Let L denotes the number of subjective functions `f : A -> B`, where set A contains 4 elementset B contains 3 elements. M denotes number of elements in the range of the function `f(x) = sec^-1(sgmx) + cosec^-1(sgn x)` where `sg n x` denotes signum function of x. And N denotes coefficient of `t^5` in `(1+t^2)^5(1+t^3)^8`. The value of `(LM+N)` is `lambda`, then the value of `lambda/19` is

Answer

To solve the problem step by step, we will calculate the values of \(L\), \(M\), and \(N\) as defined in the question, and then find the value of \(\lambda\) and finally \(\frac{\lambda}{19}\).

### Step 1: Calculate \(L\)

**Definition:** \(L\) is the number of subjective (onto) functions from set \(A\) (with 4 elements) to set \(B\) (with 3 elements).

1. **Total Functions:** The total number of functions from \(A\) to \(B\) is \(3^4 = 81\).
2. **Using the Inclusion-Exclusion Principle:**
- Let’s denote the number of functions that do not map to at least one element in \(B\).
- If we exclude one element from \(B\), we have \(2\) choices for each of the \(4\) elements in \(A\), which gives \(2^4 = 16\) functions.
- There are \(\binom{3}{1} = 3\) ways to choose which element to exclude. Therefore, we subtract \(3 \times 16 = 48\).
- If we exclude two elements from \(B\), we have only \(1\) choice for each of the \(4\) elements in \(A\), which gives \(1^4 = 1\) function.
- There are \(\binom{3}{2} = 3\) ways to choose which two elements to exclude. Therefore, we add back \(3 \times 1 = 3\).
3. **Final Calculation for \(L\):**
\[ L = 81 - 48 + 3 = 36 \]

### Step 2: Calculate \(M\)

**Definition:** \(M\) is the number of elements in the range of the function \(f(x) = \sec^{-1}(\text{sgn}(x)) + \csc^{-1}(\text{sgn}(x))\).

1. **Understanding the Signum Function:**
- The signum function \(\text{sgn}(x)\) returns:
- \(1\) if \(x > 0\)
- \(0\) if \(x = 0\)
- \(-1\) if \(x < 0\)
2. **Evaluating \(f(x)\):**
- For \(x > 0\): \(\text{sgn}(x) = 1\) gives \(f(x) = \sec^{-1}(1) + \csc^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}\).
- For \(x < 0\): \(\text{sgn}(x) = -1\) gives \(f(x) = \sec^{-1}(-1) + \csc^{-1}(-1) = \pi + (-\frac{\pi}{2}) = \frac{\pi}{2}\).
- For \(x = 0\): \(\text{sgn}(x) = 0\) is not defined for secant and cosecant.
3. **Range of \(f(x)\):** The function \(f(x)\) takes only one value \(\frac{\pi}{2}\) for \(x \neq 0\). Thus, \(M = 1\).

### Step 3: Calculate \(N\)

**Definition:** \(N\) is the coefficient of \(t^5\) in the expansion of \((1 + t^2)^5(1 + t^3)^8\).

1. **Using the Binomial Theorem:**
- The expansion of \((1 + t^2)^5\) gives terms of the form \(\binom{5}{k} t^{2k}\).
- The expansion of \((1 + t^3)^8\) gives terms of the form \(\binom{8}{m} t^{3m}\).
2. **Finding Coefficient of \(t^5\):**
- We need \(2k + 3m = 5\).
- Possible pairs \((k, m)\):
- \(k = 0, m = 1\) gives \(t^3\) from \((1 + t^3)^8\) and contributes \(\binom{5}{0} \binom{8}{1} = 1 \times 8 = 8\).
- \(k = 2, m = 1\) gives \(t^4\) from \((1 + t^2)^5\) and contributes \(\binom{5}{2} \binom{8}{1} = 10 \times 8 = 80\).
3. **Total Coefficient \(N\):**
\[ N = 8 + 80 = 88 \]

### Step 4: Calculate \(\lambda\)

Now we can compute \(\lambda\):
\[ \lambda = L \cdot M + N = 36 \cdot 1 + 88 = 36 + 88 = 124 \]

### Step 5: Calculate \(\frac{\lambda}{19}\)

Finally, we find:
\[ \frac{\lambda}{19} = \frac{124}{19} = 6.5263 \approx 6 \]

### Final Answer
The value of \(\frac{\lambda}{19}\) is approximately \(6\).