Question

If \[ f(x)=x^{3}-10x^{2}+31x-30 \] is a real valued function, then the number of values of \(c\), as stated in Rolle's theorem, that lie in the interval \((2,5)\) is

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Hint:

If \[ f(a)=f(b), \] and \(f\) is continuous on \[ [a,b] \] and differentiable on \[ (a,b), \] then Rolle's theorem guarantees at least one point \[ c\in(a,b) \] such that \[ \boxed{f'(c)=0.} \]

Answer 1

The Correct Option is C

Step 1: Verify the conditions of Rolle's theorem. Factorizing, \[ f(x) = x^{3}-10x^{2}+31x-30 = (x-2)(x-3)(x-5). \] Thus, \[ f(2)=0, \qquad f(5)=0. \] Since \(f(x)\) is a polynomial, it is continuous on \[ [2,5] \] and differentiable on \[ (2,5). \] Hence, Rolle's theorem is applicable.

Step 2: Find the values of \(c\). Differentiate: \[ f'(x) = 3x^{2}-20x+31. \] Setting \[ f'(x)=0, \] we obtain \[ 3x^{2}-20x+31=0. \] Using the quadratic formula, \[ x = \frac{20\pm\sqrt{400-372}}6 = \frac{20\pm\sqrt{28}}6 = \frac{10\pm\sqrt7}{3}. \]

Step 3: Check whether the values lie in \((2,5)\). Now, \[ \frac{10-\sqrt7}{3}\approx2.45, \] and \[ \frac{10+\sqrt7}{3}\approx4.22. \] Both belong to \[ (2,5). \] Hence, there are \[ \boxed{2} \] values of \(c\). Therefore, the correct option is \(\boxed{(C)}\).