Type I Linear Inequality
1. The number of triplets $(x, y, z)$, where $x, y, z$ are distinct non negative integers satisfying $x + y + z = 15$, is
(a) 136
(b) 114
(c) 80
(d) 92
JEE Mains 11/04/2023 Shift-I
Ans. (b) : Given, $x + y + z = 15$
Total number of non-negative integral solutions is given by:
$${}^{n+r-1}C_{r-1} = {}^{15+3-1}C_{3-1} = {}^{17}C_2 = \frac{17 \times 16}{2} = 136 \quad \text{-----(1)}$$Now, we need to subtract the cases where two or all three variables are equal:
When $x$ and $z$ are equal, then $2x + y = 15$
When $y$ and $z$ are equal, then $2y + x = 15$
When $x$ and $y$ are equal, then $2x + z = 15$
For $2x + y = 15$ (i.e. $x = z$):
If $z = 1$, then $x = 1, y = 13$
If $z = 3$, then $x = 3, y = 9$
If $z = 5$, then $x = 5, y = 5$ (All three equal: $x = y = z = 5$)
...
If $z = 13$, then $x = 13, y = 1$
If $z = 15$, then $x = 15, y = 0$
i.e., 8 solutions for $x = z$.
Similarly, for $x = y$ (where $x = 1, 3, 5, 7, 9, 11, 13, 15$), it has 8 solutions (including $x=y=z=5$).
For $y = z$ (where $y = 1, 3, 5, 7, 9, 11, 13, 15$), it has 8 solutions (including $x=y=z=5$).
Total solutions where at least two variables are equal:
$$\text{Total equal cases} = 3 \times 7 + 1 = 22 \text{ solutions}$$ $$\text{Required answer} = 136 - 22 = 114$$2. The number of integral solution $x$ of $\log_{\left(x + \frac{7}{2}\right)} \left(\frac{x - 7}{2x - 3}\right)^2 \ge 0$ is:
(a) 5
(b) 7
(c) 8
(d) 6
JEE Mains 11/04/2023 Shift-I
Ans. (d) : Given,
$$\log_{\left(x + \frac{7}{2}\right)} \left(\frac{x - 7}{2x - 3}\right)^2 \ge 0$$Feasible region:
$$x + \frac{7}{2} > 0 \implies x > -\frac{7}{2}$$And base $\neq 1$:
$$x + \frac{7}{2} \neq 1 \implies x \neq -\frac{5}{2}$$And argument domain:
$$\frac{x - 7}{2x - 3} \neq 0 \implies x \neq 7$$ $$2x - 3 \neq 0 \implies x \neq \frac{3}{2}$$Taking intersection:
$$x \in \left(-\frac{7}{2}, \infty\right) \setminus \left\{-\frac{5}{2}, \frac{3}{2}, 7\right\}$$Now, $\log_a b \ge 0$ if $(a > 1 \text{ and } b \ge 1)$ OR $(a \in (0, 1) \text{ and } b \in (0, 1])$.
Case (i): Base $a > 1 \implies x + \frac{7}{2} > 1 \implies x > -\frac{5}{2}$ and $b \ge 1$:
$$\left(\frac{x - 7}{2x - 3}\right)^2 \ge 1$$ $$(2x - 3)^2 - (x - 7)^2 \le 0$$ $$(2x - 3 + x - 7)(2x - 3 - x + 7) \le 0$$ $$(3x - 10)(x + 4) \le 0$$ $$x \in \left[-4, \frac{10}{3}\right]$$Intersection for Case (i):
$$x \in \left(-\frac{5}{2}, \frac{10}{3}\right]$$Case (ii): Base $a \in (0, 1) \implies x + \frac{7}{2} \in (0, 1)$ and $b \in (0, 1]$:
$$0 < x + \frac{7}{2} < 1 \implies -\frac{7}{2} < x < -\frac{5}{2}$$And $b \in (0, 1]$:
$$\left(\frac{x - 7}{2x - 3}\right)^2 < 1$$ $$(x - 7)^2 < (2x - 3)^2$$ $$x \in (-\infty, -4) \cup \left(\frac{10}{3}, \infty\right)$$No common values of $x$ in Case (ii).
Hence, taking the intersection with the feasible region, we get:
$$x \in \left(-\frac{5}{2}, \frac{10}{3}\right] \setminus \left\{\frac{3}{2}\right\}$$Integral values of $x$ in this range are $\{-2, -1, 0, 1, 2, 3\}$.
Number of integral values $= 6$.
Type II Number of Solution
3. A triangle is formed by X-axis, Y-axis and the line $3x + 4y = 60$. Then the number of points $P(a,b)$ which lie strictly inside the triangle, where $a$ is an integer and $b$ is a multiple of $a$, is
JEE Mains 25/01/2023 Shift-II
Ans. (31) : As $b$ is a multiple of $a$, we can write $b = ka$ where $k \in \mathbb{Z}^+$.
The required points lie on the line $y = kx$ ($k \in \mathbb{Z}$).
Since $P(a,b)$ lies strictly inside the triangle bounded by $x = 0$, $y = 0$, and $3x + 4y = 60$:
$$3x + 4y < 60$$ $$3x + 4(kx) < 60 \implies x(3 + 4k) < 60$$ $$x < \frac{60}{3 + 4k}$$For each positive integer $k$, $a$ can be any integer from $1$ to $\left\lfloor \frac{59}{3 + 4k} \right\rfloor$:
If $k = 1$, $x < \frac{60}{7} \approx 8.57 \implies a \in \{1, 2, \dots, 8\} \implies 8 \text{ internal points}$
If $k = 2$, $x < \frac{60}{11} \approx 5.45 \implies a \in \{1, 2, \dots, 5\} \implies 5 \text{ internal points}$
If $k = 3$, $x < \frac{60}{15} = 4 \implies a \in \{1, 2, 3\} \implies 3 \text{ internal points}$
If $k = 4$, $x < \frac{60}{19} \approx 3.15 \implies a \in \{1, 2, 3\} \implies 3 \text{ internal points}$
If $k = 5$, $x < \frac{60}{23} \approx 2.60 \implies a \in \{1, 2\} \implies 2 \text{ internal points}$
If $k = 6$, $x < \frac{60}{27} \approx 2.22 \implies a \in \{1, 2\} \implies 2 \text{ internal points}$
If $k = 7$, $x < \frac{60}{31} \approx 1.93 \implies a \in \{1\} \implies 1 \text{ internal point}$
If $k = 8$, $x < \frac{60}{35} \approx 1.71 \implies a \in \{1\} \implies 1 \text{ internal point}$
If $k = 9$, $x < \frac{60}{39} \approx 1.53 \implies a \in \{1\} \implies 1 \text{ internal point}$
If $k = 10$, $x < \frac{60}{43} \approx 1.39 \implies a \in \{1\} \implies 1 \text{ internal point}$
If $k = 11$, $x < \frac{60}{47} \approx 1.27 \implies a \in \{1\} \implies 1 \text{ internal point}$
If $k = 12$, $x < \frac{60}{51} \approx 1.17 \implies a \in \{1\} \implies 1 \text{ internal point}$
If $k = 13$, $x < \frac{60}{55} \approx 1.09 \implies a \in \{1\} \implies 1 \text{ internal point}$
If $k = 14$, $x < \frac{60}{59} \approx 1.01 \implies a \in \{1\} \implies 1 \text{ internal point}$
For $k \ge 15$, $x < 1$, so no positive integral solutions exist.
$$\therefore \text{Total points} = 8 + 5 + 3 + 3 + 2 + 2 + (8 \times 1) = 31 \text{ points}$$4. Number of integral solutions to the equation $x + y + z = 21$, where $x \ge 1, y \ge 3, z \ge 4$, is equal to _________.
JEE Mains 01/02/2023 Shift-II
Ans. (105) : Given,
$$x + y + z = 21, \quad x \ge 1, y \ge 3, z \ge 4$$Let substitution be:
$$\begin{cases} x = 1 + t_1, & t_1 \ge 0 \\ y = 3 + t_2, & t_2 \ge 0 \\ z = 4 + t_3, & t_3 \ge 0 \end{cases}$$Substitute these into the equation:
$$(1 + t_1) + (3 + t_2) + (4 + t_3) = 21$$ $$t_1 + t_2 + t_3 + 8 = 21$$ $$t_1 + t_2 + t_3 = 13$$Number of non-negative integral solutions is given by:
$${}^{n+r-1}C_{r-1}$$Here $n = 13$ and $r = 3$:
$${}^{13+3-1}C_{3-1} = {}^{15}C_2 = \frac{15 \times 14}{2} = 15 \times 7 = 105$$5. Number of integral solutions to the equation $x + y + z = 21$ where $x \ge 1, y \ge 3, z \ge 4$, is equal to_______.
JEE Main-01.02.2023, Shift-II
Ans. (105) : Given, number of integral solutions to the equation $x + y + z = 21$
Where $x \ge 1, y \ge 3, z \ge 4$
$$\implies (x - 1) + (y - 3) + (z - 4) = 21 - 8 = 13$$ $$\implies \text{Let, } \alpha = x - 1, \quad \beta = y - 3, \quad \gamma = z - 4$$ $$\text{Then, } \alpha + \beta + \gamma = 13 \quad (\alpha, \beta, \gamma \ge 0)$$Therefore, number of non-negative integral solutions is given by formula:
$${}^{n+r-1}C_{r-1}$$Where $n = 13$, $r = \text{number of variables} = 3$
$$\text{Number of integral solutions} = {}^{13+3-1}C_{3-1} = {}^{15}C_2$$ $${}^{15}C_2 = \frac{15!}{2! (15 - 2)!} = \frac{15!}{2! \times 13!} = \frac{15 \times 14 \times 13!}{2 \times 1 \times 13!} = 15 \times 7 = 105$$
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