Problem 89 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(q^{2}-4 p r=0, p>0\), then the domain of the function \(f(x)=\log \left\\{p x^{3}+(p+q) x^{2}+(q+r) x+r\right\\}\) is (A) \(R-\left\\{-\frac{q}{2 p}\right\\}\) (B) \(R-\left[(-\infty,-1] \cup\left\\{-\frac{q}{2 p}\right\\}\right]\) (C) \(R-\left[(-\infty,-1) \cap\left\\{-\frac{q}{2 p}\right\\}\right]\) (D) None of these

Step-by-Step Solution

Answer

The domain is option (A), \(R-\left\{-\frac{q}{2p}\right\}\).

Step 1: Analyzing the given equation

The given equation is \(q^2 - 4pr = 0\), which is the condition for a quadratic equation \(px^2 + qx + r = 0\) to have a double root. Therefore, the function inside the log has a factor that corresponds to \((x + \frac{q}{2p})^2\), because the double root is \(-\frac{q}{2p}\).

Step 2: Expressing the function in factored form

The expression inside the log, \(px^3 + (p+q)x^2 + (q+r)x + r\), can be rearranged to have a factor of \((x + \frac{q}{2p})^2\). This indicates that the expression can be rewritten as a cubic polynomial which includes the double root \(-\frac{q}{2p}\).

Step 3: Determining the domain

Given the log function \(f(x) = \log(px^3 + (p+q)x^2 + (q+r)x + r)\), the argument of the log function must be positive. The double root \(x = -\frac{q}{2p}\) results in the expression being zero. Hence, we must exclude this value from the domain. Since the cubic polynomial is continuous and \(p > 0\), \(f(x)\) is defined for all real numbers except \(-\frac{q}{2p}\).

Step 4: Conclusion

The domain of the function \(f(x) = \log(px^3 + (p+q)x^2 + (q+r)x + r)\) is all real numbers except \(-\frac{q}{2p}\). This corresponds to option (A), \(R - \left\{-\frac{q}{2p}\right\}\).

Key Concepts

Understanding Quadratic Equations

A quadratic equation takes the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients, and \(a eq 0\). This equation can have up to two solutions, known as roots. Whether it has one or two depends on its discriminant, calculated as \(b^2 - 4ac\). If the discriminant is zero, the quadratic equation has a double root or a repeated solution. By setting \(q^2 - 4pr = 0\) in our context, we determine that the quadratic equation has a double root. The specific double root here is given by \(-\frac{q}{2p}\). This root plays an essential role when determining the domain of other functions, as it causes the expression to either touch or cross the \(x\)-axis at only one point, creating a zero-value in the logarithm problem when substituted into higher-degree polynomials.

Exploring Cubic Polynomials

Cubic polynomials are equations of the form \(ax^3 + bx^2 + cx + d\), where \(a eq 0\). These equations generally have three roots, which could be real or complex. In this problem, the expression inside the logarithm is a cubic polynomial \(px^3 + (p+q)x^2 + (q+r)x + r\). When analyzing this cubic polynomial, it's essential to consider the roots because they affect the polynomial's structure and its positivity.

• The presence of a double root, identified as \(-\frac{q}{2p}\), forms a repeated factor in the polynomial, written as \((x + \frac{q}{2p})^2\).
• The additional factor, being its remaining root, ensures the polynomial crosses the \(x\)-axis again, due to its cubic nature.

Being a cubic function, it can cross the \(x\)-axis up to three times, or touch it and diverge. The configuration of roots significantly impacts the domain, especially when contained within a logarithmic function, requiring the remaining expression be positive.

Logarithmic Functions and Domain Restrictions

A logarithmic function, denoted by \(\log(g(x))\), is defined only when its argument, \(g(x)\), is positive. The question involves a function \(f(x) = \log(px^3 + (p+q)x^2 + (q+r)x + r)\), which means \(px^3 + (p+q)x^2 + (q+r)x + r\) must always be greater than zero to be valid. Here's how we handle domain restrictions for log functions:

• Logarithmic expressions take in only positive numbers because log of zero or negative numbers is undefined in the real number system.
• Since the cubic polynomial can potentially zero out at \(x = -\frac{q}{2p}\), this point is excluded from the domain.
• Because \(p > 0\), the polynomial increases in value as it moves away from this root, except at the disqualified point, ensuring other values remain positive.

Therefore, while constructing the function's domain, we must exclude \(-\frac{q}{2p}\) from all real numbers, which provides clarity on why choice (A) is correct for the domain.