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.
The problem asks for the number of real solutions to the equation \(x \left( x^2 + 3|x| + 5|x - 1| + 6|x - 2| \right) = 0\).
1. Zero-Product Property: The equation \( A \cdot B = 0 \) is true if and only if \( A = 0 \) or \( B = 0 \) (or both).
2. Properties of Absolute Value and Squares: For any real number \( a \), the following properties hold:
A sum of non-negative terms is equal to zero if and only if each term in the sum is individually equal to zero.
Step 1: Apply the Zero-Product Property to the given equation.
The equation is in the form \( A \cdot B = 0 \), where:
\[ A = x \] \[ B = x^2 + 3|x| + 5|x - 1| + 6|x - 2| \]
According to the zero-product property, the solutions can be found by setting each factor to zero. So, either \( A = 0 \) or \( B = 0 \).
Step 2: Analyze the first case, \( A = 0 \).
\[ x = 0 \]
This gives one immediate real solution.
Step 3: Analyze the second case, \( B = 0 \).
\[ x^2 + 3|x| + 5|x - 1| + 6|x - 2| = 0 \]
Step 4: Examine the individual terms of the expression in \( B \).
For any real number \( x \):
Step 5: Determine if the sum of these terms can be zero.
The expression is a sum of four non-negative terms. For the sum to be zero, each individual term must be zero simultaneously.
\[ x^2 = 0 \implies x = 0 \] \[ 3|x| = 0 \implies x = 0 \] \[ 5|x - 1| = 0 \implies x - 1 = 0 \implies x = 1 \] \[ 6|x - 2| = 0 \implies x - 2 = 0 \implies x = 2 \]
For the expression to be zero, we need \( x = 0 \), \( x = 1 \), and \( x = 2 \) to be true at the same time, which is impossible. Therefore, there is no real value of \( x \) for which the expression \( B \) is equal to zero. In fact, for any real \( x \), the expression \( B \) is always strictly positive. For example, if we test the potential solution \( x=0 \), the expression becomes \( 0^2 + 3|0| + 5|0-1| + 6|0-2| = 0 + 0 + 5 + 12 = 17 \neq 0 \).
The first factor, \( x \), gives the solution \( x = 0 \).
The second factor, \( x^2 + 3|x| + 5|x - 1| + 6|x - 2| \), can never be zero for any real \( x \).
Therefore, the only real solution to the entire equation is \( x = 0 \).
The number of real solutions is 1.
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
