Let `f (x)= (x-a) (x-b)(x-c)` be a ral vlued function where `a lt b c (a,b,c in R)` such that `f ''(alpha ) =0.` Then if `alpha in ( c _(1), c _(2)),` which one of the following is correct ?

Options

A. `alpha lt c _(1) lt b and b lt c _(2) lt c`
B. `alpha lt c _(1), c_(2) lt b`
C. `b lt c _(1), c _(2) lt c`
D. None of these

Answer

To solve the problem, we need to analyze the function \( f(x) = (x-a)(x-b)(x-c) \) and its derivatives. We are given that \( f''(\alpha) = 0 \) and that \( \alpha \) lies in the interval \( (c_1, c_2) \). Let's go through the steps to find the correct option.

### Step 1: Find the first derivative \( f'(x) \)

Using the product rule, we differentiate \( f(x) \):

\[ f'(x) = (x-a)(x-b)'(x-c) + (x-a)(x-b)(x-c)' \]

Calculating the derivatives:

\[ f'(x) = (x-a)(1)(x-c) + (x-a)(x-b)(1) \]
\[ = (x-a)(x-c) + (x-a)(x-b) \]
\[ = (x-a)((x-c) + (x-b)) \]
\[ = (x-a)(2x - (b+c)) \]

### Step 2: Find the second derivative \( f''(x) \)

Now we differentiate \( f'(x) \):

\[ f''(x) = \frac{d}{dx}[(x-a)(2x - (b+c))] \]

Using the product rule again:

\[ f''(x) = (x-a)'(2x - (b+c)) + (x-a)(2) \]
\[ = (1)(2x - (b+c)) + (x-a)(2) \]
\[ = 2x - (b+c) + 2(x-a) \]
\[ = 2x - (b+c) + 2x - 2a \]
\[ = 4x - (b+c + 2a) \]

### Step 3: Set \( f''(\alpha) = 0 \)

Now, we set the second derivative equal to zero:

\[ 4\alpha - (b+c + 2a) = 0 \]
\[ 4\alpha = b + c + 2a \]
\[ \alpha = \frac{b + c + 2a}{4} \]

### Step 4: Analyze the interval \( (c_1, c_2) \)

Given that \( \alpha \) lies in the interval \( (c_1, c_2) \), we have:

\[ c_1 < \frac{b + c + 2a}{4} < c_2 \]

### Step 5: Evaluate the options

1. **Option A**: \( \alpha < c_1 \) - Incorrect, as \( \alpha \) is in the interval \( (c_1, c_2) \).
2. **Option B**: \( \alpha < c_1 \) - Incorrect, same reasoning as above.
3. **Option C**: \( b < c_1 \) and \( c_2 < c \) - This cannot be determined as true or false without further information about \( b \) and \( c \).
4. **Option D**: This option is the only one that remains valid based on the analysis.

### Conclusion

Thus, the correct answer is **Option D**.