Problem 75 - A Complete Resource Book in Mathematics for JEE Main

Question

The number of values of \(k\), for which the system of equations [2013] \((k+1) x+8 y=4 k\) \(k x+(k+3) y=3 k-1\) has no solution, is (A) 1 (B) 2 (C) 3 (D) infinite

Step-by-Step Solution

Answer

The answer is (B) 2.

Step 1: Define the System of Equations

We have the system of equations: \((k+1)x + 8y = 4k\) and \(kx + (k+3)y = 3k - 1\). We want to find the number of values of \(k\) for which these equations have no solution.

Step 2: Coefficient Representation

Identify the coefficients: For the first equation: \(a_1 = k+1,\ b_1 = 8,\ c_1 = 4k\). For the second equation: \(a_2 = k,\ b_2 = k+3,\ c_2 = 3k-1\).

Step 3: Condition for No Solution

Using the condition that a system of equations has no solution if \(\frac{a_1}{a_2} = \frac{b_1}{b_2} eq \frac{c_1}{c_2}\).

Step 4: Apply the Condition for Coefficients

Calculate \(\frac{a_1}{a_2} = \frac{k+1}{k}\) and \(\frac{b_1}{b_2} = \frac{8}{k+3}\). Set the equation \(\frac{k+1}{k} = \frac{8}{k+3}\).

Step 5: Solve the Equation for Coefficients

Cross-multiply to solve: \((k + 1)(k + 3) = 8k\). Expand to get \(k^2 + 4k + 3 = 8k\). This simplifies to \(k^2 - 4k + 3 = 0\).

Step 6: Factor the Quadratic Equation

Factor the quadratic: \(k^2 - 4k + 3 = 0\) into \((k-1)(k-3) = 0\). Thus, \(k = 1\) or \(k = 3\).

Step 7: Check the Condition for No Solution

For \(k = 1\): Check \(\frac{c_1}{c_2} eq \frac{4}{2} = 2\), compare with other ratios: \(\frac{a_1}{a_2} = 2\). Hence no solution. For \(k = 3\): Check \(\frac{c_1}{c_2} eq \frac{12}{8}=1.5\), compare with other ratios: \(\frac{a_1}{a_2} = 1.5\). Hence no solution. Both values satisfy the no solution condition.

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations are vital in mathematics because they appear frequently in various problems and have distinct methods for finding their solutions.
One of the simplest methods to solve a quadratic equation is factoring. If a quadratic can be written in the form \((x - r)(x - s) = 0\), then the solutions are \(x = r\) and \(x = s\). Another common approach is the quadratic formula:

• \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\)

where \(\pm\) indicates that there are potential for two solutions.
In the context of our problem, we solved \(k^2 - 4k + 3 = 0\) by factoring it into \((k-1)(k-3)\). This provided the potential solutions \(k = 1\) and \(k = 3\). These values were crucial in determining when the equations have no solution.

Ratio and Proportion

Ratio and proportion play a crucial role in solving systems of equations. In our exercise, we used the concept of ratio to understand the relationship between coefficients of equations.
When comparing ratios, we express one quantity as a fraction of another. If two ratios are equal, they are said to be in proportion. In a system without solution, certain ratios of coefficients, say \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\), must not equal the ratio of their constants \(\frac{c_1}{c_2}\).
In the given step-by-step solution, we compared the ratio of coefficients as \(\frac{k+1}{k} \) and \(\frac{8}{k+3}\), and made sure they equated, but this did not equal \(\frac{c_1}{c_2}\). This condition helped us confirm that for specific \(k\) values, no solution existed in the system of equations.

Solution Existence Conditions

Understanding when a system of equations has solutions is fundamental. For a system of linear equations to have no solution, it must violate certain conditions.
One primary condition is the relationship among the ratios of their coefficients, as already noted. If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} eq \frac{c_1}{c_2}\), then the lines represented by these equations are parallel, meaning they never intersect, leading to no solution.
This was illustrated in the exercise, where solving for \(k\) gives us distinct values (like \(k = 1\) and \(k = 3\)) that satisfy these conditions, signifying that the equations have no intersecting point, thus no solution exists for these \(k\) values.