For 0

Q: Wait, why is $x=1$ always a root when the sum of coefficients is zero?

Think about the definition of a polynomial $f(x)$. If $f(1) = 0$, then $1$ is a root. The sum of coefficients is exactly $f(1)$. So, if the sum is zero, $f(1)$ is zero, and $1$ is a root. It is a beautiful, simple property that saves you from the quadratic formula!

Q: Why can we cross-multiply the inequality $\frac{c+a-2b}{a+b-2c} < 1$ without flipping the sign?

Great catch! We can only cross-multiply without flipping the sign if we are certain the denominator is positive. In this problem, we analyzed the denominator $a+b-2c$ and proved it is strictly positive because $a>c$ and $b>c$. Since it is positive, the inequality sign remains unchanged.

Q: How do I know if $b$ can be the geometric mean in the second case?

In Statement (II), we found that $b$ lies in the interval $(c, \frac{a+c}{2})$. Since the geometric mean $\sqrt{ac}$ also lies within this same interval (because $c < \sqrt{ac} < \frac{a+c}{2}$), $b$ is allowed to take the value $\sqrt{ac}$. Therefore, $b$ *may* be the geometric mean.

Question

For $0 < c < b < a$, let $(a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0$ and $\alpha 
eq 1$ be one of its root. Then, among the two statements \ (I) If $\alpha \in (-1, 0)$, then $b$ cannot be the geometric mean of $a$ and $c$ \ (II) If $\alpha \in (0, 1)$, then $b$ may be the geometric mean of $a$ and $c$

Options:

(A) Both (I) and (II) are true

(B) Neither (I) nor (II) is true

(C) Only (II) is true

(D) Only (I) is true

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Quadratic Equations Imagine you are standing before a complex quadratic equation, one that looks like a tangled mess of variables: $$(a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0$$ It is easy to feel overwhelmed. But in the world of JEE Advanced, these equations are rarely just random collections of terms. They are carefully constructed puzzles, often hiding a deep, elegant symmetry. The first step in any such problem is to look for that symmetry. Let us define our function as: $$f(x) = (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b)$$ If we sum the coefficients, we get $(a + b - 2c) + (b + c - 2a) + (c + a - 2b)$. Watch what happens: the $a$ terms, $b$ terms, and $c$ terms all cancel out perfectly to zero. This is not a coincidence; it is a signal. Since the sum of the coefficients is zero, we know immediately that $f(1) = 0$. Thus, **$x = 1$ is a guaranteed root** of this equation. ### Unlocking the Second Root Now that we have discovered one root, the problem begins to unravel. We know that for any quadratic equation $Ax^2 + Bx + C = 0$, the product of the roots is given by $\frac{C}{A}$. Let our roots be $1$ and $\alpha$. Therefore, their product is $1 \cdot \alpha = \alpha$. Using the coefficients from our equation, we find: $$\alpha = \frac{c + a - 2b}{a + b - 2c}$$ This expression for $\alpha$ is our key to unlocking the statements. But before we dive into the inequalities, we must understand the nature of our denominator. We are given the condition $0 < c < b < a$. Let us examine the denominator $a + b - 2c$. We can rewrite this as $(a - c) + (b - c)$. Since $a > c$ and $b > c$, both $(a - c)$ and $(b - c)$ are strictly positive. Thus, the denominator is guaranteed to be positive. ### The Inequality Dance: Statement (I) Statement (I) asks us to consider the case where $\alpha \in (-1, 0)$. This means $\alpha < 0$. Since our denominator is positive, for the fraction $\alpha = \frac{c + a - 2b}{a + b - 2c}$ to be negative, the numerator must be negative. So, $c + a - 2b < 0$, which simplifies to $2b > a + c$, or: $$b > \frac{a + c}{2}$$ Now, recall the Arithmetic Mean-Geometric Mean (AM-GM) inequality. For two distinct positive numbers $a$ and $c$, the Arithmetic Mean $\frac{a + c}{2}$ is strictly greater than the Geometric Mean $\sqrt{ac}$. We have just shown that $b$ is strictly greater than the Arithmetic Mean. Since the Arithmetic Mean is already greater than the Geometric Mean, it follows that $b > \sqrt{ac}$. Therefore, $b$ can never be the geometric mean of $a$ and $c$. **Statement (I) is true!** ### The Interval Analysis: Statement (II) Finally, let us look at Statement (II), where $\alpha \in (0, 1)$. This gives us two conditions. First, $\alpha > 0$, which implies $c + a - 2b > 0$, or $b < \frac{a + c}{2}$. Second, $\alpha < 1$, which means: $$\frac{c + a - 2b}{a + b - 2c} < 1$$ Because the denominator is positive, we can cross-multiply safely: $c + a - 2b < a + b - 2c$. Simplifying this, we get $3c < 3b$, or $c < b$. Combining these, we find that $b$ must lie in the interval $(c, \frac{a + c}{2})$. Now, where does the geometric mean $\sqrt{ac}$ sit? We know that $c < \sqrt{ac} < \frac{a + c}{2}$. Since the geometric mean lies entirely within the interval $(c, \frac{a + c}{2})$, and $b$ can be any value in this interval, it is entirely possible for $b$ to equal $\sqrt{ac}$. Thus, $b$ *may* be the geometric mean. **Statement (II) is also true!** ### The Final Reflection We have navigated the algebraic landscape, used the symmetry of coefficients, and applied the power of the AM-GM inequality to bound our variables. Both statements are true, leading us to the correct option. This problem is a beautiful reminder that in mathematics, especially in JEE Advanced, the most complex-looking expressions often yield to simple, fundamental principles if you approach them with patience and clarity.

Select Answer:

Visualized Solution

Analyze the Coefficients

Find the Second Root $α$

Analyze the Denominator

Analyze Statement (I): $α \in (- 1, 0)$

Geometric Mean Check for (I)

Analyze Statement (II): $α \in (0, 1)$

Upper Bound for Statement (II)

Geometric Mean Check for (II)

Final Conclusion

The Sigma Insight: Relation Between Roots and Coefficients

The Hidden Symmetry of Quadratic Equations

Unlocking the Second Root

The Inequality Dance

The Interval Analysis

The Final Reflection

Similar Questions

If $α$ and $β$ ( $α < β$ ) are the roots of the equation $x^{2} + b x + c = 0$ , where $c < 0 < b$ , then

Let $α, β$ be the roots of the equation $(x - a) (x - b) = c, c \neq = 0$ . Then the roots of the equation $(x - α) (x - β) + c = 0$ are

In the quadratic equation $a x^{2} + b x + c = 0$ , $Δ = b^{2} - 4 a c$ and $α + β, α^{2} + β^{2}, α^{3} + β^{3}$ , are in G.P. where $α, β$ are the root of $a x^{2} + b x + c = 0$ , then

Let $α, β$ be the roots of the equation $x^{2} - 2 x + 6 = 0$ and $\frac{1}{α ^{2}} + 1, \frac{1}{β ^{2}} + 1$ be the roots of the equation $x^{2} + a x + b = 0$ . Then the roots of the equation $x^{2} - (a + b - 2) x + (a + b + 2) = 0$ are :

If $α, β$ are the roots of $a x^{2} + b x + c = 0, (a \neq = 0)$ and $α + δ, β + δ$ are the roots of $A x^{2} + B x + C = 0, (A \neq = 0)$ for some constant $δ$ , then prove that $(b^{2} - 4 a c) / a^{2} = (B^{2} - 4 A C) / A^{2}$

Let $- π /6 < θ < - π /12$ . Suppose $α_{1}$ and $β_{1}$ are the roots of the equation $x^{2} - 2 x sec θ + 1 = 0$ and $α_{2}$ and $β_{2}$ are the roots of the equation $x^{2} + 2 x tan θ - 1 = 0$ . If $α_{1} > β_{1}$ and $α_{2} > β_{2}$ , then $α_{1} + β_{2}$ equals

Let $a, b, c$ be real numbers with $a \neq = 0$ and let $α, β$ be the roots of the equation $a x^{2} + b x + c = 0$ . Express the roots of $a^{3} x^{2} + ab c x + c^{3} = 0$ in terms of $α, β$ .

The value of 'a' for which one root of the quadratic equation $(a^{2} - 5 a + 3) x^{2} + (3 a - 1) x + 2 = 0$ is twice as large as the other is

If the difference between the roots of the equation $x^{2} + a x + 1 = 0$ is less than $5$ , then the set of possible values of $a$ is

The value of $a$ for which the sum of the squares of the roots of the equation $x^{2} - (a - 2) x - a - 1 = 0$ assume the least value is