A group of 40 students appeared in an examination of 3 subjects - Mathematics, Physics & Chemistry. It was found that all students passed in at least one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, at most 11 students passed in both Mathematics and Physics, at most 15 students passed in both Physics and Chemistry, at most 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is

Related Formula

Core Logic

For f(x)$f(x)$ to be defined, two main conditions must hold:

• -1 ≤ (5 - x)/(2x + 3) ≤ 1$-1 \leq \frac{5 - x}{2x + 3} \leq 1$
• 10 - x > 0$10 - x > 0$ and 10 - x ≠ 1$10 - x \neq 1$

Step 1: Solving the Logarithmic Domain

10 - x > 0 x < 10$10 - x > 0 \implies x < 10$ 10 - x ≠ 1 x ≠ 9$10 - x \neq 1 \implies x \neq 9$

Step 2: Solving the Inverse Sine Domain

-1 ≤ (5 - x)/(2x + 3) ≤ 1$-1 \leq \frac{5 - x}{2x + 3} \leq 1$ This is equivalent to |(5 - x)/(2x + 3)| ≤ 1$\left|\frac{5 - x}{2x + 3}\right| \leq 1$, which means:

Factor as difference of squares:

Multiply by -1$-1$ and flip the inequality:

So, x in (-∞, -8] [(2)/(3), ∞)$x \in (-\infty, -8] \cup \left[\frac{2}{3}, \infty\right)$.

Step 3: Intersection of Domains

Intersecting with x < 10$x < 10$ and x ≠ 9$x \neq 9$:

Domain = (-∞, -8] [(2)/(3), 10) - 9$= (-\infty, -8] \cup \left[\frac{2}{3}, 10\right) - \{9\}$

Comparing with the given form (-∞, α] [β, γ) - δ$(-\infty, \alpha] \cup [\beta, \gamma) - \{\delta\}$, we have: α = -8$\alpha = -8$, β = (2)/(3)$\beta = \frac{2}{3}$, γ = 10$\gamma = 10$, δ = 9$\delta = 9$.

Step 4: Final Value Calculation

Evaluate 6(α + β + γ + δ)$6(\alpha + \beta + \gamma + \delta)$:

Pattern Recognition

Instead of solving two separate rational inequalities for ≤ 1$\leq 1$ and ≥ -1$\geq -1$, converting the absolute value fraction into a difference of squares quickly bypasses sign analysis pitfalls, delivering the required critical points effortlessly.

Chapter Mix

Class 11 Maths: Sets, Relations and Functions Class 12 Maths: Inverse Trigonometric Functions

Related Formula

Count integer pairs (x, y)$(x, y)$ satisfying 4x² + y² < 52$4x^2 + y^2 < 52$.

Core Logic

Systematically test possible integer values of x$x$:

• For x = 0$x = 0$: y² < 52 y in 0, ± 1, , ± 7$y^2 < 52 \implies y \in \{0, \pm 1, \dots, \pm 7\}$ (15 values)
• For x = ± 1$x = \pm 1$: y² < 48 y in 0, ± 1, , ± 6$y^2 < 48 \implies y \in \{0, \pm 1, \dots, \pm 6\}$ (2 × 13 = 26$2 \times 13 = 26$ values)
• For x = ± 2$x = \pm 2$: y² < 36 y in 0, ± 1, , ± 5$y^2 < 36 \implies y \in \{0, \pm 1, \dots, \pm 5\}$ (2 × 11 = 22$2 \times 11 = 22$ values)
• For x = ± 3$x = \pm 3$: y² < 16 y in 0, ± 1, ± 2, ± 3$y^2 < 16 \implies y \in \{0, \pm 1, \pm 2, \pm 3\}$ (2 × 7 = 14$2 \times 7 = 14$ values)

Step 1: Total Sum

Total elements = 15 + 26 + 22 + 14 = 77$= 15 + 26 + 22 + 14 = 77$.

Pattern Recognition

Bound x$x$ first since its coefficient is larger, then sum valid values of y$y$ symmetrically.

Chapter Mix

Class 11 Maths: Sets and Relations

Related Formula

For logarithmic domain: argument must be strictly positive. For inverse sine domain: argument must lie in [-1, 1]$[-1, 1]$.

Core Logic

Let λ = x² - 10x + 85$\lambda = x^2 - 10x + 85$.

• Logarithmic conditions:
• λ > 0$\lambda > 0$
• 7 - ₂ λ > 0 λ < 2⁷$7 - \log_2 \lambda > 0 \implies \lambda < 2^7$
• ₅ (7 - ₂ λ) > 0 7 - ₂ λ > 1 ₂ λ < 6 λ < 64$\log_5 (7 - \log_2 \lambda) > 0 \implies 7 - \log_2 \lambda > 1 \implies \log_2 \lambda < 6 \implies \lambda < 64$

Combining: 0 < x² - 10x + 85 < 64 x² - 10x + 21 < 0 x in (3, 7)$0 < x^2 - 10x + 85 < 64 \implies x^2 - 10x + 21 < 0 \implies x \in (3, 7)$.

Step 1: Inverse Sine Domain

Step 2: Intersection of Domains

Intersection of x in (3, 7)$x \in (3, 7)$ and x in [-5, 6]$x \in [-5, 6]$ is (3, 6]$(3, 6]$. Here α = 3, β = 6 α + β = 9$\alpha = 3, \beta = 6 \implies \alpha + \beta = 9$.

Pattern Recognition

Unpack nested logarithms sequentially from outside in to determine tight bounds on inner quadratic.

Chapter Mix

Class 11 Maths: Functions and Graphs

Related Formula

Property of greatest integer function: [x+k] = [x] + k$[x+k] = [x] + k$ for integer k$k$.

Core Logic

Simplify f(x)$f(x)$:

• f(x) < 0 -2 < [x] < 3 [x] in -1, 0, 1, 2 x in [-1, 3)$f(x) < 0 \implies -2 < [x] < 3 \implies [x] \in \{-1, 0, 1, 2\} \implies x \in [-1, 3)$.

Step 1: Check Other Options

• f(x) > 0 [x] < -2$f(x) > 0 \implies [x] < -2$ or [x] > 3 x in (-∞, -2) [4, ∞)$[x] > 3 \implies x \in (-\infty, -2) \cup [4, \infty)$ (Option 1 incorrect).
• ∫₀² f(x) dx = ∫₀¹ (-6) dx + ∫₁² (-6) dx = -12$\int_{0}^{2} f(x) dx = \int_{0}^{1} (-6) dx + \int_{1}^{2} (-6) dx = -12$ (Option 3 incorrect).
• f(x) = 0 [x] = 3$f(x) = 0 \implies [x] = 3$ or [x] = -2$[x] = -2$, which has infinitely many solutions (Option 4 incorrect).

Pattern Recognition

Factorize expression in terms of [x]$[x]$ to easily check intervals for positive/negative values.

Chapter Mix

Class 11 Maths: Functions and Graphs

Related Formula

Core Logic

Evaluate 2x + y ≤ 2$2x + y \leq 2$ for all elements in A = -2, -1, 0, 1, 2, 3, 4$A = \{-2, -1, 0, 1, 2, 3, 4\}$ to find pairs (x,y) in R$(x,y) \in R$. For x = -2$x = -2$: y ≤ 6 ⇒ y in -2, -1, 0, 1, 2, 3, 4$y \leq 6 \Rightarrow y \in \{-2, -1, 0, 1, 2, 3, 4\}$ (7 pairs) For x = -1$x = -1$: y ≤ 4 ⇒ y in -2, -1, 0, 1, 2, 3, 4$y \leq 4 \Rightarrow y \in \{-2, -1, 0, 1, 2, 3, 4\}$ (7 pairs) For x = 0$x = 0$: y ≤ 2 ⇒ y in -2, -1, 0, 1, 2$y \leq 2 \Rightarrow y \in \{-2, -1, 0, 1, 2\}$ (5 pairs) For x = 1$x = 1$: y ≤ 0 ⇒ y in -2, -1, 0$y \leq 0 \Rightarrow y \in \{-2, -1, 0\}$ (3 pairs) For x = 2$x = 2$: y ≤ -2 ⇒ y in -2$y \leq -2 \Rightarrow y \in \{-2\}$ (1 pair) For x = 3$x = 3$: y ≤ -4 ⇒$y \leq -4 \Rightarrow$ None For x = 4$x = 4$: y ≤ -6 ⇒$y \leq -6 \Rightarrow$ None

Total elements in R$R$, l = 7 + 7 + 5 + 3 + 1 = 23$l = 7 + 7 + 5 + 3 + 1 = 23$.

Step 1: Calculate Minimum Additions for Reflexivity (m)

For R$R$ to be reflexive, we need (x,x) in R$(x,x) \in R$ for all x in A$x \in A$. Let's check which are missing: 2(x) + x = 3x ≤ 2$2(x) + x = 3x \leq 2$. This holds for x = -2, -1, 0$x = -2, -1, 0$. It fails for x = 1, 2, 3, 4$x = 1, 2, 3, 4$. Thus, we need to add 4 elements: (1,1), (2,2), (3,3), (4,4)$(1,1), (2,2), (3,3), (4,4)$. So, m = 4$m = 4$.

Step 2: Calculate Minimum Additions for Symmetry (n)

For R$R$ to be symmetric, if (x,y) in R$(x,y) \in R$, we must have (y,x) in R$(y,x) \in R$. Let's check elements where 2x+y ≤ 2$2x+y \leq 2$ but 2y+x > 2$2y+x > 2$. We list pairs where (x,y) in R$(x,y) \in R$ but (y,x) R$(y,x) \notin R$: For x = -2$x = -2$: ( -2, 3 )$( -2, 3 )$ and ( -2, 4 )$( -2, 4 )$ are in R$R$. Inverse (3, -2)$(3, -2)$ has 2(3)+(-2) = 4 > 2$2(3)+(-2) = 4 > 2$ (not in R$R$). Add 2 elements. For x = -1$x = -1$: (-1, 2), (-1, 3), (-1, 4)$(-1, 2), (-1, 3), (-1, 4)$ are in R$R$. Inverses: (2, -1)$(2, -1)$ has 2(2)-1=3>2$2(2)-1=3>2$; (3, -1)$(3, -1)$ has 2(3)-1=5>2$2(3)-1=5>2$; (4, -1)$(4, -1)$ has 2(4)-1=7>2$2(4)-1=7>2$. Add 3 elements. For x = 0$x = 0$: (0, 2)$(0, 2)$ is in R$R$. Inverse (2, 0)$(2, 0)$ has 2(2)+0=4>2$2(2)+0=4>2$. Add 1 element. Total pairs to add to make it symmetric, n = 2 + 3 + 1 = 6$n = 2 + 3 + 1 = 6$. (The pairs to add are (3,-2), (4,-2), (2,-1), (3,-1), (4,-1), (2,0)$(3,-2), (4,-2), (2,-1), (3,-1), (4,-1), (2,0)$).

Step 3: Final Sum

Pattern Recognition

Counting discrete relations systematically over a small finite set avoids oversight. Breaking it down by individual x$x$ constraints builds an exhaustive map.

Chapter Mix

Class 12 Maths: Relations and Functions