Question

Given that the equation $2x^2 - 2(2a+b)x+(4a^2+2b^2-2b+1)=0$ is true for exactly two real values of x, then which of the following statement(s) is INCORRECT $(a,b \in R)$

• A. $a = \frac{2}{3}, b = \frac{1}{2}$
• B. $a = -1, b = \frac{-1}{2}$
• C. $a = \frac{1}{2}, b = 1$
• D. $a = 1, b = \frac{1}{2}$

Answer 1

Answer: (C)

A, B, D

Answer 2

To determine which statement is incorrect, we first analyze the given quadratic equation: $2x^2 - 2(2a+b)x+(4a^2+2b^2-2b+1)=0$

For a quadratic equation $Ax^2 + Bx + C = 0$ to have exactly two real values of x (i.e., two distinct real roots), its discriminant $D = B^2 - 4AC$ must be strictly greater than zero ($D > 0$).

In this equation: $A = 2$ $B = -2(2a+b)$ $C = 4a^2+2b^2-2b+1$

Let's calculate the discriminant $D$: $D = [-2(2a+b)]^2 - 4(2)(4a^2+2b^2-2b+1)$ $D = 4(2a+b)^2 - 8(4a^2+2b^2-2b+1)$ $D = 4(4a^2 + 4ab + b^2) - 32a^2 - 16b^2 + 16b - 8$ $D = 16a^2 + 16ab + 4b^2 - 32a^2 - 16b^2 + 16b - 8$ $D = -16a^2 + 16ab - 12b^2 + 16b - 8$

Now, we need to analyze the condition $D > 0$: $-16a^2 + 16ab - 12b^2 + 16b - 8 > 0$ Divide the entire inequality by -4 and reverse the inequality sign: $4a^2 - 4ab + 3b^2 - 4b + 2 < 0$

Let's try to complete the square for the expression on the left side: $4a^2 - 4ab + b^2 + 2b^2 - 4b + 2 < 0$ Group the terms involving $a$ and $b$: $(4a^2 - 4ab + b^2) + (2b^2 - 4b + 2) < 0$ The first group is a perfect square: $(2a - b)^2$. The second group can be factored: $2(b^2 - 2b + 1) = 2(b - 1)^2$.

So the inequality becomes: $(2a - b)^2 + 2(b - 1)^2 < 0$

Now, let's analyze this inequality: For any real numbers $a$ and $b$: $(2a - b)^2 \ge 0$ (a square of a real number is always non-negative) $2(b - 1)^2 \ge 0$ (twice the square of a real number is always non-negative)

Therefore, the sum of these two non-negative terms, $(2a - b)^2 + 2(b - 1)^2$, must always be greater than or equal to zero. It is impossible for this sum to be strictly less than zero.

This means that the condition $D > 0$ can never be satisfied for any real values of $a$ and $b$. In other words, the premise of the question, "the equation is true for exactly two real values of x", is false for all real $a, b$.

If $D=0$, the equation has exactly one real root (a repeated root). This occurs when: $(2a - b)^2 + 2(b - 1)^2 = 0$ This is only possible if both terms are zero: $2a - b = 0 \implies b = 2a$ AND $b - 1 = 0 \implies b = 1$ Substituting $b=1$ into the first condition: $1 = 2a \implies a = 1/2$. So, if $a=1/2$ and $b=1$, then $D=0$, and the equation has exactly one real root ($x=1$). For all other values of $a$ and $b$, $D < 0$, meaning the equation has no real roots.

Since the condition "exactly two real values of x" ($D>0$) is never met, the premise of the question is false.

The only pair $(a,b)$ for which the equation has any real roots is $(1/2, 1)$. The question asks which statement about $a,b$ is INCORRECT. This means we are looking for an $(a,b)$ pair among the options that is not $(1/2, 1)$.

Hence, statements (A), (B), and (D) are incorrect as they do not match the unique $(a,b)$ pair that yields real roots. Statement (C) is the correct pair.

Explanation of the solution:

1. Calculate the discriminant $D$ of the given quadratic equation $2x^2 - 2(2a+b)x+(4a^2+2b^2-2b+1)=0$.
2. The discriminant is found to be $D = -16a^2 + 16ab - 12b^2 + 16b - 8$.
3. Rewrite $D$ by completing the square: $D = -4[(2a - b)^2 + 2(b - 1)^2]$.
4. Since $(2a - b)^2 \ge 0$ and $2(b - 1)^2 \ge 0$ for all real $a,b$, their sum $(2a - b)^2 + 2(b - 1)^2$ is always non-negative.
5. Therefore, $D = -4 \times (\text{non-negative value})$ implies $D \le 0$ for all real $a,b$.
6. The condition "exactly two real values of x" means $D > 0$, which is impossible as $D$ is always $\le 0$.
7. This implies the premise of the question is flawed. However, in such scenarios, it's common to consider the case where real roots exist, i.e., $D \ge 0$.
8. For $D \ge 0$, since $D \le 0$ is always true, the only possibility is $D=0$.
9. $D=0$ implies $(2a - b)^2 + 2(b - 1)^2 = 0$. This holds if and only if $2a - b = 0$ AND $b - 1 = 0$.
10. Solving these equations gives $b=1$ and $a=1/2$. This is the unique pair of $(a,b)$ for which the equation has real roots (specifically, one real root).
11. The question asks for the INCORRECT statement(s). This means we are looking for the options that are not $(a=1/2, b=1)$.