Question

The set of all values of k \(>\) -1, for which the equation
(3x\(^2\)+4x+3)\(^2\) - (k+1)(3x\(^2\)+4x+3)(3x\(^2\)+4x+2) + k(3x\(^2\)+4x+2)\(^2\) = 0 has real roots, is :

• A. (1, 5/2]
• B. [2, 3)
• C. [-1/2, 1)
• D. (1/2, 3/2] - \{1\}

Hint:

When an equation contains a repeated complex expression, substitution is a powerful technique. After simplifying, always relate the substituted variable back to the original variable (x) and consider the conditions for real roots, which often involves finding the range of the expression.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have a complex-looking equation and we need to find the values of k for which it has real roots for x. The key is to simplify the equation by substitution.
Step 2: Simplify the Equation:
Let the expression \(y = 3x^2 + 4x + 3\). Then the expression \(3x^2 + 4x + 2 = y - 1\).
Substituting these into the given equation:
\[ y^2 - (k+1)y(y-1) + k(y-1)^2 = 0 \] Now, expand and simplify this equation in terms of y.
\[ y^2 - (k+1)(y^2 - y) + k(y^2 - 2y + 1) = 0 \] \[ y^2 - (ky^2 - ky + y^2 - y) + (ky^2 - 2ky + k) = 0 \] \[ y^2 - ky^2 + ky - y^2 + y + ky^2 - 2ky + k = 0 \] Combine like terms:
\[ (y^2 - y^2) + (-ky^2 + ky^2) + (ky + y - 2ky) + k = 0 \] \[ y - ky + k = 0 \] \[ y(1 - k) = -k \] If \( k \neq 1 \), we can write:
\[ y = \frac{-k}{1-k} = \frac{k}{k-1} \] Step 3: Find the Range of y:
For the original equation to have real roots in x, our substituted equation \(3x^2 + 4x + 3 = y\) must have real roots for x.
Let \(f(x) = 3x^2 + 4x + 3\). This is a quadratic in x. For it to have real roots, the value of y must be within the range of f(x).
Since the coefficient of x\(^2\) is positive (3>0), the parabola opens upwards and has a minimum value.
The minimum value occurs at \( x = -\frac{b}{2a} = -\frac{4}{2(3)} = -\frac{2}{3} \).
The minimum value of f(x) is \( f(-2/3) = 3(-\frac{2}{3})^2 + 4(-\frac{2}{3}) + 3 = 3(\frac{4}{9}) - \frac{8}{3} + 3 = \frac{4}{3} - \frac{8}{3} + \frac{9}{3} = \frac{5}{3} \).
So, the range of y is \( [5/3, \infty) \).
Step 4: Solve for k:
We must have \(y \ge 5/3\).
\[ \frac{k}{k-1} \ge \frac{5}{3} \] \[ \frac{k}{k-1} - \frac{5}{3} \ge 0 \] \[ \frac{3k - 5(k-1)}{3(k-1)} \ge 0 \] \[ \frac{3k - 5k + 5}{3(k-1)} \ge 0 \] \[ \frac{-2k + 5}{k-1} \ge 0 \] Multiply by -1 and reverse the inequality sign:
\[ \frac{2k - 5}{k-1} \le 0 \] The critical points are \(k=1\) and \(k=5/2\). The expression is less than or equal to zero between these points.
Thus, \(1<k \le 5/2\). (Note that k cannot be 1 as it would make the denominator zero).
The given condition is \(k>-1\), which is satisfied by our result.
The set of values for k is (1, 5/2].