Question

Let $t$ be a real number satisfying $2t^3 - 9t^2 + \lambda = 0$ and $\lambda \in R$; then which of the following option(s) are correct?

A) If $t = x + \frac{1}{x}$ in above equation then for exactly one value of $\lambda$ the equation will have exactly 3 real and distinct solution of $x$

B) If $t = x + \frac{4}{x}$ in above equation then for exactly one value of $\lambda$ the equation will have exactly 3 real and distinct solution of $x$

C) If $t = x + \frac{1}{x}$ in above equation then for exactly 6 integral values of $\lambda$ the equation will have exactly 4 real and distinct solution of $x$

D) If $t = x + \frac{4}{x}$ in above equation then maximum positive integral value of $\lambda$ the equation will have exactly 2 real, distinct and positive solution of $x$, is 12

• A. A
• B. B
• C. C
• D. D

Answer 1

Answer: (A), (C)

A, C

Answer 2

Let $g(t) = 2t^3 - 9t^2$. The equation is $g(t) = -\lambda$. The derivative is $g'(t) = 6t^2 - 18t = 6t(t-3)$. The critical points are $t=0$ and $t=3$. The local maximum is $g(0) = 0$, and the local minimum is $g(3) = 2(3)^3 - 9(3)^2 = 54 - 81 = -27$.

The number of distinct real roots of $2t^3 - 9t^2 + \lambda = 0$ depends on $-\lambda$:

• One real root if $-\lambda > 0$ or $-\lambda < -27$ (i.e., $\lambda < 0$ or $\lambda > 27$).
• Two distinct real roots if $-\lambda = 0$ or $-\lambda = -27$ (i.e., $\lambda = 0$ or $\lambda = 27$).
• Three distinct real roots if $-27 < -\lambda < 0$ (i.e., $0 < \lambda < 27$).

For a given real value of $t$, the equation $t = x + \frac{k}{x}$ (where $k>0$) can be written as $x^2 - tx + k = 0$. The discriminant is $\Delta = t^2 - 4k$.

• If $t^2 - 4k > 0$, there are two distinct real solutions for $x$. This occurs when $t > \sqrt{4k}$ or $t < -\sqrt{4k}$.
• If $t^2 - 4k = 0$, there is exactly one real solution for $x$. This occurs when $t = \pm \sqrt{4k}$.
• If $t^2 - 4k < 0$, there are no real solutions for $x$. This occurs when $-\sqrt{4k} < t < \sqrt{4k}$.

Option A: $t = x + \frac{1}{x}$. Here $k=1$, so $\sqrt{4k}=2$.

• $t > 2$ or $t < -2$: 2 distinct real $x$ solutions.
• $t = 2$ or $t = -2$: 1 real $x$ solution.
• $-2 < t < 2$: 0 real $x$ solutions.

We want exactly 3 real and distinct solutions for $x$. This can happen if the set of distinct real roots of $t$ results in a sum of $x$ solutions equal to 3. For three distinct real roots of $t$, this sum is typically $2+1+0=3$. This requires one root $t_1$ giving 2 $x$-solutions, one root $t_2$ giving 1 $x$-solution, and one root $t_3$ giving 0 $x$-solutions. This means $t_1 \in (-\infty, -2) \cup (2, \infty)$, $t_2 \in \{-2, 2\}$, and $t_3 \in (-2, 2)$.

Consider $\lambda=20$. The equation is $2t^3 - 9t^2 + 20 = 0$. The roots are $t=2$, $t=\frac{5 \pm \sqrt{105}}{4}$.

• $t_1 = 2$: $N(t_1) = 1$ ($x=1$).
• $t_2 = \frac{5 + \sqrt{105}}{4} \approx 3.8$: $N(t_2) = 2$ (since $t_2 > 2$).
• $t_3 = \frac{5 - \sqrt{105}}{4} \approx -1.3$: $N(t_3) = 0$ (since $-2 < t_3 < 2$). The total number of distinct $x$ solutions is $1 + 2 + 0 = 3$. Thus, $\lambda=20$ yields exactly 3 real and distinct solutions for $x$. It can be shown that this is the only value of $\lambda$ for which this occurs. Option A is correct.

Option B: $t = x + \frac{4}{x}$. Here $k=4$, so $\sqrt{4k}=4$.

• $t > 4$ or $t < -4$: 2 distinct real $x$ solutions.
• $t = 4$ or $t = -4$: 1 real $x$ solution.
• $-4 < t < 4$: 0 real $x$ solutions.

We want exactly 3 real and distinct solutions for $x$. This requires a combination of $N(t)$ values summing to 3, e.g., $(2, 1, 0)$. This means one $t_i$ must be in $(-\infty, -4) \cup (4, \infty)$ (2 $x$ sols), one $t_j$ must be $\pm 4$ (1 $x$ sol), and one $t_k$ must be in $(-4, 4)$ (0 $x$ sols). If $t=4$, $\lambda=16$. The roots are $4, \frac{1 \pm \sqrt{33}}{4}$. $t_1=4 \implies N(t_1)=1$. $t_2 = \frac{1+\sqrt{33}}{4} \approx 1.67 \implies N(t_2)=0$. $t_3 = \frac{1-\sqrt{33}}{4} \approx -1.17 \implies N(t_3)=0$. Total $x$ solutions = 1. If $t=-4$, $\lambda=272$. The only real root is $t=-4 \implies N(t)=1$. Total $x$ solutions = 1. It appears there is no value of $\lambda$ for which exactly 3 distinct $x$ solutions exist. Option B is incorrect.

Option C: $t = x + \frac{1}{x}$. We need exactly 4 real and distinct solutions of $x$. This requires $N(t_1) + N(t_2) + N(t_3) = 4$. Possible combinations for distinct $t_i$ are $(2, 2, 0)$. This means two roots $t_i, t_j$ are in $(-\infty, -2) \cup (2, \infty)$, and one root $t_k$ is in $(-2, 2)$.

Consider the range $0 < \lambda < 27$ for 3 distinct real $t$ roots: $t_1 < 0 < t_2 < 3 < t_3$.

• $t_3 > 3$ always gives $N(t_3)=2$.
• We need $N(t_1) + N(t_2) = 2$.

  • Case C1: $N(t_1)=2$ and $N(t_2)=0$. $t_1 < -2$ and $t_2 \in (-2, 2)$ (and $t_2 \in (0, 3)$). So $t_2 \in (0, 2)$. This requires roots $t_1 < -2$, $t_2 \in (0, 2)$, $t_3 > 3$. We need to find $\lambda$ such that this occurs. If $\lambda=20$, roots are $2, \frac{5 \pm \sqrt{105}}{4}$. $t_1 = \frac{5-\sqrt{105}}{4} \approx -1.3 \in (-2, 2)$, so $N(t_1)=0$. $t_2=2$, $N(t_2)=1$. $t_3 = \frac{5+\sqrt{105}}{4} \approx 3.8 > 2$, so $N(t_3)=2$. Total $x$ solutions = $0+1+2=3$.

  • Case C2: $N(t_1)=0$ and $N(t_2)=2$. $t_1 \in (-2, 0)$ and $t_2 \in (-\infty, -2) \cup (2, \infty)$. Since $t_2 \in (0, 3)$, this is impossible.

Let's consider the case where the equation for $t$ has 2 distinct real roots. This happens when $\lambda=0$ or $\lambda=27$. If $\lambda=27$, the roots are $t=3$ (double root) and $t=-3/2$.

• $t_1 = 3$: $N(3)=2$ (since $3>2$).
• $t_2 = -3/2$: $N(-3/2)=0$ (since $-2 < -3/2 < 2$). Total $x$ solutions = $2+0=2$.

If $\lambda=0$, the roots are $t=0$ (double root) and $t=9/2$.

• $t_1 = 0$: $N(0)=0$.
• $t_2 = 9/2$: $N(9/2)=2$ (since $9/2 > 2$). Total $x$ solutions = $0+2=2$.

Consider the range $0 < \lambda < 27$. We have 3 distinct real roots $t_1 < 0 < t_2 < 3 < t_3$. We need $N(t_1) + N(t_2) + N(t_3) = 4$. Since $t_2 < 3$, $N(t_2)$ can only be 0. Since $t_3 > 3$, $N(t_3)$ is always 2. So we need $N(t_1) + 0 + 2 = 4$, which means $N(t_1) = 2$. This requires $t_1 < -2$. So we need the roots to satisfy $t_1 < -2$, $0 < t_2 < 3$, and $t_3 > 3$. This combination of roots occurs when $-27 < -\lambda < 0$, i.e., $0 < \lambda < 27$. We require $t_1 < -2$. The cubic $2t^3 - 9t^2 + \lambda = 0$ has a root $t_1 < -2$. This means $g(-2) = 2(-2)^3 - 9(-2)^2 = 2(-8) - 9(4) = -16 - 36 = -52$. So we need $-\lambda < g(-2)$, which means $-\lambda < -52$, or $\lambda > 52$. However, for 3 distinct real roots of $t$, we need $0 < \lambda < 27$. This implies that the scenario $N(t_1)=2$ is not possible when there are 3 distinct real roots for $t$.

Let's re-examine the condition for 4 distinct real solutions for $x$. This requires two roots of $t$ to yield 2 $x$-solutions each, and one root to yield 0 $x$-solutions. So, we need two roots $t_i, t_j$ such that $|t| > 2$, and one root $t_k$ such that $|t_k| < 2$.

Consider the case where $\lambda$ is such that there is only one real root for $t$. This happens when $\lambda < 0$ or $\lambda > 27$. If $\lambda > 27$, there is one real root $t_1$.

• If $t_1 > 2$, $N(t_1)=2$.
• If $t_1 = 2$, $N(t_1)=1$.
• If $-2 < t_1 < 2$, $N(t_1)=0$.
• If $t_1 = -2$, $N(t_1)=1$.
• If $t_1 < -2$, $N(t_1)=2$. For $\lambda > 27$, the single real root $t_1$ must be less than 0. If $\lambda=52$, $t=-2$ is a root. $N(-2)=1$. If $\lambda > 52$, then $t_1 < -2$, so $N(t_1)=2$. For example, if $\lambda=56$, $2t^3 - 9t^2 + 56 = 0$. $t=-2$ is not a root. $g(-2)=-52$. $g(0)=0$. Let's check $\lambda=56$. $2t^3 - 9t^2 + 56 = 0$. $t=-2$ is not a root. $g(t) = -56$. Since $-56 < -27$, there is one real root $t_1$. $g(-2) = -52$. $g(-3) = 2(-27) - 9(9) = -54 - 81 = -135$. So the root $t_1$ is between -2 and -3. Since $t_1 < -2$, $N(t_1)=2$. This gives 2 $x$ solutions.

Consider the case where $\lambda < 0$. There is one real root $t_1$. If $\lambda = -28$, $2t^3 - 9t^2 - 28 = 0$. $g(t) = 28$. Since $28 > 0$, there is one real root $t_1 > 3$. If $t_1 > 2$, $N(t_1)=2$. For $\lambda < 0$, the single real root $t_1$ must be greater than 0. If $\lambda=-28$, $t_1 > 3$. $N(t_1)=2$.

We need exactly 4 real and distinct solutions of $x$. This requires two roots of $t$ to give 2 $x$-solutions each. This means we need two roots $t_i, t_j$ such that $|t_i| > 2$ and $|t_j| > 2$. And we need a third root $t_k$ such that $|t_k| < 2$, which gives 0 $x$-solutions.

Let's consider the condition for 6 integral values of $\lambda$. We need $N(t_1) + N(t_2) + N(t_3) = 4$. This requires $(2, 2, 0)$. This means two roots have $|t|>2$ and one root has $|t|<2$.

Consider the range $0 < \lambda < 27$. Roots $t_1 < 0 < t_2 < 3 < t_3$. $N(t_3)=2$ since $t_3>3>2$. We need $N(t_1) + N(t_2) = 2$. Since $t_2 \in (0, 3)$, $N(t_2)$ can be 0 or 1 (if $t_2=2$). If $t_2=2$, then $\lambda=20$. Roots are $2, \frac{5 \pm \sqrt{105}}{4}$. $t_1 = \frac{5-\sqrt{105}}{4} \approx -1.3 \in (-2, 2)$, so $N(t_1)=0$. $t_2=2$, $N(t_2)=1$. $t_3 = \frac{5+\sqrt{105}}{4} \approx 3.8 > 2$, $N(t_3)=2$. Total $x$ solutions = $0+1+2=3$.

We need $N(t_1)=2$ and $N(t_2)=0$. This requires $t_1 < -2$ and $t_2 \in (-2, 2)$ (and $t_2 \in (0, 3)$). So $t_2 \in (0, 2)$. We need roots $t_1 < -2$, $t_2 \in (0, 2)$, $t_3 > 3$. This implies $\lambda$ must be such that $g(t_1) = -\lambda$, $g(t_2) = -\lambda$, $g(t_3) = -\lambda$. We need $t_1 < -2$. This means $g(t_1) < g(-2) = -52$. So $-\lambda < -52$, which means $\lambda > 52$. But for 3 distinct real roots, we need $0 < \lambda < 27$. This is a contradiction.

Let's reconsider the condition for 4 $x$-solutions. This implies two roots of $t$ result in 2 $x$-solutions each, and one root results in 0 $x$-solutions. So, two roots $t_i, t_j$ must satisfy $|t| > 2$, and one root $t_k$ must satisfy $|t_k| < 2$.

Consider the case where $\lambda$ is such that there are 3 distinct real roots for $t$. Let them be $t_1 < 0 < t_2 < 3 < t_3$. We need $N(t_1)+N(t_2)+N(t_3)=4$. $N(t_3)=2$ (since $t_3>3>2$). We need $N(t_1)+N(t_2)=2$. Since $t_2 \in (0, 3)$, $N(t_2)$ can be 0 (if $t_2 \in (0, 2)$) or 1 (if $t_2=2$). If $N(t_2)=0$, we need $N(t_1)=2$, so $t_1 < -2$. If $N(t_2)=1$, we need $N(t_1)=1$, so $t_1=-2$.

Case 1: $t_1 < -2$, $t_2 \in (0, 2)$, $t_3 > 3$. We need the cubic $2t^3 - 9t^2 + \lambda = 0$ to have these roots. The condition $t_1 < -2$ implies $-\lambda < g(-2) = -52$, so $\lambda > 52$. This contradicts the condition $0 < \lambda < 27$ for 3 distinct real roots.

Case 2: $t_1 = -2$, $t_2 \in (0, 2)$, $t_3 > 3$. If $t_1=-2$, then $\lambda = -g(-2) = 52$. For $\lambda=52$, the equation is $2t^3 - 9t^2 + 52 = 0$. The roots are $t=-2$ and the roots of $2t^2 - 13t + 26 = 0$, which has no real roots. So, for $\lambda=52$, there is only one real root $t=-2$. $N(-2)=1$. Total $x$ solutions = 1.

Let's consider the range of $\lambda$ for which we have 3 distinct real roots ($0 < \lambda < 27$). Let the roots be $t_1 < 0 < t_2 < 3 < t_3$. We want $N(t_1)+N(t_2)+N(t_3)=4$. We know $N(t_3)=2$. We need $N(t_1)+N(t_2)=2$. Since $t_2 \in (0, 3)$, $N(t_2)$ can be 0 (if $t_2 \in (0, 2)$) or 1 (if $t_2=2$). If $N(t_2)=0$, we need $N(t_1)=2$, so $t_1 < -2$. If $N(t_2)=1$, we need $N(t_1)=1$, so $t_1=-2$.

We need to find $\lambda$ such that $0 < \lambda < 27$ and either: (a) $t_1 < -2$, $t_2 \in (0, 2)$, $t_3 > 3$. (b) $t_1 = -2$, $t_2 \in (0, 2)$, $t_3 > 3$.

Condition (b) requires $t_1=-2$, which means $\lambda=52$. This is outside the range $0 < \lambda < 27$.

Consider condition (a). We need the roots to satisfy $t_1 < -2$, $t_2 \in (0, 2)$, $t_3 > 3$. The condition $t_1 < -2$ means $-\lambda < g(-2) = -52$, so $\lambda > 52$. This contradicts $0 < \lambda < 27$.

There seems to be an issue with the analysis of Option C. Let's reconsider the conditions. We need 4 distinct real solutions of $x$. This requires the sum of $N(t_i)$ to be 4. Possible combinations for 3 distinct $t$ roots: $(2, 2, 0)$. This means two roots $t_i, t_j$ must have $|t| > 2$, and one root $t_k$ must have $|t_k| < 2$.

Let's analyze the boundaries for $\lambda$ in the range $(0, 27)$. If $\lambda \to 0^+$, roots are close to $0, 0, 9/2$. $t_1 \approx 0, t_2 \approx 0, t_3 \approx 9/2$. $N(t_1)=0, N(t_2)=0, N(t_3)=2$. Total=2. If $\lambda \to 27^-$, roots are close to $3, 3, -3/2$. $t_1 \approx -3/2, t_2 \approx 3, t_3 \approx 3$. $N(t_1)=0, N(t_2)=2, N(t_3)=2$. Total=4. This happens as $\lambda \to 27^-$. Let's check $\lambda=26$. $2t^3 - 9t^2 + 26 = 0$. Roots are approximately $t_1 \approx -1.2$, $t_2 \approx 2.8$, $t_3 \approx 2.9$. $N(t_1)=0$ (since $|t_1|<2$). $N(t_2)=2$ (since $t_2>2$). $N(t_3)=2$ (since $t_3>2$). Total $x$ solutions = $0+2+2 = 4$. So, for $\lambda=26$, there are 4 real and distinct solutions of $x$.

We need exactly 6 integral values of $\lambda$. This means that for $\lambda \in \{21, 22, 23, 24, 25, 26\}$, we get 4 solutions. Let's check the roots for $\lambda=21$. $2t^3 - 9t^2 + 21 = 0$. Roots are approximately $t_1 \approx -1.1$, $t_2 \approx 2.5$, $t_3 \approx 2.6$. $N(t_1)=0$, $N(t_2)=2$, $N(t_3)=2$. Total = 4.

We need to find the range of $\lambda$ for which $t_1 < -2$, $t_2 \in (0, 2)$, $t_3 > 3$. This led to contradiction. We need to find the range of $\lambda$ for which $t_1 \in (-2, 0)$, $t_2 \in (2, 3)$, $t_3 > 3$. This gives $N(t_1)=0, N(t_2)=2, N(t_3)=2$. Total = 4. This occurs when $0 < \lambda < 27$. We need $t_2 \in (2, 3)$. The condition $t_2 > 2$ means $-\lambda > g(2) = 2(8) - 9(4) = 16 - 36 = -20$. So $\lambda < 20$. The condition $t_2 < 3$ is always true for the middle root in the range $0 < \lambda < 27$. We need $t_1 \in (-2, 0)$. The condition $t_1 > -2$ means $-\lambda > g(-2) = -52$, so $\lambda < 52$. This is always true for $0 < \lambda < 27$. The condition $t_1 < 0$ means $-\lambda < g(0) = 0$, so $\lambda > 0$. This is true for $0 < \lambda < 27$.

So, for 4 solutions, we need $t_1 \in (-2, 0)$, $t_2 \in (2, 3)$, $t_3 > 3$. This requires $0 < \lambda < 20$. The integral values of $\lambda$ are $1, 2, ..., 19$. This is 19 values.

Let's check the boundary $\lambda=27$. Roots are $3, 3, -3/2$. $N(3)=2$, $N(-3/2)=0$. Total $x$ solutions = 2.

Let's check the boundary $\lambda=0$. Roots are $0, 0, 9/2$. $N(0)=0$, $N(9/2)=2$. Total $x$ solutions = 2.

The number of $x$ solutions changes when $\lambda$ crosses values that make the $t$ roots equal to $\pm 2$. We need to find $\lambda$ such that the roots of $2t^3 - 9t^2 + \lambda = 0$ are such that two of them are greater than 2, and one is less than 2. Let $f(\lambda)$ be the number of $x$ solutions. We found $f(26)=4$. The range $0 < \lambda < 27$ gives 3 distinct real roots. The number of $x$ solutions changes when one of the roots $t_i$ becomes equal to 2 or -2. $t=2 \implies 2(8) - 9(4) + \lambda = 0 \implies 16 - 36 + \lambda = 0 \implies \lambda = 20$. $t=-2 \implies 2(-8) - 9(4) + \lambda = 0 \implies -16 - 36 + \lambda = 0 \implies \lambda = 52$.

If $\lambda = 20$: roots are $2, \frac{5 \pm \sqrt{105}}{4}$. $t_1=2 \implies N(t_1)=1$. $t_2 = \frac{5+\sqrt{105}}{4} \approx 3.8 > 2 \implies N(t_2)=2$. $t_3 = \frac{5-\sqrt{105}}{4} \approx -1.3 \in (-2, 2) \implies N(t_3)=0$. Total $x$ solutions = $1+2+0=3$.

Consider the interval $(0, 27)$. When $\lambda$ is close to 27, say $\lambda=26.9$, the roots are close to $3, 3, -1.45$. $t_1 \approx -1.45 \in (-2, 2) \implies N(t_1)=0$. $t_2 \approx 3 > 2 \implies N(t_2)=2$. $t_3 \approx 3 > 2 \implies N(t_3)=2$. Total $x$ solutions = $0+2+2=4$.

When $\lambda$ is close to 0, say $\lambda=0.1$, the roots are close to $0, 0, 4.5$. $t_1 \approx 0 \in (-2, 2) \implies N(t_1)=0$. $t_2 \approx 0 \in (-2, 2) \implies N(t_2)=0$. $t_3 \approx 4.5 > 2 \implies N(t_3)=2$. Total $x$ solutions = $0+0+2=2$.

The number of $x$ solutions is 4 when $t_1 \in (-2, 0)$, $t_2 \in (2, 3)$, $t_3 > 3$. This requires $0 < \lambda < 20$. The integral values are $1, 2, ..., 19$. (19 values)

The number of $x$ solutions is 4 when $t_1 < -2$, $t_2 \in (0, 2)$, $t_3 > 3$. This requires $\lambda > 52$. Not possible for 3 distinct $t$ roots.

The number of $x$ solutions is 4 when $t_1 \in (-2, 0)$, $t_2 \in (0, 2)$, $t_3 > 4$. This requires $t_1 \in (-2, 0)$, $t_2 \in (0, 2)$, $t_3 > 4$. The condition $t_3 > 4$ means $-\lambda < g(4) = 2(64) - 9(16) = 128 - 144 = -16$. So $\lambda > 16$. Combined with $0 < \lambda < 20$, this gives $\lambda \in (16, 20)$. Integral values: $17, 18, 19$. (3 values)

The number of $x$ solutions is 4 when $t_1 < -2$, $t_2 \in (0, 2)$, $t_3 \in (2, 3)$. This requires $t_1 < -2 \implies \lambda > 52$. Not possible.

Let's analyze the number of solutions for different $\lambda$:

• $\lambda < 0$: 1 real root $t_1 > 0$. If $t_1 > 2$, 2 $x$-sols. If $t_1=2$, 1 $x$-sol. If $t_1 \in (0, 2)$, 0 $x$-sols.
• $\lambda = 0$: Roots $0, 0, 9/2$. $N(0)=0, N(9/2)=2$. Total 2.
• $0 < \lambda < 20$: 3 distinct roots $t_1 < 0 < t_2 < 2 < t_3$. $N(t_1)=0, N(t_2)=0, N(t_3)=2$. Total 2.
• $\lambda = 20$: Roots $2, \frac{5 \pm \sqrt{105}}{4}$. $N(2)=1, N(\frac{5+\sqrt{105}}{4})=2, N(\frac{5-\sqrt{105}}{4})=0$. Total 3.
• $20 < \lambda < 27$: 3 distinct roots $t_1 < 0 < 2 < t_2 < 3 < t_3$. $N(t_1)=0, N(t_2)=2, N(t_3)=2$. Total 4. Integral values: $21, 22, 23, 24, 25, 26$. (6 values)
• $\lambda = 27$: Roots $3, 3, -3/2$. $N(3)=2, N(-3/2)=0$. Total 2.
• $27 < \lambda < 52$: 1 real root $t_1 < 0$. Since $g(-2)=-52$, and $-\lambda$ is between $-52$ and $-27$, the root $t_1$ is between -2 and 0. $N(t_1)=0$. Total 0.
• $\lambda = 52$: Root $t=-2$. $N(-2)=1$. Total 1.
• $\lambda > 52$: 1 real root $t_1 < -2$. $N(t_1)=2$. Total 2.

So, the number of $x$ solutions is 4 for $\lambda \in (20, 27)$. The integral values are $21, 22, 23, 24, 25, 26$. There are 6 such values. Option C is correct.

Option D: $t = x + \frac{4}{x}$. We need exactly 2 real, distinct and positive solutions of $x$. This requires $t > 4$. We need the cubic equation $2t^3 - 9t^2 + \lambda = 0$ to have at least one root $t > 4$. Consider $\lambda=12$. $2t^3 - 9t^2 + 12 = 0$. $g(t) = -12$. Since $-27 < -12 < 0$, there are 3 distinct real roots. $g(4) = -16$. Since $-12 > -16$, the largest root $t_3$ is greater than 4. $t_1 < 0 < t_2 < 3 < t_3$. $t_3 > 4 \implies N(t_3)=2$. $t_2 \in (0, 3) \implies N(t_2)=0$ (since $t_2<4$). $t_1 < 0 \implies N(t_1)=0$ (since $t_1 > -4$). Total $x$ solutions = 2. These solutions come from $t_3 > 4$. The equation $x^2 - t_3 x + 4 = 0$ has two distinct positive real roots since $t_3 > 4$. So for $\lambda=12$, we have 2 real, distinct and positive solutions of $x$.

We need the maximum positive integral value of $\lambda$. We need at least one root $t > 4$. This implies $-\lambda < g(4) = -16$, so $\lambda > 16$. However, if $\lambda > 27$, there is only one real root $t_1$. If $\lambda > 27$, then $t_1 < 0$. If $27 < \lambda < 52$, the root $t_1$ is in $(-2, 0)$, so $N(t_1)=0$. If $\lambda = 52$, $t_1=-2$, $N(t_1)=1$. If $\lambda > 52$, $t_1 < -2$, $N(t_1)=2$.

Let's consider the condition for 2 real, distinct and positive solutions of $x$. This requires $t > 4$. We need at least one root of $2t^3 - 9t^2 + \lambda = 0$ to be greater than 4. Let $h(t) = 2t^3 - 9t^2$. We need $h(t) = -\lambda$ for some $t > 4$. $h(4) = 2(4)^3 - 9(4)^2 = 128 - 144 = -16$. Since $h(t)$ is decreasing for $t>3$, for $t>4$, $h(t) < -16$. So we need $-\lambda < -16$, which means $\lambda > 16$. Also, for $t>4$, we need the root to be real. If $\lambda > 27$, there is only one real root $t_1$. This root must be $< 0$. So no root $t>4$. If $0 < \lambda < 27$, there are 3 real roots. $t_1 < 0 < t_2 < 3 < t_3$. We need $t_3 > 4$. This means $-\lambda < h(4) = -16$, so $\lambda > 16$. So, for $16 < \lambda < 27$, we have $t_3 > 4$, which gives 2 real, distinct and positive solutions of $x$. The integral values of $\lambda$ are $17, 18, 19, 20, 21, 22, 23, 24, 25, 26$. The maximum positive integral value is 26. Option D states the maximum is 12, which is incorrect.

Final check: Option A: Correct. $\lambda=20$ yields 3 $x$-solutions. Option B: Incorrect. No value of $\lambda$ yields 3 $x$-solutions. Option C: Correct. For $\lambda \in \{21, 22, 23, 24, 25, 26\}$, there are 4 $x$-solutions. Option D: Incorrect. The maximum integral $\lambda$ is 26.

The correct options are A and C.