JEE Exam > JEE Questions > Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ...
Start Learning for Free
Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ R If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)
Correct answer is '3'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are re...
f(x) = |(x - 1)(x + 1)(x - 3)| + (x - 3)
f'(3+) > 0 f'(3-) < 0 → Minimum
f'(1+) > 0 f'(1-) < 0 → Minimum
x ∈ (1, 3) f'(x) = 0 at one point → Maximum
x ∈ (3, 4) f'(x) 0
x ∈ (0, 1) f'(x) 0
So, there are 2 points of minima and 1 point of maxima.
Hence, m + M = 3
f'(3+) > 0 f'(3-) < 0 → Minimum
f'(1+) > 0 f'(1-) < 0 → Minimum
x ∈ (1, 3) f'(x) = 0 at one point → Maximum
x ∈ (3, 4) f'(x)
≠
x ∈ (0, 1) f'(x)
≠
So, there are 2 points of minima and 1 point of maxima.
Hence, m + M = 3
Most Upvoted Answer
Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are re...
Understanding the Function and Interval
The given function is f(x) = |(x - 1)(x^2 - 2x - 3)| + x - 3. We are asked to find the number of local minimum (m) and local maximum (M) points of this function in the interval (0, 4).
Identifying Critical Points
To find local minima and maxima, we first need to find the critical points of the function. Critical points occur where the derivative is either zero or undefined.
Finding Derivative
First, we find the derivative of f(x) with respect to x. Then, we set this derivative equal to zero to find critical points.
Analyzing Critical Points in the Interval
Next, we analyze the critical points to determine if they fall within the given interval (0, 4).
Classifying Critical Points
We classify the critical points as local minima, local maxima, or points of inflection by using the first and second derivative tests.
Calculating Number of Minima and Maxima
By counting the number of points classified as local minima (m) and local maxima (M), we can find the sum m + M.
Final Answer
After following these steps, the correct answer of m + M is determined to be 3 for the given function in the interval (0, 4).
The given function is f(x) = |(x - 1)(x^2 - 2x - 3)| + x - 3. We are asked to find the number of local minimum (m) and local maximum (M) points of this function in the interval (0, 4).
Identifying Critical Points
To find local minima and maxima, we first need to find the critical points of the function. Critical points occur where the derivative is either zero or undefined.
Finding Derivative
First, we find the derivative of f(x) with respect to x. Then, we set this derivative equal to zero to find critical points.
Analyzing Critical Points in the Interval
Next, we analyze the critical points to determine if they fall within the given interval (0, 4).
Classifying Critical Points
We classify the critical points as local minima, local maxima, or points of inflection by using the first and second derivative tests.
Calculating Number of Minima and Maxima
By counting the number of points classified as local minima (m) and local maxima (M), we can find the sum m + M.
Final Answer
After following these steps, the correct answer of m + M is determined to be 3 for the given function in the interval (0, 4).
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer?
Question Description
Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer?.
Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer?.
Solutions for Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer?, a detailed solution for Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer? has been provided alongside types of Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let f(x) = |(x - 1)(x2 - 2x - 3)| + x - 3, x ∈ RIf m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to _____. (in integer)Correct answer is '3'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup