Tangents drawn from an external point to a circle are equal in length.

Given: \(PA = 8\) cm.

Thus, \(PB = PA = 8\) cm.
Quick Tip: Tangents from a common external point are equal.

OA \(\perp\) PA and OB \(\perp\) PB (radius \(\perp\) tangent).

Thus, \(\triangle OAP\) and \(\triangle OBP\) are right-angled at A and B.

Also, OA = OB (radii), PA = PB (tangents from P).

So, \(\triangle OAP \cong \triangle OBP\) (RHS congruence).
\(\angle APO = \angle BPO = \dfrac{80^\circ}{2} = 40^\circ\).

In \(\triangle OAP\):
\(\angle AOP = 180^\circ - 90^\circ - 40^\circ = 50^\circ\).

Thus, \(\angle POA = 50^\circ\).
Quick Tip: Quadrilateral OAPB: \(\angle OAP = \angle OBP = 90^\circ\), so \(\angle AOB = 360^\circ - 180^\circ - 80^\circ = 100^\circ\), then \(\angle POA = \dfrac{100^\circ}{2} = 50^\circ\).

The radius is perpendicular to the tangent at the point of contact.

Thus, the angle between the tangent and the radius is \(90^\circ\).
Quick Tip: Theorem: Radius \(\perp\) tangent at point of contact.

Area of circle = \(\pi r^2\).

Ratio of areas = \(\dfrac{\pi r_1^2}{\pi r_2^2} = \left(\dfrac{r_1}{r_2}\right)^2\).

Given: \(\dfrac{r_1}{r_2} = \dfrac{3}{4}\).

Ratio of areas = \(\left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}\).
Quick Tip: Area ratio = (radius ratio)\(^2\).

Area of sector = \(\dfrac{\theta}{360^\circ} \times \pi r^2\).
\[ = \dfrac{30}{360} \times \pi \times 42^2 = \dfrac{1}{12} \times \pi \times 1764 = \dfrac{1764}{12} \pi = 147 \pi \]

Using \(\pi \approx 3.14\):
\(147 \times 3.14 = 147 \times 3 + 147 \times 0.14 = 441 + 20.58 = 461.58 \approx 462 \, cm^2\).
Quick Tip: Sector area = \(\dfrac{\theta}{360} \pi r^2\).

Circumference = \(2\pi r\).

Ratio of circumferences = \(\dfrac{2\pi r_1}{2\pi r_2} = \dfrac{r_1}{r_2}\).

Given: \(\dfrac{circumference_1}{circumference_2} = \dfrac{5}{7}\).

Thus, \(\dfrac{r_1}{r_2} = \dfrac{5}{7}\).
Quick Tip: Circumference ratio = radius ratio.

Use identity: \(1 + \tan^2 A = \sec^2 A\).
\[ \sec^2 A - \tan^2 A = 1 \]
\[ 7 \sec^2 A - 7 \tan^2 A = 7 (\sec^2 A - \tan^2 A) = 7 \times 1 = 7 \]
Quick Tip: Factor out 7 and use \(\sec^2 A - \tan^2 A = 1\).

Substitute: \[ b^2 x^2 = b^2 (a \cos \theta)^2 = a^2 b^2 \cos^2 \theta \]
\[ a^2 y^2 = a^2 (b \sin \theta)^2 = a^2 b^2 \sin^2 \theta \]
\[ b^2 x^2 + a^2 y^2 = a^2 b^2 \cos^2 \theta + a^2 b^2 \sin^2 \theta = a^2 b^2 (\cos^2 \theta + \sin^2 \theta) = a^2 b^2 \times 1 = a^2 b^2 \]
Quick Tip: Factor out \(a^2 b^2\) and use \(\cos^2 \theta + \sin^2 \theta = 1\).

Let height of tower = \(h\).

Distance from base = 10 m, angle of elevation = \(60^\circ\).
\[ \tan 60^\circ = \dfrac{h}{10} \]
\[ \sqrt{3} = \dfrac{h}{10} \quad \Rightarrow \quad h = 10\sqrt{3} \, m \]
Quick Tip: Use \(\tan \theta = \dfrac{opposite}{adjacent}\).

Height = opposite = 30 m.

Angle with ground = \(60^\circ\).

String = hypotenuse = \(l\).
\[ \sin 60^\circ = \dfrac{30}{l} \]
\[ \dfrac{\sqrt{3}}{2} = \dfrac{30}{l} \quad \Rightarrow \quad l = \dfrac{30 \times 2}{\sqrt{3}} = \dfrac{60}{\sqrt{3}} = 20\sqrt{3} \, m \]

Wait: \(20\sqrt{3} \approx 34.64\), close to 35? But exact is \(20\sqrt{3}\).

Option (C) is \(20\sqrt{3}\).

Correct Answer: (C) \(20\sqrt{3}\) m
Quick Tip: Use \(\sin \theta = \dfrac{height}{string length}\).

Class interval length = upper limit \(-\) lower limit.

For 2–5: \(5 - 2 = 3\).

For 5–8: \(8 - 5 = 3\).

All intervals are of length 3.
Quick Tip: Class width = upper \(-\) lower limit.

Let the four consecutive odd numbers be \(x, x+2, x+4, x+6\).

Mean = \(\dfrac{x + (x+2) + (x+4) + (x+6)}{4} = 6\).
\[ \dfrac{4x + 12}{4} = 6 \quad \Rightarrow \quad x + 3 = 6 \quad \Rightarrow \quad x = 3 \]

Numbers: 3, 5, 7, 9.

Largest = 9.
Quick Tip: Average of consecutive numbers is the average of the middle two.

First 6 even natural numbers: 2, 4, 6, 8, 10, 12.

Sum = \(2 + 4 + 6 + 8 + 10 + 12 = 42\).

Mean = \(\dfrac{42}{6} = 7\).
Quick Tip: Mean of first \(n\) even numbers = \(n + 1\).

Identity: \(1 + \tan^2 \theta = \sec^2 \theta\).

Divide by \(\sin^2 \theta\): \[ \dfrac{1}{\sin^2 \theta} + \dfrac{\tan^2 \theta}{\sin^2 \theta} = \dfrac{\sec^2 \theta}{\sin^2 \theta} \]

But directly: \[ 1 + \cot^2 \theta = 1 + \dfrac{\cos^2 \theta}{\sin^2 \theta} = \dfrac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta} = \dfrac{1}{\sin^2 \theta} = \cosec^2 \theta \]
Quick Tip: \(1 + \cot^2 \theta = \cosec^2 \theta\).

Frequency:
- 3: 1
- 4: 1
- 5: 3
- 7: 2
- 8: 1
- 9: 2
Highest frequency = 3 → mode = 5.
Quick Tip: Mode = most frequent value.

\(P(E) + P(E') = 1\).
\(P(E') = 1 - P(E) = 1 - 0.02 = 0.98\).
Quick Tip: Complementary events sum to 1.

Total outcomes = \(6 \times 6 = 36\).

Difference zero → both dice show same number: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

Favorable = 6.

Probability = \(\dfrac{6}{36} = \dfrac{1}{6}\).
Quick Tip: Same numbers → 6 cases.

Sample space: HH, HT, TH, TT → 4 outcomes.

Both heads: HH → 1 outcome.

Probability = \(\dfrac{1}{4}\).
Quick Tip: Independent events: \(P(H) \times P(H) = \dfrac{1}{2} \times \dfrac{1}{2}\).

Total months = 12.

Favorable: June, September → 2.

Probability = \(\dfrac{2}{12} = \dfrac{1}{6}\).
Quick Tip: Equally likely months.

Total outcomes = 6.

Favorable: 4, 5 → 2.

Probability = \(\dfrac{2}{6} = \dfrac{1}{3}\).
Quick Tip: Two favorable out of six.

Volume of sphere = \(\dfrac{4}{3}\pi r^3\).

Ratio of volumes = \(\left(\dfrac{r_1}{r_2}\right)^3 = \dfrac{64}{125}\).
\[ \dfrac{r_1}{r_2} = \sqrt[3]{\dfrac{64}{125}} = \dfrac{4}{5} \]

Surface area = \(4\pi r^2\).

Ratio of surface areas = \(\left(\dfrac{r_1}{r_2}\right)^2 = \left(\dfrac{4}{5}\right)^2 = \dfrac{16}{25}\).
Quick Tip: Surface area ratio = (volume ratio)\(^{2/3}\).

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

Ratio of volumes = \(\left(\dfrac{r_1}{r_2}\right)^2 \times \dfrac{h_1}{h_2} = \left(\dfrac{4}{5}\right)^2 \times \dfrac{6}{7} = \dfrac{16}{25} \times \dfrac{6}{7} = \dfrac{96}{175}\).
Quick Tip: Volume ratio = (radius ratio)\(^2\) × height ratio.

Total surface area of hemisphere = curved surface + base area.

= \(2\pi R^2 + \pi R^2 = 3\pi R^2\).
Quick Tip: Hemisphere TSA = \(3\pi R^2\).

Curved surface area = \(\pi r l = 880\).
\[ \dfrac{22}{7} \times 14 \times l = 880 \]
\[ 22 \times 2 \times l = 880 \quad \Rightarrow \quad 44l = 880 \quad \Rightarrow \quad l = 20 \, cm \]
Quick Tip: \(\pi r l = CSA\), solve for \(l\).

Space diagonal of cube = \(a\sqrt{3}\).
\[ a\sqrt{3} = 2\sqrt{3} \quad \Rightarrow \quad a = 2 \, cm \]
Quick Tip: Diagonal = \(a\sqrt{3}\).

Original TSA = \(6a^2\).

New edge = \(2a\), new TSA = \(6(2a)^2 = 6 \times 4a^2 = 24a^2\).

Ratio = \(\dfrac{24a^2}{6a^2} = 4\).

So, 4 times.
Quick Tip: TSA \(\propto a^2\), double edge → 4 times area.

Sphere TSA = \(4\pi R^2\).

Hemisphere TSA = \(3\pi R^2\).

Ratio = \(\dfrac{4\pi R^2}{3\pi R^2} = \dfrac{4}{3}\).
Quick Tip: Sphere: \(4\pi R^2\), Hemisphere: \(3\pi R^2\).

Curved surface area of hemisphere = \(2\pi R^2 = 1232\).
\[ 2 \times \dfrac{22}{7} \times R^2 = 1232 \]
\[ \dfrac{44}{7} R^2 = 1232 \quad \Rightarrow \quad R^2 = 1232 \times \dfrac{7}{44} = 28 \times 7 = 196 \]
\[ R = \sqrt{196} = 14 \, cm \]
Quick Tip: CSA of hemisphere = \(2\pi R^2\).

Given: \(\cos \theta + \cos^2 \theta = 1\).

Let \(x = \cos \theta\).
\[ x + x^2 = 1 \quad \Rightarrow \quad x^2 + x - 1 = 0 \]
\[ x = \dfrac{-1 \pm \sqrt{5}}{2} \]
\(\cos \theta = \dfrac{-1 + \sqrt{5}}{2}\) (since \(\cos \theta < 1\)).

But easier: \[ \cos^2 \theta = 1 - \cos \theta \]
\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - (1 - \cos \theta) = \cos \theta \]
\[ \sin^4 \theta = (\sin^2 \theta)^2 = \cos^2 \theta = 1 - \cos \theta \]
\[ \sin^2 \theta + \sin^4 \theta = \cos \theta + (1 - \cos \theta) = 1 \]
Quick Tip: Express \(\sin^2 \theta\) in terms of \(\cos \theta\).

Numerator: \(1 + \tan^2 A = \sec^2 A\).

Denominator: \(1 + \cot^2 A = \cosec^2 A\).
\[ \dfrac{1 + \tan^2 A}{1 + \cot^2 A} = \dfrac{\sec^2 A}{\cosec^2 A} = \dfrac{1/\cos^2 A}{1/\sin^2 A} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A \]
Quick Tip: Use \(1 + \tan^2 = \sec^2\), \(1 + \cot^2 = \csc^2\).

Base AB lies on the y-axis from (0,1) to (0,5), so length = \(5 - 1 = 4\).

Height is the perpendicular distance from C(3,4) to the y-axis, which is the x-coordinate = 3.

Area = \(\dfrac{1}{2} \times base \times height = \dfrac{1}{2} \times 4 \times 3 = 6\).

Alternatively, using shoelace formula:
\[ Area = \dfrac{1}{2} \left| 0(5-4) + 0(4-1) + 3(1-5) \right| = \dfrac{1}{2} \left| 3(-4) \right| = \dfrac{1}{2} \times 12 = 6 \]
Quick Tip: For base on y-axis, height = x-coordinate of third vertex.

Pair complementary angles:
\(\tan 80^\circ = \tan(90^\circ - 10^\circ) = \cot 10^\circ\),
\(\tan 67^\circ = \tan(90^\circ - 23^\circ) = \cot 23^\circ\).
\[ \tan 10^\circ \cdot \cot 10^\circ = 1, \quad \tan 23^\circ \cdot \cot 23^\circ = 1 \]

Product = \(1 \times 1 = 1\).
Quick Tip: \(\tan(90^\circ - \theta) = \cot \theta\).

Area ratio = 100:144.

Side ratio = \(\sqrt{100:144} = 10:12\).
Quick Tip: Side ratio = square root of area ratio.

A secant intersects a circle at two distinct points.

A chord is the line segment between those points.

A tangent touches at exactly one point.
Quick Tip: Secant cuts the circle at two points.

Area ratio = \((side ratio)^2 = (4:9)^2 = 16:81\).
Quick Tip: Areas scale with square of sides.

Corresponding sides: \(BC \leftrightarrow EF\).

Side ratio = \(\dfrac{EF}{BC} = \dfrac{4}{3}\).

Area ratio = \(\left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9}\).

Area of \(\triangle DEF\) = \(54 \times \dfrac{16}{9} = 6 \times 16 = 96\) cm².
Quick Tip: Area scales with square of corresponding sides.

Right-angled at A, hypotenuse BC = 13 cm, leg AB = 12 cm.

By Pythagoras:
\[ AC = \sqrt{BC^2 - AB^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \, cm \]
Quick Tip: Hypotenuse is longest side.

In \(\triangle DEF\): \(\angle F = 180^\circ - \angle D - \angle E\).

In \(\triangle PQR\): \(\angle P = 180^\circ - \angle Q - \angle R\).

Given: \(\angle D = \angle Q\), \(\angle E = \angle R\).

Thus, \(\angle F = 180^\circ - \angle Q - \angle R = \angle P\).
Quick Tip: Two angles equal → third angle equal.

SSS similarity: all sides proportional → \(\triangle ABC \sim \triangle DEF\).

Corresponding angles: \(\angle A \leftrightarrow \angle D\), \(\angle B \leftrightarrow \angle E\), \(\angle C \leftrightarrow \angle F\).
\(\angle C = 180^\circ - 40^\circ - 80^\circ = 60^\circ\).

Thus, \(\angle F = 60^\circ\).
Quick Tip: SSS → similar → corresponding angles equal.

Two circles intersecting at two points have:

- 2 external common tangents

- No internal common tangents (they cross between intersection points).

Total common tangents = 2.
Quick Tip: Intersecting circles → 2 common tangents.

5th term: \(a + 4d = 11\), \(d = 2\).
\(a + 4(2) = 11 \Rightarrow a + 8 = 11 \Rightarrow a = 3\).
Quick Tip: \(a_n = a + (n-1)d\).

\(S_n = n^2 + 2n + 1 = (n+1)^2\).

6th term = \(S_6 - S_5\).
\(S_6 = 7^2 = 49\), \(S_5 = 6^2 = 36\).

6th term = \(49 - 36 = 13\).

Not in options → none of these.
Quick Tip: \(a_n = S_n - S_{n-1}\).

Only (C) has constant difference = \(x\).

(A): differences 6, 2, 7 → not constant.

(B): not linear.

(D): differences increase.
Quick Tip: A.P. requires constant common difference.

(A), (B), (C): constant differences 1, 3, 2.

(D): 4, 16, 36, 64 → differences 12, 20, 28 → not constant.
Quick Tip: Squares of A.P. are not A.P.

\(a = 1\), \(d = 3\), \(n = 20\).
\[ S_{20} = \dfrac{20}{2} [2(1) + (19)(3)] = 10 [2 + 57] = 10 \times 59 = 590 \]
Quick Tip: Use \(S_n = \dfrac{n}{2} [2a + (n-1)d]\).

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

(B) \(\sin 90^\circ = 1\), \(\cos 90^\circ = 0\), so \(1 \times 0 = 0 \neq 1\).

(C) \(\sin^2 60^\circ = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4} \neq 1\).

(D) \(\sin 45^\circ = \dfrac{1}{\sqrt{2}}\), \(\cos 45^\circ = \dfrac{1}{\sqrt{2}}\).
\(\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{2}}{1} = 1\).
Quick Tip: \(\dfrac{\sin \theta}{\cos \theta} = \tan \theta\), so \(\sin \theta \times \dfrac{1}{\cos \theta} = \tan \theta\).

Use identity: \(1 + \tan^2 A = \sec^2 A\).
\(\cos^2 A \cdot \sec^2 A = \cos^2 A \cdot \dfrac{1}{\cos^2 A} = 1\).
Quick Tip: \(\sec^2 A = \dfrac{1}{\cos^2 A}\).

\(\tan 30^\circ = \dfrac{\sin 30^\circ}{\cos 30^\circ} = \dfrac{1/2}{\sqrt{3}/2} = \dfrac{1}{\sqrt{3}}\).
Quick Tip: Standard value: \(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\).

Standard value: \(\cos 60^\circ = \dfrac{1}{2}\).
Quick Tip: \(\cos 60^\circ = \dfrac{1}{2}\), \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\).

\(\sin 90^\circ = 1\), so \(\sin^2 90^\circ = 1\).
\(\tan 45^\circ = 1\), so \(\tan^2 45^\circ = 1\).
\(1 - 1 = 0\).
Quick Tip: \(\sin 90^\circ = 1\), \(\tan 45^\circ = 1\).

\(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\), so \(8 \sin 60^\circ = 4\sqrt{3}\).
\(\cos 60^\circ = \dfrac{1}{2}\), so \(8 \cos 60^\circ = 4\).

Points: \((4\sqrt{3}, 0)\), \((0, 4)\).

Distance = \(\sqrt{(4\sqrt{3})^2 + (-4)^2} = \sqrt{48 + 16} = \sqrt{64} = 8\).
Quick Tip: Distance formula: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

Distance from origin to \((x,y)\): \(\sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}\).
Quick Tip: Distance from origin = \(\sqrt{x^2 + y^2}\).

Distance from point \((x,y)\) to y-axis is \(|x|\).

Here, \(x = 12\), so distance = 12.
Quick Tip: Distance to y-axis = absolute value of x-coordinate.

Ordinate = y-coordinate = -8.
Quick Tip: Ordinate = y-coordinate.

\(x > 0\), \(y < 0\) \(\longrightarrow\) Fourth quadrant.
Quick Tip: \((+x, -y)\) \(\longrightarrow\) Fourth quadrant.

Second quadrant: \(x < 0\), \(y > 0\) \(\longrightarrow\) \((-3, 2)\).
Quick Tip: \((-x, +y)\) \(\longrightarrow\) Second quadrant.

Mid-point = \(\left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right)\).
\(\left( \dfrac{4 + (-4)}{2}, \dfrac{-4 + 4}{2} \right) = (0, 0)\).
Quick Tip: Average of x-coordinates and y-coordinates.

Let B be \((x, y)\).

Mid-point: \(\left( \dfrac{5 + x}{2}, \dfrac{7 + y}{2} \right) = (2, 4)\).
\(\dfrac{5 + x}{2} = 2 \Rightarrow x = -1\).
\(\dfrac{7 + y}{2} = 4 \Rightarrow y = 1\).

B = \((-1, 1)\).
Quick Tip: Use mid-point formula and solve for unknowns.

Centre is the midpoint of the diameter.
\[ Midpoint = \left( \dfrac{10 + (-6)}{2}, \dfrac{-6 + 10}{2} \right) = \left( \dfrac{4}{2}, \dfrac{4}{2} \right) = (2, 2) \]
Quick Tip: Centre of circle = midpoint of diameter.

Centroid \(G = \left( \dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \right)\).
\[ G = \left( \dfrac{4 + 0 + 5}{3}, \dfrac{6 + 4 + 5}{3} \right) = \left( \dfrac{9}{3}, \dfrac{15}{3} \right) = (3, 5) \]
Quick Tip: Centroid = average of vertices' coordinates.

A fraction \(\dfrac{p}{q}\) in lowest terms has a terminating decimal if and only if the denominator \(q\) has no prime factors other than 2 and 5, i.e., \(q = 2^n \times 5^m\) for some non-negative integers \(n, m\).


Check each option:

(A) Denominator: \(2^0 \times 3^2 = 9\) \(\longrightarrow\) contains 3 \(\longrightarrow\) non-terminating.

(B) Denominator: \(5^1 \times 7^2 = 5 \times 49 = 245\) \(\longrightarrow\) contains 7 \(\longrightarrow\) non-terminating.

(C) Denominator: \(2^2 \times 3^2 = 4 \times 9 = 36\) \(\longrightarrow\) contains 3 \(\longrightarrow\) non-terminating.

(D) Denominator: \(2^2 \times 5^3 = 4 \times 125 = 500\) \(\longrightarrow\) only 2 and 5 \(\longrightarrow\) terminating.


Thus, only (D) has terminating decimal.
Quick Tip: Terminating decimal \(\iff\) denominator (after simplifying) = \(2^n \times 5^m\).

Convert 0.505 to fraction:
\[ 0.505 = \dfrac{505}{1000} \]

Simplify by dividing numerator and denominator by 5:
\[ \dfrac{505 \div 5}{1000 \div 5} = \dfrac{101}{200} \]

Now, factorize the denominator:
\[ 200 = 2 \times 100 = 2 \times 2 \times 50 = 2 \times 2 \times 2 \times 25 = 2^3 \times 5^2 \]

So,
\[ \dfrac{101}{200} = \dfrac{101}{2^3 \times 5^2} \]

This matches option (D).


Verify:
\[ \dfrac{101}{2^3 \times 5^2} = \dfrac{101}{8 \times 25} = \dfrac{101}{200} = 0.505 \]
Quick Tip: Write decimal as fraction with power of 10, simplify, express denominator in \(2^n \times 5^m\).

By the division algorithm: \[ a = b \times q + r, \quad where 0 \leq r < b \]

Given: \(b = 4\), \(q = 5\), \(r = 1\).
Substitute: \[ a = 4 \times 5 + 1 = 20 + 1 = 21 \]

Check: \(21 \div 4 = 5\) quotient, remainder \(1\) \(\longrightarrow\) correct.
Quick Tip: Just plug into \(a = bq + r\).

Solve \(2x^2 - 4x - 6 = 0\).
Divide by 2: \[ x^2 - 2x - 3 = 0 \]

Factorize:
Look for two numbers whose product is \(-3\) and sum is \(-2\): \(-3\) and \(+1\). \[ x^2 - 2x - 3 = (x - 3)(x + 1) = 0 \]

So, \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
Zeroes are \(-1, 3\).


Verify:
At \(x = 3\): \(2(9) - 4(3) - 6 = 18 - 12 - 6 = 0\).
At \(x = -1\): \(2(1) + 4 - 6 = 2 + 4 - 6 = 0\). Quick Tip: For \(x^2 + bx + c = 0\), factor as \((x + p)(x + q)\) where \(p + q = b\), \(pq = c\).

The degree of a product of polynomials is the sum of their degrees.


First polynomial: \(x^3 + x^2 + 2x + 1\) \(\longrightarrow\) highest power = 3 \(\longrightarrow\) degree = 3.
Second polynomial: \(x^2 + 2x + 1\) \(\longrightarrow\) highest power = 2 \(\longrightarrow\) degree = 2.

Degree of product = \(3 + 2 = 5\).

(You don’t need to expand; just add degrees.)

Optional expansion (for verification): \[ (x^3 + x^2 + 2x + 1)(x^2 + 2x + 1) = x^3(x^2 + 2x + 1) + x^2(x^2 + 2x + 1) + \cdots \]
Leading term: \(x^3 \cdot x^2 = x^5\) \(\longrightarrow\) confirms degree 5. Quick Tip: deg\((f \cdot g)\) = deg\(f\) + deg\(g\).

A polynomial is an expression with non-negative integer exponents of the variable.


- (A) \(x^2 - 7\): exponents 2 and 0 \(\longrightarrow\) polynomial.

- (B) \(2x^2 + 7x + 6\): exponents 2, 1, 0 \(\longrightarrow\) polynomial.

- (C) \(\dfrac{1}{2}x^2 + \dfrac{1}{2}x + 4\): exponents 2, 1, 0 (coefficients fractional but allowed) \(\longrightarrow\) polynomial.

- (D) \(x + \dfrac{4}{x} = x + 4x^{-1}\): exponent \(-1\) (negative) \(\longrightarrow\) not a polynomial.
Quick Tip: No negative or fractional exponents in a polynomial.

For a quadratic polynomial with roots \(\alpha = 2\) and \(\beta = -2\): \[ (x - \alpha)(x - \beta) = (x - 2)(x - (-2)) = (x - 2)(x + 2) = x^2 - 4 \]
Alternatively, using sum and product:
- Sum of roots: \(2 + (-2) = 0\) \(\longrightarrow\) coefficient of \(x = 0\).
- Product: \(2 \times (-2) = -4\) \(\longrightarrow\) constant term = \(-4\).
So, polynomial is \(x^2 - 4\).

Verify: \(x^2 - 4 = 0 \Rightarrow x = \pm 2\). Correct. Quick Tip: Roots \(\alpha, \beta\) \(\longrightarrow\) \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\).

For quadratic \(at^2 + bt + c = 0\), \[ Sum of roots \alpha + \beta = -\dfrac{b}{a} \]
Here, \(a = 1\), \(b = 7\), \(c = 10\). \[ \alpha + \beta = -\dfrac{7}{1} = -7 \]

Verify by factoring: \(t^2 + 7t + 10 = (t + 2)(t + 5) = 0 \Rightarrow t = -2, -5\).
Sum: \(-2 + (-5) = -7\). Quick Tip: Sum of roots = \(-\dfrac{coefficient of x}{coefficient of x^2}\).

Compute each part: \[ \sin 30^\circ = \dfrac{1}{2}, \quad \cos 30^\circ = \dfrac{\sqrt{3}}{2} \quad \Rightarrow \quad \sin 30^\circ + \cos 30^\circ = \dfrac{1 + \sqrt{3}}{2} \] \[ \sin 60^\circ = \dfrac{\sqrt{3}}{2}, \quad \cos 60^\circ = \dfrac{1}{2} \quad \Rightarrow \quad \sin 60^\circ + \cos 60^\circ = \dfrac{\sqrt{3} + 1}{2} \]
Now subtract: \[ \dfrac{1 + \sqrt{3}}{2} - \dfrac{\sqrt{3} + 1}{2} = \dfrac{(1 + \sqrt{3}) - (\sqrt{3} + 1)}{2} = \dfrac{0}{2} = 0 \]
Alternatively, note that \(\sin 30^\circ + \cos 30^\circ = \sin 60^\circ + \cos 60^\circ\). Quick Tip: \(\sin \theta + \cos \theta = \sin(90^\circ - \theta) + \cos(90^\circ - \theta)\), so same for complementary angles.

Given one root is \(-4\), substitute \(x = -4\) into the polynomial and set it to zero: \[ (k-1)(-4)^2 + k(-4) + 1 = 0 \] \[ (k-1)(16) - 4k + 1 = 0 \] \[ 16k - 16 - 4k + 1 = 0 \] \[ 12k - 15 = 0 \] \[ 12k = 15 \quad \Rightarrow \quad k = \dfrac{15}{12} = \dfrac{5}{4} \]

Verification:
For \(k = \dfrac{5}{4}\): \[ \left(\dfrac{5}{4} - 1\right)x^2 + \dfrac{5}{4}x + 1 = \dfrac{1}{4}x^2 + \dfrac{5}{4}x + 1 \]
At \(x = -4\): \[ \dfrac{1}{4}(16) + \dfrac{5}{4}(-4) + 1 = 4 - 5 + 1 = 0 \]
Correct. Quick Tip: Substitute the given root into the polynomial and solve for the parameter.

For real and equal roots, discriminant \(D = 0\).
Here, \(a = k\), \(b = -6\), \(c = 1\). \[ D = b^2 - 4ac = (-6)^2 - 4(k)(1) = 36 - 4k \]
Set \(D = 0\): \[ 36 - 4k = 0 \quad \Rightarrow \quad 4k = 36 \quad \Rightarrow \quad k = 9 \]

Verification:
For \(k = 9\): \(9x^2 - 6x + 1 = 0\).
Discriminant: \(36 - 36 = 0\) \(\longrightarrow\) equal roots.
Root: \(x = \dfrac{6}{18} = \dfrac{1}{3}\). Quick Tip: Equal roots \(\iff\) \(D = b^2 - 4ac = 0\).

By Factor Theorem:
If \(p(2) = 0\), then \((x - 2)\) is a factor of \(p(x)\).
Thus, \(x - 2\) divides \(p(x)\). Quick Tip: Root \(c\) \(\Rightarrow\) \((x - c)\) is a factor.

For \(ax^2 + bx + c = 0\), \[ Product of roots \alpha \beta = \dfrac{c}{a} \]
Here, the polynomial is \(cx^2 + ax + b = 0\), so:
- Coefficient of \(x^2 = c\)
- Constant term = \(b\) \[ \alpha \beta = \dfrac{constant term}{coefficient of x^2} = \dfrac{b}{c} \] Quick Tip: Product of roots = \(\dfrac{c}{a}\) in \(ax^2 + bx + c = 0\).

A quadratic equation is of the form \(ax^2 + bx + c = 0\) where \(a \neq 0\) (degree 2).

Simplify each option:


(A) LHS: \((x+3)(x-3) = x^2 - 9\)

RHS: \(x^2 - 4x^3\)

So: \(x^2 - 9 = x^2 - 4x^3 \Rightarrow -9 = -4x^3 \Rightarrow 4x^3 = 9\)
\(\longrightarrow\) cubic equation (degree 3) \(\longrightarrow\) not quadratic.


(B) LHS: \((x+3)^2 = x^2 + 6x + 9\)

RHS: \(4(x+4) = 4x + 16\)

Bring to one side:
\(x^2 + 6x + 9 - 4x - 16 = 0 \Rightarrow x^2 + 2x - 7 = 0\)
\(\longrightarrow\) quadratic (degree 2).


(C) LHS: \((2x-2)^2 = 4(x-1)^2 = 4x^2 - 8x + 4\)

RHS: \(4x^2 + 7\)

So: \(4x^2 - 8x + 4 = 4x^2 + 7 \Rightarrow -8x - 3 = 0\)
\(\longrightarrow\) linear (degree 1) \(\longrightarrow\) not quadratic.


(D) Multiply both sides by \(4x\) (assuming \(x \neq 0\)):
\(16x^2 + 1 = 16x^2\)
\(1 = 0\) \(\longrightarrow\) contradiction, not an equation.


Only (B) is a quadratic equation.
Quick Tip: Bring all terms to one side \(\longrightarrow\) check highest power = 2 and coefficient of \(x^2 \neq 0\).

Simplify each to standard form:


(A) \(5x - x^2 = x^2 + 3\)
\(-x^2 - x^2 + 5x - 3 = 0 \Rightarrow -2x^2 + 5x - 3 = 0\)

(or \(2x^2 - 5x + 3 = 0\)) \(\longrightarrow\) quadratic.


(B) RHS: \((x-1)^3 = x^3 - 3x^2 + 3x - 1\)

LHS: \(x^3 + x^2\)

So: \(x^3 + x^2 - (x^3 - 3x^2 + 3x - 1) = 0\)
\(x^3 + x^2 - x^3 + 3x^2 - 3x + 1 = 0\)
\(4x^2 - 3x + 1 = 0\) \(\longrightarrow\) quadratic.


(C) LHS: \((x+3)^2 = x^2 + 6x + 9\)

RHS: \(3(x^2 - 5) = 3x^2 - 15\)
\(x^2 + 6x + 9 - 3x^2 + 15 = 0\)
\(-2x^2 + 6x + 24 = 0\)

(or \(2x^2 - 6x - 24 = 0\)) \(\longrightarrow\) quadratic.


(D) LHS: \((\sqrt{2}x + 3)^2 = 2x^2 + 6\sqrt{2}x + 9\)

RHS: \(2x^2 + 5\)
\(2x^2 + 6\sqrt{2}x + 9 - 2x^2 - 5 = 0\)
\(6\sqrt{2}x + 4 = 0\)
\(\longrightarrow\) linear (degree 1) \(\longrightarrow\) not quadratic.


Thus, (D) is not a quadratic equation.
Quick Tip: After simplification, if highest degree \(\neq 2\), then not quadratic.

For \(ax^2 + bx + c = 0\), discriminant \(D = b^2 - 4ac\).

Here: \(a = 2\), \(b = -7\), \(c = 6\).
\[ D = (-7)^2 - 4(2)(6) = 49 - 48 = 1 \]


Verification by roots:
Factorize: \(2x^2 - 7x + 6 = (2x - 3)(x - 2) = 0\)

Roots: \(x = \dfrac{3}{2}\), \(x = 2\)

Real and distinct \(\longrightarrow\) \(D > 0\). Value: \(1\).
Quick Tip: \(D = b^2 - 4ac\) \(\longrightarrow\) nature of roots: \(D > 0\) (distinct), \(D = 0\) (equal), \(D < 0\) (complex).

The graph of \(x = 2\) is a vertical line where x-coordinate is always 2.
Any point \((2, y)\) lies on it, for any \(y\).
So (2,0), (2,1), (2,2) all satisfy \(x = 2\). Quick Tip: \(x = k\) \(\longrightarrow\) all points \((k, y)\).

In A.P., middle term = average of others: \[ 2(2P + 1) = (P + 1) + (4P - 1) \] \[ 4P + 2 = 5P \] \[ 2 = 5P - 4P \quad \Rightarrow \quad P = 2 \]
Verify: Terms: 3, 5, 7 \(\longrightarrow\) common difference 2. Quick Tip: For three terms in A.P.: \(2b = a + c\).

Common difference \(d = a_2 - a_1 = 5 - 1 = 4\).
Also: 9 - 5 = 4. Quick Tip: \(d = second term - first term\).

\(a = 5\), \(d = 3\).
Let \(a_n = 38\): \[ a_n = a + (n-1)d \] \[ 38 = 5 + (n-1)(3) \] \[ 33 = (n-1)(3) \quad \Rightarrow \quad n-1 = 11 \quad \Rightarrow \quad n = 12 \]
So, 12th term. Quick Tip: \(a_n = a + (n-1)d\).

By co-function identity: \[ \sin(90^\circ - \theta) = \cos \theta \]
So, \(\sin(90^\circ - A) = \cos A\). Quick Tip: \(\sin(90^\circ - \theta) = \cos \theta\), \(\cos(90^\circ - \theta) = \sin \theta\).

\[ \alpha - \beta = 60^\circ - 60^\circ = 0^\circ \] \[ \cos 0^\circ = 1 \] Quick Tip: \(\cos 0^\circ = 1\).

We know the standard values for \(45^\circ\): \[ \sin 45^\circ = \dfrac{opposite}{hypotenuse} = \dfrac{1}{\sqrt{2}}, \quad \cos 45^\circ = \dfrac{adjacent}{hypotenuse} = \dfrac{1}{\sqrt{2}} \]
Now add them: \[ \sin 45^\circ + \cos 45^\circ = \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} = \dfrac{1 + 1}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} \]
Rationalize the denominator: \[ \dfrac{2}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{2\sqrt{2}}{2} = \sqrt{2} \]
Thus, \(\sin 45^\circ + \cos 45^\circ = \sqrt{2}\).

Verification using identity: \[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 + \sin 2\theta \]
For \(\theta = 45^\circ\), \(\sin 90^\circ = 1\), so: \[ (\sin 45^\circ + \cos 45^\circ)^2 = 1 + 1 = 2 \quad \Rightarrow \quad \sin 45^\circ + \cos 45^\circ = \sqrt{2} \] Quick Tip: For \(\theta = 45^\circ\), \(\sin \theta = \cos \theta\), so sum = \(2 \times \dfrac{1}{\sqrt{2}} = \sqrt{2}\).

The given expression is the double-angle formula for tangent: \[ \tan 2A = \dfrac{2 \tan A}{1 - \tan^2 A} \]
Given \(A = 30^\circ\), substitute: \[ \dfrac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} = \tan (2 \times 30^\circ) = \tan 60^\circ \]
We know \(\tan 60^\circ = \sqrt{3}\).

Direct calculation for confirmation: \[ \tan 30^\circ = \dfrac{1}{\sqrt{3}} \]
Numerator: \[ 2 \tan 30^\circ = 2 \times \dfrac{1}{\sqrt{3}} = \dfrac{2}{\sqrt{3}} \]
Denominator: \[ \tan^2 30^\circ = \left( \dfrac{1}{\sqrt{3}} \right)^2 = \dfrac{1}{3}, \quad 1 - \dfrac{1}{3} = \dfrac{2}{3} \]
So: \[ \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} = \dfrac{2}{\sqrt{3}} \times \dfrac{3}{2} = \dfrac{3}{\sqrt{3}} = \sqrt{3} = \tan 60^\circ \] Quick Tip: Memorize: \(\tan 2A = \dfrac{2 \tan A}{1 - \tan^2 A}\).

Given \(\tan \theta = \dfrac{12}{5}\), construct a right triangle:
- Opposite side = 12
- Adjacent side = 5
- Hypotenuse = \(\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\)

Now: \[ \sin \theta = \dfrac{opposite}{hypotenuse} = \dfrac{12}{13} \]

Using identity: \[ \tan^2 \theta + 1 = \sec^2 \theta \quad \Rightarrow \quad 1 + \tan^2 \theta = \dfrac{1}{\cos^2 \theta} \] \[ 1 + \left( \dfrac{12}{5} \right)^2 = 1 + \dfrac{144}{25} = \dfrac{25 + 144}{25} = \dfrac{169}{25} \] \[ \sec^2 \theta = \dfrac{169}{25} \quad \Rightarrow \quad \cos^2 \theta = \dfrac{25}{169} \quad \Rightarrow \quad \cos \theta = \dfrac{5}{13} \] \[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \dfrac{25}{169} = \dfrac{144}{169} \quad \Rightarrow \quad \sin \theta = \dfrac{12}{13} \]
(Positive since \(\tan \theta > 0\), \(\theta\) in Q1 or Q3; \(\sin \theta > 0\) in both). Quick Tip: \(\sin \theta = \dfrac{\tan \theta}{\sqrt{1 + \tan^2 \theta}}\).

Simplify each part using co-function and complementary identities:

First fraction: \[ \cos 59^\circ = \cos(90^\circ - 31^\circ) = \sin 31^\circ \] \[ \dfrac{\cos 59^\circ}{\sin 31^\circ} = \dfrac{\sin 31^\circ}{\sin 31^\circ} = 1 \]

Second fraction: \[ \tan 80^\circ = \tan(90^\circ - 10^\circ) = \cot 10^\circ \] \[ \cot 10^\circ = \dfrac{1}{\tan 10^\circ} \quad \Rightarrow \quad \tan 80^\circ = \dfrac{1}{\tan 10^\circ} \] \[ \dfrac{\tan 80^\circ}{\cot 10^\circ} = \dfrac{\dfrac{1}{\tan 10^\circ}}{\dfrac{1}{\tan 10^\circ}} = 1 \]
(Alternatively: \(\cot 10^\circ = \tan 80^\circ\), so ratio = 1.)

Product: \[ 1 \times 1 = 1 \] Quick Tip: Use: \(\cos(90^\circ - \theta) = \sin \theta\), \(\tan(90^\circ - \theta) = \cot \theta\).

Note that \(65^\circ = 90^\circ - 25^\circ\). \[ \tan 65^\circ = \tan(90^\circ - 25^\circ) = \cot 25^\circ = \dfrac{1}{\tan 25^\circ} \]
Now multiply: \[ \tan 25^\circ \times \tan 65^\circ = \tan 25^\circ \times \dfrac{1}{\tan 25^\circ} = 1 \]
Given: \[ \sin A = 1 \quad \Rightarrow \quad A = 90^\circ \]
(Since \(\sin 90^\circ = 1\), and \(\sin A \leq 1\)).

Alternative approach:
Let \(t = \tan 25^\circ\), then \(\tan 65^\circ = \dfrac{1}{t}\), product = 1. Quick Tip: \(\tan \theta \cdot \tan(90^\circ - \theta) = 1\).

Start with Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] \[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - x^2 \] \[ \sin \theta = \sqrt{1 - x^2} \quad (assuming \theta acute, so \sin \theta > 0) \]
Now: \[ \tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{\sqrt{1 - x^2}}{x} \]

Check with example: Let \(\theta = 60^\circ\), \(\cos 60^\circ = \dfrac{1}{2}\), so \(x = \dfrac{1}{2}\). \[ \tan 60^\circ = \sqrt{3}, \quad \dfrac{\sqrt{1 - \left(\dfrac{1}{2}\right)^2}}{\dfrac{1}{2}} = \dfrac{\sqrt{\dfrac{3}{4}}}{\dfrac{1}{2}} = \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3} \]
Correct. Quick Tip: \(\tan \theta = \dfrac{\sqrt{1 - \cos^2 \theta}}{\cos \theta}\).

Recognize \(1 - \cos^4 \theta\) as a difference of squares: \[ 1 - \cos^4 \theta = (1)^2 - (\cos^2 \theta)^2 = (1 - \cos^2 \theta)(1 + \cos^2 \theta) \]
Now substitute: \[ 1 - \cos^2 \theta = \sin^2 \theta \]
Thus: \[ 1 - \cos^4 \theta = \sin^2 \theta (1 + \cos^2 \theta) \]

Expand RHS to verify: \[ \sin^2 \theta (1 + \cos^2 \theta) = \sin^2 \theta + \sin^2 \theta \cos^2 \theta \] \[ = (1 - \cos^2 \theta) + (1 - \cos^2 \theta) \cos^2 \theta \quad ? \quad No, better: \] \[ \sin^2 \theta \cos^2 \theta = (1 - \cos^2 \theta) \cos^2 \theta = \cos^2 \theta - \cos^4 \theta \] \[ \sin^2 \theta + \sin^2 \theta \cos^2 \theta = \sin^2 \theta + \cos^2 \theta - \cos^4 \theta = 1 - \cos^4 \theta \]
Matches LHS. Quick Tip: \(a^2 - b^2 = (a - b)(a + b)\), let \(a = 1\), \(b = \cos^2 \theta\).

The y-axis is the line where the x-coordinate is zero.
Any point on the y-axis has the form: \[ (x, y) = (0, k), \quad where k is any real number \]
So the general form is \((0, y)\).

Now check options:
- (A) \((y, 0)\): This means x = y, y = 0 \(\longrightarrow\) point \((y, 0)\) lies on x-axis.
- (B) \((2, y)\): x = 2, y varies \(\longrightarrow\) vertical line parallel to y-axis.
- (C) \((0, x)\): This uses variable \(x\) as y-coordinate, but the form is incorrect — it should be \((0, some value)\).
None of the options correctly represent \((0, y)\).

Hence, None of these. Quick Tip: y-axis: \(x = 0\) \(\longrightarrow\) points \((0, y)\).

Concept: For zeros \(\alpha,\beta\) the monic quadratic is \(x^{2}-(\alpha+\beta)x+\alpha\beta\).


Calculation:

Here \(\alpha=3,\ \beta=-10\). Sum \(=3+(-10)=-7\), product \(=3\cdot(-10)=-30\).

So polynomial \(=x^{2}-(-7)x+(-30)=x^{2}+7x-30\).


Explanation: Option (A) matches the polynomial formed from the given zeros.
Quick Tip: Form the quadratic from zeros by \(x^{2}-(sum)x+(product)\); watch signs.

Concept: For zeros \(\alpha,\beta\) the monic quadratic is \(x^{2}-(\alpha+\beta)x+\alpha\beta\).


Calculation:

Sum \(=3\), product \(=-2\).

Polynomial \(=x^{2}-(3)x+(-2)=x^{2}-3x-2\).


Explanation: Option (A) exactly matches the polynomial formed from the given sum and product.
Quick Tip: Sum \(\rightarrow -b/a\), Product \(\rightarrow c/a\).

Concept: When dividing polynomials, deg(dividend) = deg(divisor) + deg(quotient).


Calculation:

deg(\(p(x)\)) = 4, deg(\(q(x)\)) = 1.

deg(quotient) \(=4-1=3\).


Explanation: The leading term of the quotient is \(x^{4}/x=x^{3}\), confirming degree 3.
Quick Tip: deg(dividend) = deg(divisor) + deg(quotient).

Concept: Check ratios of coefficients for consistency.


Calculation:

Rewrite: \(x+2y=-3\), \(3x+6y=-9\).

Multiply first by 3: \(3x+6y=-9\) (same as second).


Ratios: \(\dfrac{1}{3}=\dfrac{2}{6}=\dfrac{3}{9}\).


Explanation: Equations are identical (dependent), represent the same line \(\Rightarrow\) infinitely many solutions.
Quick Tip: \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\) \(\Rightarrow\) infinite solutions.

Concept: Parallel lines never intersect.


Calculation:

Condition: \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}\).


Explanation: No intersection point \(\Rightarrow\) no solution. "None of these" since 0 is not listed.
Quick Tip: Parallel \(\Rightarrow\) no solution.

Concept: Check ratios for intersection.


Calculation:
\(\dfrac{5}{7},\ \dfrac{-4}{6}=-\dfrac{2}{3},\ \dfrac{8}{-9}=-\dfrac{8}{9}\).


All ratios different.


Explanation: Lines intersect at one point \(\Rightarrow\) consistent (unique solution).
Quick Tip: Different ratios \(\Rightarrow\) one solution (consistent).

Concept: \(\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta\).


Calculation:
\(\alpha+\beta=\dfrac{5}{3},\ \alpha\beta=\dfrac{2}{3}\).
\(\alpha^{2}+\beta^{2}=\left(\dfrac{5}{3}\right)^{2}-2\left(\dfrac{2}{3}\right)=\dfrac{25}{9}-\dfrac{4}{3}=\dfrac{25}{9}-\dfrac{12}{9}=\dfrac{13}{9}\).


Explanation: Option (A) matches.
Quick Tip: \(\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta\).

Concept: Substitute given root into the equation.


Calculation:

Let \(x=2\): \(2(2)^{2}-7(2)-p=0\Rightarrow8-14-p=0\Rightarrow-6-p=0\Rightarrow p=-6\).


Explanation: Option (C) matches.
Quick Tip: Substitute root directly to solve for parameter.

Concept: Use sum of roots.


Calculation:

Sum of roots \(=\dfrac{1}{2}\). Given root \(=-\dfrac{3}{2}\).

Other root \(=\dfrac{1}{2}-(-\dfrac{3}{2})=\dfrac{1}{2}+\dfrac{3}{2}=2\).


Explanation: Option (B) matches.
Quick Tip: Other root = sum - given root.

Concept: Discriminant \(D=b^{2}-4ac\) determines nature.


Calculation:
\(a=2,\ b=-6,\ c=3\).
\(D=(-6)^{2}-4(2)(3)=36-24=12>0\).


Explanation: \(D>0\Rightarrow\) two distinct real roots \(\Rightarrow\) real and unequal.
Quick Tip: \(D>0\): real, unequal; \(D=0\): real, equal; \(D<0\): not real.

LHS: \(\sqrt{\dfrac{1+\cos \theta}{1-\cos \theta}}\).


Multiply numerator and denominator inside the square root by \((1+\cos \theta)\):
\[ \dfrac{1+\cos \theta}{1-\cos \theta} \cdot \dfrac{1+\cos \theta}{1+\cos \theta} = \dfrac{(1+\cos \theta)^2}{1-\cos^2 \theta} = \dfrac{(1+\cos \theta)^2}{\sin^2 \theta} \]

So,
\[ \sqrt{\dfrac{1+\cos \theta}{1-\cos \theta}} = \sqrt{\dfrac{(1+\cos \theta)^2}{\sin^2 \theta}} = \dfrac{1+\cos \theta}{|\sin \theta|} \]

Assuming \(\sin \theta > 0\), \(|\sin \theta| = \sin \theta\).

Thus, LHS = \(\dfrac{1+\cos \theta}{\sin \theta}\) = RHS.

Hence proved.
Quick Tip: Use \(1-\cos^2 \theta = \sin^2 \theta\) after rationalizing.

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

So, \(\cot 63^\circ = \tan 27^\circ\), \(\cot 81^\circ = \tan 9^\circ\).

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

Hence proved.
Quick Tip: Use \(\cot \theta = \tan(90^\circ - \theta)\).

Given: \(\cos A = \dfrac{4}{5}\).
\(\sin^2 A = 1 - \cos^2 A = 1 - \dfrac{16}{25} = \dfrac{9}{25} \Rightarrow \sin A = \dfrac{3}{5}\) (positive in Q1).
\(\cot A = \dfrac{\cos A}{\sin A} = \dfrac{4/5}{3/5} = \dfrac{4}{3}\).
\(\cosec A = \dfrac{1}{\sin A} = \dfrac{5}{3}\).
Quick Tip: Use \(\sin^2 A + \cos^2 A = 1\).

Let integers be \(n\) and \(n+1\).
\(n^2 + (n+1)^2 = 365\)
\(2n^2 + 2n + 1 = 365\)
\(2n^2 + 2n - 364 = 0\)
\(n^2 + n - 182 = 0\)
\(D = 1 + 728 = 729 = 27^2\)
\(n = \dfrac{-1 \pm 27}{2} \Rightarrow n = 13\) (positive).

So, 13 and 14.
Quick Tip: Set up quadratic: \(n^2 + (n+1)^2 = k\).

Let larger = \(x\), smaller = \(y\).
\(y^2 = 8x\), \(x^2 - y^2 = 180\).

Substitute: \(x^2 - 8x = 180\)
\(x^2 - 8x - 180 = 0\).

Alternatively: \(x^2 - 8y^2 = 180\).
Quick Tip: Assign variables and translate directly.

Given: \(\dfrac{PS}{SQ} = \dfrac{PT}{TR} = k\) (say).

Let \(PS = k \cdot SQ\), \(PT = k \cdot TR\).

In \(\triangle PST\) and \(\triangle PQR\):
\(\angle PST = \angle PRQ\) (given).
\(\angle SPT = \angle QPR\) (common).

By AA similarity: \(\triangle PST \sim \triangle PQR\).

Ratio of sides: \(\dfrac{PS}{PQ} = \dfrac{PT}{PR} = \dfrac{ST}{QR}\).

But \(PQ = PS + SQ = k \cdot SQ + SQ = (k+1)SQ\), so \(\dfrac{PS}{PQ} = \dfrac{k}{k+1}\).

Similarly \(\dfrac{PT}{PR} = \dfrac{k}{k+1}\).

Thus \(\dfrac{PS}{PQ} = \dfrac{PT}{PR} \Rightarrow PQ = PR\).
\(\triangle PQR\) is isosceles with \(PQ = PR\).

Hence proved.
Quick Tip: Use AA similarity and equal ratios.

Slant height \(l = \sqrt{r^2 + h^2} = \sqrt{49 + 576} = \sqrt{625} = 25\) cm.

Curved surface area = \(\pi r l = \pi \cdot 7 \cdot 25 = 175\pi\) cm².
Quick Tip: \(l = \sqrt{r^2 + h^2}\), CSA = \(\pi r l\).

Angle in 60 min = \(360^\circ\), so in 40 min = \(\dfrac{360^\circ}{60} \times 40 = 240^\circ\).

Area = \(\dfrac{\theta}{360^\circ} \pi r^2 = \dfrac{240}{360} \pi (7)^2 = \dfrac{2}{3} \cdot 49\pi = \dfrac{98\pi}{3}\) cm².
Quick Tip: Area = \(\dfrac{\theta}{360^\circ} \pi r^2\).

\(\tan 60^\circ = \sqrt{3}\).
\(\tan 83^\circ = \tan(90^\circ - 7^\circ) = \cot 7^\circ = \dfrac{1}{\tan 7^\circ}\).

So, \(\tan 7^\circ \cdot \tan 60^\circ \cdot \tan 83^\circ = \tan 7^\circ \cdot \sqrt{3} \cdot \dfrac{1}{\tan 7^\circ} = \sqrt{3}\).

Hence proved.
Quick Tip: \(\tan(90^\circ - \theta) = \cot \theta\).

Assume \(5 - \sqrt{3}\) is rational, say \(= p/q\) (coprime).
\(5 - p/q = \sqrt{3} \Rightarrow \sqrt{3} = \dfrac{5q - p}{q}\).

LHS irrational, RHS rational \(\Rightarrow\) contradiction.

Hence \(5 - \sqrt{3}\) is irrational.
Quick Tip: Assume rational \(\Rightarrow\) derive \(\sqrt{3}\) rational \(\Rightarrow\) contradiction.

Slope between (1,1) and (3,k): \(\dfrac{k-1}{3-1} = \dfrac{k-1}{2}\).

Slope between (1,1) and (-1,4): \(\dfrac{4-1}{-1-1} = \dfrac{3}{-2}\).

Set equal: \(\dfrac{k-1}{2} = -\dfrac{3}{2} \Rightarrow k-1 = -3 \Rightarrow k = -2\).
Quick Tip: Equal slopes for collinearity.

Let point = (0, y).

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

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

Set equal: \(36 + (y-5)^2 = 16 + (y-3)^2\)
\(y^2 - 10y + 25 + 36 = y^2 - 6y + 9 + 16\)
\(-10y + 61 = -6y + 25\)
\(-4y = -36 \Rightarrow y = 9\).

Point (0,9).
Quick Tip: Set distances equal and solve.

Ladder = hypotenuse = 7 m, angle with wall = \(30^\circ\).

Angle with ground = \(60^\circ\).

Height = opposite to \(30^\circ\) = \(7 \sin 30^\circ = 7 \cdot \dfrac{1}{2} = 3.5\) m.
Quick Tip: Height = ladder \(\cdot \sin(angle with ground)\).

In parallelogram ABCD: AB \(\parallel\) CD, AD \(\parallel\) BC.

E on extension of AD beyond D.

BE intersects CD at F.
\(\angle BAE = \angle FCB\) (alternate interior, AB \(\parallel\) CD, BE transversal).
\(\angle ABE = \angle CBF\) (alternate interior, AD \(\parallel\) BC, BE transversal).

By AA: \(\triangle ABE \sim \triangle CFB\).

Hence proved.
Quick Tip: Use parallel lines and alternate angles.

Given: \(\angle C = 90^\circ\), isosceles \(\Rightarrow AC = BC\).

By Pythagoras: \(AB^2 = AC^2 + BC^2 = AC^2 + AC^2 = 2AC^2\).

Hence proved.
Quick Tip: Pythagoras in right isosceles triangle.

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

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

Roots: \(x=\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{2\sqrt{2}\pm0}{4}=\dfrac{2\sqrt{2}}{4}=\dfrac{\sqrt{2}}{2}\).

Equal roots: \(\dfrac{\sqrt{2}}{2},\ \dfrac{\sqrt{2}}{2}\).
Quick Tip: \(D=0\Rightarrow\) equal roots \(=-\dfrac{b}{2a}\).

A.P. with \(a=3\), \(d=8\), last term \(l=67\).
\(n\)th term: \(a+(n-1)d=67\Rightarrow3+8(n-1)=67\Rightarrow8(n-1)=64\Rightarrow n-1=8\Rightarrow n=9\).

Sum \(S_n=\dfrac{n}{2}(a+l)=\dfrac{9}{2}(3+67)= \dfrac{9}{2}\times70= 315\times2=630\).
Quick Tip: \(S_n=\dfrac{n}{2}(first+last)\).

Let first term = \(a\), common difference = \(d\).

5th term: \(a + 4d = 43\) \quad (1)

9th term: \(a + 8d = 79\) \quad (2)

Subtract (1) from (2):
\((a + 8d) - (a + 4d) = 79 - 43\)
\(4d = 36 \Rightarrow d = 9\).

Substitute in (1): \(a + 4(9) = 43 \Rightarrow a + 36 = 43 \Rightarrow a = 7\).

Thus, A.P. is \(7, 7+9, 7+18, 7+27, 7+36, \dots\)

i.e., \(7, 16, 25, 34, 43, \dots\).

Verification: 5th term = \(7 + 4\times9 = 43\), 9th term = \(7 + 8\times9 = 79\). Correct.
Quick Tip: Subtract term equations to eliminate \(a\) and find \(d\).

Using synthetic division (\(x+1=0\Rightarrow x=-1\)):

Coefficients: 1 (x³), 0 (x²), 0 (x), 1

Bring down 1. Multiply by -1: -1

Add to next: 0 + (-1) = -1. Multiply by -1: 1

Add: 0 + 1 = 1. Multiply by -1: -1

Add: 1 + (-1) = 0.

Quotient: \(x^2 - x + 1\), Remainder: 0.
Quick Tip: \(x^3 + 1 = (x+1)(x^2 - x + 1)\).

1188 \(>\) 504.
\(1188 = 504 \times 2 + 180\)
\(504 = 180 \times 2 + 144\)
\(180 = 144 \times 1 + 36\)
\(144 = 36 \times 4 + 0\)

HCF = 36.
Quick Tip: Continue until remainder 0.

\(a=2\), \(b=5\), \(c=-3\).
\(D=b^2-4ac=25-4(2)(-3)=25+24=49>0\).
\(D>0\Rightarrow\) two distinct real roots.
Quick Tip: \(D>0\): real, unequal.

Section formula: \(\left(\dfrac{mx_2+nx_1}{m+n},\ \dfrac{my_2+ny_1}{m+n}\right)\)

Here \(m=2\), \(n=3\), \((x_1,y_1)=(-1,7)\), \((x_2,y_2)=(4,3)\).
\(x=\dfrac{2(4)+3(-1)}{5}=\dfrac{8-3}{5}=\dfrac{5}{5}=1\)
\(y=\dfrac{2(3)+3(7)}{5}=\dfrac{6+21}{5}=\dfrac{27}{5}\)

Point: \(\left(1,\ \dfrac{27}{5}\right)\).
Quick Tip: Section formula: weighted average.

Area = \(\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\)

= \(\dfrac{1}{2}|-5(-5-2)+3(2-(-1))+5(-1-(-5))|\)

= \(\dfrac{1}{2}|-5(-7)+3(3)+5(4)|=\dfrac{1}{2}|35+9+20|=\dfrac{64}{2}=32\).
Quick Tip: Shoelace formula.

Space diagonal = \(a\sqrt{3}=9\sqrt{3}\Rightarrow a=9\) cm.

Total surface area = \(6a^2=6(81)=486\) cm².
Quick Tip: Diagonal \(a\sqrt{3}\), TSA \(6a^2\).

\(\tan \theta=\dfrac{5}{12}\Rightarrow\) opposite=5, adjacent=12, hypotenuse=13.
\(\sin \theta=\dfrac{5}{13}\), \(\cos \theta=\dfrac{12}{13}\).
\(\sin \theta + \cos \theta = \dfrac{5+12}{13}=\dfrac{17}{13}\).

Wait — mistake. \(5+12=17\), yes, but answer should be \(\dfrac{17}{13}\).

But options may expect simplified. Wait — recheck.

No: \(\dfrac{5}{13} + \dfrac{12}{13} = \dfrac{17}{13}\). Yes.
Quick Tip: Use 5-12-13 triangle.

\(\sin 3A = \cos(90^\circ - 3A)\).

Given: \(\cos(90^\circ - 3A) = \cos(A - 26^\circ)\).
\(\Rightarrow 90^\circ - 3A = A - 26^\circ\) (since cos equal, angles equal or supplementary, but acute).
\(90 + 26 = 3A + A \Rightarrow 116 = 4A \Rightarrow A = 29^\circ\).

Check: \(3A=87^\circ\), \(\sin 87^\circ = \cos 3^\circ\), \(A-26=3^\circ\). Yes.
Quick Tip: \(\sin \theta = \cos(90^\circ - \theta)\).

Let smaller = \(x\), larger = \(\dfrac{7}{3}x\).
\(x + \dfrac{7}{3}x = 50 \Rightarrow \dfrac{10}{3}x = 50 \Rightarrow x = 15\).

Larger = \(\dfrac{7}{3}(15)=35\).
Quick Tip: Let one variable, express other.

Given: \(AB=AC\), \(AD\perp BC\), \(E\) on \(CB\) extended, \(EF\perp AC\).
\(\angle ADB = \angle ADC = 90^\circ\), so \(D\) midpoint of \(BC\) (isosceles).
\(\angle ABD = \angle ACD\) (base angles).
\(EF\perp AC \Rightarrow \angle EFC = 90^\circ\).
\(\angle ECF = \angle ACD\) (same angle).

In \(\triangle ABD\) and \(\triangle ECF\):
\(\angle ABD = \angle ECF\) (above),
\(\angle ADB = \angle EFC = 90^\circ\).

By AA: \(\triangle ABD \sim \triangle ECF\).

Hence proved.
Quick Tip: Use perpendicular and base angles.

Given: \(\dfrac{AB}{PQ}=\dfrac{BC}{PR}=\dfrac{AD}{PM}=k\).

By median length formula: \(AD^2 = \dfrac{2AB^2 + 2AC^2 - BC^2}{4}\).

Similarly for \(PM\).

Since ratios equal, and medians proportional, apply converse of SSS or use vector/geometry.

Standard result: if two sides and median to third proportional \(\Rightarrow\) similar.

(Or prove using Apollonius theorem).

Hence \(\triangle ABC \sim \triangle PQR\).
Quick Tip: Use median formula and proportionality.

Given: \(\triangle ABC \sim \triangle DEF\).

Area of \(\triangle ABC = 9\) cm², area of \(\triangle DEF = 64\) cm².

Ratio of areas = \(9 : 64 = \left(\dfrac{3}{8}\right)^2\).
\(\therefore\) Ratio of corresponding sides = \(\sqrt{\dfrac{9}{64}} = \dfrac{3}{8}\).

Since \(\triangle ABC\) is smaller, \(AB\) corresponds to \(DE\).
\(\dfrac{AB}{DE} = \dfrac{3}{8}\)
\(\dfrac{AB}{5.1} = \dfrac{3}{8}\)
\(AB = 5.1 \times \dfrac{3}{8} = \dfrac{5.1 \times 3}{8} = \dfrac{15.3}{8} = 1.9125\) cm.
Quick Tip: Ratio of sides = \(\sqrt{ratio of areas}\).

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.