Problem 101 - A Complete Resource Book in Mathematics for JEE Main

Question

The set of values of \(a\) for which the equation \(\left(x^{2}+x\right.\) \(+2)^{2}-(a-3)\left(x^{2}+x+2\right)\left(x^{2}+x+1\right)+(a-4)\left(x^{2}+\right.\) \(x+1)^{2}=0\) has at least one real root is (A) \(\left(5, \frac{19}{3}\right)\) (B) \(\left[5, \frac{19}{3}\right]\) (C) \(\left[5, \frac{19}{3}\right)\) (D) \(\left(5, \frac{19}{3}\right]\)

Step-by-Step Solution

Answer

The set of values of a is \( \left[5, \frac{19}{3}\right) \).

Step 1: Expression Simplification

Let's simplify the given equation. The equation can be written as: \[(x^2 + x + 2)^2 - (a-3)(x^2+x+2)(x^2+x+1) + (a-4)(x^2+x+1)^2 = 0\]Let us substitute \(t = x^2 + x\). This gives the equation in terms of \(t\):\[(t + 2)^2 - (a-3)(t + 2)(t + 1) + (a-4)(t + 1)^2 = 0\]

Step 2: Expand and Simplify the Equation

Simplify the expression:\[(t^2 + 4t + 4) - (a-3)(t^2 + 3t + 2) + (a-4)(t^2 + 2t + 1) = 0\]Simplify further:\[t^2 + 4t + 4 - (a-3)(t^2 + 3t + 2) + (a-4)(t^2 + 2t + 1)\]Distribute the constants across the terms:- Multiply out: \(-(a-3)(t^2 + 3t + 2) \rightarrow -(a-3)t^2 - 3(a-3)t - 2(a-3)\)- Multiply out: \((a-4)(t^2 + 2t + 1) \rightarrow (a-4)t^2 + 2(a-4)t + (a-4)\)

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental component of algebra that you will encounter frequently in mathematical problem-solving. Generally, they are written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. In our original exercise, the complexity was increased by involving terms that needed substitution and simplification before resolving it into a recognizable quadratic form. Understanding quadratic equations is essential because:

• They describe a wide variety of phenomena — everything from projectile motion to market equilibrium.
• Knowing how to solve them allows you to find the values of \(x\) at which the equation holds true, often referred to as the "roots" of the equation.

In our exercise, the equation was easier to handle after substituting \(t = x^2 + x\), creating a new quadratic equation in terms of \(t\). This is a classic technique that simplifies problems involving nested quadratic expressions.
Becoming comfortable with quadratic equations, their forms, and manipulation techniques like substitution is key to mastery in algebra.

Real Roots

When solving quadratic equations, one key concept you'll come across is the 'real roots'. Real roots are the values of \(x\) for which the quadratic equation equals zero, provided they can be represented on the number line. For our exercise, one of the tasks was to determine for what values of \(a\), the equation has at least one real root.In quadratic equations of the form \(ax^2 + bx + c = 0\), determining the roots often involves:

• Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Evaluating the discriminant \(b^2 - 4ac\).

The discriminant is crucial because:

• If \(b^2 - 4ac > 0\), there are two distinct real roots.
• If \(b^2 - 4ac = 0\), there is exactly one real root.
• If \(b^2 - 4ac < 0\), there are no real roots.

In our problem, simplifying and substituting the expression meant focusing on ensuring the discriminant of the resulting equation provided conditions where real roots were possible. This usually means making sure that expressions inside square roots remain positive, or zero, ensuring the solution remains in the real number domain by checking the range of parameter \(a\).

Equation Simplification

A crucial skill in solving complex equations, like the one in our exercise, is equation simplification. This involves making the expression more manageable while maintaining equality.In our case, simplification involved:

• Condensing the expression \((x^2 + x + 2)^2 - (a-3)(x^2+x+2)(x^2+x+1) + (a-4)(x^2+x+1)^2 = 0\) by substituting \(t = x^2 + x\) thus transforming it into \((t + 2)^2 - (a-3)(t + 2)(t + 1) + (a-4)(t + 1)^2 = 0\).
• Expanding each term precisely to ensure no detail in calculation is missed.
• Combining like terms and arranging the terms to get a simpler form to solve, essentially reducing the number of potential errors in solving it.

Simplification is beneficial not just for solving the particular problem but also for:

• Saving time by reducing complexity early on in a problem.
• Allowing changes in formulation that might make a tricky problem more straightforward or intuitive.

By mastering simplification techniques, you make handling elaborate mathematical expressions a smoother process, enabling you to see the path to solutions more clearly.