Question

Consider the equation $(x^2 - 4x + 6)^2 - (k + 2)(x^2 - 4x + 6) + k + 5 = 0 (k \in R)$. Which of following are correct

• A. There is no integral value of k for which given equation has exactly 4 real solutions.
• B. There is only one integral value of k for which given equation has exactly 3 real solutions.
• C. There is only one integral value of k for which given equation has exactly one real solution.
• D. The number of positive integral values of k for which given equation has no real solution are 3.

Answer 1

Answer: (A), (B), (D)

A, B, D

Answer 2

The given equation is $(x^2 - 4x + 6)^2 - (k + 2)(x^2 - 4x + 6) + k + 5 = 0$. Let $y = x^2 - 4x + 6$.

First, let's analyze the expression $y = x^2 - 4x + 6$. This is a quadratic in $x$, which can be rewritten as $y = (x - 2)^2 + 2$. The minimum value of $y$ is 2, which occurs at $x = 2$. For the equation $x^2 - 4x + 6 = y$:

1. If $y < 2$, there are no real solutions for $x$.
2. If $y = 2$, there is exactly one real solution for $x$ ($x = 2$).
3. If $y > 2$, there are exactly two distinct real solutions for $x$.

Now, substitute $y$ into the given equation, which becomes a quadratic in $y$: $y^2 - (k + 2)y + k + 5 = 0$. Let $P(y) = y^2 - (k + 2)y + k + 5$. The discriminant of this quadratic in $y$ is $D_y = (-(k + 2))^2 - 4(1)(k + 5) = (k + 2)^2 - 4(k + 5) = k^2 + 4k + 4 - 4k - 20 = k^2 - 16$.

We analyze the number of real solutions for $x$ based on the value of $k$.

Case 1: $D_y < 0$

$k^2 - 16 < 0 \implies (k - 4)(k + 4) < 0 \implies -4 < k < 4$. In this range, there are no real roots for $y$. Consequently, there are no real solutions for $x$. Number of solutions for $x = 0$.

Case 2: $D_y = 0$

$k^2 - 16 = 0 \implies k = 4$ or $k = -4$.

If $k = 4$: The quadratic in $y$ has one real root $y = \frac{k+2}{2} = \frac{4+2}{2} = 3$. Since $y = 3 > 2$, there are two distinct real solutions for $x$. Number of solutions for $x = 2$.

If $k = -4$: The quadratic in $y$ has one real root $y = \frac{k+2}{2} = \frac{-4+2}{2} = -1$. Since $y = -1 < 2$, there are no real solutions for $x$. Number of solutions for $x = 0$.

Case 3: $D_y > 0$

$k^2 - 16 > 0 \implies k < -4$ or $k > 4$. In this case, there are two distinct real roots for $y$, say $y_1$ and $y_2$.

$y_1 = \frac{k+2 - \sqrt{k^2-16}}{2}$ and $y_2 = \frac{k+2 + \sqrt{k^2-16}}{2}$.

To determine the number of solutions for $x$, we need to compare $y_1$ and $y_2$ with 2. Consider $P(2) = 2^2 - (k+2)(2) + k+5 = 4 - 2k - 4 + k + 5 = -k + 5$.

• If $P(2) > 0$ and axis of symmetry $\frac{k+2}{2} > 2$: Both roots $y_1, y_2$ are greater than 2.

  $P(2) > 0 \implies -k + 5 > 0 \implies k < 5$.

  Axis of symmetry $\frac{k+2}{2} > 2 \implies k+2 > 4 \implies k > 2$.

  Combining with $D_y > 0$ ($k < -4$ or $k > 4$): The intersection of $(k < -4 \text{ or } k > 4)$, $k < 5$, and $k > 2$ is $4 < k < 5$.

  For $k \in (4, 5)$, both $y_1 > 2$ and $y_2 > 2$. Each $y$ root gives 2 distinct $x$ solutions. So, total $2+2=4$ real solutions for $x$.

• If $P(2) = 0$: One root is 2, the other is $\ne 2$.

  $P(2) = 0 \implies -k + 5 = 0 \implies k = 5$.

  For $k=5$, $D_y = 5^2 - 16 = 9 > 0$, so roots are distinct.

  The roots are $y_1 = \frac{5+2-\sqrt{9}}{2} = \frac{7-3}{2} = 2$.

  And $y_2 = \frac{5+2+\sqrt{9}}{2} = \frac{7+3}{2} = 5$.

  For $y_1 = 2$, there is 1 real solution for $x$. For $y_2 = 5 (>2)$, there are 2 distinct real solutions for $x$. So, total $1+2=3$ real solutions for $x$.

• If $P(2) < 0$: One root is less than 2, and the other is greater than 2.

  $P(2) < 0 \implies -k + 5 < 0 \implies k > 5$.

  Combining with $D_y > 0$ ($k < -4$ or $k > 4$): The intersection is $k > 5$.

  For $k > 5$, one $y$ root is less than 2 (no $x$ solutions), and the other $y$ root is greater than 2 (2 $x$ solutions). So, total $0+2=2$ real solutions for $x$.

• If $k < -4$ (from $D_y > 0$):

  We need to check $y_1, y_2$ relative to 2.

  For $k < -4$, $k < 5$, so $P(2) = -k+5 > 0$. Also, the axis of symmetry $\frac{k+2}{2} < 2$. This means both roots $y_1, y_2$ are less than 2.

  For $y_1 < 2$ and $y_2 < 2$, there are no real solutions for $x$. So for $k < -4$, number of solutions for $x = 0$.

Summary of Number of Real Solutions for $x$:

• $k < 4$: 0 real solutions
• $k = 4$: 2 real solutions
• $4 < k < 5$: 4 real solutions
• $k = 5$: 3 real solutions
• $k > 5$: 2 real solutions

Now let's evaluate the given options:

(A) There is no integral value of k for which given equation has exactly 4 real solutions.

Exactly 4 real solutions occur when $4 < k < 5$. There are no integers in this interval. So, option (A) is correct.

(B) There is only one integral value of k for which given equation has exactly 3 real solutions.

Exactly 3 real solutions occur when $k = 5$. This is indeed one integral value. So, option (B) is correct.

(C) There is only one integral value of k for which given equation has exactly one real solution.

From our analysis, there are no values of $k$ for which the equation has exactly 1 real solution. So, option (C) is incorrect.

(D) The number of positive integral values of k for which given equation has no real solution are 3.

No real solutions occur when $k < 4$. The positive integral values of $k$ in this range are $1, 2, 3$. There are 3 such values. So, option (D) is correct.

The correct options are (A), (B), and (D).