Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
Let fx=1cosx2πx2tan2xln1+4π24πx+x2:x2π               λ:x=2π is continuous at x=2π, then the value of λ is equal to
MathematicsContinuity and DifferentiabilityJEE Main
Solution:
1256 Upvotes Verified Answer
The correct answer is: 0.5
For fx to be continuous at x=2π,
limx2πfx=f2π
Now, limx2π1-cosxtan2x2π-x2ln1+2π-x2=λ
Putting x=2π+t, we get,
limt01costt2tan2tln1+t2=λ
limt012tan2tt2t2ln1+t2=λ
So, 12×1×1=λλ=12

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.