Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

The value of c in Rolle's theorem for the function $f(x)=x^3-3x$ in the interval $[0,\sqrt 3]$ is

\[\begin{array}{1 1}(A)\;1 & (B)\;-1\\(C)\;\frac{3}{2} & (D)\;\frac{1}{3}\end{array}\]

Can you answer this question?

1 Answer

0 votes
  • Rolle's Theorem :Let $f$ be a real valued function defined once close interval $[a,b]$,such that
  • $\quad(i)$ it is continuous on the closed interval $[a,b]$.
  • $\quad(ii)$ it is differentiable in the open interval $(a,b)$.
  • $\quad(iii)$ $f(a)=f(b)$,then there exists a real numbers $c\in (a,b)$ such that $f'(c)=0$
Step 1:
$f(x)=x^3-3x\quad[0,\sqrt 3]$
Since $c$ lies between the given interval $[0,\sqrt 3]$ for $f'(c)=0$
Step 2:
$\Rightarrow c=\pm 1$
But only $+1$ lies in the interval $[0,\sqrt 3]$
Hence the correct option is $A$
answered Jul 4, 2013 by sreemathi.v
edited Jul 5, 2013 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App