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Relations and Functions - Functions

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

A function ff from a set AA to a set BB is a specific type of relation where every element aAa \in A has exactly one unique image bBb \in B, denoted as f(a)=bf(a) = b. Visually, this is represented by an arrow diagram where every point in the starting set (Domain) has exactly one outgoing arrow, and no point has multiple arrows originating from it.

The domain of a function f:ABf: A \rightarrow B is the set AA, while the codomain is the set BB. The range is the set of all actual outputs or images, defined as {f(x):xA}\{f(x) : x \in A\}. On a Cartesian coordinate system, the domain is represented by the span of the graph along the xx-axis, while the range is represented by the span along the yy-axis.

A Real-Valued Function is a function where the range is a subset of the set of real numbers R\mathbb{R}. If the domain is also R\mathbb{R} or a subset of it, it is called a Real Function. The graph of such a function is a curve or line drawn on the xyxy-plane.

The Identity Function f:RRf: \mathbb{R} \rightarrow \mathbb{R} is defined by f(x)=xf(x) = x for each xRx \in \mathbb{R}. Its graph is a straight line passing through the origin, bisecting the first and third quadrants at a 4545^{\circ} angle to both axes.

The Modulus Function f(x)=xf(x) = |x| is defined as xx if x0x \geq 0 and x-x if x<0x < 0. Visually, the graph forms a characteristic 'V' shape with the vertex at the origin (0,0)(0,0), where the right arm is the line y=xy = x and the left arm is the line y=xy = -x.

The Greatest Integer Function f(x)=[x]f(x) = [x] assumes the value of the greatest integer less than or equal to xx. Its graph is often called a 'step function' or 'staircase graph' because it consists of horizontal line segments of length 1, where the left endpoint is included (closed circle) and the right endpoint is excluded (open circle).

The Signum Function f(x)f(x) returns 11 if x>0x > 0, 00 if x=0x = 0, and 1-1 if x<0x < 0. Geometrically, this looks like two disjoint horizontal rays: one at y=1y=1 for positive xx and one at y=1y=-1 for negative xx, with a single isolated point at the origin (0,0)(0,0).

Algebra of Functions: For two real functions f:XRf: X \rightarrow \mathbb{R} and g:XRg: X \rightarrow \mathbb{R}, we define addition as (f+g)(x)=f(x)+g(x)(f+g)(x) = f(x) + g(x), subtraction as (fg)(x)=f(x)g(x)(f-g)(x) = f(x) - g(x), and multiplication as (fg)(x)=f(x)g(x)(fg)(x) = f(x)g(x). For division, (fg)(x)=f(x)g(x)(\frac{f}{g})(x) = \frac{f(x)}{g(x)}, provided g(x)0g(x) \neq 0.

📐Formulae

Identity Function: f(x)=xf(x) = x

Constant Function: f(x)=cf(x) = c, where cc is a constant

Modulus Function: f(x)=x={x,x0x,x<0f(x) = |x| = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases}

Signum Function: f(x)={1,x>00,x=01,x<0=xx,x0f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases} = \frac{|x|}{x}, x \neq 0

Greatest Integer Function: f(x)=[x]f(x) = [x], where nx<n+1[x]=nn \leq x < n+1 \Rightarrow [x] = n

Domain of fg\frac{f}{g}: DfDg{x:g(x)=0}D_f \cap D_g - \{x : g(x) = 0\}

Domain of f(x)\sqrt{f(x)}: {xDf:f(x)0}\{x \in D_f : f(x) \geq 0\}

💡Examples

Problem 1:

Find the domain and range of the real function f(x)=9x2f(x) = \sqrt{9 - x^2}.

Solution:

  1. For f(x)f(x) to be defined as a real function, the expression inside the square root must be non-negative: 9x209 - x^2 \geq 0. \ 2. Factoring the inequality: (3x)(3+x)0(3 - x)(3 + x) \geq 0. This implies 3x3-3 \leq x \leq 3. So, Domain =[3,3]= [-3, 3]. \ 3. To find the range, let y=9x2y = \sqrt{9 - x^2}. Since square roots are non-negative, y0y \geq 0. \ 4. Squaring both sides: y2=9x2x2=9y2y^2 = 9 - x^2 \Rightarrow x^2 = 9 - y^2. \ 5. Since x20x^2 \geq 0, then 9y20y293y39 - y^2 \geq 0 \Rightarrow y^2 \leq 9 \Rightarrow -3 \leq y \leq 3. \ 6. Combining y0y \geq 0 and 3y3-3 \leq y \leq 3, we get 0y30 \leq y \leq 3. So, Range =[0,3]= [0, 3].

Explanation:

We determine the domain by ensuring the radicand of the square root is non-negative. For the range, we solve for xx in terms of yy and apply the constraints of the square root's output.

Problem 2:

Let f(x)=x+1f(x) = x + 1 and g(x)=2x3g(x) = 2x - 3. Find (f+g)(x)(f + g)(x), (fg)(x)(f - g)(x), and (fg)(x)(\frac{f}{g})(x).

Solution:

  1. Addition: (f+g)(x)=f(x)+g(x)=(x+1)+(2x3)=3x2(f + g)(x) = f(x) + g(x) = (x + 1) + (2x - 3) = 3x - 2. \ 2. Subtraction: (fg)(x)=f(x)g(x)=(x+1)(2x3)=x+12x+3=x+4(f - g)(x) = f(x) - g(x) = (x + 1) - (2x - 3) = x + 1 - 2x + 3 = -x + 4. \ 3. Division: (fg)(x)=f(x)g(x)=x+12x3(\frac{f}{g})(x) = \frac{f(x)}{g(x)} = \frac{x + 1}{2x - 3}. \ 4. Condition for division: The denominator cannot be zero, so 2x30x322x - 3 \neq 0 \Rightarrow x \neq \frac{3}{2}.

Explanation:

This demonstrates the algebraic operations on functions. For addition and subtraction, we combine like terms. For division, we must explicitly state the restriction on the domain where the divisor is zero.