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Punjab Board Class 11 Mathematics

Chapter 3: Theory of Quadratic Functions — Class 11 Maths Notes

Quadratic functions: roots, discriminant, vertex and axis of symmetry, with graph-based reasoning fully worked. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 3.

Exercise solutions — Chapter 3

Every exercise of Theory of Quadratic Functions solved step by step.

Exercise 3.1 Exercise 3.2

Mathematics 11 (Punjab Board)

Chapter 3: Theory of Quadratic Functions — Theory

Definitions, key results and methods for Punjab Board Class 11 Mathematics, Chapter 3: Theory of Quadratic Functions.

The Quadratic Function

A quadratic function has the standard form \(f(x) = ax^2 + bx + c\) with \(a \neq 0\). Its graph is a parabola: agar \(a > 0\) to upward khulti hai, agar \(a < 0\) to downward.

\[ f(x) = ax^2 + bx + c, \qquad a \neq 0 \]

The vertex \((h, k)\) is the turning point of the parabola, and the axis of symmetry is the vertical line \(x = h\), where:

\[ h = -\frac{b}{2a}, \qquad k = f(h) \]

Key point: The sign of \(a\) alone decides up vs down; the vertex formula \(-\frac{b}{2a}\) gives the exact turning point — sign error yahan sab se common hai.

Completing the Square: Vertex Form

Completing the square rewrites the quadratic as \(f(x) = a(x-h)^2 + k\), which shows the vertex \((h, k)\) directly:

\[ ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right) \]

Example: \(x^2 + 6x + 13 = (x+3)^2 - 9 + 13 = (x+3)^2 + 4\), so the vertex is \((-3, 4)\).

Key point: Jab \(a \neq 1\) ho to pehle \(a\) common factor nikaal lein, phir bracket ke andar square complete karen.

Maximum and Minimum Values

Since \((x-h)^2 \ge 0\) always, the vertex gives the extreme value of the function:

\[ a > 0 \implies \text{minimum } k \text{ at } x = h \qquad a < 0 \implies \text{maximum } k \text{ at } x = h \]

Example: \(f(x) = -x^2 + 8x + 13 = -(x-4)^2 + 29\) has maximum value \(29\) at \(x = 4\).

Key point: "Maximum/minimum value" means the \(y\)-value \(k\) at the vertex — students often answer with the \(x\)-value \(h\) by mistake. Dono likhen: value \(k\) at \(x = h\).

Analyzing a Quadratic by Sketching

To sketch \(y = ax^2 + bx + c\): (1) note whether it opens up or down from the sign of \(a\); (2) find the y-intercept \((0, c)\); (3) find x-intercepts by solving \(ax^2 + bx + c = 0\) (0, 1 or 2 points); (4) plot the vertex \(\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)\); (5) join with a smooth symmetric curve.

The domain of every quadratic function is \(\mathbb{R}\). The range starts (or ends) at the vertex: \([k, \infty)\) if \(a > 0\), \((-\infty, k]\) if \(a < 0\).

Key point: Symmetry about \(x = h\) is a free check — agar sketch symmetric na lage to vertex ya intercepts mein ghalti hai.

Inverse of a Quadratic Function

A quadratic is not one-to-one on all of \(\mathbb{R}\) (it fails the horizontal line test), so it has no inverse there. Lekin agar domain ko vertex ke ek side tak restrict kar dein (jahan function strictly increasing ya decreasing ho), to inverse exist karta hai.

Method: write \(y = f(x)\), interchange \(x\) and \(y\), solve the resulting quadratic for \(y\) with the quadratic formula, and keep the branch whose range matches the restricted domain.

\[ x = ay^2 + by + c \implies y = \frac{-b \pm \sqrt{b^2 - 4a(c-x)}}{2a} \quad (\text{choose the correct } \pm) \]

Domain of \(f^{-1}\) = range of \(f\); range of \(f^{-1}\) = (restricted) domain of \(f\).

Key point: The \(\pm\) choice is decided by the restricted domain — wrong branch ka matlab inverse ka range original domain se match nahi karega.

Absolute Value

The absolute value \(|x|\) is the distance of \(x\) from zero:

\[ |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases} \]

Absolute value equations: \(|A| = B\) (with \(B \ge 0\)) splits into \(A = B\) or \(A = -B\). Also \(|A| = |B|\) splits into \(A = B\) or \(A = -B\).

Absolute value inequalities: \(|A| < B\) becomes the double inequality \(-B < A < B\); \(|A| > B\) becomes \(A < -B\) or \(A > B\).

\[ |A| < B \iff -B < A < B \qquad |A| > B \iff A < -B \text{ or } A > B \]

Key point: \(|A| \le B\) likhne se pehle check karen ke \(B \ge 0\) hai — agar right side negative ho to koi solution nahi.

Equations Reducible to the Quadratic Equation

Rational equations: multiply both sides by the LCD to clear denominators, solve the resulting quadratic, then reject any root that makes an original denominator zero.

Radical equations: isolate one radical, square both sides, repeat if a radical remains. Squaring can introduce extraneous roots, so har root ko original equation mein rakh kar verify karna lazmi hai.

Substitution: for equations like \(3x^2 + 15x - 2\sqrt{x^2+5x+1} = 2\), set \(y = \sqrt{x^2+5x+1}\) (so \(y \ge 0\)), solve for \(y\) first, then for \(x\).

Key point: Extraneous roots are the examiner's favourite trap in radical equations — verification is part of the solution, not optional.

Real-World Quadratic Models

Area, profit, motion and cost problems often reduce to a quadratic equation: define the unknown, translate the words into an equation, solve, and reject answers that make no physical sense (negative lengths, negative time).

"At least" / "at most" conditions become quadratic inequalities: solve the equality for boundary points, then test which side satisfies the inequality.

Key point: Word problem ka aakhri step hamesha sanity check hai — negative sheep ya negative time ka jawab reject karen.

Mathematics 11 (Punjab Board)

Solved Examples

Worked textbook examples for Punjab Board Class 11 Mathematics, Chapter 3: Theory of Quadratic Functions.

Example 1: Sketch and analyze \(y = -x^2 - 2x + 3\).

Sketch ke liye chaar cheezein chahiye: direction (sign of \(a\)), y-intercept, x-intercepts, aur vertex. Phir increasing/decreasing intervals vertex se tay hote hain.

Here \(a = -1 < 0\), so the parabola opens downward.

y-intercept: \(x = 0 \implies y = 3\). Point: \((0, 3)\).

x-intercepts: set \(y = 0\):

\[ -x^2 - 2x + 3 = 0 \implies x^2 + 2x - 3 = 0 \implies (x+3)(x-1) = 0 \implies x = -3, 1 \]

Points: \((-3, 0)\) and \((1, 0)\).

Vertex: using \(h = -\frac{b}{2a}\):

\[ h = -\frac{-2}{2(-1)} = -1, \qquad k = -(-1)^2 - 2(-1) + 3 = -1 + 2 + 3 = 4 \]

Vertex \((-1, 4)\) is a maximum point. The function increases on \((-\infty, -1)\) and decreases on \((-1, \infty)\). Domain \(= \mathbb{R}\), Range \(= (-\infty, 4]\).

Answer: Vertex \((-1, 4)\) (maximum); intercepts \((0,3), (-3,0), (1,0)\); increasing on \((-\infty,-1)\), decreasing on \((-1,\infty)\)

Example 2: Find the maximum or minimum value of \(f(x) = -2x^2 + 4x + 3\) by completing the square.

Pehle \(-2\) common factor nikaal kar bracket ke andar square complete karenge. Phir \(a < 0\) hone ki wajah se vertex maximum dega.

\[ f(x) = -2x^2 + 4x + 3 = -2(x^2 - 2x) + 3 \]
\[ = -2\left[(x-1)^2 - 1\right] + 3 = -2(x-1)^2 + 2 + 3 = -2(x-1)^2 + 5 \]

Since \(-2(x-1)^2 \le 0\) for all \(x\), the largest value occurs when \((x-1)^2 = 0\), i.e. at \(x = 1\).

Answer: Maximum value \(5\) at \(x = 1\)

Example 3: Find the maximum or minimum value of \(f(x) = x^2 - 2x - 3\).

Yahan \(a = 1 > 0\), is liye minimum value hogi. Completing square se vertex form banayenge.

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

Since \((x-1)^2 \ge 0\), the smallest value is \(-4\), attained at \(x = 1\).

Answer: Minimum value \(-4\) at \(x = 1\)

Example 4: Find the inverse of \(f(x) = x^2 + 4x + 3\), \(x \ge -2\). Also find its domain and range.

Domain vertex ke right side tak restricted hai (\(x \ge -2\)), is liye function one-to-one hai aur inverse exist karta hai. \(y = f(x)\) likh kar \(x\) aur \(y\) interchange karenge, phir quadratic formula se \(y\) solve karke sahi branch choose karenge.

Let \(y = x^2 + 4x + 3\). Interchanging \(x\) and \(y\): \(x = y^2 + 4y + 3\), i.e. \(y^2 + 4y + (3 - x) = 0\). Apply the quadratic formula in \(y\):

\[ y = \frac{-4 \pm \sqrt{16 - 4(3-x)}}{2} = \frac{-4 \pm \sqrt{4 + 4x}}{2} = \frac{-4 \pm 2\sqrt{x+1}}{2} = -2 \pm \sqrt{x+1} \]

The range of \(f^{-1}\) must equal the (restricted) domain of \(f\), which is \([-2, \infty)\). Since \(-2 - \sqrt{x+1} \le -2\) and \(-2 + \sqrt{x+1} \ge -2\), we keep the \(+\) branch:

\[ f^{-1}(x) = -2 + \sqrt{x+1} \]

Check: \(f(-2) = 4 - 8 + 3 = -1\), and \(f\) increases for \(x \ge -2\), so Range of \(f = [-1, \infty) =\) Domain of \(f^{-1}\). Range of \(f^{-1} = [-2, \infty)\).

Answer: \(f^{-1}(x) = -2 + \sqrt{x+1}\); Domain of \(f^{-1} = [-1, \infty)\), Range of \(f^{-1} = [-2, \infty)\)

Common mistake: Keeping the wrong \(\pm\) branch. The branch must give outputs in the original restricted domain \(x \ge -2\).

Example 5: Solve \(|x^2 - 4| = 5\).

\(|A| = B\) ka matlab \(A = B\) ya \(A = -B\). Dono cases solve karke real roots rakhenge.

Case 1: \(x^2 - 4 = 5 \implies x^2 = 9 \implies x = \pm 3\).

Case 2: \(x^2 - 4 = -5 \implies x^2 = -1\), which has no real solution.

Answer: \(\{-3, 3\}\)

Example 6: Solve \(|x^2 - 6x - 4| < 3\).

\(|A| < 3\) ko double inequality \(-3 < A < 3\) mein badalte hain. Phir dono quadratic inequalities alag solve karke unka intersection lete hain.

\[ -3 < x^2 - 6x - 4 < 3 \]

Right part: \(x^2 - 6x - 4 < 3 \implies x^2 - 6x - 7 < 0\). Roots of \(x^2 - 6x - 7 = 0\) are \(x = \frac{6 \pm \sqrt{36+28}}{2} = \frac{6 \pm 8}{2}\), i.e. \(x = -1\) and \(x = 7\). The upward parabola is negative between its roots: \(-1 < x < 7\).

Left part: \(x^2 - 6x - 4 > -3 \implies x^2 - 6x - 1 > 0\). Roots of \(x^2 - 6x - 1 = 0\) are \(x = \frac{6 \pm \sqrt{36+4}}{2} = 3 \pm \sqrt{10} \approx -0.162\) and \(6.162\). The upward parabola is positive outside its roots: \(x < 3 - \sqrt{10}\) or \(x > 3 + \sqrt{10}\).

Intersection: combine \((-1, 7)\) with \(\big((-\infty, 3-\sqrt{10}) \cup (3+\sqrt{10}, \infty)\big)\):

\[ (-1,\, 3-\sqrt{10}) \cup (3+\sqrt{10},\, 7) \]

Check: at \(x = -0.5\), \(|0.25 + 3 - 4| = 0.75 < 3\) ✓; at \(x = 0\), \(|{-4}| = 4 \not< 3\) ✓ (correctly excluded).

Answer: \((-1, 3-\sqrt{10}) \cup (3+\sqrt{10}, 7)\)

Example 7: Solve \(\frac{1}{x} + \frac{2}{x+1} = 1\), \(x \neq 0\), \(x \neq -1\).

Dono taraf \(x(x+1)\) se multiply karke denominators khatam karenge. Jo quadratic bane use solve karke check karenge ke koi root excluded values \(0\) ya \(-1\) to nahi.

Multiply both sides by \(x(x+1)\):

\[ (x+1) + 2x = x(x+1) \implies 3x + 1 = x^2 + x \implies x^2 - 2x - 1 = 0 \]

Apply the quadratic formula:

\[ x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]

Neither \(1 + \sqrt{2}\) nor \(1 - \sqrt{2}\) equals \(0\) or \(-1\), so both are valid.

Answer: \(\{1 + \sqrt{2},\, 1 - \sqrt{2}\}\)

Example 8: Solve \(\sqrt{x+8} + \sqrt{x+3} = \sqrt{12x+13}\).

Dono taraf square karenge, phir bacha hua radical isolate karke dobara square karenge. Squaring extraneous roots paida kar sakti hai, is liye har candidate ko original equation mein verify karenge.

Square both sides:

\[ (x+8) + (x+3) + 2\sqrt{(x+8)(x+3)} = 12x + 13 \]
\[ 2x + 11 + 2\sqrt{x^2+11x+24} = 12x + 13 \implies 2\sqrt{x^2+11x+24} = 10x + 2 \]
\[ \sqrt{x^2+11x+24} = 5x + 1 \]

Square again:

\[ x^2 + 11x + 24 = 25x^2 + 10x + 1 \implies 24x^2 - x - 23 = 0 \]
\[ x = \frac{1 \pm \sqrt{1 + 2208}}{48} = \frac{1 \pm 47}{48} \implies x = 1 \text{ or } x = -\frac{23}{24} \]

Verification: For \(x = 1\): \(\sqrt{9} + \sqrt{4} = 3 + 2 = 5\) and \(\sqrt{25} = 5\) ✓. For \(x = -\frac{23}{24}\): the step \(\sqrt{x^2+11x+24} = 5x+1\) needs \(5x+1 \ge 0\), but \(5\left(-\frac{23}{24}\right) + 1 = -\frac{91}{24} < 0\) ✗ — extraneous, rejected.

Answer: \(\{1\}\)

Common mistake: Accepting \(x = -\frac{23}{24}\) without checking. A square root can never equal a negative number — that step alone rejects it.

Example 9: The length of a room is 3 metres greater than its breadth. If the area of the room is 180 square metres, find the length and the breadth.

Breadth ko \(x\) farz karke length \(x+3\) likhenge. Area ka formula \(x(x+3) = 180\) ek quadratic equation dega; negative root reject kar denge kyunki length negative nahi ho sakti.

Let breadth \(= x\) m, so length \(= (x+3)\) m. Then:

\[ x(x+3) = 180 \implies x^2 + 3x - 180 = 0 \implies (x+15)(x-12) = 0 \]

So \(x = -15\) or \(x = 12\). Breadth cannot be negative, so \(x = 12\), and length \(= 12 + 3 = 15\).

Answer: Breadth \(12\) m, Length \(15\) m

Example 10: A company's weekly profit (in thousands of dollars) is \(P(x) = -x^2 + 2x + 3\), where \(x\) is laptops produced (in hundreds). Find the production levels where the company makes at least \$4,000 profit.

"\$4,000" ka matlab \(P(x) \ge 4\) (thousands mein). Inequality ko standard form mein laa kar solve karenge.

\[ -x^2 + 2x + 3 \ge 4 \implies -x^2 + 2x - 1 \ge 0 \implies x^2 - 2x + 1 \le 0 \implies (x-1)^2 \le 0 \]

A square is never negative, so \((x-1)^2 \le 0\) forces \((x-1)^2 = 0\), i.e. \(x = 1\). Indeed \(P(1) = -1 + 2 + 3 = 4\), the maximum profit.

Answer: Exactly \(x = 1\) (i.e. 100 laptops per week), giving the maximum profit of \$4,000

Mathematics 11 (Punjab Board)

MCQs: Chapter 3 – Theory of Quadratic Functions

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

  1. The vertex of the parabola \(y = x^2 - 4x\) is

    1. \((2, -4)\)
    2. \((-2, 4)\)
    3. \((4, -2)\)
    4. \((0, -4)\)
    \(h = -\frac{-4}{2} = 2\), \(k = 4 - 8 = -4\) — option A.

  2. The minimum value of \(f(x) = x^2 + 6x + 13\) is

    1. \(13\)
    2. \(-3\)
    3. \(22\)
    4. \(4\)
    \((x+3)^2 + 4 \ge 4\), minimum \(4\) at \(x = -3\) — option D.

  3. The maximum value of \(f(x) = -x^2 + 8x + 13\) is

    1. \(29\)
    2. \(13\)
    3. \(4\)
    4. \(21\)
    \(-(x-4)^2 + 29 \le 29\), maximum \(29\) at \(x = 4\) — option A.

  4. The axis of symmetry of \(y = -2x^2 + 4x + 3\) is

    1. \(x = 2\)
    2. \(x = 1\)
    3. \(x = -1\)
    4. \(x = 4\)
    \(x = -\frac{b}{2a} = -\frac{4}{-4} = 1\) — option B.

  5. If \(|x^2 + 1| = 5\), then \(x =\)

    1. \(\pm\sqrt{6}\)
    2. \(\pm 3\)
    3. \(\pm 2\)
    4. \(\pm\sqrt{5}\)
    \(x^2 + 1 = 5 \implies x^2 = 4\); \(x^2 + 1 = -5\) ka koi real hal nahi — option C.

  6. The solution set of \(|x^2 - 4| < 5\) is

    1. \(\{-3, 3\}\)
    2. \((-5, 5)\)
    3. \([-3, 3]\)
    4. \((-3, 3)\)
    \(-5 < x^2 - 4 < 5 \implies x^2 < 9\), yani poora interval \((-3, 3)\) — option D.

  7. \(|A| = |B|\) implies

    1. \(A = B\) only
    2. \(A = -B\) only
    3. \(A = B\) or \(A = -B\)
    4. \(A^2 = B^2\) only
    Dono cases possible hain — option C.

  8. Extraneous roots can appear when we

    1. factor a quadratic
    2. complete the square
    3. multiply both sides by 1
    4. square both sides of an equation
    Squaring false roots paida kar sakti hai, is liye verification lazmi hai — option D.

  9. The solution of \(\frac{1}{x} + \frac{2}{x+1} = 1\) (\(x \neq 0, -1\)) is

    1. \(1 \pm \sqrt{2}\)
    2. \(1, 2\)
    3. \(\pm\sqrt{2}\)
    4. \(-1, 0\)
    \(x^2 - 2x - 1 = 0\) se \(x = 1 \pm \sqrt{2}\) — option A.

  10. Solving \(\sqrt{x+5} - \sqrt{x-3} = 2\) gives \(x =\)

    1. \(3\)
    2. \(4\)
    3. \(5\)
    4. \(1\)
    Square karne par \(\sqrt{x-3} = 1\), yani \(x = 4\); verify bhi hota hai — option B.

  11. The roots of \(2x^2 - 3x + 1 = 0\) are

    1. \(1, 2\)
    2. \(\frac{1}{2}, 1\)
    3. \(-1, -\frac{1}{2}\)
    4. \(\frac{3}{2}, 1\)
    \((2x-1)(x-1) = 0\) — option B.

  12. For \(f(x) = x^2 - 3\) with \(x \le 0\), \(f^{-1}(x) =\)

    1. \(\sqrt{x+3}\)
    2. \(\sqrt{x-3}\)
    3. \(-\sqrt{x-3}\)
    4. \(-\sqrt{x+3}\)
    \(y = \pm\sqrt{x+3}\) mein se woh branch jiska range \((-\infty, 0]\) ho — option D.

  13. The domain of \(f^{-1}\) for \(f(x) = x^2 - 3\), \(x \le 0\), is

    1. \((-\infty, 0]\)
    2. \([0, \infty)\)
    3. \(\mathbb{R}\)
    4. \([-3, \infty)\)
    Inverse ka domain = original ki range \(= [-3, \infty)\) — option D.

  14. The solution of \(|2x^2 - 3x - 5| < 4\) is

    1. a single interval
    2. a union of two open intervals
    3. the empty set
    4. all real numbers
    Intersection se \(\left(-\frac{3}{2}, \frac{3-\sqrt{17}}{4}\right) \cup \left(\frac{3+\sqrt{17}}{4}, 3\right)\) — option B.

  15. \(f(x) = -2x^2 - x + 21\) attains its maximum at

    1. \(x = \frac{1}{4}\)
    2. \(x = -\frac{1}{4}\)
    3. \(x = \frac{1}{2}\)
    4. \(x = -\frac{1}{2}\)
    \(h = -\frac{b}{2a} = -\frac{-1}{-4} = -\frac{1}{4}\) — option B (sign mat bhoolen).

  16. The minimum value of \(f(x) = x^2 + 5x + 4\) is

    1. \(-4\) at \(x = -2\)
    2. \(4\) at \(x = \frac{5}{2}\)
    3. \(-\frac{9}{4}\) at \(x = -\frac{5}{2}\)
    4. \(-\frac{25}{4}\) at \(x = 0\)
    Completing square: \(\left(x+\frac{5}{2}\right)^2 - \frac{9}{4}\) — option C.

  17. The parabola \(y = -x^2 - 2x + 3\) opens

    1. upward
    2. downward
    3. to the left
    4. to the right
    \(a = -1 < 0\) — downward — option B.

  18. The y-intercept of \(y = -x^2 - 2x + 3\) is

    1. \((3, 0)\)
    2. \((0, -3)\)
    3. \((0, 3)\)
    4. \((-1, 4)\)
    \(x = 0\) rakhne par \(y = 3\) — option C.

  19. \(|x^2 - 6x + 8| = 4\) gives \(x =\)

    1. \(2, 4\)
    2. \(\pm 2\)
    3. \(3 \pm \sqrt{5}\)
    4. \(3 \pm \sqrt{3}\)
    \(x^2 - 6x + 4 = 0\) se \(x = 3 \pm \sqrt{5}\); negative case ka koi real hal nahi — option C.

  20. The extraneous root when solving \(\sqrt{x+7} + \sqrt{x+2} = \sqrt{6x+13}\) is

    1. \(2\)
    2. \(-\frac{5}{3}\)
    3. \(0\)
    4. \(1\)
    \(x = -\frac{5}{3}\) verify nahi hota (square root negative ke barabar) — option B.

  21. For \(P(x) = -x^2 + 2x + 3\) (profit in thousands of dollars), the maximum profit occurs at

    1. \(x = 2\)
    2. \(x = 1\)
    3. \(x = 0\)
    4. \(x = 3\)
    \(P(x) = -(x-1)^2 + 4\), maximum \$4,000 at \(x = 1\) — option B.

  22. Solving \(|x^2 - 5x + 6| \le x + 2\) first requires

    1. \(x \le -2\)
    2. \(x \ge -2\)
    3. \(x > 0\)
    4. no restriction
    \(|A| \le B\) ke liye \(B \ge 0\) chahiye, yani \(x + 2 \ge 0\) — option B.

Mathematics 11 (Punjab Board)

Chapter 3: Theory of Quadratic Functions — Formula Sheet

Quick-revision formulas for Punjab Board Class 11 Mathematics, Chapter 3: Theory of Quadratic Functions.

Quadratic Function Basics

\[ f(x) = ax^2 + bx + c, \quad a \neq 0 \]
\[ \text{Vertex form: } f(x) = a(x-h)^2 + k \]
\[ h = -\frac{b}{2a}, \qquad k = f(h), \qquad \text{axis: } x = h \]
\[ a > 0 \implies \text{min } k \text{ at } x = h \qquad a < 0 \implies \text{max } k \text{ at } x = h \]
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]

Inverse of a Quadratic (Restricted Domain)

\[ y = ax^2 + bx + c \;\xrightarrow{\text{swap}}\; x = ay^2 + by + c \]
\[ y = \frac{-b \pm \sqrt{b^2 - 4a(c-x)}}{2a} \quad (\text{keep the branch matching the restricted domain}) \]
\[ \text{Dom}(f^{-1}) = \text{Range}(f), \qquad \text{Range}(f^{-1}) = \text{Dom}(f) \]

Absolute Value

\[ |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases} \]
\[ |A| = B \implies A = \pm B \qquad |A| = |B| \implies A = \pm B \]
\[ |A| < B \iff -B < A < B \]
\[ |A| > B \iff A < -B \text{ or } A > B \]

Equations Reducible to Quadratic

\[ \text{Rational: multiply by LCD, reject roots making any denominator zero} \]
\[ \text{Radical: isolate, square, repeat — then verify every root (extraneous roots possible)} \]
\[ \text{Substitution: } y = \sqrt{\text{quadratic}} \ge 0 \]

Completing the Square (Reference)

\[ x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 \]
\[ ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right) \]

Chapter 3 FAQs

What is covered in Chapter 3 (Theory of Quadratic Functions)?
This chapter covers Quadratic functions, roots, discriminant, vertex, axis of symmetry, graphs. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.
How many marks does Chapter 3 carry in the Punjab Board paper?
Chapter 3 (Theory of Quadratic Functions) carries approximately 13 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 3?
Use the exercise index on this page — every exercise of Chapter 3 has its own page with complete step-by-step solutions.