Concept: Co-function identity: \(\sin(90^\circ - \theta) = \cos \theta\).

Calculation: \[ \sin(90^\circ - A) = \cos A. \]

Explanation: Standard complementary angle identity.
Quick Tip: \(\sin(90^\circ - \theta) = \cos \theta\), \(\cos(90^\circ - \theta) = \sin \theta\).

Concept: \(\cos(\alpha - \beta) = \cos 0^\circ = 1\) when \(\alpha = \beta\).

Calculation: \[ \alpha - \beta = 60^\circ - 60^\circ = 0^\circ, \quad \cos 0^\circ = 1. \]

Explanation: Equal angles give zero difference.
Quick Tip: \(\cos 0^\circ = 1\), always.

Concept: At \(45^\circ\), \(\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}\).

Calculation: \[ \sin 45^\circ + \cos 45^\circ = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}. \]

Explanation: Both values equal, sum is \(\sqrt{2}\).
Quick Tip: For \(45^\circ\), \(\sin = \cos = \frac{1}{\sqrt{2}}\).

Concept: Double-angle formula: \(\tan 2A = \dfrac{2 \tan A}{1 - \tan^2 A}\).

Calculation: \[ \tan 30^\circ = \frac{1}{\sqrt{3}}, \quad \tan^2 30^\circ = \frac{1}{3} \] \[ \dfrac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \dfrac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{2} = \sqrt{3} = \tan 60^\circ \]
But \(2 \tan 60^\circ = 2\sqrt{3}\), no.
Wait: The expression is \(\tan 2A = \tan 60^\circ = \sqrt{3}\).
But option (C) says \(2 \tan 60^\circ\), which is wrong.
Correct: equals \(\tan 60^\circ\).
Wait: Let's see options.
Wait: \(\dfrac{2 \tan A}{1 - \tan^2 A} = \tan 2A = \tan 60^\circ = \sqrt{3}\).
So (B).

Wait: But earlier said (C). Mistake.
No: (B) is \(\tan 60^\circ\), yes.

Explanation: Identity gives \(\tan 60^\circ\).
Quick Tip: Recognize \(\tan 2\theta\) formula.

Concept: \(\tan \theta = \frac{opp}{adj}\), hypotenuse = \(\sqrt{12^2 + 5^2} = 13\).

Calculation: \[ \sin \theta = \frac{opp}{hyp} = \frac{12}{13}. \]

Explanation: 5-12-13 right triangle.
Quick Tip: Use Pythagorean triplet: \(5^2 + 12^2 = 25 + 144 = 169 = 13^2\).

Concept: Use co-function and reciprocal identities.

Calculation: \[ \cos 59^\circ = \sin(90^\circ - 59^\circ) = \sin 31^\circ \] \[ \dfrac{\cos 59^\circ}{\sin 31^\circ} = \dfrac{\sin 31^\circ}{\sin 31^\circ} = 1 \] \[ \tan 80^\circ = \tan(90^\circ - 10^\circ) = \cot 10^\circ \] \[ \dfrac{\tan 80^\circ}{\cot 10^\circ} = \dfrac{\cot 10^\circ}{\cot 10^\circ} = 1 \] \[ 1 \times 1 = 1. \]

Explanation: Each fraction simplifies to 1.
Quick Tip: \(\cos(90^\circ - \theta) = \sin \theta\), \(\tan(90^\circ - \theta) = \cot \theta\).

Concept: \(\tan(90^\circ - \theta) = \cot \theta\), so \(\tan \theta \cdot \tan(90^\circ - \theta) = 1\).

Calculation: \[ \tan 65^\circ = \tan(90^\circ - 25^\circ) = \cot 25^\circ \] \[ \tan 25^\circ \cdot \cot 25^\circ = 1 = \sin 90^\circ. \]

Explanation: Product is 1, which is \(\sin 90^\circ\).
Quick Tip: Complementary angles: \(\tan \theta \cdot \cot \theta = 1\).

Concept: \(\sin^2 \theta = 1 - \cos^2 \theta\), \(\tan \theta = \dfrac{\sin \theta}{\cos \theta}\).

Calculation: \[ \sin \theta = \sqrt{1 - x^2}, \quad \tan \theta = \dfrac{\sqrt{1 - x^2}}{x}. \]

Explanation: Direct from Pythagorean identity.
Quick Tip: \(\tan \theta = \dfrac{\sqrt{1 - \cos^2 \theta}}{\cos \theta}\).

Concept: \(a^2 - b^2 = (a - b)(a + b)\).

Calculation: \[ 1 - \cos^4 \theta = (1 - \cos^2 \theta)(1 + \cos^2 \theta) = \sin^2 \theta (1 + \cos^2 \theta). \]

Explanation: Difference of squares.
Quick Tip: \(1 - \cos^4 \theta = (1 - \cos^2 \theta)(1 + \cos^2 \theta)\).

Concept: Point on y-axis has \(x = 0\), so form \((0, y)\).

Calculation: \((2, y)\) has \(x = 2\), not on y-axis.

Explanation: Only \((0, y)\) lies on y-axis.
Quick Tip: Y-axis: \(x = 0\).

Concept: For equal roots, \(D = b^2 - 4ac = 0\).

Calculation: \(a = k\), \(b = -6\), \(c = 1\), \[ (-6)^2 - 4(k)(1) = 0 \quad \Rightarrow \quad 36 - 4k = 0 \quad \Rightarrow \quad k = 9. \]

Explanation: Discriminant zero when \(k = 9\).
Quick Tip: \(D = 0 \Rightarrow b^2 = 4ac\).

Concept: If \(r\) is a zero, then \((x - r)\) is a factor.

Calculation:
Zero = 2 → factor = \(x - 2\).

Explanation: Factor theorem.
Quick Tip: \(p(r) = 0 \Rightarrow (x - r)\) divides \(p(x)\).

Concept: For \(ax^2 + bx + c = 0\), product of roots = \(\frac{c}{a}\).

Calculation:
Here, \(a = c\), \(b = a\), \(c = b\), \[ \alpha \beta = \frac{b}{c}. \]

Explanation: Standard quadratic formula.
Quick Tip: Product = \(\frac{constant term}{leading coefficient}\).

Concept: Quadratic has highest degree 2.

Calculation:
(A): RHS has \(x^3\) → cubic.
(B): Expand: \(x^2 + 6x + 9 = 4x + 16\) → \(x^2 + 2x - 7 = 0\) → quadratic.
(C): LHS degree 2, RHS degree 2, but simplify: \(4(x-1)^2 = 4x^2 + 7\) → not equal.
(D): Multiply by \(4x\): \(16x^2 + 1 = 16x^2\) → \(1 = 0\), contradiction.

Explanation: Only (B) reduces to quadratic.
Quick Tip: Expand and bring to standard form.

Concept: A quadratic equation has highest degree 2 after simplification.

Calculation:
(A): \[ 5x - x^2 = x^2 + 3 \quad \Rightarrow \quad -2x^2 + 5x - 3 = 0 \quad \Rightarrow \quad 2x^2 - 5x + 3 = 0 \]
→ Quadratic (degree 2).

(B): \[ x^3 - x^2 = (x-1)^3 = x^3 - 3x^2 + 3x - 1 \] \[ x^3 - x^2 - x^3 + 3x^2 - 3x + 1 = 0 \quad \Rightarrow \quad 2x^2 - 3x + 1 = 0 \]
→ Quadratic (degree 2).

(C): \[ (x+3)^2 = 3(x^2 - 5) \] \[ x^2 + 6x + 9 = 3x^2 - 15 \] \[ 0 = 3x^2 - 15 - x^2 - 6x - 9 \quad \Rightarrow \quad 2x^2 - 6x - 24 = 0 \]
→ Quadratic (degree 2).

(D): \[ (\sqrt{2}x + 3)^2 = 2x^2 + 5 \] \[ 2x^2 + 6\sqrt{2}x + 9 = 2x^2 + 5 \] \[ 6\sqrt{2}x + 9 - 5 = 0 \quad \Rightarrow \quad 6\sqrt{2}x + 4 = 0 \]
→ Linear (degree 1).

Explanation: Only (D) simplifies to a linear equation.
Quick Tip: Simplify each equation completely to check the degree.

Concept: \(D = b^2 - 4ac\).

Calculation: \(a = 2\), \(b = -7\), \(c = 6\), \[ D = (-7)^2 - 4(2)(6) = 49 - 48 = 1 \quad (Wait) \]
Wait: 49 - 48 = 1.
So (A).

Wait: But said (D). Mistake.
Wait: Recheck: \(4 \times 2 \times 6 = 48\), yes. \(49 - 48 = 1\).
Correct: (A) 1.

Explanation: \(D = 1\).
Quick Tip: \(D = b^2 - 4ac\).

Concept: \(x = 2\) is vertical line, all points with \(x = 2\).

Calculation:
Any point \((2, y)\) lies on it.

Explanation: All points with first coordinate 2.
Quick Tip: Vertical line: constant \(x\).

Concept: In A.P., middle term = average of others.

Calculation: \[ 2(2P + 1) = (P + 1) + (4P - 1) \] \[ 4P + 2 = 5P \quad \Rightarrow \quad 2 = P \]
Wait: \(5P - P = 4P\), no:
Left: \(4P + 2\), right: \(P + 1 + 4P - 1 = 5P\). \(4P + 2 = 5P\) → \(2 = P\).
So (B) 2.

Wait: But said 3.
Wait: Common difference:
Second - first = third - second \((2P + 1) - (P + 1) = (4P - 1) - (2P + 1)\) \(P = 2P - 2\) \(P = 2\).
Yes, \(P = 2\).

Explanation: \(P = 2\).
Quick Tip: Use \(2b = a + c\).

Concept: \(d = a_2 - a_1\).

Calculation: \[ 5 - 1 = 4. \]

Explanation: Constant difference 4.
Quick Tip: Subtract consecutive terms.

Concept: \(a_n = a + (n-1)d\).

Calculation: \(a = 5\), \(d = 3\), \[ 38 = 5 + (n-1)3 \] \[ 33 = (n-1)3 \quad \Rightarrow \quad n-1 = 11 \quad \Rightarrow \quad n = 12. \]

Explanation: 12th term.
Quick Tip: Solve \(a_n = a + (n-1)d\).

Concept: Area = \(\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\).

Calculation: \[ Area = \frac{1}{2} |0(5-4) + 0(4-1) + 3(1-5)| = \frac{1}{2} |3(-4)| = \frac{1}{2} \times 12 = 6. \]

Explanation: Base \(AB = 4\) units (vertical), height = 3 units → area = \(\frac{1}{2} \times 4 \times 3 = 6\).
Quick Tip: If two points have same \(x\), base is vertical.

Concept: \(\tan(90^\circ - \theta) = \cot \theta\).

Calculation: \[ \tan 80^\circ = \cot 10^\circ, \quad \tan 67^\circ = \cot 23^\circ \] \[ \tan 10^\circ \cdot \tan 23^\circ \cdot \cot 10^\circ \cdot \cot 23^\circ = 1 \cdot 1 = 1. \]

Explanation: Complementary pairs multiply to 1.
Quick Tip: Pair \(\tan \theta\) and \(\tan(90^\circ - \theta)\).

Concept: Ratio of sides = \(\sqrt{ratio of areas}\).

Calculation: \[ \sqrt{\dfrac{100}{144}} = \dfrac{10}{12}. \]

Explanation: Simplify square root of ratio.
Quick Tip: Area ratio → side ratio = square root.

Concept: Secant intersects circle at two points; tangent at one; chord is segment inside.

Explanation: By definition.
Quick Tip: Secant: 2 points, Tangent: 1 point.

Concept: Area ratio = (side ratio)\(^2\).

Calculation: \[ \left(\dfrac{4}{9}\right)^2 = \dfrac{16}{81}. \]

Explanation: Square the side ratio.
Quick Tip: Area scales with square of linear dimensions.

Concept: Area ratio = (side ratio)\(^2\).

Calculation: \[ \dfrac{Area_{\triangle DEF}}{Area_{\triangle ABC}} = \left(\dfrac{EF}{BC}\right)^2 = \left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9} \] \[ Area_{\triangle DEF} = 54 \times \dfrac{16}{9} = 6 \times 16 = 96 cm^2. \]

Explanation: Scale area by square of side ratio.
Quick Tip: Identify corresponding sides.

Concept: Pythagoras theorem.

Calculation: \[ AC = \sqrt{BC^2 - AB^2} = \sqrt{169 - 144} = \sqrt{25} = 5 cm. \]

Explanation: 5-12-13 right triangle.
Quick Tip: Memorize 5-12-13 triplet.

Concept: Sum of angles = \(180^\circ\).

Calculation: \[ \angle F = 180^\circ - \angle D - \angle E, \quad \angle P = 180^\circ - \angle Q - \angle R \]
Given \(\angle D = \angle Q\), \(\angle E = \angle R\), \[ \angle F = 180^\circ - \angle Q - \angle R = \angle P. \]

Explanation: Third angles equal.
Quick Tip: Two angles equal → third angle equal.

Concept: SSS similarity → corresponding angles equal.

Calculation: \(\angle C = 180^\circ - 40^\circ - 80^\circ = 60^\circ\).
Corresponding to \(\angle F\) (since \(CA \leftrightarrow DF\)).

Explanation: \(\angle C = \angle F = 60^\circ\).
Quick Tip: SSS → angles opposite corresponding sides.

Concept: Intersecting circles → 2 common tangents (both external or one each).

Explanation: Two external tangents.
Quick Tip: Intersecting: 2 tangents; separate: 4; one inside: 2.

Concept: Volume ratio = \((r_1/r_2)^3\), surface area = \((r_1/r_2)^2\).

Calculation: \[ \left(\dfrac{r_1}{r_2}\right)^3 = \dfrac{64}{125} \quad \Rightarrow \quad \dfrac{r_1}{r_2} = \dfrac{4}{5} \] \[ Surface area ratio = \left(\dfrac{4}{5}\right)^2 = \dfrac{16}{25}. \]

Explanation: Square the cube root of volume ratio.
Quick Tip: Surface area ratio = \(\sqrt[3]{volume ratio}^2\).

Concept: Volume of cylinder = \(\pi r^2 h\).

Calculation:
Let the two cylinders be I and II.
Given: \[ \dfrac{r_I}{r_{II}} = \dfrac{4}{5}, \quad \dfrac{h_I}{h_{II}} = \dfrac{6}{7} \]
Ratio of volumes: \[ \dfrac{V_I}{V_{II}} = \dfrac{\pi r_I^2 h_I}{\pi r_{II}^2 h_{II}} = \left(\dfrac{r_I}{r_{II}}\right)^2 \cdot \dfrac{h_I}{h_{II}} = \left(\dfrac{4}{5}\right)^2 \cdot \dfrac{6}{7} \] \[ = \dfrac{16}{25} \cdot \dfrac{6}{7} = \dfrac{16 \times 6}{25 \times 7} = \dfrac{96}{175} \]
Thus, \[ V_I : V_{II} = 96 : 175. \]

Explanation: Volume ratio = (ratio of radii squared) × (ratio of heights).
Quick Tip: Volume ∝ \(r^2 h\) → multiply square of radius ratio and height ratio.

Concept: TSA = curved + base = \(2\pi R^2 + \pi R^2 = 3\pi R^2\).

Explanation: Curved surface + flat base.
Quick Tip: Hemisphere TSA = \(3\pi R^2\), CSA = \(2\pi R^2\).

Concept: CSA = \(\pi r l\).

Calculation: \[ \pi \times 14 \times l = 880 \quad \Rightarrow \quad l = \dfrac{880}{14\pi} = \dfrac{880 \div 44}{14 \div 44 \times \pi} \quad wait: \] \[ \dfrac{880}{14 \times \pi} = \dfrac{880 \div 44}{14 \div 44 \times \pi} = 20 / \pi \times \pi = 20 cm. \]

Explanation: \(l = \dfrac{CSA}{\pi r}\).
Quick Tip: Isolate \(l\) from \(\pi r l\).

Concept: Diagonal = \(a\sqrt{3}\).

Calculation: \[ a\sqrt{3} = 2\sqrt{3} \quad \Rightarrow \quad a = 2 cm. \]

Explanation: Divide by \(\sqrt{3}\).
Quick Tip: Cube diagonal = \(a\sqrt{3}\).

Concept: TSA = \(6a^2\), new = \(6(2a)^2 = 24a^2\).

Calculation: \[ \dfrac{24a^2}{6a^2} = 4. \]

Explanation: Area scales with square of linear dimensions.
Quick Tip: Linear scale factor \(k\) → area \(k^2\).

Concept: Sphere TSA = \(4\pi R^2\), hemisphere = \(3\pi R^2\).

Calculation: \[ \dfrac{4\pi R^2}{3\pi R^2} = \dfrac{4}{3}. \]

Explanation: Direct ratio.
Quick Tip: Sphere: \(4\pi R^2\), hemisphere: \(3\pi R^2\).

Concept: CSA of hemisphere = \(2\pi R^2\).

Calculation: \[ 2\pi R^2 = 1232 \quad \Rightarrow \quad R^2 = \dfrac{1232}{2\pi} = \dfrac{616}{\pi} \] \[ R = \sqrt{\dfrac{616}{\pi}} = \sqrt{196} = 14 cm \quad (since 196\pi / \pi = 196). \]

Explanation: \(R = \sqrt{\dfrac{CSA}{2\pi}}\).
Quick Tip: Hemisphere CSA = \(2\pi R^2\).

Concept: \(\cos^2 \theta = 1 - \sin^2 \theta\).

Calculation: \[ \cos \theta + \cos^2 \theta = 1 \quad \Rightarrow \quad \cos \theta = 1 - \cos^2 \theta = \sin^2 \theta \] \[ \cos^2 \theta = \sin^4 \theta \]
Let \(u = \sin^2 \theta\), \[ u^2 = 1 - u \quad \Rightarrow \quad u^2 + u - 1 = 0 \]
But directly: \[ \sin^2 \theta + \sin^4 \theta = \cos \theta + \cos^2 \theta = 1. \]

Explanation: Given equation shows sum equals 1.
Quick Tip: Substitute \(\cos \theta = \sin^2 \theta\).

Concept: \(1 + \tan^2 \theta = \sec^2 \theta\), \(1 + \cot^2 \theta = \csc^2 \theta\).

Calculation: \[ \dfrac{\sec^2 A}{\csc^2 A} = \dfrac{1/\cos^2 A}{1/\sin^2 A} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A. \]

Explanation: Simplify using reciprocal identities.
Quick Tip: \(\sec^2 / \csc^2 = \tan^2\).

Concept: Decimal terminates if denominator (after simplifying) has prime factors only 2 and/or 5.

Calculation:
(A): \(3^2\) → non-terminating.
(B): \(7^2\) → non-terminating.
(C): \(3^2\) → non-terminating.
(D): \(2^2 \times 5^3\) → terminating.

Explanation: Only (D) has denominator of form \(2^n 5^m\).
Quick Tip: Check prime factors of denominator.

Concept: \(0.505 = \dfrac{505}{1000}\). Simplify and match denominator.

Calculation: \[ 0.505 = \dfrac{505}{1000} = \dfrac{101 \times 5}{200 \times 5} = \dfrac{101}{200} = \dfrac{101}{2^3 \times 5^2}. \]

Explanation: \(1000 = 2^3 \times 5^3\), cancel one 5.
Quick Tip: Multiply numerator and denominator by 1000 for three decimal places.

Calculation: \[ a = 4 \times 5 + 1 = 20 + 1 = 21. \]

Explanation: Direct substitution.
Quick Tip: \(a = bq + r\).

Calculation: \[ 2x^2 - 4x - 6 = 0 \quad \Rightarrow \quad x^2 - 2x - 3 = 0 \] \[ (x - 3)(x + 1) = 0 \quad \Rightarrow \quad x = 3, -1. \]

Explanation: Factorize or use quadratic formula.
Quick Tip: Divide by 2 to simplify.

Concept: Degree of product = sum of degrees.

Calculation:
Degree of first: 3, second: 2 → total = 5.

Explanation: Leading terms: \(x^3 \cdot x^2 = x^5\).
Quick Tip: Add degrees of factors.

Concept: Polynomial has non-negative integer exponents.

Calculation:
(D): \(x^{-1}\) → not polynomial.

Explanation: Negative exponent.
Quick Tip: All exponents ≥ 0.

Calculation:
Sum = 0, product = -4 → \(x^2 - 4 = 0\).

Explanation: \((x-2)(x+2)\).
Quick Tip: Roots \(r, -r\) → \(x^2 - r^2\).

Concept: Sum of roots = \(-\frac{b}{a}\).

Calculation: \[ \alpha + \beta = -7. \]

Explanation: For \(t^2 + 7t + 10\), \(b = 7\), \(a = 1\).
Quick Tip: Sum = \(-\frac{b}{a}\).

Calculation: \[ \sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2} \quad \Rightarrow \quad \frac{1 + \sqrt{3}}{2} \] \[ \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2} \quad \Rightarrow \quad \frac{\sqrt{3} + 1}{2} \]
Difference = 0.

Explanation: Both sums are equal.
Quick Tip: \(\sin \theta + \cos \theta = \sin(90^\circ - \theta) + \cos(90^\circ - \theta)\).

Calculation: \[ (k-1)(-4)^2 + k(-4) + 1 = 0 \] \[ 16(k-1) - 4k + 1 = 0 \quad \Rightarrow \quad 16k - 16 - 4k + 1 = 0 \] \[ 12k - 15 = 0 \quad \Rightarrow \quad k = \frac{15}{12} = \frac{5}{4} \quad wait \]
Wait: \(k = \frac{15}{12} = \frac{5}{4}\)? But answer is negative.
Wait: Plug \(x = -4\): \[ (k-1)(16) + k(-4) + 1 = 0 \] \[ 16k - 16 - 4k + 1 = 0 \quad \Rightarrow \quad 12k - 15 = 0 \quad \Rightarrow \quad k = \frac{15}{12} = \frac{5}{4} \]
But option (B). Wait: Question says -4, but calculation gives positive.
Wait: Let's verify:
For \(k = \frac{5}{4}\): \[ \left(\frac{5}{4} - 1\right)x^2 + \frac{5}{4}x + 1 = \frac{1}{4}x^2 + \frac{5}{4}x + 1 \]
At \(x = -4\): \[ \frac{1}{4}(16) + \frac{5}{4}(-4) + 1 = 4 - 5 + 1 = 0 \quad \]
So \(k = \frac{5}{4}\).

Correct: (B) \(\frac{5}{4}\).

Explanation: Substitute zero into polynomial.
Quick Tip: \(p(root) = 0\).

Concept: Tangents from external point are equal.

Calculation: \[ PA = PB = 8 cm. \]

Explanation: Theorem of equal tangents.
Quick Tip: \(PA = PB\).

Concept: Quadrilateral OAPB: \(\angle OAP = \angle OBP = 90^\circ\), \(\angle APB = 80^\circ\).

Calculation: \[ \angle AOP + \angle BOP = 360^\circ - 90^\circ - 90^\circ - 80^\circ = 100^\circ \]
Since OA = OB (radii), \(\triangle AOP \cong \triangle BOP\) → \(\angle AOP = \angle BOP = 50^\circ\).

Explanation: \(\angle POA = 50^\circ\).
Quick Tip: Tangent ⊥ radius.

Concept: Tangent is perpendicular to radius at point of contact.

Explanation: Theorem.
Quick Tip: Radius ⊥ tangent.

Calculation: \[ \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}. \]

Explanation: Area ∝ \(r^2\).
Quick Tip: Square the radius ratio.

Calculation: \[ Area = \dfrac{30^\circ}{360^\circ} \pi (42)^2 = \dfrac{1}{12} \pi \times 1764 = \dfrac{1764 \pi}{12} = 147 \pi \approx 147 \times 3.14 = 461.58 \approx 462. \]

Explanation: \(\dfrac{\theta}{360} \pi r^2\).
Quick Tip: Use \(\pi \approx 3.14\) or \(22/7\).

Concept: \(C = 2\pi r\) → \(C \propto r\).

Calculation: \[ \dfrac{r_1}{r_2} = \dfrac{C_1}{C_2} = \dfrac{5}{7}. \]

Explanation: Direct proportion.
Quick Tip: Circumference ratio = radius ratio.

Calculation: \[ 7(\sec^2 A - \tan^2 A) = 7(1) = 7. \]

Explanation: Identity \(\sec^2 - \tan^2 = 1\).
Quick Tip: Factor out 7.

Calculation: \[ b^2 (a \cos \theta)^2 + a^2 (b \sin \theta)^2 = a^2 b^2 \cos^2 \theta + a^2 b^2 \sin^2 \theta = a^2 b^2 (\cos^2 \theta + \sin^2 \theta) = a^2 b^2. \]

Explanation: Factor out \(a^2 b^2\).
Quick Tip: Use \(\cos^2 + \sin^2 = 1\).

Calculation: \[ \tan 60^\circ = \sqrt{3} = \dfrac{h}{10} \quad \Rightarrow \quad h = 10\sqrt{3} m. \]

Explanation: Opposite over adjacent.
Quick Tip: \(\tan \theta = \dfrac{opp}{adj}\).

Calculation:
Height = opposite = 30 m, angle with ground = \(60^\circ\), hypotenuse = string length \(l\). \[ \sin 60^\circ = \dfrac{30}{l} \quad \Rightarrow \quad \dfrac{\sqrt{3}}{2} = \dfrac{30}{l} \quad \Rightarrow \quad l = \dfrac{60}{\sqrt{3}} = 20\sqrt{3} m. \]

Correct: (C) \(20\sqrt{3}\) m.

Explanation: \(\sin \theta = \dfrac{opp}{hyp}\).
Quick Tip: Angle with ground → use sin for height.

The general formula for the \(n\)th term of an A.P. is: \[ a_n = a + (n-1)d \]
where \(a\) is the first term and \(d\) is the common difference.

Given:
- 5th term: \(a_5 = 11\)
- Common difference: \(d = 2\)

Substitute into the formula: \[ a_5 = a + (5-1)d \quad \Rightarrow \quad 11 = a + 4 \times 2 \] \[ 11 = a + 8 \] \[ a = 11 - 8 = 3 \]

Verification:
- 1st term: 3
- 2nd term: \(3 + 2 = 5\)
- 3rd term: \(5 + 2 = 7\)
- 4th term: \(7 + 2 = 9\)
- 5th term: \(9 + 2 = 11\) (correct)

Thus, the first term is 3.

Explanation: Subtract \(4d\) from the 5th term to find the first term.
Quick Tip: To go backwards from \(n\)th term to 1st term, subtract \((n-1)d\).

The sum of the first \(n\) terms is given as: \[ S_n = n^2 + 2n + 1 = (n+1)^2 \]

The \(n\)th term is found using: \[ a_n = S_n - S_{n-1} \]

For the 6th term: \[ a_6 = S_6 - S_5 \]

Calculate: \[ S_6 = (6+1)^2 = 7^2 = 49 \] \[ S_5 = (5+1)^2 = 6^2 = 36 \] \[ a_6 = 49 - 36 = 13 \]

General formula for \(a_n\): \[ a_n = (n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1 \] \[ a_6 = 2(6) + 1 = 13 \]

Since 13 is not in the options (A) 29, (B) 19, (C) 15, the answer is (D) none of these.

Explanation: The sum \(S_n = (n+1)^2\) gives \(a_n = 2n + 1\), so 6th term is 13.
Quick Tip: For any A.P., \(a_n = S_n - S_{n-1}\).

An A.P. requires a constant common difference \(d\).

- (A): Differences:
\(7-1 = 6\), \(9-7 = 2\), \(16-9 = 7\) → not constant → not A.P.

- (B): Differences:
\(x^3 - x^2 = x^2(x-1)\), \(x^4 - x^3 = x^3(x-1)\), etc. → not constant → not A.P.

- (C): Differences:
\(2x - x = x\), \(3x - 2x = x\), \(4x - 3x = x\) → constant \(d = x\) → A.P.

- (D): Differences:
\(16 - 4 = 12\), \(36 - 16 = 20\), \(64 - 36 = 28\) → not constant → not A.P.

Only (C) has a constant difference.

Explanation: Check consecutive differences; only (C) satisfies.
Quick Tip: Compute at least first three differences.

Check common difference:

- (A): \(d = 1\) → A.P.
- (B): \(d = 3\) → A.P.
- (C): \(d = 2\) → A.P.
- (D): Sequence: \(4, 16, 36, 64\)
Differences:
\(16-4 = 12\), \(36-16 = 20\), \(64-36 = 28\) → increasing → not A.P.

Only (D) fails.

Explanation: Squares of even numbers form a quadratic sequence.
Quick Tip: Second differences are constant only for quadratic sequences.

Given:
- First term: \(a = 1\)
- Common difference: \(d = 3\)
- Number of terms: \(n = 20\)

Formula for sum: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] \[ S_{20} = \frac{20}{2} [2(1) + (19)(3)] = 10 [2 + 57] = 10 \times 59 = 590 \]

Alternative method: \[ S_n = \frac{n}{2} (a + l), \quad l = 1 + 19 \times 3 = 58 \] \[ S_{20} = \frac{20}{2} (1 + 58) = 10 \times 59 = 590 \]

Verification:
- 20th term: \(1 + 19 \times 3 = 58\)
- Average of first and last: \(\frac{1+58}{2} = 29.5\)
- Sum: \(20 \times 29.5 = 590\)

Explanation: Both methods confirm 590.
Quick Tip: Use \(S_n = \frac{n}{2} (a + l)\) for quick check.

Evaluate each:

- (A):
\(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\cos 60^\circ = \frac{1}{2}\)
\(\left(\frac{\sqrt{3}}{2}\right)^2 + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} \neq 1\)

- (B):
\(\sin 90^\circ = 1\), \(\cos 90^\circ = 0\)
\(1 \times 0 = 0 \neq 1\)

- (C):
\(\sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \neq 1\)

- (D):
\(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\)
\(\frac{1}{\sqrt{2}} \times \frac{1}{\frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2}} \times \sqrt{2} = 1\)

Only (D) equals 1.

Explanation: (D) simplifies to \(\tan 45^\circ = 1\).
Quick Tip: \(\frac{\sin \theta}{\cos \theta} = \tan \theta\).

Use identity: \[ 1 + \tan^2 A = \sec^2 A \] \[ \cos^2 A (1 + \tan^2 A) = \cos^2 A \cdot \sec^2 A = \cos^2 A \cdot \frac{1}{\cos^2 A} = 1 \]

Alternative: \[ \tan^2 A = \frac{\sin^2 A}{\cos^2 A} \quad \Rightarrow \quad 1 + \tan^2 A = 1 + \frac{\sin^2 A}{\cos^2 A} = \frac{\cos^2 A + \sin^2 A}{\cos^2 A} = \frac{1}{\cos^2 A} = \sec^2 A \]
Same result.

Explanation: Direct application of Pythagorean identity.
Quick Tip: Memorize: \(1 + \tan^2 \theta = \sec^2 \theta\).

In a 30-60-90 triangle:
- Opposite to 30°: 1
- Adjacent: \(\sqrt{3}\)
- Hypotenuse: 2
\[ \tan 30^\circ = \frac{opposite}{adjacent} = \frac{1}{\sqrt{3}} \]
Rationalize: \[ \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3} \]
Both forms are acceptable, but option is \(\frac{1}{\sqrt{3}}\).

Explanation: Standard value from unit triangle.
Quick Tip: 30°: \(\sin = \frac{1}{2}\), \(\cos = \frac{\sqrt{3}}{2}\), \(\tan = \frac{1}{\sqrt{3}}\).

30-60-90 triangle sides:
- Opposite 30°: 1
- Opposite 60°: \(\sqrt{3}\)
- Hypotenuse: 2
\[ \cos 60^\circ = \frac{adjacent}{hypotenuse} = \frac{1}{2} \]

Or: \[ \cos 60^\circ = \frac{\sqrt{3}}{2} \quad (incorrect for 60°) \quad No, \cos 60^\circ = \frac{1}{2} \]

Explanation: Adjacent to 60° is half the hypotenuse.
Quick Tip: 60°: \(\cos = \frac{1}{2}\), \(\sin = \frac{\sqrt{3}}{2}\).

\[ \sin 90^\circ = 1 \quad \Rightarrow \quad \sin^2 90^\circ = 1 \] \[ \tan 45^\circ = 1 \quad \Rightarrow \quad \tan^2 45^\circ = 1 \] \[ 1 - 1 = 0 \]

Explanation: Both terms are 1.
Quick Tip: \(\sin 90^\circ = 1\), \(\tan 45^\circ = 1\).

Roots: 3 and -10
Sum of roots: \(3 + (-10) = -7\)
Product: \(3 \times (-10) = -30\)

Standard form: \[ x^2 - (sum)x + product = 0 \] \[ x^2 - (-7)x + (-30) = x^2 + 7x - 30 = 0 \]

Verification: \[ (x - 3)(x + 10) = x^2 + 10x - 3x - 30 = x^2 + 7x - 30 \]

Matches (A).

Explanation: Sum is negative → positive linear coefficient.
Quick Tip: Roots \(r_1, r_2\): \((x - r_1)(x - r_2)\).

Sum = 3 → coefficient of \(x\) = -3
Product = -2 → constant term = -2
\[ x^2 - (sum)x + product = x^2 - 3x - 2 \]

Matches (A).

Explanation: Standard quadratic form.
Quick Tip: Sum = \(-b\), Product = \(c\).

Degree of \(p(x)\): 4
Degree of \(q(x)\): 1

Degree of quotient: \[ \deg(p) - \deg(q) = 4 - 1 = 3 \]

Long division: \[ x^4 - 2x^3 + 17x^2 - 4x + 30 \div (x + 2) \]
Leading term: \(x^3\) → quotient starts with \(x^3\) → degree 3.

Explanation: Polynomial division rule.
Quick Tip: Subtract degrees of dividend and divisor.

Second equation: \[ 3x + 6y + 9 = 3(x + 2y + 3) = 0 \]
So: \[ 3(x + 2y + 3) = 0 \quad \Rightarrow \quad x + 2y + 3 = 0 \]
Both equations are identical → same line → infinitely many solutions.

Explanation: Dependent system.
Quick Tip: If one equation is a multiple of the other → infinite solutions.

Parallel lines never intersect → no common point → no solution.

Condition: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

Explanation: Inconsistent system.
Quick Tip: Parallel → no solution.

Check ratios: \[ \frac{5}{7} \approx 0.714, \quad \frac{-4}{6} = -0.667, \quad \frac{8}{-9} \approx -0.889 \]
All ratios different → unique solution → consistent.

Explanation: Intersect at one point.
Quick Tip: If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), then consistent.

Sum: \(\alpha + \beta = \frac{5}{3}\)
Product: \(\alpha \beta = \frac{2}{3}\)
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{3}\right)^2 - 2 \cdot \frac{2}{3} \] \[ = \frac{25}{9} - \frac{4}{3} = \frac{25}{9} - \frac{12}{9} = \frac{13}{9} \]

Explanation: Use identity.
Quick Tip: \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\).

Substitute \(x = 2\): \[ 2(2)^2 - 7(2) - p = 0 \] \[ 8 - 14 - p = 0 \quad \Rightarrow \quad -6 - p = 0 \quad \Rightarrow \quad p = -6 \]

Verification: \[ 2x^2 - 7x + 6 = 0 \quad (p = -6 \Rightarrow +6) \] \[ (x-2)(2x-3) = 0 \quad \Rightarrow \quad x = 2, \frac{3}{2} \]
Root 2 confirmed.

Explanation: Direct substitution.
Quick Tip: \(p(root) = 0\).

Sum of roots: \[ \alpha + \beta = \frac{1}{2} \]
One root: \(\alpha = -\frac{3}{2}\)
Other root: \[ \beta = \frac{1}{2} - (-\frac{3}{2}) = \frac{1}{2} + \frac{3}{2} = 2 \]

Factorize: \[ 2x^2 - x - 6 = (2x + 3)(x - 2) = 0 \]
Roots: \(x = -\frac{3}{2}\), \(x = 2\)

Explanation: Sum of roots = \(-\frac{b}{a}\).
Quick Tip: Other root = sum − given root.

Discriminant: \[ D = b^2 - 4ac = (-6)^2 - 4(2)(3) = 36 - 24 = 12 \] \[ D = 12 > 0 \quad \Rightarrow \quad two distinct real roots \]

Roots: \[ x = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2} \]
Both real and unequal.

Explanation: \(D > 0\) → real and distinct.
Quick Tip: - \(D > 0\): real, unequal - \(D = 0\): real, equal - \(D < 0\): not real

Class interval length (or class width) is the difference between the upper and lower boundaries of any class.

Given classes:
- 2–5 → upper limit = 5, lower limit = 2 \[ Width = 5 - 2 = 3 \]
- 5–8 → \(8 - 5 = 3\)
- 8–11 → \(11 - 8 = 3\)

All classes have the same width of 3.

Note: These are exclusive classes (since 5 is the start of the next class), so the width is still \(5 - 2 = 3\).

Verification:
The sequence of lower limits: 2, 5, 8, ... → common difference = 3 → class width = 3.

Explanation: Class width = upper limit − lower limit = 3.
Quick Tip: Class width = upper boundary − lower boundary.

Let the four consecutive odd numbers be: \[ x, \, x+2, \, x+4, \, x+6 \]
where \(x\) is the first odd number.

Their mean is 6: \[ Mean = \frac{x + (x+2) + (x+4) + (x+6)}{4} = 6 \] \[ \frac{4x + 12}{4} = 6 \] \[ x + 3 = 6 \quad \Rightarrow \quad x = 3 \]

So the numbers are: \[ 3, \, 5, \, 7, \, 9 \]

Largest number = 9

Verification:
Sum = \(3 + 5 + 7 + 9 = 24\)
Mean = \(24 \div 4 = 6\) (correct)

Explanation: The mean of four consecutive odd numbers is the average of the second and third terms, which equals 6 → third term = 7, so largest = 9.
Quick Tip: For even number of terms, mean = average of middle two terms.

First 6 even natural numbers: \[ 2, 4, 6, 8, 10, 12 \]

Sum: \[ 2 + 4 + 6 + 8 + 10 + 12 = 42 \]
(or use formula: sum of first \(n\) even numbers = \(n(n+1)\) → \(6 \times 7 = 42\))

Mean: \[ \frac{42}{6} = 7 \]

Alternative:
This is an A.P. with \(a=2\), \(d=2\), \(n=6\) \[ Mean = \frac{first + last}{2} = \frac{2 + 12}{2} = 7 \]

Explanation: Mean of first \(n\) even numbers = \(n+1\) → for \(n=6\), mean = 7.
Quick Tip: Mean of first \(n\) even numbers = \(n+1\).

Recall the identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \]
Divide both sides by \(\sin^2 \theta\): \[ 1 + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta} \] \[ 1 + \cot^2 \theta = \csc^2 \theta \]

Alternative: \[ \cot \theta = \frac{\cos \theta}{\sin \theta} \quad \Rightarrow \quad \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \] \[ 1 + \cot^2 \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta} = \csc^2 \theta \]

Explanation: Pythagorean identity in terms of cotangent.
Quick Tip: Memorize: \(1 + \cot^2 \theta = \csc^2 \theta\).

Data: 8, 7, 9, 3, 9, 5, 4, 5, 7, 5

Frequency table:
\begin{tabular{|c|c|
\hline
Number & Frequency

\hline
3 & 1

4 & 1

5 & 3

7 & 2

8 & 1

9 & 2

\hline
\end{tabular

Mode = value with highest frequency = 5 (appears 3 times)

Explanation: Mode is the most frequent observation.
Quick Tip: Count occurrences → highest count = mode.

By complement rule: \[ P(E') = 1 - P(E) \] \[ P(E') = 1 - 0.02 = 0.98 \]

Explanation: Probability of complement is 1 minus probability of event.
Quick Tip: \(P(not E) = 1 - P(E)\).

Total outcomes when two dice are thrown: \(6 \times 6 = 36\)

Favorable outcomes (difference = 0 → both dice show same number): \[ (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \]
Number of favorable = 6

Probability: \[ P = \frac{favorable}{total} = \frac{6}{36} = \frac{1}{6} \]

Explanation: Only when both dice are equal → 6 cases out of 36.
Quick Tip: Same number on both dice → 6 outcomes.

Sample space for two coins: \[ \{HH, HT, TH, TT\} \quad (4 outcomes) \]

Favorable: both heads → \(HH\) → 1 outcome

Probability: \[ P(both heads) = \frac{1}{4} \]

Alternative: \[ P(H on first) = \frac{1}{2}, \quad P(H on second) = \frac{1}{2} \] \[ P(both) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

Explanation: Independent events → multiply probabilities.
Quick Tip: For two coins: \(P(HH) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).

Total months in a year = 12

Favorable: June, September → 2 months

Probability: \[ P = \frac{favorable}{total} = \frac{2}{12} = \frac{1}{6} \]

Explanation: Two specific months out of twelve.
Quick Tip: Equal likelihood → count favorable months.

Total outcomes on a die: 6 (1, 2, 3, 4, 5, 6)

Favorable: 4 or 5 → 2 outcomes

Probability: \[ P(4 or 5) = \frac{2}{6} = \frac{1}{3} \]

Alternative: \[ P(4) = \frac{1}{6}, \quad P(5) = \frac{1}{6} \quad \Rightarrow \quad P(4 or 5) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \]

Explanation: Mutually exclusive events → add probabilities.
Quick Tip: For "or" with equal probability: \(\frac{number of favorable}{total}\).

First, compute the values: \[ \sin 60^\circ = \dfrac{\sqrt{3}}{2}, \quad \cos 60^\circ = \dfrac{1}{2} \]

Points: \[ A = \left(8 \cdot \dfrac{\sqrt{3}}{2}, \, 0\right) = (4\sqrt{3}, \, 0) \] \[ B = \left(0, \, 8 \cdot \dfrac{1}{2}\right) = (0, \, 4) \]

Distance formula: \[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] \[ AB = \sqrt{(0 - 4\sqrt{3})^2 + (4 - 0)^2} = \sqrt{(4\sqrt{3})^2 + 4^2} = \sqrt{48 + 16} = \sqrt{64} = 8 \]

Verification:
- Horizontal distance: \(4\sqrt{3} \approx 6.928\)

- Vertical distance: 4

- Hypotenuse: \(\sqrt{(6.928)^2 + 4^2} \approx \sqrt{48 + 16} = \sqrt{64} = 8\)

Explanation: The points lie on axes; distance simplifies to \(\sqrt{(4\sqrt{3})^2 + 4^2} = 8\).
Quick Tip: Substitute trig values early.

Distance from origin \((0,0)\) to point \((x,y)\): \[ OP = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} \]

This is the distance formula from origin.

Example:
- Point (3,4): \(OP = \sqrt{9 + 16} = 5\)
- Point (1,1): \(OP = \sqrt{1 + 1} = \sqrt{2}\)

Explanation: Pythagorean theorem in coordinate plane.
Quick Tip: Distance from origin = \(\sqrt{x^2 + y^2}\).

Distance from a point \((x, y)\) to the y-axis is the absolute value of the x-coordinate.

Point: \((12, 14)\) → x = 12 \[ Distance = |12| = 12 \]

Geometric meaning:
- y-axis is the line \(x = 0\)
- Horizontal distance from (12,14) to (0,14) = 12 units

Explanation: Only x-coordinate matters for y-axis distance.
Quick Tip: Distance to y-axis = |x|.

In a point \((x, y)\):
- Abscissa (x-coordinate) = x
- Ordinate (y-coordinate) = y

Given point: \((-6, -8)\) \[ Ordinate = -8 \]

Explanation: Ordinate is the y-value.
Quick Tip: Ordinate = y-coordinate.

Quadrant rules:
- I: \(x > 0\), \(y > 0\)
- II: \(x < 0\), \(y > 0\)
- III: \(x < 0\), \(y < 0\)
- IV: \(x > 0\), \(y < 0\)

Point: \((3, -4)\)
- \(x = 3 > 0\)
- \(y = -4 < 0\)

→ Fourth quadrant

Explanation: Positive x, negative y → IV.
Quick Tip: Sign of (x, y): (+, −) → IV.

Second quadrant: \(x < 0\), \(y > 0\)

Check:
- (A) (3,2): \(x > 0\), \(y > 0\) → I
- (B) (-3,2): \(x < 0\), \(y > 0\) → II
- (C) (3,-2): \(x > 0\), \(y < 0\) → IV
- (D) (-3,-2): \(x < 0\), \(y < 0\) → III

Only (B) is in second quadrant.

Explanation: Negative x, positive y.
Quick Tip: II: left and above origin.

Mid-point formula: \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

Points: \(A(4, -4)\), \(B(-4, 4)\) \[ x = \frac{4 + (-4)}{2} = \frac{0}{2} = 0 \] \[ y = \frac{-4 + 4}{2} = \frac{0}{2} = 0 \] \[ Mid-point = (0, 0) \]

Verification:
- Average of x: \(\frac{4 + (-4)}{2} = 0\)
- Average of y: \(\frac{-4 + 4}{2} = 0\)

Explanation: The points are symmetric about origin → midpoint is origin.
Quick Tip: Mid-point = average of coordinates.

Let B be \((x, y)\).
Mid-point: \[ \left( \frac{5 + x}{2}, \frac{7 + y}{2} \right) = (2, 4) \]

Solve: \[ \frac{5 + x}{2} = 2 \quad \Rightarrow \quad 5 + x = 4 \quad \Rightarrow \quad x = -1 \] \[ \frac{7 + y}{2} = 4 \quad \Rightarrow \quad 7 + y = 8 \quad \Rightarrow \quad y = 1 \] \[ B = (-1, 1) \]

Verification:
Mid-point of \(A(5,7)\) and \(B(-1,1)\): \[ x = \frac{5 + (-1)}{2} = 2, \quad y = \frac{7 + 1}{2} = 4 \quad \checkmark \]

Explanation: Use mid-point formula and solve for unknown.
Quick Tip: Let B = (x,y) → set up equations.

Centre of a circle is the mid-point of any diameter.

Endpoints: \(A(10, -6)\), \(B(-6, 10)\) \[ x = \frac{10 + (-6)}{2} = \frac{4}{2} = 2 \] \[ y = \frac{-6 + 10}{2} = \frac{4}{2} = 2 \] \[ Centre = (2, 2) \]

Verification:
Distance from centre to A: \[ \sqrt{(10-2)^2 + (-6-2)^2} = \sqrt{64 + 64} = \sqrt{128} \]
To B: \[ \sqrt{(-6-2)^2 + (10-2)^2} = \sqrt{64 + 64} = \sqrt{128} \]
Equal → correct centre.

Explanation: Mid-point of diameter = centre.
Quick Tip: Centre = mid-point of diameter.

Centroid formula for triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\): \[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

Vertices:
- \(A(4,6)\)
- \(B(0,4)\)
- \(C(5,5)\)
\[ x = \frac{4 + 0 + 5}{3} = \frac{9}{3} = 3 \] \[ y = \frac{6 + 4 + 5}{3} = \frac{15}{3} = 5 \quad wait! \]
Wait: \(6 + 4 + 5 = 15\), \(15 \div 3 = 5\)? But answer is (3,4)?

Wait: Recheck: \[ y: 6 + 4 + 5 = 15 \quad \Rightarrow \quad \frac{15}{3} = 5 \]
But option (B) is (3,4) → mistake?

Wait: Let's list again:
- (4,6) → y=6
- (0,4) → y=4
- (5,5) → y=5
Sum of y: \(6 + 4 + 5 = 15\) \[ \frac{15}{3} = 5 \]
So centroid = (3, 5) → (D)

Wait: But earlier said (B) → error.

Correct calculation: \[ x = \frac{4+0+5}{3} = 3, \quad y = \frac{6+4+5}{3} = 5 \] \[ \Rightarrow (3, 5) \]

Correct Answer: (D) (3,5)

Explanation: Average of x-coordinates and y-coordinates.
Quick Tip: Centroid = \(\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)\).

Concept: Curved Surface Area (C.S.A.) of a cone = \(\pi r l\), where \(r\) is radius and \(l\) is slant height.


Step 1: Find slant height using Pythagoras theorem: \(l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49+576} = \sqrt{625} = 25 cm\).


Step 2: Compute C.S.A.: \(\pi r l = \pi \times 7 \times 25 = 175\pi cm^2\).


Explanation: Curved surface area uses slant height, not vertical height.
Quick Tip: Always calculate slant height first using \(l=\sqrt{r^2+h^2}\) for cones.

Concept: Area swept by minute hand = area of sector of a circle: \(A = \frac{\theta}{360} \pi r^2\), where \(\theta\) in degrees.


Step 1: Find angle swept in 40 minutes: 1 minute → 6° (360°/60). So 40 minutes → \(40 \times 6 = 240^\circ\).


Step 2: Compute area: \(A = \frac{240}{360} \pi (7^2) = \frac{2}{3} \pi (49) = \frac{98\pi}{3} cm^2 \approx 102.67 cm^2\).


Explanation: Angle of sector proportional to time fraction.
Quick Tip: Area of sector = \(\frac{\theta}{360} \pi r^2\), convert time to angle first.

Concept: Use complementary angle identity: \(\tan(90^\circ - \theta) = \cot \theta\).


Calculation:
\(\tan 7^\circ \cdot \tan 60^\circ \cdot \tan 83^\circ = \tan 7^\circ \cdot \sqrt{3} \cdot \tan (90^\circ - 7^\circ) = \tan 7^\circ \cdot \sqrt{3} \cdot \cot 7^\circ = \sqrt{3}\).


Explanation: \(\tan\theta \cdot \cot\theta =1\), so product simplifies to \(\sqrt{3}\).
Quick Tip: Use complementary angle identity: \(\tan(90^\circ - \theta) = \cot \theta\) to simplify products.

Concept: Section formula (internal division): \((x,y) = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)\).


Calculation:
\(m:n = 2:3\), \(P(x_1,y_1)=(-1,7)\), \(Q(x_2,y_2)=(4,-3)\)
\(x = \frac{3*4 + 2*(-1)}{3+2} = \frac{12-2}{5} = \frac{10}{5} = 2\)
\(y = \frac{3*(-3) + 2*7}{5} = \frac{-9+14}{5} = \frac{5}{5}=1\)


Point = (2,1).


Explanation: Substitute correctly into section formula.
Quick Tip: Internal division formula: \(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\).

Concept: Area of triangle with vertices \((x_1,y_1),(x_2,y_2),(x_3,y_3)\):
\(A = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\)


Calculation:
\(A = \frac{1}{2} |(-5)((-5)-2) + 3*(2 -(-1)) +5*((-1)-(-5))|\)
\(= \frac{1}{2} |(-5)*(-7) + 3*(3) + 5*(4)| = \frac{1}{2} |35 + 9 + 20| = \frac{1}{2}*64 = 32\)


Area = 32 sq. units.


Explanation: Use determinant formula for area of triangle.
Quick Tip: Area formula: \(A = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\); signs do not matter because of absolute value.

Concept: Diagonal of a cube \(d = a\sqrt{3}\) where \(a\) is edge. Total Surface Area (TSA) = \(6a^2\).


Step 1: Find edge length: \(a = \frac{d}{\sqrt{3}} = \frac{9\sqrt{3}}{\sqrt{3}} = 9~cm\).


Step 2: TSA = \(6a^2 = 6*9^2 = 6*81 = 486~cm^2\).


Explanation: Cube surface area formula uses edge length.
Quick Tip: Cube diagonal \(d = a\sqrt{3}\), TSA = \(6a^2\), always find edge first.

Concept: Quadratic formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).


Given \(a=2, b=-2\sqrt{2}, c=1\).

Discriminant \(D = b^2 - 4ac = (-2\sqrt{2})^2 - 4*2*1 = 8 - 8 = 0\).


Since \(D=0\), roots are real and equal: \(x = \frac{-b}{2a} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}\).


Explanation: Zero discriminant → repeated root.
Quick Tip: Check discriminant first; \(D=0\) implies equal roots.

Concept: Sum of n-term A.P.: \(S_n = \frac{n}{2}[2a + (n-1)d]\).


Step 1: Identify first term \(a=3\), common difference \(d=8\).

Step 2: Find number of terms \(n\): \(l = a + (n-1)d = 67 \Rightarrow 3 + (n-1)*8 = 67 \Rightarrow 8(n-1)=64 \Rightarrow n=9\).


Step 3: Compute sum: \(S_9 = \frac{9}{2}[2*3 + (9-1)*8] = \frac{9}{2}[6 + 64] = \frac{9}{2}*70 = 315\)

Correction: Check calculation: \(2*3 + 8*8 = 6 + 64 = 70\), \(70*9/2 = 315\) → Correct Answer: 315.


Explanation: Use formula carefully.
Quick Tip: Identify \(a\), \(d\), \(n\) carefully before applying \(S_n = \frac{n}{2}[2a + (n-1)d]\).

Concept: \(n\)-th term of A.P.: \(T_n = a + (n-1)d\).


Given: \(T_5 = a + 4d = 43\), \(T_9 = a + 8d = 79\)

Subtract equations: \((a+8d) - (a+4d) = 79 - 43 \Rightarrow 4d = 36 \Rightarrow d=9\)

Then \(a + 4*9 = 43 \Rightarrow a = 43 -36 =7\)


A.P.: 7,16,25,34,43,52,61,70,...


Explanation: Solve system of two equations for \(a\) and \(d\).
Quick Tip: Use \(T_n = a + (n-1)d\), set up equations and solve for \(a\) and \(d\).

Concept: Let consecutive integers be \(n\) and \(n+1\). Then \(n^2 + (n+1)^2 = 365\).


Step 1: \(n^2 + n^2 + 2n +1 = 365 \Rightarrow 2n^2 +2n +1=365 \Rightarrow 2n^2+2n-364=0 \Rightarrow n^2 + n -182=0\)


Step 2: Solve quadratic: \(n = \frac{-1 \pm \sqrt{1+728}}{2} = \frac{-1 \pm 27}{2}\)


Positive root: \(n = \frac{26}{2} = 13\) or \(n = -14\) (ignore negative)


Then consecutive integers: 12 and 13 (check sum: \(12^2+13^2=144+169=313\)?) Correction: Recalculate.

\(2n^2 + 2n +1 = 365 \Rightarrow 2n^2 + 2n -364 =0 \Rightarrow n^2 + n -182=0\)


Discriminant \(D=1+728=729\), \(\sqrt{729}=27\)

\(n = (-1 +27)/2 =26/2=13\), \((-1-27)/2=-28/2=-14\) → ignore


Then consecutive integers: 13 and 14 → sum of squares: \(13^2 +14^2 = 169+196=365\)


Explanation: Let integers be \(n\) and \(n+1\), form quadratic from sum of squares.
Quick Tip: Let consecutive integers be \(n\) and \(n+1\), write equation for sum of squares and solve quadratic.

Concept: Let larger number = \(x\), smaller number = \(y\).


Step 1: Difference of squares: \(y^2 - x^2 = 180\)

Step 2: Square of smaller number is 8 times larger: \(y^2 = 8x\)


Equations: \(y^2 - x^2 =180\), \(y^2 = 8x\)


Explanation: Translate word statement into algebraic equations.
Quick Tip: Let unknowns = x (larger), y (smaller), write equations step by step from statements.

Concept: Use basic proportionality theorem and angle properties in triangles.


Step 1: \(\frac{PS}{SQ}=\frac{PT}{TR} \Rightarrow ST \parallel QR\) (by converse of basic proportionality theorem).

Step 2: \(\angle PST = \angle PRQ \Rightarrow \angle PST = \angle STQ\) (alternate interior angles)

Step 3: Thus \(\triangle PST \sim \triangle PQR \Rightarrow PS/PT = PQ/PR\)

Step 4: Given ratios equal → \(PQ = PR\)


Explanation: Triangle is isosceles with sides opposite equal angles equal.
Quick Tip: Use similarity and proportionality of line segments and angles to prove isosceles property.

Concept: Use right-angle triangle similarity criteria (AA criterion).


Step 1: \(\angle ADB = \angle EFC = 90^\circ\) (given perpendiculars)

Step 2: \(\angle ABD = \angle CEF\) (alternate angles, corresponding sides)

Step 3: By AA similarity criterion, \(\triangle ABD \sim \triangle ECF\)


Explanation: Two angles equal → triangles similar.
Quick Tip: Use AA similarity rule for right-angle triangles with perpendiculars.

Concept: If sides and corresponding medians are proportional, then triangles are similar (SSS criterion for triangles using medians).


Step 1: \(AB/PQ = BC/PR = AD/PM = k\) (proportionality constant)

Step 2: Triangles with sides proportional are similar → \(\triangle ABC \sim \triangle PQR\)


Explanation: Proportional sides and corresponding median → similarity.
Quick Tip: Use properties of medians and proportionality to establish similarity (SSS).

Concept: Areas of similar triangles proportional to square of corresponding sides: \(\frac{Area_1}{Area_2} = \left(\frac{AB}{DE}\right)^2\).


Step 1: \(\frac{9}{64} = \left(\frac{AB}{5.1}\right)^2 \Rightarrow \frac{3}{8} = \frac{AB}{5.1}\)

Step 2: \(AB = 5.1 \cdot \frac{3}{8} = 1.9125\) cm? Correction: \(\sqrt{9/64} = 3/8\)
\(AB = 5.1 * 3/8 = 15.3/8 = 1.9125\) cm


Explanation: Take square root of area ratio first, then multiply by corresponding side.
Quick Tip: For similar triangles, side ratio = square root of area ratio.

Concept: Use Pythagorean identity \(\sin^2\theta = 1-\cos^2\theta\).


LHS: \(\sqrt{\frac{1+\cos\theta}{1-\cos\theta}} = \sqrt{\frac{1+\cos\theta}{(1-\cos\theta)} \cdot \frac{1+\cos\theta}{1+\cos\theta}} = \sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\theta}}\)
\(= \sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}} = \frac{1+\cos\theta}{\sin\theta} = RHS\)


Explanation: Rationalize denominator and apply \(\sin^2\theta + \cos^2\theta=1\).
Quick Tip: For trigonometric identities, try rationalizing or multiplying by conjugate to simplify.

Use complementary angle property: \(\cot \theta = \tan(90^\circ - \theta)\)

\(\cot 63^\circ \cdot \cot 81^\circ = \tan(27^\circ) \cdot \tan(9^\circ)\)


Hence \(\tan 9^\circ \cdot \tan 27^\circ = \cot 63^\circ \cdot \cot 81^\circ\).


Explanation: Use \(\cot \theta = \tan(90^\circ-\theta)\) for angle conversions.
Quick Tip: Complementary angle conversions often simplify trigonometric products.

Concept: \(\sin^2 A + \cos^2 A =1\), \(\cot A = \frac{\cos A}{\sin A}\), \(\csc A = \frac{1}{\sin A}\).


Step 1: \(\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}\)

Step 2: \(\cot A = \frac{\cos A}{\sin A} = \frac{4/5}{3/5} = \frac{4}{3}\)

Step 3: \(\csc A = \frac{1}{\sin A} = \frac{1}{3/5} = \frac{5}{3}\)
Quick Tip: Use \(\sin^2 A + \cos^2 A = 1\) to find unknown trigonometric ratios.

Concept: Ladder forms right triangle with wall. Height \(h = L \cdot \sin \theta\) if angle with wall is given.


Step 1: \(h = 7 \cdot \sin 30^\circ = 7 \cdot \frac{1}{2} = 3.5~m\)


Explanation: Use basic right triangle trigonometry.
Quick Tip: Height from ladder: \(h = L \sin\theta\), base from wall: \(b = L \cos\theta\).

Concept: Use AA criterion for similarity in triangles.


Step 1: \(\angle AEB = \angle CFB\) (vertically opposite angles)

Step 2: \(\angle ABE = \angle CBF\) (alternate interior angles as BE || CF)

Step 3: Hence by AA criterion, \(\triangle ABE \sim \triangle CFB\)
Quick Tip: Look for vertical and alternate interior angles to apply AA similarity.

Concept: In right triangle, Pythagoras theorem: hypotenuse\(^2 = sum of squares of legs\).


Step 1: Let legs be \(AC = BC\), hypotenuse \(AB\).

Step 2: \(AB^2 = AC^2 + BC^2 = AC^2 + AC^2 = 2 AC^2\)
Quick Tip: For isosceles right triangle: hypotenuse = leg \(\times \sqrt{2}\).

Concept: \(\tan \theta = \frac{\sin\theta}{\cos\theta}\), use Pythagorean theorem.


Step 1: Let \(\sin \theta = 5k\), \(\cos \theta = 12k\), then \((5k)^2 + (12k)^2 = 1\)

Step 2: \(25 k^2 + 144 k^2 = 1 \Rightarrow 169 k^2 =1 \Rightarrow k = \frac{1}{13}\)

Step 3: \(\sin \theta + \cos \theta = 5/13 + 12/13 = 17/13\)
Quick Tip: Express \(\sin\) and \(\cos\) in terms of \(\tan\) and Pythagorean identity.

Concept: \(\sin \theta = \cos(90^\circ - \theta)\)


Step 1: \(\sin 3A = \cos(A-26^\circ) \Rightarrow 3A = 90^\circ - (A-26^\circ)\)

Step 2: \(3A = 116^\circ - A \Rightarrow 4A = 116^\circ \Rightarrow A = 29^\circ\)
Quick Tip: Use \(\sin \theta = \cos (90^\circ - \theta)\) to convert and solve easily.

Let smaller number = \(x\), larger = \(7x/3\)
\(x + 7x/3 = 50 \Rightarrow 10x/3 = 50 \Rightarrow x = 15\)

Then larger = \(7*15/3 = 35\) ✅ Wait sum = 15+35=50 ✅ Correct.
Quick Tip: Translate "multiple of other" into algebraic equation to solve easily.

Concept: Sum/difference of a rational and an irrational number is irrational.


Step 1: \(5\) is rational, \(\sqrt{3}\) is irrational

Step 2: \(5 - \sqrt{3}\) = rational - irrational = irrational
Quick Tip: Rational ± Irrational = Irrational; use to prove such expressions.

Concept: Three points collinear if slope between any two pairs is same.


Slope (1,1)-(3,k) = \(\frac{k-1}{3-1} = \frac{k-1}{2}\)

Slope (1,1)-(-1,4) = \(\frac{4-1}{-1-1} = \frac{3}{-2} = -3/2\)

Set equal: \((k-1)/2 = -3/2 \Rightarrow k-1 = -3 \Rightarrow k = -2\) ✅ Wait slope check: (-1,4)-(1,1): (1-4)/(3-1)=? Better: slope = (4-1)/(-1-1)=3/-2=-3/2 ✅ correct. (3,k)-(1,1): (k-1)/(3-1)=(k-1)/2=-3/2 => k-1=-3 => k=-2 ✅ Correct.
Quick Tip: Use slope equality condition for collinearity of three points.

Let point on y-axis = \((0,y)\)


Distance to (6,5) = \(\sqrt{(0-6)^2 + (y-5)^2} = \sqrt{36 + (y-5)^2}\)


Distance to (-4,3) = \(\sqrt{(-4-0)^2 + (3-y)^2} = \sqrt{16 + (y-3)^2}\)


Equate squares: \(36 + (y-5)^2 = 16 + (y-3)^2 \Rightarrow (y-5)^2 - (y-3)^2 = -20\)

\((y^2-10y+25)-(y^2-6y+9)=-20 \Rightarrow -4y +16=-20 \Rightarrow y=9\) Wait recall\(:\) \(-4y\) +16=\(-20\) \(=>\) \(-4y\)=\(-36\) \(=>\) y=9 Correct. Check distance: (0,9) to (6,5)=√(36+16)=√52, (0,9) to (-4,3)=√(16+36)=√52 Correct.
Quick Tip: Use distance formula and equate distances to find points equidistant from two points.

Concept: Factorization \(x^3+1=(x+1)(x^2 - x +1)\)

Step 1: Divide using long division: \(x^3+1 ÷ (x+1)\)

Step 2: Quotient = \(x^2 - x + 1\), remainder = 0
Quick Tip: Use sum of cubes formula: \(a^3+b^3=(a+b)(a^2-ab+b^2)\) for division.

Step 1: \(1188 ÷ 504 = 2\) remainder \(180\)

Step 2: \(504 ÷ 180 = 2\) remainder \(144\)

Step 3: \(180 ÷ 144 =1\) remainder \(36\)

Step 4: \(144 ÷ 36 =4\) remainder \(0\)

HCF = 36 Wait recalc: 504 ÷ 180 =2 rem 144 180 ÷ 144=1 rem 36 144 ÷36=4 rem 0 Correct. HCF=36 But 504 and 1188 ÷36? 36*14=504 36*33=1188 Correct. HCF=36 Correct.
Quick Tip: Apply Euclidean algorithm: divide larger number by smaller, replace and repeat until remainder 0.

Concept: \(D = b^2 - 4ac\), nature of roots: \(D>0\) real unequal, \(D=0\) real equal, \(D<0\) non-real


Step 1: \(a=2\), \(b=5\), \(c=-3\)

Step 2: \(D = b^2-4ac = 25 - 4*2*(-3) = 25 +24 = 49\)

Step 3: Since \(D>0\), roots are real and unequal
Quick Tip: Use \(D = b^2-4ac\) to determine the nature of roots before solving quadratic.

Concept: Solve for \(y\), plot two points per line, draw, find intersection.

Calculation:

Line 1: \(x + 3y = 6\) → \(y = \dfrac{6-x}{3}\)
- \(x=0\): \(y=2\) → \((0,2)\)
- \(x=6\): \(y=0\) → \((6,0)\)

Line 2: \(2x - 3y = 12\) → \(y = \dfrac{2x-12}{3}\)
- \(x=6\): \(y=0\) → \((6,0)\)
- \(x=0\): \(y=-4\) → \((0,-4)\)

Algebraically:
Add equations: \[ (x + 3y) + (2x - 3y) = 6 + 12 \quad \Rightarrow \quad 3x = 18 \quad \Rightarrow \quad x = 6 \]
Substitute in first: \(6 + 3y = 6\) → \(y = 0\).

Explanation: Graphs intersect at \((6, 0)\). Solution: \(x=6\), \(y=0\).
Quick Tip: Plot x-intercept and y-intercept for quick graphing.

Concept: SAS similarity criterion.

Calculation:
Let \(\triangle ABC\), \(\triangle DEF\).
Given: \(\angle A = \angle D\), \[ \dfrac{AB}{DE} = \dfrac{AC}{DF} = k \quad (say) \]
Construct \(\triangle AD'E'\) on \(DE\) such that \(AD' = AB\), \(AE' = AC\).
Then \(\triangle AD'E' \cong \triangle ABC\) (SAS).
But \(D'E' \parallel BC\) (by construction and equal sides). \(\Rightarrow \angle AD'E' = \angle ABC\) (corresponding), \(\angle AE'D = \angle ACB\) (corresponding).
Thus, \(\angle ABC = \angle DEF\), \(\angle ACB = \angle DFE\).
So \(\triangle ABC \sim \triangle DEF\) by AAA.

Explanation: Equal angle and proportional including sides imply other angles equal via parallel lines.
Quick Tip: Use SAS to construct congruent triangle, then use parallel lines.

Concept: Let number be \(10x + y\). Then: \[ 10x + y = 4(x + y), \quad 10x + y = 2xy \]

Calculation: \[ 10x + y = 4x + 4y \quad \Rightarrow \quad 6x = 3y \quad \Rightarrow \quad y = 2x \quad (1) \] \[ 10x + y = 2xy \quad \Rightarrow \quad 10x + 2x = 2x(2x) \quad \Rightarrow \quad 12x = 4x^2 \] \[ 4x^2 - 12x = 0 \quad \Rightarrow \quad 4x(x - 3) = 0 \quad \Rightarrow \quad x = 3 \] \(y = 2(3) = 6\).
Number: \(36\).
But check:
Sum = 9, 4×9=36
Product = 18, 2×18=36
Wait: \(36 = 36\), yes.
But earlier said 24. Let’s check 24:
Sum=6, 4×6=24
Product=8, 2×8=16 ≠24
So 36 is correct.

Explanation: Number is \(36\).
Quick Tip: Let digits be \(x, y\); form two equations from conditions.

Concept: Use section formula or ruler division.

Calculation:
Total parts = \(5 + 8 = 13\).
Length of each part = \(\dfrac{7.6}{13} \approx 0.5846\) cm.
First part (5 parts): \(5 \times 0.5846 \approx 2.923 \approx 2.9\) cm
Second part (8 parts): \(8 \times 0.5846 \approx 4.677 \approx 4.7\) cm

Using formula:
Point dividing \(AB = 7.6\) cm in \(5:8\): \[ Position = \dfrac{5 \cdot 7.6 + 8 \cdot 0}{13} = \dfrac{38}{13} \approx 2.923 cm from A \]

Explanation: Parts measure \(2.9\) cm and \(4.7\) cm.
Quick Tip: Total parts = sum of ratio; divide length accordingly.

Concept: Rationalize LHS and simplify.

Calculation:
Let \(a = \sec\theta\), \(b = \tan\theta\).
LHS: \[ \dfrac{a - b}{a + b} \cdot \dfrac{a - b}{a - b} = \dfrac{(a - b)^2}{a^2 - b^2} \]
But \(a^2 - b^2 = \sec^2\theta - \tan^2\theta = 1\), \[ \Rightarrow \dfrac{(a - b)^2}{1} = (\sec\theta - \tan\theta)^2 \] \[ = \sec^2\theta - 2\sec\theta\tan\theta + \tan^2\theta \] \[ = (\sec^2\theta + \tan^2\theta) - 2\sec\theta\tan\theta \] \[ = (1 + \tan^2\theta + \tan^2\theta) - 2\sec\theta\tan\theta = 1 + 2\tan^2\theta - 2\sec\theta\tan\theta = RHS \]

Explanation: Rationalizing and using identity \(sec^2 - tan^2 = 1\) proves equality.
Quick Tip: Multiply numerator and denominator by conjugate of denominator.

Concept: \(C = 2\pi r\). Sum of circumferences = \(2\pi(r_1 + r_2)\).

Calculation: \[ C_1 = 2\pi(19), \quad C_2 = 2\pi(9) \] \[ C_1 + C_2 = 2\pi(19 + 9) = 2\pi(28) \]
New circle: \(2\pi r = 2\pi(28)\) → \(r = 28\) cm.

Explanation: Radius is sum of given radii.
Quick Tip: Factor out \(2\pi\): sum of radii gives new radius.

Concept: Mean of grouped data = \(\dfrac{\sum f_i x_i}{\sum f_i}\), where \(x_i\) is the class mark.

Calculation:
Class marks (\(x_i\)): \[ \dfrac{11+13}{2} = 12, \quad \dfrac{13+15}{2} = 14, \quad 16, \quad 18, \quad 20, \quad 22, \quad 24 \]

Now compute \(f_i x_i\):

\[ \bar{x} = \dfrac{1152}{64} = 18 \]
Wait: \(64 \times 18 = 1152\), yes.
But recheck sum:
84 + 84 = 168
168 + 144 = 312
312 + 234 = 546
546 + 400 = 946
946 + 110 = 1056
1056 + 96 = 1152. Yes. \(N = 7+6+9+13+20+5+4 = 64\). \[ \bar{x} = \dfrac{1152}{64} = 18 \]

Wait: But earlier said 17.7 — mistake. \(1152 \div 64\):
64 × 18 = 1152 → 18.

Explanation: Mean = \(18\).
Quick Tip: Class mark = \(\dfrac{lower + upper}{2}\); verify \(\sum f_i x_i\) by addition.

Concept: Curved surface area = \(\pi l (r_1 + r_2)\), where perimeters give \(2\pi r_1, 2\pi r_2\).

Calculation:
Let perimeters: \(P_1 = 18\), \(P_2 = 6\), slant height \(l = 4\). \[ r_1 = \dfrac{18}{2\pi}, \quad r_2 = \dfrac{6}{2\pi} \] \[ r_1 + r_2 = \dfrac{18 + 6}{2\pi} = \dfrac{24}{2\pi} = \dfrac{12}{\pi} \]
Curved surface area: \[ \pi \cdot 4 \cdot \dfrac{12}{\pi} = 4 \times 12 = 48 cm^2 \]
But wait: units? \(\pi\) cancels: \(48\) (no \(\pi\))?
No: \[ \pi l (r_1 + r_2) = \pi \cdot 4 \cdot \dfrac{12}{\pi} = 48 \]
But standard formula uses perimeter, not radius sum:
Actually, correct formula: \[ CSA = \dfrac{1}{2} \times (P_1 + P_2) \times l \] \[ = \dfrac{1}{2} (18 + 6) \times 4 = \dfrac{1}{2} \times 24 \times 4 = 48 cm^2 \]
But many textbooks write \(\pi(r_1 + r_2)l\), but here perimeters given, so use average perimeter × slant height.

Explanation: CSA = \(\dfrac{1}{2} (P_1 + P_2) l = 48\) cm².
Quick Tip: For frustum: CSA = average circumference × slant height.