Analyze the equation:
\[ x(x^2 + 3x) + |x - 1| + 6|k - 2| = 0 \implies x^3 + 3x^2 + |x - 1| + C = 0. \]
Substitute \( C = 6|k - 2| \):
\[ x^3 + 3x^2 + |x - 1| + C = 0. \]
Consider cases for absolute value:
Case 1: \( x \geq 1 \):
\[ x^3 + 3x^2 + (x - 1) + C = 0 \implies x^3 + 3x^2 + x + (C - 1) = 0. \]
Case 2: \( x < 1 \):
\[ x^3 + 3x^2 + (-x + 1) + C = 0 \implies x^3 + 3x^2 - x + (C + 1) = 0. \]
Finding the number of real solutions: The cubic functions yield at most one real root due to their monotonically increasing nature.
Evaluate the function: Since the cubic functions are monotonic, we conclude: There is one real solution in each case.
Thus, the number of real solutions is: 1.
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)