Let `f:R->[1,oo)` be defined as `f(x)=log_10(sqrt(3x^2-4x+k+1)+10)` If f(x) is surjective then k =

Options

A. `k=(1)/(3)`
B. `klt(1)/(3)`
C. `kgt(1)/(3)`
D. `k=1`

Explanation

To determine the value of \( k \) such that the function \( f(x) = \log_{10}(\sqrt{3x^2 - 4x + k + 1} + 10) \) is surjective from \( \mathbb{R} \) to \( [1, \infty) \), we need to ensure that the expression inside the logarithm can take on all values greater than or equal to 1.

### Step 1: Set the condition for surjectivity
For \( f(x) \) to be surjective, the argument of the logarithm must be at least 1. Thus, we need:
\[ \sqrt{3x^2 - 4x + k + 1} + 10 \geq 10 \]
This simplifies to:
\[ \sqrt{3x^2 - 4x + k + 1} \geq 0 \]
This condition is always satisfied since the square root is non-negative.

### Step 2: Determine when the expression inside the square root is non-negative
Next, we need to ensure that the expression inside the square root is non-negative:
\[ 3x^2 - 4x + k + 1 \geq 0 \]
This is a quadratic inequality. For this quadratic to be non-negative for all \( x \), its discriminant must be less than or equal to zero.

### Step 3: Calculate the discriminant
The discriminant \( D \) of the quadratic \( 3x^2 - 4x + (k + 1) \) is given by:
\[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 3 \cdot (k + 1) \]
Calculating this gives:
\[ D = 16 - 12(k + 1) \]
Setting the discriminant less than or equal to zero for the quadratic to be non-negative:
\[ 16 - 12(k + 1) \leq 0 \]

### Step 4: Solve the inequality
Rearranging the inequality:
\[ 16 \leq 12(k + 1) \]
Dividing both sides by 12:
\[ \frac{16}{12} \leq k + 1 \]
Simplifying gives:
\[ \frac{4}{3} \leq k + 1 \]
Subtracting 1 from both sides:
\[ \frac{4}{3} - 1 \leq k \]
This simplifies to:
\[ \frac{4}{3} - \frac{3}{3} \leq k \]
Thus:
\[ \frac{1}{3} \leq k \]

### Step 5: Conclusion
For \( f(x) \) to be surjective, the minimum value of \( k \) must be \( \frac{1}{3} \). Therefore, the value of \( k \) that makes \( f(x) \) surjective is:
\[ k = \frac{1}{3} \]