Question

\(f:\mathbb{R}\rightarrow\mathbb{R}\) is increasing in \((-\infty,0)\cup(1,\infty)\) and decreasing in \((0,1)\). If \(f(2)=6\) and the continuous curve \(y=f(x)\) passes through \((0,0)\), then

• A. A point \(a\in(0,2)\) such that \(f'(a)=3\)
• B. A point \(a\in(0,2)\) such that \(f'(a)=0\)
• C. \(f'(0)<0\)
• D. \(f'(1)>0\)

Hint:

Whenever values of a function at two points are given, immediately think of the Mean Value Theorem: \[ f'(c)=\frac{f(b)-f(a)}{b-a}. \] This theorem is frequently used in objective-type calculus questions.

Answer 1

The Correct Option is A

Concept: The Mean Value Theorem (MVT) states that if a function is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists a point \(c\in(a,b)\) such that \[ f'(c)=\frac{f(b)-f(a)}{b-a}. \] The given information about increasing and decreasing intervals helps us understand the shape of the curve.

Step 1: Use the given points.
The curve passes through \((0,0)\), therefore \[ f(0)=0. \] Also, \[ f(2)=6. \] Hence \[ \frac{f(2)-f(0)}{2-0} = \frac{6-0}{2} = 3. \]

Step 2: Apply Mean Value Theorem.
Since the function is continuous and differentiable on the interval under consideration, by MVT there exists some \[ a\in(0,2) \] such that \[ f'(a) = \frac{f(2)-f(0)}{2-0} = 3. \] Therefore, \[ f'(a)=3. \]

Step 3: Examine the remaining options.
Since the function is decreasing in \((0,1)\), \[ f'(x)\le 0 \] in that interval. Since the function is increasing in \((1,\infty)\), \[ f'(x)\ge0 \] for \(x>1\). The information given does not guarantee that \[ f'(a)=0 \] for some \(a\in(0,2)\). Similarly, no definite conclusions can be made about \(f'(0)\) or \(f'(1)\).

Step 4: Choose the correct option.
The only statement that must be true is \[ \boxed{\exists\,a\in(0,2)\text{ such that }f'(a)=3}. \] Hence option (A) is correct.