If the equation `5arctan(x^(2)+x+k)+3arc cot(x^(2)+x+k)=2 pi,` has two distinct solutions,then the range of `k,` is

Options

A. `(0 , 5/4]`
B. ` ( - infty, 5/4)`
C. `(5/4 , infty)`
D. `( - infty, 5/4]`

Answer

To solve the equation \( 5 \arctan(x^2 + x + k) + 3 \text{arc cot}(x^2 + x + k) = 2\pi \) for the values of \( k \) that allow for two distinct solutions, we can follow these steps:

### Step 1: Simplify the Equation
We know that:
\[ \text{arc cot}(y) = \frac{\pi}{2} - \arctan(y) \]
Thus, we can rewrite the equation as:
\[ 5 \arctan(x^2 + x + k) + 3\left(\frac{\pi}{2} - \arctan(x^2 + x + k)\right) = 2\pi \]
This simplifies to:
\[ 5 \arctan(x^2 + x + k) + \frac{3\pi}{2} - 3 \arctan(x^2 + x + k) = 2\pi \]
Combining like terms gives:
\[ (5 - 3) \arctan(x^2 + x + k) + \frac{3\pi}{2} = 2\pi \]
Thus, we have:
\[ 2 \arctan(x^2 + x + k) = 2\pi - \frac{3\pi}{2} \]
This simplifies to:
\[ 2 \arctan(x^2 + x + k) = \frac{\pi}{2} \]

### Step 2: Solve for \(\arctan\)
Dividing both sides by 2:
\[ \arctan(x^2 + x + k) = \frac{\pi}{4} \]
Taking the tangent of both sides gives:
\[ x^2 + x + k = \tan\left(\frac{\pi}{4}\right) = 1 \]
Thus, we have:
\[ x^2 + x + k = 1 \]

### Step 3: Rearranging the Equation
Rearranging gives:
\[ x^2 + x + (k - 1) = 0 \]

### Step 4: Determine Conditions for Distinct Solutions
For the quadratic equation \( x^2 + x + (k - 1) = 0 \) to have two distinct solutions, the discriminant must be positive:
\[ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (k - 1) > 0 \]
This simplifies to:
\[ 1 - 4(k - 1) > 0 \]
\[ 1 - 4k + 4 > 0 \]
\[ 5 - 4k > 0 \]
\[ 4k < 5 \]
\[ k < \frac{5}{4} \]

### Step 5: Determine the Lower Bound
Next, we find if there’s a lower bound for \( k \). The quadratic will have real solutions for all \( k \) since it can be zero or negative. However, we need to ensure that the quadratic does not become a perfect square (which would give only one solution). The condition for a perfect square is:
\[ D = 0 \Rightarrow 1 - 4(k - 1) = 0 \]
Solving gives:
\[ 1 - 4k + 4 = 0 \Rightarrow 5 - 4k = 0 \Rightarrow k = \frac{5}{4} \]
Thus, the quadratic has two distinct solutions for:
\[ k < \frac{5}{4} \]

### Final Range
Since there is no lower limit derived from the discriminant, we can conclude that \( k \) can take any value less than \( \frac{5}{4} \).

### Conclusion
The range of \( k \) for which the equation has two distinct solutions is:
\[ k < \frac{5}{4} \]