Question

If a function \(f(x)\) is continuous in the closed interval \([a, b]\) and the first derivative \(f'(x)\) exists in the open interval \((a, b)\), then according to the Lagrange's Mean Value Theorem: \[ \frac{f(b) - f(a)}{b - a} = f'(c) \] If \(a = 0, b = 1.5\), and \(f(x) = x(x - 1)(x - 2)\), then the value(s) of \(c\) in \([a, b]\) is/are:

• A. 0.50
• B. 0.75
• C. 1.00
• D. 1.50

Hint:

Always check endpoints when applying LMVT. The valid \(c\) lies strictly within the interval \((a, b)\).

Answer 1

The Correct Option is A

Step 1: Function evaluation.
Given: \[ f(x) = x(x-1)(x-2) = x^3 - 3x^2 + 2x \] 

Step 2: Apply LMVT formula.
\[ \frac{f(b) - f(a)}{b - a} = f'(c) \] Here, \(a=0\), \(b=1.5\). \[ f(0) = 0(0-1)(0-2) = 0 \] \[ f(1.5) = (1.5)(0.5)(-0.5) = -0.375 \] So, \[ \frac{f(1.5) - f(0)}{1.5 - 0} = \frac{-0.375}{1.5} = -0.25 \] Thus, \[ f'(c) = -0.25 \] 

Step 3: Differentiate.
\[ f(x) = x^3 - 3x^2 + 2x \quad \Rightarrow \quad f'(x) = 3x^2 - 6x + 2 \] We require: \[ 3c^2 - 6c + 2 = -0.25 \] \[ 3c^2 - 6c + 2.25 = 0 \] 

Step 4: Solve quadratic.
\[ c^2 - 2c + 0.75 = 0 \] \[ c = \frac{2 \pm \sqrt{4 - 3}}{2} = \frac{2 \pm 1}{2} \] \[ c = \frac{1}{2}, \; \frac{3}{2} \] But since \(c \in (0, 1.5)\), valid values: \[ c = 0.5 \quad \text{and} \quad c = 1.5 \; (\text{endpoint not allowed in LMVT}) \] Hence, only \(c = 0.5\). 

Step 5: Verify again.
Wait — check calculation again. Actually: \[ 3c^2 - 6c + 2 = -0.25 \quad \Rightarrow \quad 3c^2 - 6c + 2.25 = 0 \] Divide by 3: \[ c^2 - 2c + 0.75 = 0 \] Roots: \[ c = \frac{2 \pm \sqrt{4 - 3}}{2} = \frac{2 \pm 1}{2} = 0.5, \, 1.5 \] Since \(c\) must lie in \((0, 1.5)\), the only valid solution is: \[ c = 0.5 \] 

Final Answer: \[ \boxed{0.50} \]