If the domain of the function f(x) = ⁻¹((2x - 5)/(11 - 3x)) + ⁻¹(2x² - 3x + 1)$f(x) = \cos^{-1}\left(\frac{2x - 5}{11 - 3x}\right) + \sin^{-1}(2x^2 - 3x + 1)$ is the interval [α, β]$[\alpha, \beta]$ , then α + 2β$\alpha + 2\beta$ is equal to :

Core Logic

Let's first determine the domain of f(x)$f(x)$. For logarithmic expressions, the argument must be strictly positive:

Hence, x in (-3, -1)$x \in (-3, -1)$, which gives α = -3$\alpha = -3$ and β = -1$\beta = -1$.

Step 1: Finding the domain of g(x)

For the function g(x) = ⁻¹((7x+10)/(x-2))$g(x) = \sin^{-1}\left(\frac{7x+10}{x-2}\right)$, the argument must lie within [-1, 1]$[-1, 1]$:

Let's break this into two separate inequalities:

Inequality A: (7x+10)/(x-2) ≥ -1 (7x+10+x-2)/(x-2) ≥ 0 (8x+8)/(x-2) ≥ 0 x in (-∞, -1] (2, ∞)$\frac{7x+10}{x-2} \ge -1 \implies \frac{7x+10+x-2}{x-2} \ge 0 \implies \frac{8x+8}{x-2} \ge 0 \implies x \in (-\infty, -1] \cup (2, \infty)$

Inequality B: (7x+10)/(x-2) ≤ 1 (7x+10-x+2)/(x-2) ≤ 0 (6x+12)/(x-2) ≤ 0 x in [-2, 2)$\frac{7x+10}{x-2} \le 1 \implies \frac{7x+10-x+2}{x-2} \le 0 \implies \frac{6x+12}{x-2} \le 0 \implies x \in [-2, 2)$

Taking the intersection of both intervals:

x in [-2, -1]$x \in [-2, -1]$

Thus, γ = -2$\gamma = -2$ and δ = -1$\delta = -1$.

Step 2: Computing the final sum of squares

Now we calculate α² + β² + γ² + δ²$\alpha^2 + \beta^2 + \gamma^2 + \delta^2$:

Pattern Recognition

For nested logs, start from the outermost log condition and work your way inward step-by-step. Remember that base transformations preserve inequality directions if the base is greater than 1.

Chapter Mix

Class 12 Mathematics: Relations and Functions

Related Formula

For a composite function f(g(x))$f(g(x))$:

Core Logic

Substitute g(x) = (2 - 3x)/(1 - x)$g(x) = \frac{2 - 3x}{1 - x}$ into f(x)$f(x)$:

For the domain interval [2, 4]$[2, 4]$, calculate the boundary values since the function is monotonic:

Thus, the range [α, β] = [(1)/(2), (29)/(56)]$[\alpha, \beta] = \left[\frac{1}{2}, \frac{29}{56}\right]$.

Step 1: Calculate the Difference

Pattern Recognition

When dealing with composite functions of linear fractions, simplify algebraically first. If the resulting function has no vertical asymptote in the specified interval, it is monotonic, and the extreme values occur exactly at the endpoints.

Chapter Mix

Class 11 Mathematics: Sets, Relations and Functions Class 12 Mathematics: Relations and Functions

Related Formula

The number of one-one (injective) functions from a set D$D$ with n(D)$n(D)$ elements to a set C$C$ with n(C)$n(C)$ elements is given by:

Core Logic

Step 1: Find the number of elements in set D = A B$D = A \cap B$. Solving the equations of circle A$A$ and ellipse B$B$ simultaneously: From A$A$, y² = 25 - x²$y^2 = 25 - x^2$. Substitute this into B$B$:

Correspondingly, y = ± √(119)2√(2)$y = \pm\frac{\sqrt{119}}{2\sqrt{2}}$. Thus, there are exactly 4 distinct intersection points, so n(D) = 4$n(D) = 4$.

Step 1: Count elements in Set C

Set C$C$ consists of integral lattice points (x,y)$(x,y)$ inside or on the circle x² + y² ≤ 4$x^2 + y^2 \le 4$: Possible integer pairs are: (0,0), (±1,0), (0,±1), (±2,0), (0,±2), (±1,±1)$(0,0), (\pm1,0), (0,\pm1), (\pm2,0), (0,\pm2), (\pm1,\pm1)$. Counting them yields n(C) = 13$n(C) = 13$ elements.

Step 2: Calculate Injective Functions

The total number of one-one functions from D$D$ to C$C$ is:

Pattern Recognition

The intersection of a concentric circle and ellipse always yields 4 points if they cross completely. Break down the problem by counting the cardinality of domain and codomain independently before applying permutation formulas.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 12 Mathematics: Relations and Functions

Related Formula

Sum of first n$n$ squares:

Core Logic

From functional relation f(x) - f((x)/(2)) = (x)/(2)$f(x) - f\left(\frac{x}{2}\right) = \frac{x}{2}$. Write a telescoping sequence by scaling variable down:

$\dots$

Step 1: Evaluate the Limit Definition

Summing all equations creates a telescoping sum on the left side:

Taking the limit as n → ∞$n \to \infty$:

Step 2: Final Sum Evaluation

We need to compute Σr=1¹⁰ G(r²) = Σr=1¹⁰ r²$\sum_{r=1}^{10} G(r^2) = \sum_{r=1}^{10} r^2$:

Pattern Recognition

Linear iterative arguments of type f(2x)-f(x)=x$f(2x)-f(x)=x$ naturally condense into geometric progression properties via geometric series limits. Always look for telescoping patterns in infinite limits of difference terms.

Chapter Mix

Class 11 Mathematics: Sequence and Series Class 12 Mathematics: Relations and Functions

Related Formula

An ellipse equation is structured as:

Core Logic

Analyzing set A$A$:

Thus, region A$A$ forms a rectangle bounded between x in [-3, 5]$x \in [-3, 5]$ and y in [-1, 11]$y \in [-1, 11]$.

Analyzing set B$B$:

Dividing by 144:

This represents the interior and boundary of an ellipse centered at (2, 6)$(2, 6)$ with semi-minor axis a = 3$a = 3$ and semi-major axis b = 4$b = 4$.

Step 1: Spatial Inclusion Check

Let's check the extreme horizontal and vertical extents of the ellipse B$B$: Horizontal extent: x in [2 - 3, 2 + 3] = [-1, 5]$x \in [2 - 3, 2 + 3] = [-1, 5]$ Vertical extent: y in [6 - 4, 6 + 4] = [2, 10]$y \in [6 - 4, 6 + 4] = [2, 10]$

Comparing with the boundaries of rectangle A$A$ (x in [-3, 5]$x \in [-3, 5]$ and y in [-1, 11]$y \in [-1, 11]$):

Since all points of the ellipse lie perfectly inside the rectangular region, we conclusively find that B A$B \subset A$.

Pattern Recognition

A bounding box check (finding h ± a$h \pm a$ and k ± b$k \pm b$) for conics is the fastest analytical shortcut to verify set inclusion without plotting extensive coordinates.

Chapter Mix

Class 11 Mathematics: Sets, Relations and Functions Class 11 Mathematics: Conic Sections