Punjab Board Class 11 Mathematics

Chapter 2: Functions and Graphs — Class 11 Maths Notes

Functions and graphs: domain and range, inverse functions and graph transformations, explained with worked examples. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 2.

Exercise solutions — Chapter 2

Every exercise of Functions and Graphs solved step by step.

Exercise 2.1 Exercise 2.2

Mathematics 11 (Punjab Board)

Chapter 2: Functions and Graphs — Theory

Definitions, key results and graph facts for Punjab Board Class 11 Mathematics, Chapter 2: Functions and Graphs.

What Is a Function?

A relation pairs inputs with outputs. A function is a special relation: har input ka exactly ek output hota hai — no input is allowed to produce two different outputs.

If \(A\) and \(B\) are sets, a function \(f\) from \(A\) to \(B\) (written \(f: A \to B\)) assigns to every element \(x\) in \(A\) exactly one element \(y\) in \(B\). We write \(y = f(x)\), and call \(x\) the independent variable and \(y\) the dependent variable.

f: A \to B, \qquad y = f(x)

Domain is the set of all allowed inputs. Co-domain is the set we declare as the target. Range is the set of outputs actually produced. The range is always a subset of the co-domain.

Key point: On a graph, the vertical line test decides: agar koi vertical line graph ko ek se zyada points par kaate, to woh graph kisi function ka nahi hai.

One-to-One, Onto and Bijective Functions

One-to-one (injective): different inputs give different outputs. Formally, \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\). Example: \(f(x) = 3x - 5\) is one-to-one, but \(g(x) = x^2 - 3x\) is not, because \(g(0) = 0 = g(3)\).

f \text{ is one-to-one if } f(x_1) = f(x_2) \implies x_1 = x_2

Onto (surjective): the range equals the co-domain — co-domain ka har element kisi na kisi input ka output hai. Example: \(f(x) = 3x - 5\) with co-domain \(\mathbb{R}\) is onto, but \(f(x) = \frac{2x-3}{x+1}\) with co-domain \(\mathbb{R}\) is not onto, because the value \(y = 2\) is never attained.

Bijective: a function that is both one-to-one and onto. Only a bijective function has an inverse function.

Key point: To prove one-to-one, start from \(f(x_1) = f(x_2)\) and derive \(x_1 = x_2\). To disprove, one counter-example pair (like \(0\) and \(3\) for \(x^2-3x\)) is enough.

Domain and Range of Standard Functions

Polynomial functions (linear, quadratic, cubic) are defined for every real number: domain \(\mathbb{R}\).

Rational functions: denominator kabhi zero nahi ho sakta. Exclude every \(x\) that makes the denominator zero. To find the range, write \(y = f(x)\), solve for \(x\) in terms of \(y\), and exclude the \(y\)-values that make the expression impossible.

\text{For } f(x) = \frac{x+2}{3-x}: \quad \text{Domain } = \mathbb{R} - \{3\}, \quad \text{Range } = \mathbb{R} - \{-1\}

Even roots (square root): the quantity inside must be non-negative. Odd roots (cube root) accept every real number.

Absolute value \(|x - a|\) is defined for all real \(x\) and is never negative, so its range starts at \(0\).

Key point: Domain restrictions are the first thing to write down — zero denominator and negative radicand are the two traps examiners set most often.

The Difference Quotient

The expression \(\frac{f(a+h) - f(a)}{h}\) (for \(h \neq 0\)) measures the average rate of change of \(f\) between \(a\) and \(a+h\). Is mein pehle \(f(a+h)\) aur \(f(a)\) nikaal kar subtract karte hain, phir \(h\) se divide karke simplify karte hain.

\text{Difference quotient} = \frac{f(a+h) - f(a)}{h}, \qquad h \neq 0

For a linear function the result is just the slope; for \(\sin x\) and \(\tan x\) the standard sum-to-product identities simplify it.

Key point: Expand \(f(a+h)\) fully before subtracting \(f(a)\) — most sign errors happen when the subtraction bracket is skipped.

Inverse Functions

If \(f\) is bijective, its inverse \(f^{-1}\) undoes it: \(f^{-1}\) takes each output back to the input that produced it. The domain of \(f^{-1}\) is the range of \(f\), and the range of \(f^{-1}\) is the domain of \(f\) — domain aur range aapas mein swap ho jate hain.

f(f^{-1}(x)) = x, \qquad f^{-1}(f(x)) = x

A function that is not one-to-one (like \(x^2\) on all of \(\mathbb{R}\)) has no inverse unless its domain is restricted to a part where it is one-to-one.

Key point: No bijective, no inverse. Always check one-to-one and onto before writing \(f^{-1}\).

Intercepts and Intersection of Graphs

The y-intercept is where the graph meets the y-axis: put \(x = 0\). The x-intercept is where it meets the x-axis: put \(y = 0\) and solve for \(x\).

Two graphs \(y = f(x)\) and \(y = g(x)\) intersect where \(f(x) = g(x)\). Is equation ko solve karke \(x\)-values nikaalte hain, phir kisi ek function mein rakh kar \(y\) nikaalte hain.

f(x) = g(x) \quad \text{gives the x-coordinates of intersection points}

Line with line: one intersection point. Line with parabola: zero, one (tangent — repeated root of the quadratic) or two points.

Key point: A repeated (double) root means the line just touches the curve — tangent case, sirf ek intersection point.

Graphs of Square Root and Cube Root Functions

The basic square root graph \(y = \sqrt{x}\) starts at the origin \((0,0)\) and curves upward to the right. Its domain is \(x \ge 0\) and its range is \(y \ge 0\).

Transformations shift or flip this shape: \(y = \sqrt{x} + 5\) moves it 5 units up (starting point \((0,5)\)); \(y = -\sqrt{x+1} + 2\) moves it left 1, up 2, and reflects it downward (starting point \((-1,2)\), range \(y \le 2\)); \(y = a\sqrt{x}\) stretches it vertically.

The cube root graph \(y = \sqrt[3]{x}\) is defined for all real \(x\), passes through the origin, and has an S-shape — negative \(x\) par negative \(y\), positive \(x\) par positive \(y\).

Key point: For graphing, first write domain and range, then build a small table with easy values (perfect squares for square roots), then join the points with a smooth curve.

Exponential Growth and Decay Models

Quantities that grow (or decay) in proportion to their current size follow the exponential model \(P(t) = P_0 e^{rt}\), where \(P_0\) is the initial amount, \(r\) the rate per unit time, and \(t\) the time.

P(t) = P_0 e^{rt}, \qquad \text{doubling time } t = \frac{\ln 2}{r}

Radioactive decay with half-life \(h\) follows \(Q(t) = Q_0\left(\frac{1}{2}\right)^{t/h}\). After each half-life the quantity is halved.

Key point: Doubling time depends only on the rate \(r\), not on the starting amount \(P_0\).

Mathematics 11 (Punjab Board)

Solved Examples

Worked textbook examples for Punjab Board Class 11 Mathematics, Chapter 2: Functions and Graphs.

Example 1: Given \(f(x) = x^3 - 2x^2 + 4x - 1\), find (i) \(f(0)\) (ii) \(f(1)\) (iii) \(f(-2)\) (iv) \(f(1+x)\) (v) \(f\left(\frac{1}{x}\right)\)

Is example mein function ki value nikalni hai — jahan \(x\) likha hai wahan di gayi value rakh kar simplify karenge. \(f(1+x)\) mein poora bracket \((1+x)\) substitute hoga, is liye expansion carefully karenge.

(i) \(f(0) = 0^3 - 2(0)^2 + 4(0) - 1 = -1\)

(ii) \(f(1) = 1^3 - 2(1)^2 + 4(1) - 1 = 1 - 2 + 4 - 1 = 2\)

(iii) \(f(-2) = (-2)^3 - 2(-2)^2 + 4(-2) - 1 = -8 - 8 - 8 - 1 = -25\)

(iv) Substitute \(x = 1+x\) and expand using \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\) and \((a+b)^2 = a^2 + 2ab + b^2\):

\[ f(1+x) = (1+x)^3 - 2(1+x)^2 + 4(1+x) - 1 \]

\[ = (1 + 3x + 3x^2 + x^3) - 2(1 + 2x + x^2) + (4 + 4x) - 1 \]

\[ = 1 + 3x + 3x^2 + x^3 - 2 - 4x - 2x^2 + 4 + 4x - 1 \]

\[ = x^3 + x^2 + 3x + 2 \]

(v) \(f\left(\frac{1}{x}\right) = \frac{1}{x^3} - \frac{2}{x^2} + \frac{4}{x} - 1\), valid for \(x \neq 0\).

Answer: (i) \(-1\) (ii) \(2\) (iii) \(-25\) (iv) \(x^3 + x^2 + 3x + 2\) (v) \(\frac{1}{x^3} - \frac{2}{x^2} + \frac{4}{x} - 1\), \(x \neq 0\)

Common mistake: In \(f(-2)\), students write \(-2^2 = -4\) instead of \((-2)^2 = 4\). Bracket lagana na bhoolen.

Example 2: Find the domain and range of \(f(x) = x^2\).

Har real number ka square nikal sakta hai, is liye domain mein koi restriction nahi. Lekin kisi bhi real number ka square kabhi negative nahi hota — yahi range tay karega.

Every real number \(x\) can be squared, so there is no restriction on the input.

For every real \(x\), \(x^2 \ge 0\), and every non-negative number is the square of some real number. Hence the outputs cover exactly \([0, \infty)\).

Answer: Domain \(= \mathbb{R}\), Range \(= [0, \infty)\)

Example 3: Find the domain and range of \(f(x) = \frac{x}{x^2 - 4}\).

Rational function mein denominator zero nahi ho sakta — yahan se domain milega. Range ke liye \(y = f(x)\) likh kar \(x\) ko \(y\) ke terms mein solve karenge, phir dekhenge ke \(y\) ki kaunsi values par \(x\) real rehta hai.

Domain: \(x^2 - 4 \neq 0 \implies x \neq \pm 2\). So Domain \(= \mathbb{R} - \{-2, 2\}\).

Range: Let \(y = \frac{x}{x^2-4}\). Cross-multiplying:

yx^2 - 4y = x \implies yx^2 - x - 4y = 0

This is quadratic in \(x\). If \(y = 0\), the equation gives \(x = 0\), which is valid, so \(0\) is in the range. If \(y \neq 0\), \(x\) is real exactly when the discriminant is non-negative:

\Delta = (-1)^2 - 4(y)(-4y) = 1 + 16y^2 > 0 \quad \text{for every real } y

Since \(1 + 16y^2\) is always positive, \(x\) is real for every real \(y\).

Answer: Domain \(= \mathbb{R} - \{-2, 2\}\), Range \(= \mathbb{R}\)

Example 4: Find the domain and range of \(f(x) = \sqrt{x^2 - 9}\).

Square root ke andar wali quantity negative nahi ho sakti. Is liye \(x^2 - 9 \ge 0\) solve karke domain nikaalte hain. Square root ka result hamesha non-negative hota hai — yahi range dega.

Domain: \(x^2 - 9 \ge 0 \implies (x-3)(x+3) \ge 0 \implies x \le -3\) or \(x \ge 3\).

Range: The square root of any non-negative number is non-negative, and as \(|x|\) grows, \(\sqrt{x^2-9}\) takes every value \(\ge 0\).

Answer: Domain \(= (-\infty, -3] \cup [3, \infty)\), Range \(= [0, \infty)\)

Example 5: Show that \(f(x) = x + 1\), with domain and co-domain both \(\mathbb{R}\), is bijective.

Bijective prove karne ke liye do cheezein dikhani hain: one-to-one aur onto. One-to-one ke liye \(f(x_1) = f(x_2)\) se \(x_1 = x_2\) nikaalte hain; onto ke liye har \(y\) ke liye ek \(x\) dhoondhte hain.

One-to-one: Suppose \(f(x_1) = f(x_2)\). Then:

x_1 + 1 = x_2 + 1 \implies x_1 = x_2

So \(f\) is one-to-one.

Onto: For any real \(y\) in the co-domain, choose \(x = y - 1\) (which is real). Then \(f(x) = (y-1) + 1 = y\). Every \(y\) has a pre-image, so \(f\) is onto.

Since \(f\) is both one-to-one and onto, it is bijective.

Answer: \(f(x) = x+1\) is bijective on \(\mathbb{R}\).

Example 6: Show that \(f(x) = x^2 - 2\), with domain and co-domain both \(\mathbb{R}\), is neither one-to-one nor onto.

Disprove karne ke liye sirf ek counter-example kaafi hai. One-to-one ke liye do different inputs dhoondhenge jinka output same ho; onto ke liye co-domain ka ek aisa element jiska koi pre-image na ho.

Not one-to-one: \(f(2) = 4 - 2 = 2\) and \(f(-2) = 4 - 2 = 2\). Here \(2 \neq -2\) but \(f(2) = f(-2)\), so \(f\) is not one-to-one.

Not onto: Since \(x^2 \ge 0\) for all real \(x\), we have \(f(x) = x^2 - 2 \ge -2\). The number \(-3\) lies in the co-domain \(\mathbb{R}\) but \(f(x) = -3\) would need \(x^2 = -1\), which has no real solution. So \(-3\) has no pre-image, and \(f\) is not onto.

Answer: \(f(x) = x^2 - 2\) is neither one-to-one nor onto on \(\mathbb{R}\).

Example 7: Find the points of intersection of the line \(y = 2x + 6\) with the coordinate axes graphically.

Axes se intersection ka matlab x-intercept aur y-intercept. Y-intercept ke liye \(x = 0\), x-intercept ke liye \(y = 0\) rakh kar solve karenge, phir points ki table bana kar line draw karenge.

y-intercept: put \(x = 0\): \(y = 2(0) + 6 = 6\). Point: \((0, 6)\).

x-intercept: put \(y = 0\): \(0 = 2x + 6 \implies x = -3\). Point: \((-3, 0)\).

Table of values for plotting:

\(x\)	\(-3\)	\(-2\)	\(-1\)	\(0\)
\(y = 2x+6\)	\(0\)	\(2\)	\(4\)	\(6\)

Joining these points with a straight line, the graph cuts the x-axis at \((-3, 0)\) and the y-axis at \((0, 6)\).

Answer: \((-3, 0)\) and \((0, 6)\)

Example 8: Find the point of intersection of \(y = 3x + 2\) and \(y = -x + 6\) graphically.

Do lines ka intersection woh point hai jahan dono ki \(y\)-values barabar hon. Is liye \(3x + 2 = -x + 6\) solve karke \(x\) nikaalte hain, phir \(y\) nikaalte hain. Table bana kar dono lines plot karne par wahi point milega.

Set the two expressions for \(y\) equal:

3x + 2 = -x + 6 \implies 4x = 4 \implies x = 1

Put \(x = 1\) in \(y = 3x + 2\): \(y = 3(1) + 2 = 5\). (Check: \(y = -1 + 6 = 5\). ✓)

Tables for plotting:

\(x\)	\(0\)	\(1\)	\(2\)
\(y = 3x+2\)	\(2\)	\(5\)	\(8\)
\(y = -x+6\)	\(6\)	\(5\)	\(4\)

Both lines pass through \((1, 5)\), which is the point of intersection.

Answer: \((1, 5)\)

Example 9: Solve the linear function \(y = -x + 3\) and the quadratic function \(y = x^2 - 6x + 3\) graphically.

Line aur parabola ka intersection nikalne ke liye dono ko barabar rakhte hain — ek quadratic equation banegi jiske roots intersection points ke \(x\)-coordinates honge.

Set \(-x + 3 = x^2 - 6x + 3\):

x^2 - 5x = 0 \implies x(x - 5) = 0 \implies x = 0 \text{ or } x = 5

For \(x = 0\): \(y = -0 + 3 = 3\). For \(x = 5\): \(y = -5 + 3 = -2\). (Check on the parabola: \(0 - 0 + 3 = 3\) ✓, \(25 - 30 + 3 = -2\) ✓.)

For the graph: the line has intercepts \((0, 3)\) and \((3, 0)\). The parabola \(y = x^2 - 6x + 3\) has y-intercept \((0, 3)\), vertex at \(h = \frac{6}{2} = 3\), \(k = 9 - 18 + 3 = -6\), i.e. \((3, -6)\), and x-intercepts at \(x = 3 \pm \sqrt{6} \approx 0.55\) and \(5.45\). Plotting both, they meet at \((0, 3)\) and \((5, -2)\).

Answer: \((0, 3)\) and \((5, -2)\)

Example 10: Graph the square root function \(y = 2\sqrt{x} + 1\).

Pehle domain aur range likhenge, phir aisi \(x\)-values ki table banayenge jinke square root easy hon (0, 1, 4, 9), aur points ko smooth curve se join karenge.

Domain: \(x \ge 0\) (the square root of a negative number is not real). Range: since \(\sqrt{x} \ge 0\), \(y = 2\sqrt{x} + 1 \ge 1\).

\(x\)	\(0\)	\(1\)	\(4\)	\(9\)
\(y = 2\sqrt{x}+1\)	\(1\)	\(3\)	\(5\)	\(7\)

The graph starts at \((0, 1)\) and curves upward to the right — the basic \(\sqrt{x}\) shape, stretched vertically by 2 and shifted up 1 unit.

Answer: Curve starting at \((0,1)\), rising to the right; Domain \(x \ge 0\), Range \(y \ge 1\)

Example 11: Graph the cube root function \(y = \sqrt[3]{x} - 1\).

Cube root har real number ke liye defined hai, is liye domain aur range dono \(\mathbb{R}\) honge. Perfect cubes (-8, -1, 0, 1, 8) table ke liye best hain.

Domain: all real numbers. Range: all real numbers.

\(x\)	\(-8\)	\(-1\)	\(0\)	\(1\)	\(8\)
\(y = \sqrt[3]{x}-1\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)

Joining the points gives the S-shaped cube root curve, shifted 1 unit down, passing through \((0,-1)\) and \((1,0)\).

Answer: S-shaped curve through \((0,-1)\); Domain \(= \mathbb{R}\), Range \(= \mathbb{R}\)

Example 12: A stock valued at Rs. 5,000 grows at 5% per year with \(P(t) = P_0 e^{rt}\). (i) Find its value after 10 years. (ii) After how many years will it double?

Exponential growth model mein given values substitute karenge. Doubling ke liye \(P(t) = 2P_0\) rakh kar natural log lete hain — \(P_0\) cancel ho jayega.

Given \(P_0 = 5000\), \(r = 0.05\).

(i) Using the exponential growth formula \(P(t) = P_0 e^{rt}\):

P(10) = 5000 \cdot e^{0.05 \times 10} = 5000 \cdot e^{0.5}

With \(e^{0.5} \approx 1.6487\): \(P(10) \approx 5000 \times 1.6487 = 8243.6\).

(ii) Set \(P(t) = 2P_0\) and take the natural logarithm of both sides:

2P_0 = P_0 e^{rt} \implies 2 = e^{rt} \implies \ln 2 = rt \implies t = \frac{\ln 2}{r}

t = \frac{0.6931}{0.05} \approx 13.86

Answer: (i) Rs. 8,244 (approx.) (ii) About 13.86 years (≈ 14 years)

Example 13: Pollutant concentration in a lake is \(C(t) = \frac{100}{\sqrt{t+1}}\) ppm, where \(t\) is days since introduction. (i) Find the concentration after 4 days. (ii) After how many days will it drop below 10 ppm?

Part (i) mein sirf \(t = 4\) substitute karna hai. Part (ii) mein inequality \(C(t) < 10\) solve karenge — dono taraf positive hon to square karna safe hai.

(i)

C(4) = \frac{100}{\sqrt{4+1}} = \frac{100}{\sqrt{5}} \approx \frac{100}{2.236} \approx 44.72

(ii) Solve \(\frac{100}{\sqrt{t+1}} < 10\). Both sides are positive, so cross-multiply and square:

100 < 10\sqrt{t+1} \implies 10 < \sqrt{t+1} \implies 100 < t + 1 \implies t > 99

Answer: (i) ≈ 44.72 ppm (ii) After 99 days

Mathematics 11 (Punjab Board)

MCQs: Chapter 2 – Functions and Graphs

Choose the correct option for each question. The answer key is checked against the worked solutions.

If \(f(x) = x^2 - 1\), then \(f(-3) =\)
1. \(10\)
2. \(8\)
3. \(-10\)
4. \(0\)
\((-3)^2 - 1 = 9 - 1 = 8\), is liye option B durust hai.
The domain of \(f(x) = \sqrt{2x + 3}\) is
1. \(x \ge -\frac{3}{2}\)
2. \(x > -\frac{3}{2}\)
3. all real numbers
4. \(x \ge 0\)
Square root ke andar \(2x + 3 \ge 0\) chahiye, yani \(x \ge -\frac{3}{2}\) — option A.
If \(f(x) = 4x + 7\), then \(\frac{f(a+h) - f(a)}{h} =\)
1. \(4h\)
2. \(4\)
3. \(4a\)
4. \(7\)
Numerator \(4h\) banta hai, \(h\) se divide karne par \(4\) — option B.
A function \(f\) is one-to-one if
1. \(f(x_1) = f(x_2) \implies x_1 = x_2\)
2. range = co-domain
3. every \(y\) has two pre-images
4. domain = co-domain
Yeh one-to-one ki definition hai — option A.
The range of \(g(x) = |x - 5|\) is
1. \(\mathbb{R}\)
2. \([0, \infty)\)
3. \((-\infty, 0]\)
4. \((5, \infty)\)
Absolute value kabhi negative nahi hoti, minimum \(0\) — option B.
The x-intercept of \(y = 2x - 1\) is
1. \((0, -1)\)
2. \(\left(\frac{1}{2}, 0\right)\)
3. \(\left(-\frac{1}{2}, 0\right)\)
4. \((1, 0)\)
\(y = 0\) rakhne par \(x = \frac{1}{2}\) — option B. \((0,-1)\) y-intercept hai.
The domain of \(f(x) = \frac{x}{x^2 - 4}\) is
1. \(\mathbb{R}\)
2. \(\mathbb{R} - \{-2, 2\}\)
3. \(\mathbb{R} - \{2\}\)
4. \([-2, 2]\)
Denominator \(x^2 - 4 = 0\) par \(x = \pm 2\), dono exclude — option B.
The function \(f(x) = 3x - 5\) with \(f: \mathbb{R} \to \mathbb{R}\) is
1. one-to-one only
2. onto only
3. bijective
4. neither
Linear function dono properties rakhti hai — one-to-one bhi, onto bhi — option C.
The function \(g(x) = x^2 - 3x\) with \(g: \mathbb{R} \to \mathbb{R}\) is
1. injective
2. surjective
3. neither injective nor surjective
4. bijective
\(g(0) = g(3) = 0\) (not injective) aur range \([-\frac{9}{4}, \infty) \neq \mathbb{R}\) (not surjective) — option C.
The range of \(y = \sqrt{x} + 5\) is
1. \([0, \infty)\)
2. \([5, \infty)\)
3. \(\mathbb{R}\)
4. \((5, \infty)\)
\(\sqrt{x} \ge 0\) mein \(5\) add karne par \(y \ge 5\) — option B.
The graphs of \(y = 3x + 2\) and \(y = -x + 6\) intersect at
1. \((1, 5)\)
2. \((5, 1)\)
3. \((2, 8)\)
4. \((1, 7)\)
\(3x + 2 = -x + 6 \implies x = 1, y = 5\) — option A.
The domain of \(y = \sqrt[3]{2x + 1}\) is
1. \(x \ge 0\)
2. all real numbers
3. \(x \ge -\frac{1}{2}\)
4. \([0, \infty)\)
Cube root har real number ke liye defined hai — option B.
The function \(f(x) = e^x\) with \(f: \mathbb{R} \to \mathbb{R}^+\) is
1. one-to-one only
2. onto only
3. bijective
4. neither
Domain \(\mathbb{R}\) par har positive \(y\) ka pre-image \(x = \ln y\) hai — dono properties, option C.
The y-intercept of \(y = -5x + 10\) is
1. \((2, 0)\)
2. \((0, 10)\)
3. \((0, 2)\)
4. \((10, 0)\)
\(x = 0\) rakhne par \(y = 10\) — option B. \((2,0)\) x-intercept hai.
If \(f\) is bijective, then its inverse \(f^{-1}\)
1. swaps the domain and range of \(f\)
2. is always a linear function
3. does not exist
4. equals \(f\)
Inverse mein domain aur range interchange ho jate hain — option A.
If \(f(x) = x^3 - ax^2 + bx + 1\) with \(f(2) = -3\) and \(f(-1) = 0\), then \(a =\)
1. \(2\)
2. \(-2\)
3. \(1\)
4. \(-1\)
Equations \(2a - b = 6\) aur \(a = -b\) se \(a = 2\) — option A.
The range of \(f(x) = x^2\) is
1. \(\mathbb{R}\)
2. \([0, \infty)\)
3. \((-\infty, 0]\)
4. \(\{0\}\)
Square kabhi negative nahi hota — option B.
The graphs of \(y = 2x + 5\) and \(y = -x + 5\) intersect at
1. \((0, 5)\)
2. \((5, 0)\)
3. \((0, -5)\)
4. \((2, 9)\)
\(2x + 5 = -x + 5 \implies x = 0, y = 5\) — option A.
\(f(x) = \frac{2x-3}{x+1}\) is not onto \(\mathbb{R}\) because
1. \(x = -1\) is excluded from the domain
2. \(y = 2\) is never attained
3. it is not one-to-one
4. its domain is empty
Range \(\mathbb{R} - \{2\}\) hai, co-domain \(\mathbb{R}\) — \(y = 2\) miss hota hai, option B.
The doubling time for \(P(t) = P_0 e^{0.05t}\) is
1. \(\frac{\ln 2}{0.05}\)
2. \(\frac{0.05}{\ln 2}\)
3. \(e^{0.05}\)
4. \(20\) years
\(2 = e^{rt}\) se \(\ln 2 = rt\), yani \(t = \frac{\ln 2}{r}\) — option A.
The vertical line test is used to check whether
1. a function is one-to-one
2. a graph represents a function
3. a function is onto
4. a function is continuous
Agar vertical line graph ko ek se zyada baar kaate to function nahi — option B.
The range of \(g(x) = 5 - x\) is
1. \(\mathbb{R}\)
2. \([0, \infty)\)
3. \((-\infty, 5]\)
4. \([5, \infty)\)
Linear function har real value leti hai — option A.

Mathematics 11 (Punjab Board)

Chapter 2: Functions and Graphs — Formula Sheet

Quick-revision formulas for Punjab Board Class 11 Mathematics, Chapter 2: Functions and Graphs.

Function Basics

\[ f: A \to B, \qquad y = f(x) \]

\[ \text{One-to-one: } f(x_1) = f(x_2) \implies x_1 = x_2 \]

\[ \text{Onto: Range} = \text{Co-domain} \]

\[ \text{Inverse: } f(f^{-1}(x)) = x, \quad \text{Dom}(f^{-1}) = \text{Range}(f) \]

\[ \text{Difference quotient: } \frac{f(a+h) - f(a)}{h} \]

Domain Rules

\[ \text{Rational } \frac{p(x)}{q(x)}: \quad q(x) \neq 0 \]

\[ \text{Even root } \sqrt{g(x)}: \quad g(x) \ge 0 \]

\[ \text{Odd root } \sqrt[3]{g(x)}: \quad \text{all real } x \]

\[ \text{Polynomial: Domain} = \mathbb{R} \]

Graphs: Intercepts and Intersections

\[ \text{y-intercept: set } x = 0 \qquad \text{x-intercept: set } y = 0 \]

\[ \text{Intersection of } f \text{ and } g: \text{ solve } f(x) = g(x) \]

\[ \text{Vertex of } y = ax^2 + bx + c: \quad h = -\frac{b}{2a} \]

Root Function Graphs

\[ y = \sqrt{x}: \text{ Domain } x \ge 0,\; \text{Range } y \ge 0,\; \text{starts at } (0,0) \]

\[ y = \sqrt{x} + k: \text{ shifts up by } k \qquad y = -\sqrt{x}: \text{ reflects downward} \]

\[ y = \sqrt[3]{x}: \text{ Domain } \mathbb{R},\; \text{Range } \mathbb{R},\; \text{S-shape through } (0,0) \]

Exponential Models

\[ P(t) = P_0 e^{rt} \]

\[ \text{Doubling time: } t = \frac{\ln 2}{r} \]

\[ \text{Decay with half-life } h: \quad Q(t) = Q_0\left(\frac{1}{2}\right)^{t/h} \]

Geometry as Functions

\[ \text{Square: } A = \frac{P^2}{16} \]

\[ \text{Circle: } C = 2\sqrt{\pi A} \]

\[ \text{Cube: } S = 6V^{2/3} \]

Trig Identities Used in This Chapter

\[ \sin C - \sin D = 2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right) \]

\[ \sin A\cos B - \cos A\sin B = \sin(A-B) \]

Chapter 2 FAQs

What is covered in Chapter 2 (Functions and Graphs)?

This chapter covers Relations, functions, domain and range, inverse functions, graph transformations. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.

How many marks does Chapter 2 carry in the Punjab Board paper?

Chapter 2 (Functions and Graphs) carries approximately 7 marks in the Punjab Board Class 11 Mathematics paper.

Are these notes for the new PCTB syllabus?

Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.

Where are the exercise solutions for Chapter 2?

Use the exercise index on this page — every exercise of Chapter 2 has its own page with complete step-by-step solutions.