From \( a + 5b = 42 \), where \( a, b \in \mathbb{N} \), we have:
\[ a = 42 - 5b. \]
Since \( a > 0 \), we require:\[ 42 - 5b > 0 \implies b < \frac{42}{5} \implies b \leq 8. \]
The possible values of \( (a, b) \) are:
\[ (37, 1), (32, 2), (27, 3), (22, 4), (17, 5), (12, 6), (7, 7), (2, 8). \]
Thus, \( m = 8 \).
The sum is:
\[ \sum_{n=1}^{8}(1 - i^n). \]
For \( n \geq 4 \), \( i^n \) repeats every 4 terms:
\[ i^1 = i, \; i^2 = -1, \; i^3 = -i, \; i^4 = 1. \]
Compute:
\[ \sum_{n=1}^{8}(1 - i^n) = (1 - i) + (1 - (-1)) + (1 - (-i)) + (1 - 1) + (1 - i) + (1 - (-1)) + (1 - (-i)) + (1 - 1). \]
Simplify:
\[ = (1 - i) + 2 + (1 + i) + 0 + (1 - i) + 2 + (1 + i) + 0 = 5 - i + i = 5. \]
Thus:
\[ x + y = 5, \; m = 8, \; m + x + y = 8 + 5 - 1 = 12. \]
Final Answer: 12.
The speed-density relation on a one-way, single lane road is shown in the figure, where speed \( u \) is in km/hour and density \( k \) is in vehicles/km. The maximum flow (in vehicles/hour) on this road is
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32