Search for: For f(x)=lnx2+α7x Roll’s theorem is applicable on [3, 4] with c∈(3,4) such that f′(c)=0. The value of f′′(c) isFor f(x)=lnx2+α7x Roll's theorem is applicable on [3, 4] with c∈(3,4) such that f′(c)=0. The value of f′′(c) isA112B−112C16D−16 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:Roll's theorem is applicable to f(x)=lnx2+α7x on [3,4]. ∴ f(3)=f(4)⇒ ln9+α21=ln16+α28⇒ 9+α21=16+α28⇒36+4α=48+3α⇒α=12Now, f(x)=lnx2+α7x⇒ f(x)=lnx2+12−ln7x⇒ f′(x)=2xx2+12−1x=x2−12xx2+12∴ f′(x)=0⇒x2−12=0⇒x=±12∴ c=12Now, f′(x)=x2−12xx2+12xx2+12⇒f′′(x)=2x2x2+12−x2−12x2+12+2x2x2x2+122⇒f′′(x)=−x4+48x2+144x2x2+122⇒f′′(12)=112 Post navigationPrevious: ∫xlog1+1xdx =f(x)⋅logc(x+1)+g(x)logex +Lx+C, then 4L=Next: ∫tan xtan 2xtan 3x dx is equal to Related content 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 NEET Crash Course – NEET Crash Course 2023 JEE Advanced Crash Course – JEE Advanced Crash Course 2023