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

We find the domains of the two constituent terms separately and then find their intersection.

Step 1: Find the domain of the first term

For the first term to be defined:

Step 2: Find the domain of the second term

For the second term to be defined:

x + |x| > 0$x + |x| > 0$

• If x ≥ 0$x \ge 0$: x + x = 2x > 0 x > 0$x + x = 2x > 0 \implies x > 0$.
• If x < 0$x < 0$: x - x = 0 0$x - x = 0 \ngtr 0$.

Thus, the domain of the second term is:

Step 3: Find intersection and calculate the final expression

Intersecting domains (1) and (2):

Comparing this with (a, b)$(a, b)$ gives a = 0$a = 0$ and b = 5$b = 5$.

Now calculate the value:

Pattern Recognition

Modulus domain constraint: The function x + |x|$x + |x|$ is non-zero only for positive values of x$x$. This is a standard math trick that collapses complex domains down to x > 0$x > 0$ instantly.

Chapter Mix

Class 11 Mathematics: Relations and Functions

Related Formula

Definition of properties of binary relations:

• Reflexive: (f, f) in R f(0) = f(1)$(f, f) \in R \iff f(0) = f(1)$
• Symmetric: (f, g) in R (g, f) in R$(f, g) \in R \implies (g, f) \in R$
• Transitive: (f, g) in R and (g, h) in R (f, h) in R$(f, g) \in R \text{ and } (g, h) \in R \implies (f, h) \in R$

Core Logic

Evaluate reflexivity, symmetry, and transitivity sequentially by plugging standard arbitrary function values into the condition definition.

Step 1: Reflexivity Audit

For (f,f) in R$(f,f) \in R$, we require f(0) = f(1)$f(0) = f(1)$ and f(1) = f(0)$f(1) = f(0)$. This holds true only for functions whose values at 0$0$ and 1$1$ are identical. Since it does not hold true for all possible functions mapping Z → Z$\mathbb{Z} \to \mathbb{Z}$ (e.g., f(x)=x$f(x)=x$), R$R$ is not reflexive.

Step 2: Symmetry Audit

Assume (f,g) in R f(0) = g(1)$(f,g) \in R \implies f(0) = g(1)$ and f(1) = g(0)$f(1) = g(0)$. To check if (g,f) in R$(g,f) \in R$, check its matching constraints: g(0) = f(1)$g(0) = f(1)$ and g(1) = f(0)$g(1) = f(0)$. Both statements are perfectly identical to our assumption. Therefore, R$R$ is symmetric.

Step 3: Transitivity Audit

Assume (f,g) in R f(0) = g(1), f(1) = g(0)$(f,g) \in R \implies f(0) = g(1), f(1) = g(0)$. Assume (g,h) in R g(0) = h(1), g(1) = h(0)$(g,h) \in R \implies g(0) = h(1), g(1) = h(0)$. For (f,h) in R$(f,h) \in R$, we need f(0) = h(1)$f(0) = h(1)$ and f(1) = h(0)$f(1) = h(0)$. From assumptions: f(0) = g(1) = h(0)$f(0) = g(1) = h(0)$ and f(1) = g(0) = h(1)$f(1) = g(0) = h(1)$. This means f(0) = h(0)$f(0) = h(0)$ and f(1) = h(1)$f(1) = h(1)$, which does not necessarily satisfy f(0)=h(1)$f(0)=h(1)$. Hence, R$R$ is not transitive.

Pattern Recognition

The relation swaps indices 0$0$ and 1$1$. Swapping twice returns you to the original position, which visually justifies why symmetry holds trivially, while transitivity creates a cyclic dependency that fails standard property constraints.

Chapter Mix

Class 12 Mathematics: Relations and Functions

Related Formula

For a logarithmic term b(g(x))$\log_b(g(x))$ to be defined:

• g(x) > 0$g(x) > 0$
• b > 0, b ≠ 1$b > 0, b \neq 1$

Core Logic

Let's set defining inequalities sequentially:

• Inside the outer logarithm:

Since base is 4 > 1$4 > 1$:

Step 1: Finding bounds for inner logarithmic term

• Inside the inner logarithm:

Step 2: Intersection of regions

Taking the intersection of (1) and (2):

This gives:

Calculating the sum:

Pattern Recognition

Logarithmic domains must check arguments from the innermost level to the outermost level. Remember that bases >1$>1$ maintain inequality direction upon exponentiation, while bases <1$<1$ reverse it.

Chapter Mix

Class 11 Mathematics: Relations and Functions

Related Formula

A relation R$R$ on set A$A$ is:

• Reflexive: If (x, x) in R$(x, x) \in R$ for all x in A$x \in A$.
• Symmetric: If (x, y) in R (y, x) in R$(x, y) \in R \implies (y, x) \in R$.

Core Logic

Let's find the explicit set R$R$ using y = x, 1$y = \max\{x, 1\}$:

• x = -2 y = 1 (-2, 1) in R$x = -2 \implies y = 1 \implies (-2, 1) \in R$
• x = -1 y = 1 (-1, 1) in R$x = -1 \implies y = 1 \implies (-1, 1) \in R$
• x = 0 y = 1 (0, 1) in R$x = 0 \implies y = 1 \implies (0, 1) \in R$
• x = 1 y = 1 (1, 1) in R$x = 1 \implies y = 1 \implies (1, 1) \in R$
• x = 2 y = 2 (2, 2) in R$x = 2 \implies y = 2 \implies (2, 2) \in R$
• x = 3 y = 3 (3, 3) in R$x = 3 \implies y = 3 \implies (3, 3) \in R$

Thus, l = 6$l = 6$ elements.

Step 1: Making the Relation Reflexive

For R$R$ to be reflexive, it must contain all elements (x, x)$(x, x)$ where x in A = -2, -1, 0, 1, 2, 3$x \in A = \{-2, -1, 0, 1, 2, 3\}$. Currently, R$R$ has (1,1), (2,2), (3,3)$(1,1), (2,2), (3,3)$. Missing elements: (-2, -2), (-1, -1), (0, 0)$(-2, -2), (-1, -1), (0, 0)$.

Thus, minimum number of elements to add: m = 3$m = 3$

Step 2: Making the Relation Symmetric

For R$R$ to be symmetric, if (x, y) in R$(x, y) \in R$ and x ≠ y$x \neq y$, then (y, x)$(y, x)$ must also be in R$R$.

• (-2, 1) in R$(-2, 1) \in R \implies$ need (1, -2)$(1, -2)$
• (-1, 1) in R$(-1, 1) \in R \implies$ need (1, -1)$(1, -1)$
• (0, 1) in R$(0, 1) \in R \implies$ need (1, 0)$(1, 0)$

Missing elements to form symmetric pairs: (1, -2), (1, -1), (1, 0)${(1, -2), (1, -1), (1, 0)}$.

Thus, minimum number of elements to add: n = 3$n = 3$

Calculating total:

Pattern Recognition

To quickly solve counting tasks of elements in relations: Write down the pairs explicitly since the set size is small (|A|=6$|A|=6$). Count the elements already satisfying standard relations, then subtract from |A|$|A|$ to find missing diagonal terms for reflexivity.

Chapter Mix

Class 12 Mathematics: Relations and Functions

Related Formula

A functional equation relates the values of a function at different arguments. We can find values by substituting symmetric inputs that map to each other (e.g., x$x$ and (24)/(x)$\frac{24}{x}$).

Core Logic

Given:

Step 1: Substitution of values

Substitute x = 3$x = 3$:

Substitute x = 8$x = 8$:

Step 2: Linear combination of equations

Add equations (2) and (3) directly:

Pattern Recognition

Instead of solving for the general function f(x)$f(x)$ (which is also easy by substitution: replace x → 24/x$x \to 24/x$), look at the symmetric nature of the target expression f(3) + f(8)$f(3) + f(8)$. Direct addition of symmetric systems avoids resolving the individual values and saves time.

Chapter Mix

Class 11 Mathematics: Relations and Functions