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For n ≥ 2, let fn: R → R be given by fn(x) = xn sin x. Then at x = 0, fn has a
- a)local maximum if n is even
- b)local maximum if n is odd
- c)local minimum if n is even
- d)local minimum if n is odd
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
For n≥ 2, let fn: R → R be given by fn(x) = xn sin x. Then...
Explanation:
To determine whether the function f_n(x) = x^n sin(x) has a local maximum or minimum at x = 0, we need to examine the behavior of the function near x = 0.
First Derivative Test:
Using the first derivative test, we can determine whether f_n(x) has a local maximum or minimum at x = 0 by examining the sign of f'_n(x) near x = 0.
f'_n(x) = n x^(n-1) sin(x) + x^n cos(x)
Odd and Even:
Now, we need to consider whether n is even or odd.
If n is even, then n-1 is odd, and x^(n-1) is positive near x = 0.
If n is odd, then n-1 is even, and x^(n-1) is negative near x = 0.
Positive and Negative:
Next, we need to consider the sign of sin(x) and cos(x) near x = 0.
sin(x) is positive for x > 0 and negative for x < 0.="" />
cos(x) is positive for x > 0 and negative for x < 0.="" />
Conclusion:
Therefore, if n is even, then f'_n(x) is positive for x > 0 and negative for x < 0="" near="" x="0." this="" means="" that="" f_n(x)="" has="" a="" local="" minimum="" at="" x="0." />
If n is odd, then f'_n(x) is negative for x > 0 and positive for x < 0="" near="" x="0." this="" means="" that="" f_n(x)="" has="" a="" local="" maximum="" at="" x="0." />
So, in this case, the correct answer is option (D) local minimum if n is odd.
To determine whether the function f_n(x) = x^n sin(x) has a local maximum or minimum at x = 0, we need to examine the behavior of the function near x = 0.
First Derivative Test:
Using the first derivative test, we can determine whether f_n(x) has a local maximum or minimum at x = 0 by examining the sign of f'_n(x) near x = 0.
f'_n(x) = n x^(n-1) sin(x) + x^n cos(x)
Odd and Even:
Now, we need to consider whether n is even or odd.
If n is even, then n-1 is odd, and x^(n-1) is positive near x = 0.
If n is odd, then n-1 is even, and x^(n-1) is negative near x = 0.
Positive and Negative:
Next, we need to consider the sign of sin(x) and cos(x) near x = 0.
sin(x) is positive for x > 0 and negative for x < 0.="" />
cos(x) is positive for x > 0 and negative for x < 0.="" />
Conclusion:
Therefore, if n is even, then f'_n(x) is positive for x > 0 and negative for x < 0="" near="" x="0." this="" means="" that="" f_n(x)="" has="" a="" local="" minimum="" at="" x="0." />
If n is odd, then f'_n(x) is negative for x > 0 and positive for x < 0="" near="" x="0." this="" means="" that="" f_n(x)="" has="" a="" local="" maximum="" at="" x="0." />
So, in this case, the correct answer is option (D) local minimum if n is odd.
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For n≥ 2, let fn: R → R be given by fn(x) = xn sin x. Then at x = 0, fn has aa)local maximum if n is evenb)local maximum if n is oddc)local minimum if n is evend)local minimum if n is oddCorrect answer is option 'D'. Can you explain this answer?
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For n≥ 2, let fn: R → R be given by fn(x) = xn sin x. Then at x = 0, fn has aa)local maximum if n is evenb)local maximum if n is oddc)local minimum if n is evend)local minimum if n is oddCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about For n≥ 2, let fn: R → R be given by fn(x) = xn sin x. Then at x = 0, fn has aa)local maximum if n is evenb)local maximum if n is oddc)local minimum if n is evend)local minimum if n is oddCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For n≥ 2, let fn: R → R be given by fn(x) = xn sin x. Then at x = 0, fn has aa)local maximum if n is evenb)local maximum if n is oddc)local minimum if n is evend)local minimum if n is oddCorrect answer is option 'D'. Can you explain this answer?.
For n≥ 2, let fn: R → R be given by fn(x) = xn sin x. Then at x = 0, fn has aa)local maximum if n is evenb)local maximum if n is oddc)local minimum if n is evend)local minimum if n is oddCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about For n≥ 2, let fn: R → R be given by fn(x) = xn sin x. Then at x = 0, fn has aa)local maximum if n is evenb)local maximum if n is oddc)local minimum if n is evend)local minimum if n is oddCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For n≥ 2, let fn: R → R be given by fn(x) = xn sin x. Then at x = 0, fn has aa)local maximum if n is evenb)local maximum if n is oddc)local minimum if n is evend)local minimum if n is oddCorrect answer is option 'D'. Can you explain this answer?.
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