Let `f (x)` be a continuous function (define for all x) which satisfies `f ^(3) (x)-5 f ^(2) (x)+ 10f (x) -12 ge 0, f ^(2) (x) + 3 ge 0 and f ^(2) (x) -5f(x)+ 6 le 0`
If distinct positive number `b_(1), b _(2) and b _(3)` ar in G.P. then `f (1)+ ln b _1), f (2) + ln b _(2), f (3)+ ln b _(3)` are in :

Options

A. A.P.
B. G.P.
C. H. P.
D. A. G. P.

Answer

To solve the problem, we need to analyze the given conditions for the continuous function \( f(x) \) and determine the nature of the expressions involving \( f(1) + \ln b_1 \), \( f(2) + \ln b_2 \), and \( f(3) + \ln b_3 \).

### Step 1: Analyze the inequalities involving \( f(x) \)

We are given three inequalities:

1. \( f^{(3)}(x) - 5f^{(2)}(x) + 10f(x) - 12 \geq 0 \)
2. \( f^{(2)}(x) + 3 \geq 0 \)
3. \( f^{(2)}(x) - 5f(x) + 6 \leq 0 \)

Let’s denote \( f(x) = t \). Then, we can rewrite the inequalities in terms of \( t \).

### Step 2: Solve the first inequality

The first inequality can be rewritten as:
\[ t^3 - 5t^2 + 10t - 12 \geq 0 \]

To find the roots of the polynomial \( t^3 - 5t^2 + 10t - 12 = 0 \), we can use the Rational Root Theorem or synthetic division. Testing \( t = 2 \):
\[ 2^3 - 5(2^2) + 10(2) - 12 = 8 - 20 + 20 - 12 = -4 \quad \text{(not a root)} \]
Testing \( t = 3 \):
\[ 3^3 - 5(3^2) + 10(3) - 12 = 27 - 45 + 30 - 12 = 0 \quad \text{(is a root)} \]

Now we can factor \( t - 3 \) out of the polynomial:
\[ t^3 - 5t^2 + 10t - 12 = (t - 3)(t^2 - 2t + 4) \]

The quadratic \( t^2 - 2t + 4 \) has a discriminant of:
\[ D = (-2)^2 - 4(1)(4) = 4 - 16 = -12 < 0 \]
This means \( t^2 - 2t + 4 > 0 \) for all \( t \). Thus, the first inequality holds for \( t \geq 3 \).

### Step 3: Solve the second inequality

The second inequality is:
\[ f^{(2)}(x) + 3 \geq 0 \implies f^{(2)}(x) \geq -3 \]
Since \( f^{(2)}(x) \) is always non-negative, this condition is satisfied.

### Step 4: Solve the third inequality

The third inequality is:
\[ f^{(2)}(x) - 5f(x) + 6 \leq 0 \]
Substituting \( t \):
\[ t^2 - 5t + 6 \leq 0 \]
Factoring gives:
\[ (t - 2)(t - 3) \leq 0 \]
The roots are \( t = 2 \) and \( t = 3 \). The solution to this inequality is:
\[ 2 \leq t \leq 3 \]

### Step 5: Combine results

From the first inequality, we found \( t \geq 3 \). From the third inequality, we found \( 2 \leq t \leq 3 \). The only value that satisfies both inequalities is:
\[ t = 3 \]
Thus, \( f(x) = 3 \) for all \( x \).

### Step 6: Analyze the logarithmic expressions

Now we need to analyze the expressions:
\[ f(1) + \ln b_1, \quad f(2) + \ln b_2, \quad f(3) + \ln b_3 \]
Substituting \( f(x) = 3 \):
\[ 3 + \ln b_1, \quad 3 + \ln b_2, \quad 3 + \ln b_3 \]

### Step 7: Determine the nature of the expressions

Since \( b_1, b_2, b_3 \) are in geometric progression (G.P.), we have:
\[ \frac{b_2}{b_1} = \frac{b_3}{b_2} \]
This implies:
\[ \ln b_2 - \ln b_1 = \ln b_3 - \ln b_2 \]
Thus, the differences:
\[ (3 + \ln b_2) - (3 + \ln b_1) = \ln b_2 - \ln b_1 \]
\[ (3 + \ln b_3) - (3 + \ln b_2) = \ln b_3 - \ln b_2 \]
are equal, indicating that:
\[ f(1) + \ln b_1, f(2) + \ln b_2, f(3) + \ln b_3 \]
are in arithmetic progression (A.P.).

### Final Answer:
The expressions \( f(1) + \ln b_1, f(2) + \ln b_2, f(3) + \ln b_3 \) are in **A.P.**.