Let f, g: (1, ∞) → R$f, g: (1, \infty) \to \mathbb{R}$ be defined as f(x) = (2x + 3)/(5x + 2)$f(x) = \frac{2x + 3}{5x + 2}$ and g(x) = (2 - 3x)/(1 - x)$g(x) = \frac{2 - 3x}{1 - x}$. If the range of the function f(g(x))$f(g(x))$ on the interval [2, 4]$[2, 4]$ is [α, β]$[\alpha, \beta]$, then (1)/(β - α)$\frac{1}{\beta - \alpha}$ is equal to

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