All Courses
Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity Get the App Login Join for Free
Sign in Open App
JEE Exam  >  JEE Questions  >   A continuous, even periodic function f with ... Start Learning for Free
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?
Clear all your JEE doubts with EduRev
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))
Free Test
FREE
Start Free Test
Start Free Test
Community Answer
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.
Explore Courses for JEE exam

Similar JEE Doubts

Not differentiable then S 2019 won is an empty set to equals minus 2 minus 1 0 1 2 3 equals minus 2 minus 1 1 2 4 equals minus 2 2 ANS 2 solution FX is not differentiable at negative 2 minus 1 and 2 s equals minus 2 minus 1 0 1 2 qu 21 let F be a differentiable function such than FXX greater than 0 and F14 then 2019 1 exists and equals 4/7 2 exists and equals 4 3 does not exist 4 exists and equals 0 ANS 2 solutions reply equals FX solution of differential equation 22 let's denote the greatest integer less than or equal to 10 spend 2019 equals pi plus 1 11 equals 0 and 1 solution just since it does not exist You 23 2019-102 like four for one ANS 4 solution huge 24 let F equals RB differentiable at CR and F C equals 0 If GX equals FX then at X equals C G is 20191 not differentiable if FC equals 0 to differentiable if FC03 differentiable if FC equals 04 not differentiable ANS 3 solution Q25 is continuous at x equals 0 than the ordered pair PQ is equal to 2019 ANS 3 solution Q26 LED FRRB a continuously differentiable function such that F2 equals 6 and F2 equals 148 2019 1 18/24 3/12/436 ANS 1 solution Q27 2019/1424 square root 2 3/8 square root 248 A for solution Q28 2019 ANS 2 solution huge 29 if the function has to find on is continuous then K is equal to 2019 1 2 2 1 have 3 1 4 1 / square Benz FX is continuous then Q30 Let FX equals 15 minus X 10 X are then the set of all values of X at This is not differentiable is 2019-15 to 10 15 3 5 10 15 24 10 ANS 1 solution Since FX equals 15 minus 10X GX equals FFX equals 15 minus 10 minus 15 minus 10X equals 15 10X 5 then the points were function GX is non-differential or 10 X equals 0 and 10 X equals 5X equals 10 and X 10 equals plus minus 5 X equals 10 and X equals where it's denotes the greatest integer function then 2019 one F is continuous at X equals 4 ANI One solution Q32 if the function is continuous at x equals 5 then the value of A B is 2019 ANS 4 solution function is continuous at x equals 5 LHL equals RHL5 minus pi B plus 3 equals 5 minus pi A plus 1 Q3 and 1914 square root 22 square root 232 square root 244 ANS 1 solution Q34 let FRRB a differential function satisfying F3 plus F2 equals 0 then is equal to 2019 1 1 2 E 1 3 4 E2ANS 1 solution Q35 if FRR is a differentiable function and F2 equals 6 then 2019/124F222F230412F2 ANS 4 solution Using L hospital rule and Live NetSerum we get putting X = 2 2 F2 F2 = 12 F2 F2 = 6 Q36 2019/18/33 seconds 4/3 AMS One solution GX = FX + FX Then in the Interval - 2/2 G is one differentiable at all points two not continuous three not differentiable at two points four not differentiable at one point ANS 4 solution GX is non-differentiable at x equals 1 GX is not differentiable at one point Q38 let KB the set of all real values of x were the function fx = sin x minus x plus 2x minus pi cos x is not differentiable Then the set k is equal to 2019 empty set 2 pi 3 0 4 0 pi ANS 1 solution Then function f x is differentiable for l Related: JEE Main Previous year questions (2016-20): Limits, Continuity and Differentiability?

Top Courses for JEEView all

Top Courses for JEE

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?
Question Description
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. 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 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.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
SaveJoin the Discussion on the App for FREE
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Get App
© 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