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

For functional equations with inversion, substituting x → (1)/(x)$x \to \frac{1}{x}$ establishes a solvable system of algebraic equations to isolate f(x)$f(x)$ directly.

Core Logic

The given equation is:

Substitute x → (1)/(x)$x \to \frac{1}{x}$ in equation (1):

Step 1: Eliminate f(1/x)$f(1/x)$

Multiply equation (2) by 6$6$ and add it to equation (1):

Divide across by -35$-35$:

Step 2: Evaluate the Limit

We are given that the following limit evaluates to a finite constant β$\beta$:

For the limit to be a finite value, the coefficient of (1)/(x)$\frac{1}{x}$ must vanish completely:

When α = 3$\alpha = 3$, the limit simplifies directly to the constant term:

Step 3: Calculate Final Value

Substitute the determined parameters α$\alpha$ and β$\beta$:

Pattern Recognition

In limit problems involving fractional components where x → 0$x \to 0$, any term like (1)/(x)$\frac{1}{x}$ or higher negative powers must have a net coefficient of zero to guarantee existence of a finite limit value.

Chapter Mix

Class 11 Mathematics: Functions Class 11 Mathematics: Limits and Derivatives

Related Formula

For a function to be onto, its Codomain must equal its Range.

Core Logic

First, find range A$A$ for f(x) = 2x³ - 15x² + 36x + 7$f(x) = 2x^3 - 15x^2 + 36x + 7$ over [0, 3]$[0, 3]$. Differentiating f(x)$f(x)$:

Critical points are x=2$x=2$ and x=3$x=3$. Evaluate f(x)$f(x)$ at boundary and critical points:

• f(0) = 7$f(0) = 7$
• f(2) = 2(8) - 15(4) + 36(2) + 7 = 16 - 60 + 72 + 7 = 35$f(2) = 2(8) - 15(4) + 36(2) + 7 = 16 - 60 + 72 + 7 = 35$
• f(3) = 2(27) - 15(9) + 36(3) + 7 = 54 - 135 + 108 + 7 = 34$f(3) = 2(27) - 15(9) + 36(3) + 7 = 54 - 135 + 108 + 7 = 34$

Thus, Range A = [7, 35]$A = [7, 35]$.

Step 1: Find Range B for g(x)

Now look at g(x) = x²⁰²⁵x²⁰²⁵+1 = 1 - 1x²⁰²⁵+1$g(x) = \frac{x^{2025}}{x^{2025}+1} = 1 - \frac{1}{x^{2025}+1}$ over [0, ∞)$[0, \infty)$.

• At x = 0$x = 0$, g(0) = 0$g(0) = 0$.
• As x → ∞$x \to \infty$, g(x) → 1$g(x) \to 1$.

Since g(x)$g(x)$ is continuous and monotonically strictly increasing, Range B = [0, 1)$B = [0, 1)$.

Step 2: Find the Integer Count of Union Set S

S = x in Z : x in A or x in B = Z (A B)$S = \{x \in \mathbb{Z} : x \in A \text{ or } x \in B\} = \mathbb{Z} \cap (A \cup B)$

The integers in this set are:

• From [0, 1)$[0, 1)$: x = 0$x = 0$
• From [7, 35]$[7, 35]$: x = 7, 8, 9, , 35$x = 7, 8, 9, \dots, 35$

Total number of integers n(S)$n(S)$:

Pattern Recognition

The condition 'or' means union. Be careful not to include integers between 1$1$ and 6$6$ since they are not present in either continuous range segment.

Chapter Mix

Class 12 Mathematics: Functions Class 12 Mathematics: Application of Derivatives

Related Formula

The domain of ⁻¹(y)$\sec^{-1}(y)$ is given by |y| ≥ 1$|y| \ge 1$, which means:

Core Logic

For f(x) = ⁻¹(2[x]+1)$f(x) = \sec^{-1}(2[x]+1)$ to be defined:

Step 1: Solve individual inequalities

Case 1:

This holds true for all x < 0$x < 0$, i.e., x in (-∞, 0)$x \in (-\infty, 0)$.

Case 2:

This holds true for all x ≥ 0$x \ge 0$, i.e., x in [0, ∞)$x \in [0, \infty)$.

Step 2: Take Union of the Solutions

Pattern Recognition

Since [x]$[x]$ covers all integer values and 2[x]+1$2[x]+1$ forms all odd integer values, the expression inside ⁻¹$\sec^{-1}$ is always a non-zero integer. Non-zero integers always have absolute value ≥ 1$\ge 1$. Hence, it is valid for all real numbers.

Chapter Mix

Class 11 Mathematics: Functions Class 12 Mathematics: Inverse Trigonometric Functions

Related Formula

Standard result for functional equation of a polynomial satisfying f(x)f(1/x) = f(x) + f(1/x)$f(x)f(1/x) = f(x) + f(1/x)$:

Core Logic

Given that f(x)$f(x)$ is a polynomial of degree 2, the identity implies:

We are given the range is bounded above: (-∞, 1]$(-\infty, 1]$.

• For 1 + x²$1 + x^2$, the range is [1, ∞)$[1, \infty)$.
• For 1 - x²$1 - x^2$, the range is (-∞, 1]$(-\infty, 1]$.

Therefore, the correct functional form is f(x) = 1 - x²$f(x) = 1 - x^2$.

Step 1: Solve for K

Given condition: f(K) = -2K$f(K) = -2K$

Let the roots of this equation be K₁$K_1$ and K₂$K_2$. From quadratic properties (Vieta's formulas): K₁ + K₂ = 2$K_1 + K_2 = 2$

Step 2: Calculate Sum of Squares

We need the sum of squares of the values of K:

Pattern Recognition

The functional equation f(x)f(1/x)=f(x)+f(1/x)$f(x)f(1/x)=f(x)+f(1/x)$ uniquely forces polynomials to be 1 ± xⁿ$1 \pm x^n$. Remembering this shortcut saves valuable time required to derive the template from general coefficients.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations Class 12 Mathematics: Functions

Related Formula

Core Logic

To show R$R$ is an equivalence relation, verify reflexive, symmetric, and transitive properties sequentially.

Step 1: Reflexive Property

For any x in [0, π/2)$x \in [0, \pi/2)$:

Step 2: Symmetric Property

If xRy ² x - ² y = 1$xRy \implies \sec^2 x - \tan^2 y = 1$. Using identities: (1 + ² x) - ( ² y - 1) = 1 ² y - ² x = 1 yRx (Symmetric)$(1 + \tan^2 x) - (\sec^2 y - 1) = 1 \implies \sec^2 y - \tan^2 x = 1 \implies yRx \quad \text{(Symmetric)}$

Step 3: Transitive Property

If$

Step 3: Transitive Property

If $xRyand$ and $yRz \implies \sec^2 x - \tan^2 y = 1and ² y - ² z = 1. Adding both equations: