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A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal to
- a)2π − 3
- b)3 − 2π
- c)2π + 3
- d)3 − π
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A continuous, even periodic function f with period 8 is such that f(0...
f(x) is given to be an even, periodic function with period equal to 8.
⇒ f(x + 8) = f(x)
1. f(−5) = f(3) = 2
2. f(20) = f(12) = f(4) = 3
3. f(−10) = f(−2) = f(2) = 1
4. f(17) = f(9) = f(1) = −2
f(−5) + f(20) + cos−1(f(−10)) + f(17) = 2 + 3 + cos−1(1)−2 = 3
tan−1(tan(3)) = tan−1(tan(3 − π)) = 3 − π
(using tan−1(tanx) = x − π if x ∈ (π/2, π))
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A continuous, even periodic function f with period 8 is such that f(0...
Given Information:
- The function f is continuous and periodic with a period of 8.
- f(0) = 0, f(1) = -2, f(2) = 1, f(3) = 2, f(4) = 3.
Solution:
We are given a periodic function f with a period of 8. This means that the function repeats its values every 8 units. We can use this information to find the values of f for other inputs.
Finding f(-5):
Since the period of f is 8, we can write -5 as -5 + 8 = 3. Therefore, f(-5) = f(3) = 2.
Finding f(20):
Similarly, we can write 20 as 20 - 8 = 12. Therefore, f(20) = f(12) = f(4) = 3.
Finding cos^-1(f(-10)):
We are given that the function f is continuous, which means that f(-10) is also equal to f(6) (since -10 + 8 = 6). Therefore, we need to find cos^-1(f(6)).
Finding f(6):
We can write 6 as 6 - 8 = -2. Therefore, f(6) = f(-2).
Finding cos^-1(f(-2)):
We are given that the function f is continuous, which means that f(-2) is also equal to f(6) (since -2 + 8 = 6). Therefore, we need to find cos^-1(f(6)).
Finding f(6):
We can write 6 as 6 - 8 = -2. Therefore, f(6) = f(-2).
Finding tan^-1(tan{f(-5) f(20) cos^-1(f(-10)) f(17)}):
We can now substitute the values we found into the given expression:
tan^-1(tan{2 3 cos^-1(f(6)) f(17)})
Since f(6) = f(-2), we can substitute f(-2) for f(6):
tan^-1(tan{2 3 cos^-1(f(-2)) f(17)})
Since f(17) = f(1), we can substitute f(1) for f(17):
tan^-1(tan{2 3 cos^-1(f(-2)) (-2)})
Now, using the identity tan^-1(tan(x)) = x for -π/2 < x="" />< π/2,="" we="" can="" simplify="" the="" />
tan^-1(tan{2 3 cos^-1(f(-2)) (-2)}) = 2 3 cos^-1(f(-2)) (-2)
= 3 cos^-1(f(-2)) (-2)
= 3 (-2)
= -6
Therefore, the value of tan^-1(tan{f(-5) f(20) cos^-1(f(-10)) f(17)}) is equal to -6.
The correct answer is option D) 3 - π.
- The function f is continuous and periodic with a period of 8.
- f(0) = 0, f(1) = -2, f(2) = 1, f(3) = 2, f(4) = 3.
Solution:
We are given a periodic function f with a period of 8. This means that the function repeats its values every 8 units. We can use this information to find the values of f for other inputs.
Finding f(-5):
Since the period of f is 8, we can write -5 as -5 + 8 = 3. Therefore, f(-5) = f(3) = 2.
Finding f(20):
Similarly, we can write 20 as 20 - 8 = 12. Therefore, f(20) = f(12) = f(4) = 3.
Finding cos^-1(f(-10)):
We are given that the function f is continuous, which means that f(-10) is also equal to f(6) (since -10 + 8 = 6). Therefore, we need to find cos^-1(f(6)).
Finding f(6):
We can write 6 as 6 - 8 = -2. Therefore, f(6) = f(-2).
Finding cos^-1(f(-2)):
We are given that the function f is continuous, which means that f(-2) is also equal to f(6) (since -2 + 8 = 6). Therefore, we need to find cos^-1(f(6)).
Finding f(6):
We can write 6 as 6 - 8 = -2. Therefore, f(6) = f(-2).
Finding tan^-1(tan{f(-5) f(20) cos^-1(f(-10)) f(17)}):
We can now substitute the values we found into the given expression:
tan^-1(tan{2 3 cos^-1(f(6)) f(17)})
Since f(6) = f(-2), we can substitute f(-2) for f(6):
tan^-1(tan{2 3 cos^-1(f(-2)) f(17)})
Since f(17) = f(1), we can substitute f(1) for f(17):
tan^-1(tan{2 3 cos^-1(f(-2)) (-2)})
Now, using the identity tan^-1(tan(x)) = x for -π/2 < x="" />< π/2,="" we="" can="" simplify="" the="" />
tan^-1(tan{2 3 cos^-1(f(-2)) (-2)}) = 2 3 cos^-1(f(-2)) (-2)
= 3 cos^-1(f(-2)) (-2)
= 3 (-2)
= -6
Therefore, the value of tan^-1(tan{f(-5) f(20) cos^-1(f(-10)) f(17)}) is equal to -6.
The correct answer is option D) 3 - π.
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A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer?
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A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer?.
A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal toa)2π − 3b)3 − 2πc)2π + 3d)3 − πCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
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