Search for: If y=f(x)=1u2+u+1 where u=1x−1 , then the function is discontinuous at x =If y=f(x)=1u2+u+1 where u=1x−1 , then the function is discontinuous at x =A1B1/2C2D-2 Register to Get Free Mock Test and Study Material +91 Verify OTP Code (required) I agree to the terms and conditions and privacy policy. Solution:u=1x−1 is not defined at x=1. Hence,y=f(x)=1u2+u+1 is discontinuous at x=1.Post navigationPrevious: If ey+xy=e, then the value of d2ydx2 for x=0, is ke2 then k=Next: If f(x)=x−4|x−4|+a,x<4a+b, x=4x−4|x−4|+b,x>4 Then, f(x) is continuous at x=4 when a2+b2=Related content JEE Advanced 2023 NEET Rank Assurance Program | NEET Crash Course 2023 JEE Main 2023 Question Papers with Solutions JEE Main 2024 Syllabus Best Books for JEE Main 2024 JEE Advanced 2024: Exam date, Syllabus, Eligibility Criteria JEE Main 2024: Exam dates, Syllabus, Eligibility Criteria JEE 2024: Exam Date, Syllabus, Eligibility Criteria NCERT Solutions For Class 6 Maths Data Handling Exercise 9.3 JEE Crash Course – JEE Crash Course 2023
