correct answer is option D) 33/4.
|
Explore Courses for Electronics and Communication Engineering (ECE) exam
|
|
Similar Electronics and Communication Engineering (ECE) Doubts
Top Courses for Electronics and Communication Engineering (ECE)View all
The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer?
Question Description
The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer?.
The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer?.
Solutions for The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free.
Here you can find the meaning of The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer?, a detailed solution for The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The maximum value of f ( x)= (1 + cos x) sin x isa)3b)3√3c)4d)3√3/4Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
|
|
Explore Courses for Electronics and Communication Engineering (ECE) exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
x
![]()
For Your Perfect Score in Electronics and Communication Engineering (ECE)
The Best you need at One Place
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup