Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The constant c of Rolle's theorem for the function <img alt="f(x)=(x-a)^m(x-b)^n in\, \, [a, b]" src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3D%28x-a%29%5Em%28x-b%29%5En%20in%

The constant c of Rolle's theorem for the function f(x)=(x-a)^m(x-b)^n in\, \, [a, b] where m, n are positive integers, is

Option: 1

\frac{a+b}{2}


Option: 2

\frac{m a+n b}{m+n}


Option: 3

\frac{m b+n a}{m+n}


Option: 4

\text { None of these }


Answers (1)

best_answer

 f(x) is continuous on [a, b] and derivable on (a, b).
\mathrm{\begin{aligned} & f^{\prime}(x)=(x-a)^m n(x-b)^{n-1}+m(x-a)^{m-1}(x-b)^n \\ & =(x-a)^{m-1}(x-b)^{n-1}(n x-n a+m x-m b) \end{aligned} }
And f(a)=0=f(b).
By Rolle's theorem, \mathrm{\exists c \in(a, b) } such that  \mathrm{ f^{\prime}(c)=0 }
\mathrm{\begin{aligned} & \Rightarrow \quad(c-a)^{m-1}(c-b)^{n-1}(n c-n a+m c-m b)=0 \\ & \Rightarrow \quad(m+n) c=m b+n a \Rightarrow c=\frac{m b+n a}{m+n} \end{aligned} }

Posted by

Rakesh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks
anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule
anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question