Let `lambda^("*")` be the largest value of `lambda` for which the function `f_(lambda)(x) = 4lambdax^(3)-36lambdax^(2) +36 x +48` is increasing for all `x in R` . Then `f_(lambda)"*"(1)+f_(lambda)"*"(-1)` is equal to :

Options

A. 36
B. 48
C. 64
D. 72

Answer

To solve the problem, we need to find the largest value of \( \lambda \) such that the function

\[ f_{\lambda}(x) = 4\lambda x^3 - 36\lambda x^2 + 36x + 48 \]

is increasing for all \( x \in \mathbb{R} \).

### Step 1: Find the derivative of \( f_{\lambda}(x) \)

To determine when the function is increasing, we first compute its derivative:

\[ f'_{\lambda}(x) = \frac{d}{dx}(4\lambda x^3 - 36\lambda x^2 + 36x + 48) \]

Calculating the derivative:

\[ f'_{\lambda}(x) = 12\lambda x^2 - 72\lambda x + 36 \]

### Step 2: Set the derivative greater than or equal to zero

For \( f_{\lambda}(x) \) to be increasing for all \( x \), we need:

\[ f'_{\lambda}(x) \geq 0 \quad \text{for all } x \in \mathbb{R} \]

This means the quadratic \( 12\lambda x^2 - 72\lambda x + 36 \) must not have any real roots. Therefore, we require its discriminant to be less than or equal to zero.

### Step 3: Calculate the discriminant

The discriminant \( D \) of the quadratic \( ax^2 + bx + c \) is given by:

\[ D = b^2 - 4ac \]

For our quadratic:

- \( a = 12\lambda \)
- \( b = -72\lambda \)
- \( c = 36 \)

Calculating the discriminant:

\[ D = (-72\lambda)^2 - 4(12\lambda)(36) = 5184\lambda^2 - 1728\lambda \]

### Step 4: Set the discriminant less than or equal to zero

We need:

\[ 5184\lambda^2 - 1728\lambda \leq 0 \]

Factoring out \( 1728\lambda \):

\[ 1728\lambda(3\lambda - 1) \leq 0 \]

### Step 5: Solve the inequality

This inequality holds when:

1. \( \lambda = 0 \)
2. \( 3\lambda - 1 = 0 \) which gives \( \lambda = \frac{1}{3} \)

The critical points are \( \lambda = 0 \) and \( \lambda = \frac{1}{3} \). Testing intervals, we find:

- For \( \lambda < 0 \): The expression is positive.
- For \( 0 < \lambda < \frac{1}{3} \): The expression is negative.
- For \( \lambda > \frac{1}{3} \): The expression is positive.

Thus, the solution to the inequality is:

\[ 0 \leq \lambda \leq \frac{1}{3} \]

The largest value of \( \lambda \) for which \( f_{\lambda}(x) \) is increasing for all \( x \) is:

\[ \lambda^* = \frac{1}{3} \]

### Step 6: Calculate \( f_{\lambda^*}(1) + f_{\lambda^*}(-1) \)

Now we need to compute:

\[ f_{\lambda^*}(1) + f_{\lambda^*}(-1) \]

Substituting \( \lambda = \frac{1}{3} \):

\[ f_{\frac{1}{3}}(1) = 4 \cdot \frac{1}{3} \cdot 1^3 - 36 \cdot \frac{1}{3} \cdot 1^2 + 36 \cdot 1 + 48 \]

Calculating each term:

\[ = \frac{4}{3} - 12 + 36 + 48 = \frac{4}{3} + 72 - 12 = \frac{4}{3} + 60 = \frac{4 + 180}{3} = \frac{184}{3} \]

Now for \( f_{\frac{1}{3}}(-1) \):

\[ f_{\frac{1}{3}}(-1) = 4 \cdot \frac{1}{3} \cdot (-1)^3 - 36 \cdot \frac{1}{3} \cdot (-1)^2 + 36 \cdot (-1) + 48 \]

Calculating each term:

\[ = -\frac{4}{3} - 12 - 36 + 48 = -\frac{4}{3} + 0 = -\frac{4}{3} \]

### Step 7: Combine results

Adding these results together:

\[ f_{\frac{1}{3}}(1) + f_{\frac{1}{3}}(-1) = \frac{184}{3} - \frac{4}{3} = \frac{180}{3} = 60 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{72} \]