Concept: The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is found using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \]
This measures the straight-line (Euclidean) distance between two coordinate points.


Calculation:

Given points are \(A(8\sin60^\circ, 0)\) and \(B(0, 8\cos60^\circ)\).

We know: \(\sin60^\circ = \dfrac{\sqrt{3}}{2}\) and \(\cos60^\circ = \dfrac{1}{2}\).

Hence, \(A(4\sqrt{3}, 0)\) and \(B(0, 4)\).


Now applying the distance formula: \[ AB = \sqrt{(4\sqrt{3} - 0)^2 + (0 - 4)^2} \] \[ = \sqrt{(16 \times 3) + 16} = \sqrt{64} = 8. \]


Explanation:
Thus, the distance between the two points is \(8\) units. The trigonometric values simplify the coordinates, and substituting them carefully gives the correct result.
Quick Tip: When solving coordinate geometry questions, always begin by simplifying trigonometric terms like \(\sin\) and \(\cos\) before applying any formula.
Write coordinates clearly and label them as \((x_1, y_1)\) and \((x_2, y_2)\) to avoid confusion.
After substituting, square and add each term carefully to prevent sign errors.
Finally, check if the result is a perfect square to simplify the final answer neatly.

Concept: The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \]
Here, the origin \(O(0,0)\) and point \(P(x,y)\) are given.


Calculation:

Using the distance formula, \[ OP = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}. \]


Explanation: The distance from the origin to any point \((x, y)\) is always \(\sqrt{x^2 + y^2}\).
Quick Tip: Always remember that the distance from the origin to any point \((x, y)\) is obtained by applying Pythagoras’ theorem in coordinate geometry.
It represents the length of the hypotenuse of a right triangle formed by x- and y-axes.
This result is frequently used to write equations of circles centered at the origin.

Concept: The distance of a point \((x, y)\) from the y-axis is given by the absolute value of its x-coordinate.


Calculation:

For the point \((12, 14)\), distance from the y-axis = \(|x| = |12| = 12.\)


Explanation: The y-axis is the line \(x = 0\), so horizontal distance from any point to it is simply the magnitude of the x-coordinate.
Quick Tip: Distance from the y-axis depends only on the x-coordinate; distance from the x-axis depends only on the y-coordinate.
For point \((x,y)\), distance to y-axis = \(|x|\) and to x-axis = \(|y|\).
Quickly scan coordinates to answer such questions without lengthy calculations.

Concept: In coordinate geometry, the x-coordinate is called the abscissa and the y-coordinate is called the ordinate.


Calculation:

Given point = \((-6, -8)\). Hence, the ordinate = y-coordinate = \(-8\).


Explanation: The ordinate indicates the vertical position of a point on the coordinate plane and is the second component in \((x,y)\).
Quick Tip: Remember: for a point \((x,y)\), \(x\) = abscissa, \(y\) = ordinate.
Label coordinates clearly to avoid mixing up x and y.
This basic terminology appears frequently in coordinate geometry questions.

Concept: A point \((x, y)\) lies in a specific quadrant based on the signs of x and y coordinates.

First: (+, +), Second: (-, +), Third: (-, -), Fourth: (+, -).


Calculation:

Given point \((3, -4)\) has x positive and y negative.

Therefore, it lies in the Fourth Quadrant.
Quick Tip: Memorize quadrant sign order: I → (+,+), II → (-,+), III → (-,-), IV → (+,-).
A quick sketch of axes helps if you're unsure.
Use the signs of coordinates to identify the quadrant instantly.

Concept: In the second quadrant, x is negative and y is positive.


Calculation:

Among the given options, \((-3,2)\) satisfies x < 0 and y > 0, so it lies in the second quadrant.
Quick Tip: Second quadrant points have negative x and positive y.
Check signs of coordinates rather than plotting if options are clear.
A small mental map of quadrants speeds up answers.

Concept: The midpoint of a line joining points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \[ M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right). \]


Calculation:

Substitute: \((x_1, y_1) = (4, -4)\), \((x_2, y_2) = (-4, 4)\).
\[ M = \left(\frac{4 + (-4)}{2}, \frac{-4 + 4}{2}\right) = (0, 0). \]
Quick Tip: Apply midpoint formula by averaging x-coordinates and y-coordinates separately.
If points are negatives of each other, midpoint is the origin (0,0).
Midpoint problems often simplify if symmetry is recognized early.

Concept: If \(M(x_m, y_m)\) is the midpoint of \(AB\), then \[ x_m = \frac{x_A + x_B}{2}, \quad y_m = \frac{y_A + y_B}{2}. \]


Calculation:

Given: \(M(2,4)\), \(A(5,7)\). Let \(B(x, y)\). Then, \[ 2 = \frac{5 + x}{2} \Rightarrow 4 = 5 + x \Rightarrow x = -1. \] \[ 4 = \frac{7 + y}{2} \Rightarrow 8 = 7 + y \Rightarrow y = 1. \]
So \(B(-1,1)\).
Quick Tip: To find the other endpoint, multiply midpoint coordinates by 2 and subtract the known endpoint coordinates.
Work coordinate-wise (x and y separately) to avoid mistakes.
Always replug found coordinates into midpoint formula to verify.

Concept: The centre of a circle is the midpoint of the line joining the endpoints of its diameter.


Calculation:

Let \(A(10,-6)\), \(B(-6,10)\). Then, \[ C = \left(\frac{10 + (-6)}{2}, \frac{-6 + 10}{2}\right) = \left(\frac{4}{2}, \frac{4}{2}\right) = (2,2). \]
Quick Tip: The centre of a circle with endpoints of diameter given is always their midpoint.
Compute midpoint using averages of coordinates.
This is a direct application of the midpoint formula—no need for distance formula here.

Concept: The centroid of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) is given by \[ G\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right). \]


Calculation:

Substitute: \((4,6), (0,4), (5,5)\).
\[ G = \left(\frac{4+0+5}{3}, \frac{6+4+5}{3}\right) = \left(\frac{9}{3}, \frac{15}{3}\right) = (3,5). \]
Quick Tip: Centroid = average (mean) of the vertex coordinates; compute x-mean and y-mean separately.
Centroid lies inside the triangle and divides medians in a 2:1 ratio.
Use centroid formula for problems on centres and balancing points.

Concept: The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) can be found using the determinant formula: \[ Area = \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|. \]


Calculation:

Substitute: \(A(0,1)\), \(B(0,5)\), \(C(3,4)\).
\[ Area = \frac{1}{2}\left|0(5-4) + 0(4-1) + 3(1-5)\right| = \frac{1}{2}\left|0 + 0 -12\right| = \frac{12}{2} = 6. \]


Explanation: Two vertices share the same x-coordinate (x=0), so the base can be taken along the y-axis and height measured horizontally, simplifying the determinant calculation.
Quick Tip: Use the coordinate (determinant) formula for area to avoid geometry-based constructions.
If two points have same x (or y), area calculation simplifies—use that symmetry.
Always take absolute value to ensure a positive area result.

Concept: Use complementary-angle identities: \(\tan(90^\circ - \theta) = \cot\theta = \dfrac{1}{\tan\theta}\).


Calculation:

Note that \(\tan 80^\circ = \tan(90^\circ - 10^\circ) = \cot 10^\circ = \dfrac{1}{\tan 10^\circ}\).

Also \(\tan 67^\circ = \tan(90^\circ - 23^\circ) = \cot 23^\circ = \dfrac{1}{\tan 23^\circ}\).

Therefore, \[ \tan 10^\circ \tan 23^\circ \tan 80^\circ \tan 67^\circ = \tan 10^\circ \tan 23^\circ \cdot \frac{1}{\tan 10^\circ} \cdot \frac{1}{\tan 23^\circ} = 1. \]
Quick Tip: Look for complementary-angle pairs (sum to \(90^\circ\)) to convert tangents to cotangents.
Cotangent is the reciprocal of tangent, which often cancels products neatly.
Recognizing these identities saves computation and reduces errors in trigonometric products.

Concept: For two similar figures (here triangles), the ratio of their areas equals the square of the ratio of corresponding sides. That is, if side ratio = \(k\), then area ratio = \(k^2\).


Detailed Calculation and Reasoning:

We are given area ratio: \(\dfrac{Area_1}{Area_2} = \dfrac{100}{144}\). Let the ratio of corresponding sides be \(\dfrac{s_1}{s_2} = k\). Then \(k^2 = \dfrac{100}{144}\).


Take the positive square root (side lengths are positive): \[ k = \sqrt{\dfrac{100}{144}} = \dfrac{\sqrt{100}}{\sqrt{144}} = \dfrac{10}{12}. \]
Simplify if desired: \(\dfrac{10}{12} = \dfrac{5}{6}\), but among the options \(10:12\) is listed directly.


Check: Squaring \(10:12\) gives \(100:144\), which matches the given area ratio. This confirms the result.
Quick Tip: Remember area ratio = (side ratio)\(^2\) for similar shapes.
Always take the positive square root when converting area ratio to side ratio (sides are lengths).
Simplify the ratio only if required by options—both forms may be equivalent.

Concepts and Definitions:

- A \emph{tangent is a line that touches the circle at exactly one point.

- A \emph{chord is a line segment whose endpoints lie on the circle.

- A \emph{secant is a line that intersects the circle in two distinct points; equivalently, it is the extension of a chord to a full line.


Detailed Explanation:

If a line meets the circle at two distinct points \(A\) and \(B\), then the segment \(AB\) joining these two points is a chord. Extending that segment indefinitely gives the secant line. The question asks for the term for the \emph{line that meets the circle at two points, which by definition is a \emph{secant.


Comparison to other options:

- It is not a tangent because a tangent meets the circle only once.

- It is not just the chord (option 1), because chord refers to the segment between the two points, whereas the question names the line that intersects the circle in two points (secant).
Quick Tip: Visualize: tangent touches once, secant cuts through (twice), chord is the inside segment between the two cut points.
A secant contains a chord as part of it — remember this hierarchy to avoid confusion.
Tangents are perpendicular to the radius at point of contact; secants have no such perpendicularity property globally.

Concept: For similar triangles, area ratio \(=\) (side ratio)\(^2\).


Step-by-step Calculation:

Given side ratio \(\dfrac{s_1}{s_2} = \dfrac{4}{9}\). Then area ratio is \[ \left(\frac{4}{9}\right)^2 = \frac{16}{81}. \]

Interpretation and Check:

This means if one triangle has side lengths scaled by factor \(4/9\) relative to the other, its area scales down by \((4/9)^2\). Option (2) directly matches this computed ratio.
Quick Tip: Square the side ratio to get the area ratio for similar figures.
Keep track of which triangle corresponds to numerator/denominator to avoid inversions.
For quick recall: sides → power 1, areas → power 2, volumes → power 3.

Concept: For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides. Use that to relate given areas and sides.


Step-by-step:

1. Side ratio: \(\dfrac{BC}{EF} = \dfrac{3}{4}\). This is the ratio of the corresponding sides of triangle ABC to triangle DEF.

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

3. Substitute the known area: \(\dfrac{54}{Area_{DEF}} = \dfrac{9}{16}.\)

4. Solve for \(Area_{DEF}\): multiply both sides by \(Area_{DEF}\) and divide by \(9/16\): \[ Area_{DEF} = \frac{54 \times 16}{9} = 54 \times \frac{16}{9} = 54 \times \frac{16}{9}. \]
Compute: \(54 \div 9 = 6\), then \(6 \times 16 = 96\). So \(Area_{DEF} = 96~cm^2.\)


Sanity check: Since triangle DEF has larger corresponding side (4 vs 3), its area should be larger than 54 — 96 is larger, consistent with expectation.
Quick Tip: Write the proportion \(\dfrac{Area_1}{Area_2} = (\dfrac{side_1}{side_2})^2\) and solve—cross-multiplication avoids mistakes.
Check direction: if side ratio > 1 then area of the corresponding triangle must be larger.
Keep units consistent and compute stepwise to reduce arithmetic errors.

Concept: In a right-angled triangle (here \(\angle A = 90^\circ\)), the Pythagorean theorem relates the sides: if the hypotenuse is \(BC\) and the legs are \(AB\) and \(AC\), then \(BC^2 = AB^2 + AC^2\).


Detailed Steps:

Given: \(BC = 13\) (hypotenuse), \(AB = 12\). Plug into Pythagoras: \[ 13^2 = 12^2 + AC^2. \]
Compute squares: \(169 = 144 + AC^2\). Rearranging: \(AC^2 = 169 - 144 = 25\). Taking positive square root (length): \(AC = \sqrt{25} = 5\) cm.


Interpretation: This is the well-known Pythagorean triple \(5\!-\!12\!-\!13\).
Quick Tip: Recognize common Pythagorean triplets (3–4–5, 5–12–13, 7–24–25) to answer quickly.
Ensure you identify the hypotenuse correctly (the side opposite the right angle) before applying the theorem.
Always take the positive square root for side lengths.

Concept: Sum of interior angles in any triangle is \(180^\circ\). If two angles of one triangle equal two angles of another, the third angles must also be equal (AA → similarity).


Detailed Reasoning:

Given: \(\angle D = \angle Q\) and \(\angle E = \angle R\). Consider the third angles in each triangle: in \(\triangle DEF\), \(\angle F = 180^\circ - (\angle D + \angle E)\). In \(\triangle PQR\), \(\angle P = 180^\circ - (\angle Q + \angle R)\). Substitute the equalities: since \(\angle D = \angle Q\) and \(\angle E = \angle R\), we get \(\angle F = 180^\circ - (\angle Q + \angle R) = \angle P\).


Conclusion: \(\angle F = \angle P\), which is option (1). This also establishes the triangles are similar by the AA criterion.
Quick Tip: AA similarity: if two angles of one triangle equal two angles of another, the triangles are similar and the third angles match automatically.
Use the angle-sum property (\(180^\circ\)) to prove equality of the third angles formally.
Label corresponding vertices to avoid mismatching corresponding angles.

Concept: Equality of all three corresponding side ratios implies triangles are similar (S-S-S similarity), so corresponding angles are equal. Use angle-sum property to find the missing angle in triangle ABC, then map to triangle DEF.


Detailed Steps:

1. In \(\triangle ABC\), angles given: \(\angle A = 40^\circ\), \(\angle B = 80^\circ\). Sum of angles is \(180^\circ\).

2. Compute \(\angle C = 180^\circ - (40^\circ + 80^\circ) = 180^\circ - 120^\circ = 60^\circ.\)

3. Because triangles are similar, \(\angle F\) corresponds to \(\angle C\) (match order of letters if needed: given side ratios AB/DE = BC/EF = CA/DF suggests \(A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F\)). Therefore \(\angle F = \angle C = 60^\circ.\)


Cross-check: Option (3) matches this computed value.
Quick Tip: When side ratios for all three pairs are equal, triangles are similar — map vertices in the same order to identify corresponding angles.
Compute the third angle using \(180^\circ\) and then transfer it to the corresponding angle in the similar triangle.
Careful labeling avoids mismatches in correspondence.

Concept: The number of common tangents to two circles depends on their relative position: disjoint (outside each other), touching externally, intersecting, touching internally, or one inside the other.


Detailed Reasoning by Cases:

- If two circles are separate (no intersection), there are four common tangents (two direct/external and two transverse/internal).

- If they touch externally (one point), there are three common tangents (two external, one common internal becomes coincident at point of contact).

- If they intersect at two points (the present case), there are exactly two common tangents, both external. The internal tangents cross the line of centers between intersection points and do not exist as distinct tangents in this case.

- If they touch internally (one circle inside the other tangent internally), there is exactly one common tangent.

- If one circle is strictly inside the other with no touch, there are no common tangents.


Thus for two circles that intersect (two common points), the correct number of common tangents is 2 (the two external tangents).
Quick Tip: Memorize the relation: intersecting → 2, externally tangent → 3, internally tangent → 1, disjoint outside → 4, one inside the other → 0.
Sketching the circles and possible tangents clarifies the count in exam conditions.
Use the line of centers and nature of intersection to reason about tangents.

Concept: A fundamental property of tangents from an external point is that the two tangent segments drawn from that external point to the circle are equal in length. This follows from congruent right triangles formed by radii to the points of contact.


Detailed Explanation:

1. Let the circle have centre \(O\) and the tangency points be \(A\) and \(B\). Then \(OA\) and \(OB\) are radii and are perpendicular to tangents \(PA\) and \(PB\) respectively. So \(\triangle OAP\) and \(\triangle OBP\) are right triangles.

2. \(OA = OB\) (radii), and \(OP\) is common to both triangles. Thus by RHS (right-angle, hypotenuse, side) congruence, \(\triangle OAP \cong \triangle OBP\).

3. By congruence, corresponding sides \(PA\) and \(PB\) are equal. Given \(PA = 8\) cm, therefore \(PB = 8\) cm.


Alternate quick reasoning: Tangents from an external point are equal—memorized theorem.
Quick Tip: Use the tangent-length theorem: tangents from the same external point are equal.
If needed, prove via congruent right triangles formed with radii to points of contact.
This property helps in many geometry problems involving tangents and power of a point.

Concepts and Key Observations:

- \(PA\) and \(PB\) are tangents from the external point \(P\) to the circle with centre \(O\).

- Radii \(OA\) and \(OB\) are perpendicular to tangents at points of contact: \(OA \perp PA\) and \(OB \perp PB\).

- The line \(OP\) (joining centre to external point) bisects the angle \(\angle APB\) because the two right triangles \(\triangle OAP\) and \(\triangle OBP\) are congruent (radii equal and \(OP\) common). Thus \(\angle AOP\) and \(\angle POB\) are equal in measure when considered at point \(P\).


Detailed Angle Relationship and Calculation:

1. Consider angle \(\angle APB = 80^\circ\). This is the angle between the two tangent segments from \(P\).

2. The line \(OP\) is the external bisector of \(\angle APB\) for the tangents because triangles \(\triangle OAP\) and \(\triangle OBP\) are congruent, implying \(\angle APO = \angle OPB = \tfrac{1}{2}\angle APB\). Thus each of these half-angles is \(40^\circ\).

3. The central angle \(\angle AOB\) subtended by chord \(AB\) at the centre is related to the angle between tangents: specifically, \(\angle AOB = 180^\circ - \angle APB\). (Reason: quadrilateral \(AO P B\) has two right angles at \(A\) and \(B\), sum of angles in quadrilateral gives this relation; alternatively, using properties of tangent-chord and exterior angles one can derive the same.) For this question we need \(\angle POA\), which is half of \(\angle AOB\) when measured from \(OP\) to each radius. But the simpler direct route is to note that \(OP\) bisects \(\angle AOB\) externally with respect to the tangents; consequently \(\angle POA = \tfrac{1}{2}\angle APB\).

4. Hence \(\angle POA = \dfrac{1}{2} \times 80^\circ = 40^\circ.\)


Alternate concise justification: Since tangents from P are equal, \(OP\) bisects \(\angle APB\), so \(\angle POA = \tfrac{1}{2}\angle APB = 40^\circ\).
Quick Tip: Know these standard facts: OP bisects angle between tangents; angle between tangents = supplementary to central angle subtending the chord of contact.
Often it’s quickest to recall that \(\angle POA = \tfrac{1}{2}\angle APB\) for tangents drawn from an external point.
Draw radii and mark right angles at points of contact to visualize congruent triangles.

Concept: The tangent to a circle is a straight line that touches the circle at exactly one point. The radius drawn to the point of contact is always perpendicular to the tangent at that point.


Explanation:

If \(O\) is the centre of the circle, and \(P\) is the point of contact where the tangent touches the circle, then the line \(OP\) (radius) meets the tangent \(PT\) such that \[ \angle OPT = 90^\circ. \]
This is a fundamental property of circles derived from the definition of a tangent — it touches the circle at one point and does not cut it, meaning the radius is perpendicular at that contact point.


Therefore, the required angle is \(90^\circ\).
Quick Tip: Always remember: a tangent touches a circle at only one point, and the radius at that point is perpendicular to it.
This property is frequently used in geometry, especially when solving problems involving tangents and circles.
It also helps in finding lengths and angles in tangential triangles or when two tangents are drawn from an external point.
Keep in mind the 90° rule — it is a guaranteed geometric relationship for every tangent.

Concept: The area of a circle is given by the formula \(A = \pi r^2\). The ratio of the areas of two circles is equal to the square of the ratio of their radii.


Calculation:

Let the radii of two circles be \(r_1\) and \(r_2\). Given, \[ r_1 : r_2 = 3 : 4 \]
Then, \[ Ratio of areas = r_1^2 : r_2^2 = 3^2 : 4^2 = 9 : 16. \]

Explanation:

Thus, when radii are in ratio \(3:4\), the areas are in ratio \(9:16\), since area increases with the square of the radius.
Quick Tip: When comparing geometrical figures, remember that ratios of areas depend on the \emph{square} of the linear ratio (like radius, side, or diameter).
For circles, \(A \propto r^2\), for squares \(A \propto side^2\), and for spheres, surface area \(\propto r^2\).
Always square the ratio of radii to find the ratio of areas.

Concept: The area of a sector of a circle is given by \[ A = \frac{\theta}{360^\circ} \times \pi r^2, \]
where \(\theta\) is the central angle and \(r\) is the radius of the circle.


Calculation:

Given, \(r = 42~cm\) and \(\theta = 30^\circ\).
\[ A = \frac{30}{360} \times \pi \times 42^2 \] \[ A = \frac{1}{12} \times \pi \times 1764 = 147\pi. \]
Taking \(\pi = \frac{22}{7}\), \[ A = 147 \times \frac{22}{7} = 462~cm^2. \]

Explanation:

The formula proportionally relates the angle to the entire circle. Since \(30^\circ\) is one-twelfth of a full circle, the area of the sector is one-twelfth of the total area of the circle.
Quick Tip: Always express the central angle in degrees when using the formula \(\frac{\theta}{360}\pi r^2\).
For smaller sectors, note that the area scales directly with the angle.
Use \(\pi = \frac{22}{7}\) for exact numeric answers and \(\pi = 3.14\) for approximate ones.
Visualize a circle divided into 360 equal parts to understand the sector proportion.

Concept: Circumference of a circle is \(C = 2\pi r\). Thus circumference is directly proportional to the radius. If \(C_1:C_2 = 5:7\), then \(2\pi r_1 : 2\pi r_2 = 5:7\). Cancelling the common factor \(2\pi\) gives \(r_1:r_2 = 5:7\).


Detailed Reasoning:

Let circumferences be \(C_1 = 2\pi r_1\) and \(C_2 = 2\pi r_2\). Given \(C_1:C_2 = 5:7\). Therefore \[ \frac{2\pi r_1}{2\pi r_2} = \frac{5}{7} \implies \frac{r_1}{r_2} = \frac{5}{7}. \]
Thus the ratio of radii is \(5:7\).
Quick Tip: Circumference and radius are linearly related (\(C \propto r\)), so their ratios are identical.
For area, remember to square the radius ratio.
Cancel common constants early (like \(2\pi\)) to simplify calculations.

Concept: Use the trigonometric identity \(\sec^2 A - \tan^2 A = 1\). Multiply both sides by 7.


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

Explanation: The identity follows from \(\sec^2A = 1 + \tan^2A\). Subtracting \(\tan^2A\) gives 1, hence multiplication by 7 yields 7.
Quick Tip: Memorize basic Pythagorean trig identities: \(\sec^2A = 1 + \tan^2A\), \(\csc^2A = 1 + \cot^2A\).
Look for patterns like \(\sec^2 - \tan^2\) which simplify directly to 1.
Factor common numeric multipliers early to shorten work.

Concept: Substitute given expressions for \(x\) and \(y\) and use the identity \(\sin^2\theta+\cos^2\theta=1\).


Detailed Calculation:

Given \(x = a\cos\theta\) and \(y = b\sin\theta\). Compute: \[ b^2x^2 + a^2y^2 = b^2(a^2\cos^2\theta) + a^2(b^2\sin^2\theta) \] \[ = a^2b^2(\cos^2\theta + \sin^2\theta) = a^2b^2 \times 1 = a^2b^2. \]

Explanation: Factoring \(a^2b^2\) and using \(\sin^2 + \cos^2 = 1\) gives the result directly.
Quick Tip: When expressions involve \(a\cos\theta\) and \(b\sin\theta\), try factoring common \(a^2b^2\) or apply \(\sin^2+\cos^2=1\).
Substitution followed by factoring is an efficient approach for such algebraic-trigonometric expressions.
Double-check algebraic factoring to avoid sign or power errors.

Concept: For angle of elevation \(\theta\), height \(h\) and horizontal distance \(d\) satisfy \(\tan\theta = \dfrac{h}{d}\). Here \(\theta = 60^\circ\), \(d = 10\).


Detailed Calculation:
\[ \tan 60^\circ = \sqrt{3} = \frac{h}{10} \Rightarrow h = 10\sqrt{3}~m. \]

Explanation: Using basic right-triangle trigonometry with tangent yields the tower height directly.
Quick Tip: Remember: \(\tan\theta = \dfrac{opposite}{adjacent}\). For \(\theta=60^\circ\), \(\tan60=\sqrt{3}\).
Always identify which side is opposite (height) and adjacent (horizontal distance) relative to the angle of elevation.
Use standard trig values for 30°, 45°, 60° to compute quickly.

Concept: In the right triangle formed by the kite, the vertical height is opposite to the angle the string makes with the horizontal. If string length = \(s\) and height = \(h\), then \(h = s\sin\theta\). Given \(h = 30\), \(\theta = 60^\circ\).


Detailed Calculation:
\[ 30 = s\sin60^\circ = s \times \frac{\sqrt{3}}{2} \Rightarrow s = \frac{30 \times 2}{\sqrt{3}} = \frac{60}{\sqrt{3}} = 20\sqrt{3}. \]

Explanation: Use sine because the height is opposite the given angle with the horizontal. Rationalize/express simply to get \(20\sqrt{3}\) m.
Quick Tip: Decide whether to use \(\sin\), \(\cos\) or \(\tan\) based on which sides (height/string/distance) are known.
For angle with horizontal, vertical height = string × \(\sin\) (angle).
Keep radicals in simplest form for final answers.

Concept: A quadratic \(ax^2+bx+c=0\) has real and equal roots when its discriminant \(D = b^2 - 4ac = 0\). Here \(a=k\), \(b=-6\), \(c=1\).


Detailed Calculation:

Compute discriminant: \[ D = (-6)^2 - 4(k)(1) = 36 - 4k. \]
Set \(D = 0\) for equal roots: \[ 36 - 4k = 0 \Rightarrow 4k = 36 \Rightarrow k = 9. \]

Verification: With \(k=9\), equation becomes \(9x^2 - 6x + 1 = 0\) which has discriminant zero, hence real and equal roots.
Quick Tip: Use discriminant \(b^2 - 4ac\): = 0 for equal roots, >0 for real distinct, <0 for complex.
Substitute coefficients carefully (watch sign of b).
This test is quick for parameter problems in quadratics.

Concept: If \(r\) is a zero (root) of polynomial \(p(x)\), then \((x-r)\) is a factor of \(p(x)\). This is the Factor Theorem.


Detailed Reasoning:

Given one zero is \(2\). By Factor Theorem, \((x-2)\) divides \(p(x)\) exactly, so \(x-2\) is a factor. None of the other listed linear factors correspond to the root 2.
Quick Tip: Factor Theorem: \(p(r)=0 \iff (x-r)\) is a factor of \(p(x)\).
When given a root, immediately write the corresponding linear factor \((x-r)\).
This is useful for factorization and synthetic division tasks.

Concept: For a quadratic \(cx^2 + ax + b\), the product of its roots \(\alpha \cdot \beta = \frac{c}{b}\)? Let's check carefully.


Detailed Reasoning:

Standard formula: For quadratic \(ax^2 + bx + c = 0\), product of roots = \(c/a\) and sum of roots = \(-b/a\). Here, \(cx^2 + ax + b = 0\), so comparing with standard form \(ax^2 + bx + c\), \(a \to c, b \to a, c \to b\). Hence, \[ \alpha \cdot \beta = \frac{c}{a} = \frac{b}{c}. \]

Explanation: Using the standard relation carefully ensures correct identification of coefficients.
Quick Tip: Always compare the given quadratic with standard form \(ax^2+bx+c=0\) carefully.
Remember: product of roots = constant term / leading coefficient.
Check the positions of coefficients to avoid mistakes with non-standard ordering.

Concept: A quadratic equation has degree 2. Expand each option to identify degree.


Detailed Reasoning:

Option 1: \((x+3)(x-3) = x^2-9\), but RHS \(x^2-4x^3\) has degree 3 → not quadratic.

Option 2: \((x+3)^2 = x^2 +6x +9\), RHS \(4(x+4) = 4x +16\), subtract \(4x+16\) to bring to standard form: \(x^2 +2x -7=0\) → degree 2 → quadratic.

Option 3: \((2x-2)^2 = 4x^2-8x+4\), RHS \(4x^2+7\), subtract RHS: \(4x^2-8x+4 - 4x^2 -7 = -8x -3=0\) → degree 1 → not quadratic.

Option 4: \(4x + 1/(4x) = 4x\) → contains \(1/x\) term → not polynomial → not quadratic.


Hence, correct option is 2.
Quick Tip: Expand all expressions to bring to standard form \(ax^2+bx+c=0\).
Check the highest power of \(x\) after simplification to determine degree.
Ignore equations containing non-polynomial terms like \(1/x\).

Concept: A quadratic equation has degree 2. Expand each option to check degree.


Detailed Reasoning:

Option 1: \(5x-x^2=x^2+3 \implies -2x^2+5x-3=0\) → degree 2 → quadratic.

Option 2: \(x^3-x^2 = (x-1)^3 = x^3-3x^2+3x-1 \implies x^3-x^2 - x^3 +3x^2 -3x +1=0 \implies 2x^2-3x+1=0\) → degree 2 after simplification? Wait carefully: LHS: \(x^3 - x^2\), RHS: \(x^3 -3x^2 +3x -1\), subtract RHS: \(x^3 - x^2 - (x^3-3x^2+3x-1) = 2x^2 -3x +1\) → degree 2 → actually quadratic. Hmm, let's check carefully.

Yes, after simplification, it becomes \(2x^2-3x+1=0\) → degree 2 → quadratic.

Option 3: \((x+3)^2 - 3(x^2-5) = x^2+6x+9 -3x^2 +15 = -2x^2 +6x +24=0\) → degree 2 → quadratic.

Option 4: \((\sqrt{2}x+3)^2 -2x^2 -5 = 2x^2+6\sqrt{2}x +9 -2x^2 -5 =6\sqrt{2}x +4=0\) → degree 1 → not quadratic.


Hence, correct option is 4.
Quick Tip: After moving all terms to one side, the highest power of \(x\) determines degree.
Quadratic → degree 2, cubic → degree 3, linear → degree 1.
Carefully expand and simplify before deciding degree.

Concept: Discriminant \(D = b^2 - 4ac\) for quadratic \(ax^2+bx+c=0\). Here \(a=2\), \(b=-7\), \(c=6\).


Detailed Calculation:
\[ D = (-7)^2 - 4(2)(6) = 49 - 48 = 1. \]

Explanation: Discriminant is positive, hence roots are real and distinct.
Quick Tip: Discriminant tells nature of roots: \(D>0\) → real & distinct, \(D=0\) → real & equal, \(D<0\) → complex.
Carefully identify coefficients \(a\), \(b\), \(c\) from standard form.
Square negative numbers correctly when computing \(b^2\).

Concept: Graph of \(x = 2\) is a vertical line passing through \(x=2\). All points on this line have x-coordinate 2.


Explanation: Points \((2,0)\), \((2,1)\), \((2,2)\) all satisfy \(x=2\), hence all lie on the graph.
Quick Tip: Vertical lines have equation \(x = k\); all points on them share the same x-coordinate.
Horizontal lines have equation \(y = k\); all points share the same y-coordinate.
Check coordinates to determine if point lies on a line.

Concept: For three numbers in A.P., middle term ×2 = sum of first and last.


Detailed Calculation:
\[ 2(2P+1) = (P+1) + (4P-1) \implies 4P+2 = 5P \implies P=2. \]

Explanation: Applying A.P. property directly gives the value of P.
Quick Tip: Check the A.P. property: \(2 \times\) middle term = sum of extremes.
Substitute carefully and solve simple linear equations.

Concept: Common difference \(d\) of A.P. = difference between any consecutive terms.


Calculation: \(d = 5-1 =4\).


Explanation: Difference between consecutive terms is constant in A.P., hence \(d=4\).
Quick Tip: Subtract consecutive terms to find common difference.
Check multiple consecutive pairs if necessary to verify A.P.

Concept: \(n\)th term of A.P. \(a_n = a + (n-1)d\). Here \(a=5\), \(d=3\).


Calculation: \[ 5 + (n-1)3 = 38 \implies 3(n-1)=33 \implies n-1=11 \implies n=12. \]

Explanation: 12th term of the sequence is 38.
Quick Tip: Use formula \(a_n = a + (n-1)d\) and solve for \(n\).
Check arithmetic carefully for correct term identification.

Concept: Volume \(V \propto r^3\), Surface area \(S \propto r^2\).


Calculation: \[ \frac{V_1}{V_2} = \frac{r_1^3}{r_2^3} = \frac{64}{125} \implies \frac{r_1}{r_2} = \frac{4}{5}. \] \[ \frac{S_1}{S_2} = \frac{r_1^2}{r_2^2} = \left(\frac{4}{5}\right)^2 = \frac{16}{25}. \]

Explanation: Surface area ratio is square of radii ratio derived from volume ratio.
Quick Tip: For spheres: \(V \propto r^3\), \(S \propto r^2\).
Volume ratio → cube root → radius ratio → square → surface area ratio.

Concept: Volume of cylinder \(V = \pi r^2 h\). Given \(r_1:r_2 = 4:5\), \(h_1:h_2 = 6:7\).


Calculation: \[ \frac{V_1}{V_2} = \frac{r_1^2 h_1}{r_2^2 h_2} = \frac{4^2 \cdot 6}{5^2 \cdot 7} = \frac{16 \cdot 6}{25 \cdot 7} = \frac{96}{175}. \]

Explanation: Cylinder volume ratio depends on square of radius ratio multiplied by height ratio.
Quick Tip: Volume of cylinder: \(V=\pi r^2 h\).
Square the radius ratio, multiply by height ratio to get volume ratio.
Always simplify fractions carefully.

Concept: Total surface area (TSA) of a hemisphere includes the curved surface area (CSA) and base area.


Detailed Calculation:

Curved surface area of hemisphere = \(2\pi R^2\), base area = \(\pi R^2\).

Total surface area = \(2\pi R^2 + \pi R^2 = 3\pi R^2\).


Explanation: TSA accounts for both the curved portion and flat circular base.
Quick Tip: Hemisphere TSA = CSA + base area.
CSA of hemisphere = \(2\pi R^2\), base = \(\pi R^2\).
Always include the base when calculating TSA.

Concept: Curved surface area (CSA) of cone = \(\pi r l\), where \(r\) = radius, \(l\) = slant height.


Calculation:
\[ \pi r l = 880 \implies 3.14 \times 14 \times l = 880 \implies l = \frac{880}{43.96} \approx 20 cm. \]

Explanation: Solving the CSA formula gives the slant height directly.
Quick Tip: CSA of cone = \(\pi r l\).
Given CSA and radius, divide CSA by \(\pi r\) to get slant height.

Concept: Diagonal of cube \(d = a\sqrt{3}\), where \(a\) = edge of cube.


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

Explanation: Diagonal formula relates cube edge to space diagonal.
Quick Tip: Diagonal of cube = \(a\sqrt{3}\).
Given diagonal, divide by \(\sqrt{3}\) to find edge.

Concept: TSA of cube = \(6a^2\), where \(a\) = edge.


Calculation:

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

Old TSA = \(6a^2\), ratio = \(24a^2 : 6a^2 = 4:1\).


Explanation: Doubling edge increases surface area by factor of square of 2.
Quick Tip: TSA depends on square of edge.
Doubling edge → surface area × 4.

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


Calculation:
\[ \frac{TSA of sphere}{TSA of hemisphere} = \frac{4\pi R^2}{3\pi R^2} = \frac{4}{3}. \]

Explanation: Ratio is simple division of TSA values.
Quick Tip: TSA sphere = \(4\pi R^2\), TSA hemisphere = \(3\pi R^2\).
For ratio, divide sphere TSA by hemisphere TSA.

Concept: Curved surface area of hemisphere = \(2\pi R^2\).


Calculation:
\[ 2\pi R^2 = 1232 \implies R^2 = \frac{1232}{2 \cdot 3.14} \approx 196 \implies R \approx 14 cm. \]

Explanation: Solve CSA formula to find radius.
Quick Tip: CSA hemisphere = \(2\pi R^2\).
Divide given CSA by \(2\pi\) and take square root for radius.

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


Calculation:

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

Oops! Wait, let's double-check: \(\cos\theta + \cos^2\theta=1\) → yes \(\cos\theta + \cos^2\theta =1\), so \(\sin^2\theta = 1-\cos^2\theta = 1-(1-\cos\theta)=\cos\theta\) → \(\sin^4\theta = (\cos\theta)^2 = \cos^2\theta\) → sum = \(\cos\theta+\cos^2\theta=1\).

Corrected: sum = 1.
Quick Tip: Use identity \(\sin^2\theta = 1 - \cos^2\theta\) carefully.
Substitute given equation to simplify expressions.

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


Calculation:
\[ \frac{1+\tan^2A}{1+\cot^2A} = \frac{\sec^2A}{\csc^2A} = \frac{1/\cos^2A}{1/\sin^2A} = \frac{\sin^2A}{\cos^2A} = \tan^2A? \]

Wait, carefully: \(\frac{\sec^2A}{\csc^2A} = \frac{1/\cos^2A}{1/\sin^2A} = \frac{\sin^2A}{\cos^2A} = \tan^2A\) → correct. So correct answer should be \(\tan^2A\) (4). Quick Tip: Use \(1+\tan^2A = \sec^2A\), \(1+\cot^2A = \csc^2A\).
Simplify step by step to avoid mistakes.

Concept: Complementary angle identity: \(\sin(90^\circ - A) = \cos A\).


Explanation: Direct trigonometric identity.
Quick Tip: Complementary angles: \(\sin(90^\circ - \theta) = \cos \theta\), \(\cos(90^\circ - \theta) = \sin \theta\).
Memorize these fundamental identities.

Concept: \(\cos(\alpha - \beta) = \cos 0 = 1\), since \(\alpha = \beta\).


Explanation: Cosine of zero degrees is 1.
Quick Tip: Use formula \(\cos(A-B)\) and substitute values carefully.
If angles are equal, \(\cos(A-A) = 1\).

Concept: \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\).


Calculation:
\[ \sin\theta+\cos\theta = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}. \]

Explanation: Sum of sine and cosine for 45° is simply doubled value of either.
Quick Tip: For special angles like \(30^\circ, 45^\circ, 60^\circ\), memorize sine and cosine values.
Adding or subtracting these values often appears in questions.

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


Calculation:
\[ \tan 2A = \tan 2(30^\circ) = \tan 60^\circ. \]

Explanation: Use tangent double angle identity directly.
Quick Tip: Remember identity: \(\tan 2A = \frac{2\tan A}{1-\tan^2A}\).
Substitute angle carefully for exact value.

Concept: \(\tan\theta = \frac{opposite}{adjacent}\). Use Pythagoras to find hypotenuse.


Calculation:

Let opposite = 12, adjacent = 5, hypotenuse \(= \sqrt{12^2+5^2} = \sqrt{144+25} = \sqrt{169} = 13\).
\[ \sin\theta = \frac{opposite}{hypotenuse} = \frac{12}{13}. \]

Explanation: Use triangle definition for sine from tangent ratio.
Quick Tip: Draw a right triangle when given tangent or cotangent.
Use Pythagoras to compute other trigonometric ratios.

Concept: Use complementary angle identities: \(\cos 59^\circ = \sin 31^\circ\), \(\tan 80^\circ = \cot 10^\circ\).


Calculation:
\[ \frac{\cos 59^\circ}{\sin 31^\circ} \times \frac{\tan 80^\circ}{\cot 10^\circ} = \frac{\sin 31^\circ}{\sin 31^\circ} \times \frac{\cot 10^\circ}{\cot 10^\circ} = 1. \]

Explanation: Simplify each term using basic trigonometric properties.
Quick Tip: Check if angles are complementary or co-functions.
Use identities: \(\cos(90^\circ-\theta)=\sin\theta\), \(\tan(90^\circ-\theta)=\cot\theta\).

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


Calculation:
\[ \tan 25^\circ \cdot \tan 65^\circ = 1 \implies \sin A = 1 \implies A = 90^\circ? \]

Check carefully: \(\sin A = 1\) → \(A = 90^\circ\). So correct answer should be (3) \(90^\circ\). Quick Tip: Use complementary angle identity: \(\tan\theta \cdot \tan(90^\circ-\theta)=1\).
Check whether the final value equals 1 for sine or cosine carefully.

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


Calculation:
\[ \tan \theta = \frac{\sqrt{1-x^2}}{x}. \]

Explanation: Use Pythagorean identity to find sine from cosine.
Quick Tip: Pythagorean identity: \(\sin^2\theta + \cos^2\theta = 1\).
Express unknown trig ratios in terms of given ones.

Concept: Factor using difference of squares: \(1-\cos^4\theta = (1-\cos^2\theta)(1+\cos^2\theta)\).


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

Explanation: Use \(1-\cos^2\theta = \sin^2\theta\).
Quick Tip: Always try difference of squares factorization for expressions like \(1-\cos^4\theta\).
Then apply Pythagorean identity to simplify further.

Concept: Points on y-axis have x-coordinate 0, y can be any value.


Calculation:
\[ Point on y-axis: (0, y) or (0, x) as per given notation. \]

Explanation: X-coordinate = 0 for all points on y-axis.
Quick Tip: For points on axes: x-axis → y=0, y-axis → x=0.
Always check which coordinate is fixed and which is variable.

Concept: If α and β are zeros, polynomial = \((x-\alpha)(x-\beta)\).


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

Explanation: Expand (x - zero1)(x - zero2) to find quadratic polynomial.
Quick Tip: Use factorization formula: \((x-zero1)(x-zero2)\) for any quadratic with given zeros.

Concept: For quadratic \(x^2 - (sum~of~zeros)x + (product~of~zeros)\).


Calculation:

Sum of zeros = 3 → coefficient of x = -3.

Product of zeros = -2 → constant term = -2.
\[ Quadratic: x^2 - 3x - 2 \]

Explanation: Directly use standard relation between zeros and coefficients.
Quick Tip: For quadratic polynomial with zeros α, β: \(x^2-(\alpha+\beta)x + \alpha\beta\).
Check signs carefully based on sum and product.

Concept: Degree of the quotient = degree of dividend - degree of divisor.


Calculation:

Degree of \(p(x)\) = 4, degree of \(q(x)\) = 1.

Degree of quotient = \(4 - 1 = 3\).

Explanation: Subtract the degree of divisor from dividend to find quotient degree.
Quick Tip: Degree of polynomial quotient = degree(dividend) - degree(divisor).
Remember, this is valid if divisor degree > 0.

Concept: Two equations are proportional if coefficients are multiples → infinitely many solutions.


Calculation:
\(3(x+2y+3) = 3x + 6y + 9\) → same as second equation.


Explanation: Since second equation is a multiple of the first, the lines coincide → infinitely many solutions.
Quick Tip: Check ratios of coefficients.
If ratios of \(x, y, \) and constant terms are equal → infinitely many solutions.

Concept: Parallel lines have same slope but different intercept → no intersection → no solution.


Calculation:

Slope = m, y-intercepts different → lines never meet.

Explanation: Parallel linear equations → inconsistent → no solution.
Quick Tip: Compare slopes: equal slopes, different intercepts → no solution.
Equal slopes, equal intercepts → infinitely many solutions.

Concept: Solve by elimination or substitution to check if unique solution exists.


Calculation:

Solve \(5x-4y=-8\) and \(7x+6y=9\).

Multiply first by 3, second by 2: \(15x-12y=-24\), \(14x+12y=18\).

Add: \(29x=-6 \implies x=-6/29\), then \(y=(-8-5x)/(-4)=(-8+30/29)/(-4) = (-202/29)/(-4)= 202/(116)=101/58\).

Explanation: Unique solution exists → equations are consistent.
Quick Tip: Use elimination/substitution to check consistency.
Consistent → at least one solution; dependent → infinitely many; inconsistent → no solution.

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


Calculation:

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

Explanation: Use standard relations between roots and coefficients.
Quick Tip: Formulas: \(\alpha+\beta=-\frac{b}{a}\), \(\alpha\beta=\frac{c}{a}\).
\(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\).

Concept: Substitute the root into equation to find unknown constant.


Calculation:
\[ 2(2)^2 - 7(2) - p = 0 \implies 8-14-p=0 \implies -6-p=0 \implies p=-6. \]

Check: Correct value is p=-6 → matches calculation.

Explanation: Always plug in known root to solve for unknown.
Quick Tip: Substitute given root into equation directly.
Solve for unknown coefficient.

Concept: Sum of roots formula: \(\alpha + \beta = -\frac{b}{a}\), product: \(\alpha \beta = \frac{c}{a}\).


Calculation:

Here, \(a=2\), \(b=-1\), \(c=-6\). Sum of roots \(\alpha+\beta = \frac{1}{2}\).

Given \(\alpha = -\frac{3}{2}\), then \(\beta = \frac{1}{2} - (-3/2) = 2\).

Explanation: Use sum of roots to find the other root when one root is known.
Quick Tip: For quadratic \(ax^2+bx+c=0\), sum = \(-b/a\), product = \(c/a\).
Substitute known root to calculate the other root.

Concept: Nature of roots depends on discriminant \(D=b^2-4ac\).


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

Explanation: Positive discriminant → roots are real and unequal.
Quick Tip: Check discriminant: \(D>0\) → real and unequal, \(D=0\) → real and equal, \(D<0\) → non-real.

Concept: Class interval = Upper limit - Lower limit.


Calculation:
\(2-5 \implies 5-2=3\), \(5-8 \implies 8-5=3\), \(8-11 \implies 11-8=3\).

Explanation: All intervals are equal → length = 3.
Quick Tip: Subtract lower class boundary from upper class boundary to get class interval.

Concept: Consecutive odd numbers: \(n, n+2, n+4, n+6\). Mean = sum/4.


Calculation:
\[ \frac{n + (n+2) + (n+4) + (n+6)}{4} = 6 \implies \frac{4n+12}{4}=6 \implies n+3=6 \implies n=3 \]
Largest number = \(n+6=3+6=9\).

Explanation: For consecutive odd numbers, difference = 2. Sum = 4 numbers → divide by 4 for mean.
Quick Tip: For k consecutive odd numbers: first = n, then n+2, n+4,...
Use mean = sum/number of terms to find any number.

Concept: Mean = Sum of numbers / Number of numbers.


Calculation:

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

Sum = 2+4+6+8+10+12 = 42.

Mean = 42/6 = 7.

Explanation: Mean is the arithmetic average of the numbers.
Quick Tip: Add all given numbers and divide by the count to find the mean.

Concept: Pythagorean identity: \(1+\cot^2\theta = \csc^2\theta\).


Calculation:

Using \(\cot^2\theta + 1 = \csc^2\theta\).

Explanation: This is a standard trigonometric identity.
Quick Tip: Remember the Pythagorean identities: \(1+\tan^2\theta = \sec^2\theta\), \(1+\cot^2\theta = \csc^2\theta\).

Concept: Mode is the value appearing most frequently.


Calculation:

Count of each number: 3→1, 4→1, 5→3, 7→2, 8→1, 9→2.

Most frequent = 5 (occurs 3 times).

Explanation: Mode is the number with highest frequency.
Quick Tip: Count occurrences of each number; the highest frequency is the mode.

Concept: Complementary probability: \(P(E') = 1-P(E)\).


Calculation:
\(P(E') = 1-0.02 = 0.98\).

Explanation: Probability of the event not occurring = 1 minus probability of event occurring.
Quick Tip: Use \(P(E')=1-P(E)\) for complementary events.

Concept: Difference = 0 → numbers are equal.


Calculation:

Total outcomes = 6×6 = 36.

Equal numbers: (1,1),(2,2),...,(6,6) = 6 outcomes.

Probability = 6/36 = 1/6.

Explanation: Equal numbers = zero difference.
Quick Tip: When finding probability of "same number" on dice, count outcomes with equal faces and divide by total outcomes.

Concept: Two coins, independent events, probability of both heads = P(H)×P(H).


Calculation:

P(H) = 1/2 each coin.

Probability = 1/2 × 1/2 = 1/4.

Explanation: Multiply probabilities of independent events.
Quick Tip: For independent events, probability of both occurring = product of individual probabilities.

Concept: Total months = 12. Favorable = June, September = 2.


Calculation:

Probability = 2/12 = 1/6.

Explanation: Count favorable months / total months.
Quick Tip: When selecting one item at random, probability = favorable outcomes / total outcomes.

Concept: Favorable outcomes = 4,5 → 2 outcomes. Total outcomes = 6.


Calculation:

Probability = 2/6 = 1/3.

Explanation: Count favorable faces on the die / total faces.
Quick Tip: Probability = number of favorable outcomes / total outcomes.

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


Calculation:

5th term = a + 4d = 11.

d = 2 → a + 4(2) = 11 → a + 8 = 11 → a = 3.

Explanation: Use n-th term formula to find first term.
Quick Tip: n-th term of A.P.: \(a_n = a + (n-1)d\). Substitute values to find unknowns.

Concept: \(n\)-th term \(t_n = S_n - S_{n-1}\).


Calculation:
\(S_n = n^2 + 2n +1\), \(S_5 = 25 + 10 +1=36\).
\(S_6 = 36 + 12 +1=49\)? Wait calculate correctly: \(S_6 = 36 + 12 +1? = 49\) (check)
\(S_6 = 6^2 +2*6 +1 = 36+12+1=49\), \(S_5=25+10+1=36\), \(t_6 = S_6 - S_5 = 49-36 = 13\).

Actually correct 6th term = 13. The options given may differ. Use \(t_n=S_n-S_{n-1}\).

Explanation: Use formula \(t_n = S_n - S_{n-1}\) for n-th term from sum.
Quick Tip: To find n-th term from sum: \(t_n = S_n - S_{n-1}\). Always calculate carefully.

Concept: A.P. → difference between consecutive terms is constant.


Calculation:

Option 3: Differences: 2x-x = x, 3x-2x=x, 4x-3x=x → constant difference x → A.P.

Explanation: Only option with constant difference is in A.P.
Quick Tip: Check consecutive differences. Constant difference → A.P.

Concept: Difference must be constant for A.P.


Calculation:

Option 4: Differences: 16-4=12, 36-16=20 → not constant → not A.P.

Explanation: Squares of even numbers are not in A.P.
Quick Tip: Subtract consecutive terms; if difference not constant → not A.P.

Concept: Sum of n terms: \(S_n = n/2 [2a + (n-1)d]\).


Calculation:

a=1, d=3, n=20
\(S_{20}=20/2 [2*1 + 19*3] =10[2+57]=10*59=590\) → wait options: correct sum = 590


Yes, correct sum = 590. So option (3) 590.

Explanation: Use sum formula carefully.
Quick Tip: Check a, d, n carefully, apply formula: \(S_n = n/2[2a + (n-1)d]\).

Concept: \(\sin 45^\circ = \cos 45^\circ = 1/\sqrt{2}\)


Calculation:
\(\sin 45^\circ * (1/\cos 45^\circ) = (1/\sqrt{2})*(\sqrt{2}/1) =1\)

Explanation: Trigonometric simplification.
Quick Tip: Check trigonometric values at special angles and simplify fractions.

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


Calculation:
\(\cos^2A*(1+\tan^2A) = \cos^2A * \sec^2A = \cos^2A * (1/\cos^2A) =1\)

Explanation: Apply identity \(1+\tan^2A = \sec^2A\).
Quick Tip: Use trigonometric identities carefully; simplify step by step.

Concept: Standard trigonometric value.


Calculation:
\(\tan 30^\circ = 1/\sqrt{3}\).

Explanation: Use the standard values of \(\tan\) at special angles.
Quick Tip: Remember \(\tan 30^\circ = 1/\sqrt{3}, \tan 45^\circ =1, \tan 60^\circ = \sqrt{3}\).

Concept: Standard trigonometric value for \(\cos 60^\circ\).


Calculation:

From the unit circle or equilateral triangle, \(\cos 60^\circ = \frac{1}{2}\).


Explanation: \(\cos 60^\circ = \frac{adjacent}{hypotenuse} = \frac{1}{2}\) in a 30-60-90 triangle.
Quick Tip: Memorize standard angles: \(0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ\).

Concept: Use standard values: \(\sin 90^\circ = 1\), \(\tan 45^\circ = 1\).


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


Explanation: Both terms are 1, so difference is 0.
Quick Tip: Direct substitution of known values simplifies trigonometric expressions.

Concept: A fraction \(\frac{p}{q}\) in lowest terms has terminating decimal if \(q\) has prime factors only 2 and/or 5.


Calculation:

(A) Denominator: \(3^2\) → non-terminating.
(B) Denominator: \(5 \times 7^2\) → has 7 → non-terminating.
(C) Denominator: \(2^2 \times 3^2\) → has 3 → non-terminating.
(D) Denominator: \(2^2 \times 5^3\) → only 2 and 5 → terminating.


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

Concept: \(0.505 = \frac{505}{1000}\). Simplify and express with denominator \(2^n 5^m\).


Calculation:
\(0.505 = \frac{505}{1000}\). \(1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3\). \(\frac{505}{1000} = \frac{505}{2^3 \times 5^3}\).
But 505 = 5 × 101, so \(\frac{5 \times 101}{2^3 \times 5^3} = \frac{101}{2^3 \times 5^2}\).
Wait: Cancel one 5: \(\frac{101}{2^3 \times 5^{3-1}} = \frac{101}{2^3 \times 5^2}\).
But option (D). Wait, recheck. \(0.505 = 505/1000\), but 505 ÷ 5 = 101, 1000 ÷ 5 = 200? No.
Better: \(0.505 = \frac{505}{1000} = \frac{101}{200}\), since divide numerator and denominator by 5. \(200 = 2^3 \times 5^2\)? 200 = 2 × 100 = 2 × \(2^2\) × \(5^2\) = \(2^3\) × \(5^2\). Yes.
But \(\frac{101}{200} = \frac{101}{2^3 \times 5^2}\), which is (D).
Wait, but 0.505 = 505/1000 = 101/200, yes.
But let's check value: 101/200 = 0.505, yes.
Now options:
(A) 101/(2×25) = 101/50 = 2.02, no.
(B) 101/(2×125)=101/250≈0.404, no.
(C) 101/(4×25)=101/100=1.01, no.
(D) 101/(8×25)=101/200=0.505, yes.

Explanation: \(0.505 = \frac{101}{200} = \frac{101}{2^3 \times 5^2}\).
Quick Tip: Multiply numerator and denominator by power of 10 to eliminate decimal, then simplify.

Concept: Division algorithm: \(a = bq + r\), where \(0 \leq r < b\).


Calculation:
\(a = 4 \times 5 + 1 = 20 + 1 = 21\).


Explanation: Dividend = divisor × quotient + remainder.
Quick Tip: Direct substitution into \(a = bq + r\).

Concept: Solve \(2x^2 - 4x - 6 = 0\) by factoring or quadratic formula.


Calculation:

Divide by 2: \(x^2 - 2x - 3 = 0\).
Factor: \((x - 3)(x + 1) = 0\). \(x = 3\) or \(x = -1\).


Explanation: Roots are \(x = 3, -1\).
Quick Tip: Divide by leading coefficient to simplify factoring.

Concept: Degree of product = sum of degrees of factors.


Calculation:

First polynomial: degree 3,
Second: degree 2,
Product: degree \(3 + 2 = 5\).


Explanation: Leading term: \(x^3 \cdot x^2 = x^5\).
Quick Tip: Add degrees when multiplying polynomials.

Concept: Polynomial has non-negative integer exponents.


Calculation:

(A), (B), (C): all terms have integer exponents ≥ 0.
(D): \(\frac{4}{x} = 4x^{-1}\) → negative exponent → not a polynomial.


Explanation: Negative exponent violates polynomial definition.
Quick Tip: Check for negative or fractional exponents.

Concept: For roots \(\alpha, \beta\), polynomial is \((x - \alpha)(x - \beta)\).


Calculation:

Roots 2, -2: \((x - 2)(x + 2) = x^2 - 4\).
Matches option (B).


Explanation: \(x^2 - 4 = (x-2)(x+2)\), roots 2 and -2.
Quick Tip: Sum of roots = 0 → no \(x\) term → \(x^2 + c\).

Concept: For \(at^2 + bt + c = 0\), sum of roots = \(-\frac{b}{a}\).


Calculation:

Here, \(a = 1\), \(b = 7\), \(\alpha + \beta = -\frac{7}{1} = -7\).


Explanation: Sum of zeroes is negative of coefficient of \(t\).
Quick Tip: Sum of roots = \(-\frac{b}{a}\), product = \(\frac{c}{a}\).

Concept: Use standard values.


Calculation:
\(\sin 30^\circ = \frac{1}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), sum = \(\frac{1 + \sqrt{3}}{2}\). \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\cos 60^\circ = \frac{1}{2}\), sum = \(\frac{\sqrt{3} + 1}{2}\).
Difference: \(\frac{1 + \sqrt{3}}{2} - \frac{1 + \sqrt{3}}{2} = 0\).


Explanation: Both expressions are equal.
Quick Tip: Group terms to spot identical expressions.

Concept: Substitute root \(x = -4\) into the polynomial and set to zero.


Calculation:
\((k-1)(-4)^2 + k(-4) + 1 = 0\), \((k-1)(16) - 4k + 1 = 0\), \(16k - 16 - 4k + 1 = 0\), \(12k - 15 = 0\), \(12k = 15\), \(k = \frac{15}{12} = \frac{5}{4}\).
Wait: But option (B).
Wait: Recompute: \(16(k - 1) - 4k + 1 = 0\), \(16k - 16 - 4k + 1 = 0\), \(12k - 15 = 0\), \(k = \frac{15}{12} = \frac{5}{4}\).
But answer should be negative? Wait, let's check by plugging.
If \(k = \frac{5}{4}\), then coefficient: \(k-1 = \frac{5}{4} - 1 = \frac{1}{4}\).
Polynomial: \(\frac{1}{4} x^2 + \frac{5}{4} x + 1\).
At \(x = -4\): \(\frac{1}{4}(16) + \frac{5}{4}(-4) + 1 = 4 - 5 + 1 = 0\). Yes.
But options include negative. Wait, question says one zero is -4.
It works for \(k = \frac{5}{4}\).
But answer marked as (A) negative. Wait, perhaps typo.
Wait: Let's use product of roots.
Roots: -4 and \(\beta\), product = \(\frac{1}{k-1}\).
Sum = \(-4 + \beta = -\frac{k}{k-1}\).
But easier: substitute.
We got \(k = \frac{5}{4}\).
But let's see if \(k = -\frac{5}{4}\): \(k-1 = -\frac{5}{4} -1 = -\frac{9}{4}\),
At x=-4: \(-\frac{9}{4}(16) + (-\frac{5}{4})(-4) +1 = -36 + 5 +1 = -30 \neq 0\). No.
So correct is \(k = \frac{5}{4}\), option (B).
But earlier said (A). Wait, mistake in initial.
Wait: Let's solve again. \((k-1)(16) -4k +1 =0\),
16k -16 -4k +1=0,
12k -15=0,
k=15/12=5/4. Yes.

Explanation: Substituting \(x = -4\) gives \(k = \frac{5}{4}\).
Quick Tip: Substitute the given root directly into the polynomial.

Concept: Section formula for internal division in ratio \(m:n\): \[ \left( \dfrac{m x_2 + n x_1}{m+n}, \dfrac{m y_2 + n y_1}{m+n} \right). \]

Calculation:

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

Explanation: The point is \(\left(1, \dfrac{27}{5}\right)\).
Quick Tip: Section formula gives weighted average; ratio \(m:n\) means \(m\) parts toward second point, \(n\) parts toward first.

Concept: Area using shoelace formula: \[ \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|. \]

Calculation:

Let \(A(-5,-1)\), \(B(3,5)\), \(C(5,2)\). \[ Area = \frac{1}{2} \left| -5(5-2) + 3(2-(-1)) + 5(-1-5) \right| \] \[ = \frac{1}{2} \left| -15 + 9 - 30 \right| = \frac{1}{2} \left| -36 \right| = 18. \]

Explanation: Shoelace formula yields area \(18\) square units.
Quick Tip: List coordinates in order and repeat first at end for shoelace method.

Concept: Space diagonal of cube = \(a\sqrt{3}\), surface area = \(6a^2\).

Calculation:

Given \(a\sqrt{3} = 9\sqrt{3}\), \[ a = 9 cm. \]
Surface area = \(6 \times 9^2 = 6 \times 81 = 486\) cm².

Explanation: Side length \(9\) cm gives total surface area \(486\) cm².
Quick Tip: Memorize cube formulas: diagonal \(a\sqrt{3}\), surface \(6a^2\), volume \(a^3\).

Concept: Form right triangle with opposite = 5, adjacent = 12.
Hypotenuse = \(\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\).

Calculation: \[ \sin \theta = \dfrac{5}{13}, \quad \cos \theta = \dfrac{12}{13}, \] \[ \sin \theta + \cos \theta = \dfrac{5}{13} + \dfrac{12}{13} = \dfrac{17}{13}. \]

Explanation: Sum of sine and cosine is \(\dfrac{17}{13}\).
Quick Tip: Use 5-12-13 Pythagorean triplet for quick recognition.

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

Calculation: \[ \sin 3A = \cos(90^\circ - 3A), \] \[ A - 26^\circ = 90^\circ - 3A \quad (since \cos is co-function of \sin), \] \[ A + 3A = 90^\circ + 26^\circ, \quad 4A = 116^\circ, \quad A = 29^\circ. \]
Verification: \(3A = 87^\circ\) (acute), \(\sin 87^\circ = \cos 3^\circ\), and \(A - 26^\circ = 3^\circ\).

Explanation: \(A = 29^\circ\) satisfies the equation.
Quick Tip: Rewrite using \(\sin \theta = \cos(90^\circ - \theta)\) and equate arguments.

Concept: Let the smaller number be \(x\), then the other is \(\dfrac{7}{3}x\). Their sum is 50.

Calculation: \[ x + \dfrac{7}{3}x = 50 \] \[ \dfrac{3x + 7x}{3} = 50 \quad \Rightarrow \quad \dfrac{10x}{3} = 50 \] \[ 10x = 150 \quad \Rightarrow \quad x = 15 \]
Other number = \(\dfrac{7}{3} \times 15 = 35\).

Explanation: The numbers are \(15\) and \(35\).
Quick Tip: Use one variable and express the second in terms of it using the given ratio.

Concept: The ladder forms a right triangle with the wall. Height = opposite side to \(30^\circ\).

Calculation: \[ \sin 30^\circ = \dfrac{height}{ladder} \quad \Rightarrow \quad \dfrac{1}{2} = \dfrac{h}{7} \] \[ h = \dfrac{7}{2} = 3.5 m \quad or \quad \dfrac{7}{2} m \]
But wait: angle with wall is \(30^\circ\), so angle with ground is \(60^\circ\).
Ladder is hypotenuse. Angle with wall = \(30^\circ\) → angle with ground = \(60^\circ\).
Height = adjacent to \(30^\circ\) = \(\cos 30^\circ \times 7\). \[ \cos 30^\circ = \dfrac{\sqrt{3}}{2}, \quad h = 7 \times \dfrac{\sqrt{3}}{2} = \dfrac{7\sqrt{3}}{2} m. \]

Explanation: Height is \(\dfrac{7\sqrt{3}}{2}\) m.
Quick Tip: Identify the angle with the wall carefully; use \(\cos\) for adjacent (height).

Concept: Use properties of parallelogram and alternate angles.

Calculation:

In \(\parallelogram ABCD\): \(AB \parallel CD\), \(AD \parallel BC\). \(E\) is on extension of \(AD\) beyond \(D\), so \(A-D-E\). \(BE\) intersects \(CD\) at \(F\).
\(\angle BAE = \angle DCF\) (alternate interior angles, \(AB \parallel CD\), transversal \(AE\)). \(\angle ABE = \angle CBF\) (alternate interior angles, \(AB \parallel CD\), transversal \(BE\)).
Thus, \(\angle BAE = \angle DCF\), \(\angle ABE = \angle CBF\).
By AA similarity: \(\triangle ABE \sim \triangle CBF\).
(Note: \(\triangle CFB\) is same as \(\triangle CBF\)).

Explanation: Two pairs of alternate angles equal imply similarity.
Quick Tip: In parallelograms, parallel lines create alternate interior angles with transversals.

Concept: Isosceles right triangle: \(AC = BC\), \(\angle C = 90^\circ\).

Calculation:
By Pythagoras: \[ AB^2 = AC^2 + BC^2 \]
Since \(AC = BC\), \[ AB^2 = AC^2 + AC^2 = 2AC^2. \]

Explanation: Hypotenuse squared equals twice leg squared.
Quick Tip: In 45-45-90 triangle, hypotenuse = \(leg \sqrt{2}\), so squared: \(2 \times leg^2\).

Concept: \(AB = AC\), \(AD \perp BC\) (altitude and median), \(E\) on \(CB\) extended, \(EF \perp AC\).


Calculation:
1. \(AB = AC \Rightarrow \angle ABC = \angle ACB\).


2. \(AD \perp BC \Rightarrow \angle ADB = \angle ADC = 90^\circ\).


3. \(EF \perp AC \Rightarrow \angle EFC = 90^\circ\).


4. \(\angle BAC = \angle FAC\) (common).


5. In \(\triangle ABD\) and \(\triangle ECF\):
\(\angle BAD = \angle FAC\) (part of \(\angle BAC\)),
\(\angle ABD = \angle FCE\) (corresponding angles, \(AB \parallel EF\)? Wait).


Actually:
Since \(AB = AC\), \(AD\) is median, altitude, angle bisector. \(E\) on \(CB\) extended, so \(C\) between \(B\) and \(E\). \(EF \perp AC\).


Consider: \(\angle BAD = \angle CAF\) (since \(AD\) bisects \(\angle BAC\)). \(\angle ABD = \angle FCE\) (alternate angles? Need parallel).
Note: \(EF \perp AC\), \(AD \perp BC\).

Better: \(\angle ADB = \angle EFC = 90^\circ\). \(\angle BAD = \angle CAF\) (angle bisector).
Thus, by AA: \(\triangle ABD \sim \triangle ECF\).

Explanation: Right angles and equal angles from angle bisector prove similarity.
Quick Tip: In isosceles triangle, altitude to base is also angle bisector.

Concept: Medians divide triangles; use Apollonius's theorem or vector geometry.

Calculation:
Given: \[ \dfrac{AB}{PQ} = \dfrac{BC}{PR} = \dfrac{AD}{PM} = k \quad (say) \]
Let \(D, M\) be midpoints of \(BC, QR\).
By median length formula: \[ AD^2 = \dfrac{2AB^2 + 2AC^2 - BC^2}{4} \]
But easier: use vector or coordinate geometry.
Place \(\triangle PQR\) and scale.
Since two sides and median proportional, use converse of midline theorem.
Actually, standard result: if two sides and median to third side proportional, triangles similar.
Proof via SSS: need to show \(AC/QR = k\).
Use Apollonius: \[ 4AD^2 = 2AB^2 + 2AC^2 - BC^2 \] \[ 4PM^2 = 2PQ^2 + 2PR^2 - QR^2 \]
Substitute proportional: \[ 4(k \cdot PM)^2 = 2(k \cdot PQ)^2 + 2AC^2 - (k \cdot PR)^2 \]
Wait, better:
Given \(AB = k \cdot PQ\), \(BC = k \cdot PR\), \(AD = k \cdot PM\).
Then \[ 4 (k PM)^2 = 2 (k PQ)^2 + 2 AC^2 - (k PR)^2 \] \[ 4 k^2 PM^2 = 2 k^2 PQ^2 + 2 AC^2 - k^2 PR^2 \]
Divide by \(k^2\): \[ 4 PM^2 = 2 PQ^2 + \dfrac{2 AC^2}{k^2} - PR^2 \]
But from \(\triangle PQR\): \[ 4 PM^2 = 2 PQ^2 + 2 QR^2 - PR^2 \]
So \[ 2 PQ^2 + 2 QR^2 - PR^2 = 2 PQ^2 + \dfrac{2 AC^2}{k^2} - PR^2 \] \[ 2 QR^2 = \dfrac{2 AC^2}{k^2} \quad \Rightarrow \quad AC^2 = k^2 QR^2 \quad \Rightarrow \quad AC = k \cdot QR \]
Thus, \(AB = k PQ\), \(BC = k PR\), \(AC = k QR\).
By SSS: \(\triangle ABC \sim \triangle PQR\).

Explanation: Proportionality extends to third side via median formula, hence SSS similarity.
Quick Tip: Use median length formula to relate third side.

Concept: Ratio of areas = square of ratio of corresponding sides.

Calculation: \[ \dfrac{Area \triangle ABC}{Area \triangle DEF} = \dfrac{9}{64} = \left( \dfrac{AB}{DE} \right)^2 \] \[ \dfrac{AB}{5.1} = \sqrt{\dfrac{9}{64}} = \dfrac{3}{8} \] \[ AB = 5.1 \times \dfrac{3}{8} = \dfrac{51}{10} \times \dfrac{3}{8} = \dfrac{153}{80} = \dfrac{23}{8} = 2.875 cm. \]

Explanation: \(AB = \dfrac{23}{8}\) cm.
Quick Tip: Area ratio → side ratio = square root.

Concept: Polynomial long division or synthetic division.

Calculation:
Using synthetic division (\(x = -1\)):
Coefficients of \(x^3 + 0x^2 + 0x + 1\): \[ \begin{array}{r|rrrr} -1 & 1 & 0 & 0 & 1
& & -1 & 1 & -1
\hline & 1 & -1 & 1 & 0
\end{array} \]
Quotient: \(x^2 - x + 1\), remainder \(0\).
Or factor: \(x^3 + 1 = (x+1)(x^2 - x + 1)\).

Explanation: Quotient is \(x^2 - x + 1\).
Quick Tip: Use \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\).

Concept: Euclid: \(\gcd(a,b) = \gcd(b, a \mod b)\).

Calculation: \[ 1188 = 2 \times 504 + 180 \] \[ 504 = 2 \times 180 + 144 \] \[ 180 = 1 \times 144 + 36 \] \[ 144 = 4 \times 36 + 0 \]
HCF = \(36\).

Explanation: Repeated division yields HCF \(36\).
Quick Tip: Continue until remainder is zero; last non-zero remainder is HCF.

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

Calculation: \(a = 2\), \(b = 5\), \(c = -3\), \[ D = 5^2 - 4(2)(-3) = 25 + 24 = 49 > 0. \]
Since \(D > 0\), roots are real and distinct.

Explanation: \(D = 49\), two distinct real roots.
Quick Tip: \(D > 0\): two real distinct; \(D = 0\): equal; \(D < 0\): not real.

Concept: Curved surface area of cone = \(\pi r l\), where \(l = \sqrt{r^2 + h^2}\).

Calculation: \[ l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 cm. \]
Curved surface area = \(\pi \times 7 \times 25 = 175\pi\) cm².

Explanation: Slant height is \(25\) cm, so area is \(175\pi\) cm².
Quick Tip: Recognize 7-24-25 as a Pythagorean triplet.

Concept: Area swept = sector area. In 60 min, hand sweeps \(360^\circ\). In 40 min: \(\dfrac{40}{60} \times 360^\circ = 240^\circ\).

Calculation: \[ Area = \dfrac{240^\circ}{360^\circ} \times \pi r^2 = \dfrac{2}{3} \pi (7)^2 = \dfrac{2}{3} \times 49\pi = \dfrac{98\pi}{3} cm². \]
Wait: 240/360 = 2/3, yes. But 98π/3 ≈ 102.6, but answer has 1026/9?
Wait: Recheck.
40 min → angle = \(6^\circ \times 40 = 240^\circ\), yes. \[ \dfrac{240}{360} \pi 49 = \dfrac{2}{3} \times 49\pi = \dfrac{98\pi}{3}. \]
But 1026/9 = 114, no. Perhaps mistake.
Wait: Minute hand in 60 min → full circle. Yes.
Area = \(\dfrac{40}{60} \pi r^2 = \dfrac{2}{3} \times 49\pi = \dfrac{98\pi}{3}\).
But 98/3 = 32.666π, but let's see.
Perhaps they want in terms of 1026/9? No.
98π/3 = (98 × 3)π / 9 = 294π / 9? No.
Wait: Perhaps 40 min from 12.
Correct: \(\dfrac{98\pi}{3}\) cm².

Explanation: Sector angle \(240^\circ\), area \(\dfrac{98\pi}{3}\) cm².
Quick Tip: Minute hand: \(6^\circ\) per minute.

Concept: Use identity \(\tan(90^\circ - \theta) = \cot \theta\), and \(\tan 60^\circ = \sqrt{3}\).

Calculation: \[ \tan 83^\circ = \tan(90^\circ - 7^\circ) = \cot 7^\circ. \]
So, \[ \tan 7^\circ \cdot \tan 60^\circ \cdot \tan 83^\circ = \tan 7^\circ \cdot \sqrt{3} \cdot \cot 7^\circ = \sqrt{3} \cdot (\tan 7^\circ \cdot \dfrac{1}{\tan 7^\circ}) = \sqrt{3} \cdot 1 = \sqrt{3}. \]

Explanation: Pairing \(\tan 7^\circ\) and \(\tan 83^\circ = \cot 7^\circ\) gives 1, leaving \(\sqrt{3}\).
Quick Tip: Use \(\tan(90^\circ - \theta) = \cot \theta\) for complementary angles.

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

Calculation: \(a = 2\), \(b = -2\sqrt{2}\), \(c = 1\), \[ D = (-2\sqrt{2})^2 - 4(2)(1) = 8 - 8 = 0. \] \[ x = \dfrac{2\sqrt{2}}{4} = \dfrac{\sqrt{2}}{2}. \]
Double root: \(x = \dfrac{\sqrt{2}}{2}\).

Wait: \(D = 0\), yes.
But let's confirm: \[ 2\left(\dfrac{\sqrt{2}}{2}\right)^2 - 2\sqrt{2} \cdot \dfrac{\sqrt{2}}{2} + 1 = 2 \cdot \dfrac{2}{4} - 2 \cdot 1 + 1 = 1 - 2 + 1 = 0. \]
Yes.

Explanation: Discriminant zero, repeated root \(\dfrac{\sqrt{2}}{2}\).
Quick Tip: When \(D = 0\), roots are equal: \(x = -\dfrac{b}{2a}\).

Concept: A.P. with first term \(a = 3\), common difference \(d = 8\), last term \(l = 67\).

Calculation:
Number of terms: \[ 67 = 3 + (n-1)8 \quad \Rightarrow \quad 64 = (n-1)8 \quad \Rightarrow \quad n-1 = 8 \quad \Rightarrow \quad n = 9. \]
Sum: \[ S_n = \dfrac{n}{2} (a + l) = \dfrac{9}{2} (3 + 67) = \dfrac{9}{2} \times 70 = 9 \times 35 = 315. \]
Wait: Mistake.
Wait: Sequence: 3, 11, 19, 27, 35, 43, 51, 59, 67. Yes, 9 terms.


Sum = \(\dfrac{9}{2} (3 + 67) = \dfrac{9 \times 70}{2} = 315\).


But earlier said 620. Wait, recheck last term.


67 - 3 = 64, 64/8 = 8, n = 9, yes.
3+11+19+27+35+43+51+59+67:


Pair: (3+67)=70, (11+59)=70, (19+51)=70, (27+43)=70, middle 35.


4 pairs × 70 = 280, +35 = 315.


Wait, but perhaps question has different last term.


Wait: Let's see: 3,11,19,... common difference 8.


8×8=64, 3+64=67, yes.


Sum = 315.


But earlier said 620. Perhaps typo.


Wait: Maybe up to 83 or something.


Wait: 67 is correct. Perhaps answer is 315.


But let's accept 315. Explanation: 9 terms, sum \(315\).
Quick Tip: Pair first and last terms for quick sum.

Concept: \(a_5 = a + 4d = 43\), \(a_9 = a + 8d = 79\).

Calculation:
Subtract: \[ (a + 8d) - (a + 4d) = 79 - 43 \quad \Rightarrow \quad 4d = 36 \quad \Rightarrow \quad d = 9. \] \[ a + 4(9) = 43 \quad \Rightarrow \quad a + 36 = 43 \quad \Rightarrow \quad a = 7. \]
Wait: But 7 + 4×9 = 7+36=43, yes.
But 7 + 8×9 = 7+72=79, yes.
A.P.: \(7, 16, 25, 34, 43, \ldots\).


Wait: But answer said 11,19,...
Wait: Let's check.


If a=11, d=8: 11+4\times8=11+32=43, 11+8\times8=11+64=75 ≠79. No.


Wait: Given 5th=43, 9th=79.


Difference 4d=36, d=9, a=43-36=7. Yes.


A.P.: 7, 16, 25, 34, 43, 52, 61, 70, 79, ...


Explanation: First term 7, common difference 9.
Quick Tip: Subtract term equations to find \(d\).

Concept: Use half-angle identities.

Calculation:
LHS: \[ \sqrt{\dfrac{1+\cos \theta}{1-\cos \theta}} \cdot \dfrac{1+\cos \theta}{1+\cos \theta} = \sqrt{\dfrac{(1+\cos \theta)^2}{(1-\cos \theta)(1+\cos \theta)}} = \sqrt{\dfrac{(1+\cos \theta)^2}{1 - \cos^2 \theta}} = \dfrac{1+\cos \theta}{\sin \theta}. \]
(RHS).

Explanation: Rationalizing inside square root simplifies to RHS.
Quick Tip: Multiply numerator and denominator inside square root by conjugate.

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

Calculation:
RHS: \(\cot 63^\circ = \tan 27^\circ\), \(\cot 81^\circ = \tan 9^\circ\).
So, \[ \cot 63^\circ \cdot \cot 81^\circ = \tan 27^\circ \cdot \tan 9^\circ = LHS. \]

Explanation: Complementary angles convert cot to tan.
Quick Tip: \(\cot \theta = \tan(90^\circ - \theta)\).

Concept: \(\sin^2 A + \cos^2 A = 1\).

Calculation: \[ \cos A = \dfrac{4}{5} \quad \Rightarrow \quad \sin A = \sqrt{1 - \left(\dfrac{4}{5}\right)^2} = \sqrt{1 - \dfrac{16}{25}} = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}. \] \[ \cot A = \dfrac{\cos A}{\sin A} = \dfrac{4/5}{3/5} = 4, \quad \operatorname{cosec} A = \dfrac{1}{\sin A} = \dfrac{5}{3}. \]

Explanation: Using identity, \(\sin A = \dfrac{3}{5}\), hence ratios follow.
Quick Tip: 3-4-5 triangle: \(\sin = 3/5\), \(\cos = 4/5\).

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

Calculation: \[ n^2 + n^2 + 2n + 1 = 365 \quad \Rightarrow \quad 2n^2 + 2n + 1 = 365 \] \[ 2n^2 + 2n - 364 = 0 \quad \Rightarrow \quad n^2 + n - 182 = 0 \] \[ D = 1 + 728 = 729 = 27^2, \quad n = \dfrac{-1 \pm 27}{2}. \] \(n = 13\) (positive).

Explanation: \(13^2 + 14^2 = 169 + 196 = 365\).
Quick Tip: Expand \((n+1)^2 = n^2 + 2n + 1\).

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

Calculation: \[ x^2 - 8x = 180 \quad \Rightarrow \quad x^2 - 8x - 180 = 0. \]
Or substitute: \[ x^2 - y^2 = 180, \quad y^2 = 8x \quad \Rightarrow \quad x^2 - 8x = 180. \]

Explanation: Equation: \(x^2 - 8x - 180 = 0\).
Quick Tip: Define variables clearly: larger and smaller.

Concept: Use similar triangles and ratio.

Calculation:
Given \(\dfrac{PS}{SQ} = \dfrac{PT}{TR} = k\) (say).
So \(PS = k \cdot SQ\), \(PQ = PS + SQ = k SQ + SQ = (k+1)SQ\).
Similarly for \(PR\).
Now \(\angle PST = \angle PRQ\).
In \(\triangle PST\) and \(\triangle PQR\): \(\angle SPT = \angle QPR\) (common), \(\angle PST = \angle PRQ\) (given).
By AA: \(\triangle PST \sim \triangle PQR\).
Thus, \[ \dfrac{PS}{PQ} = \dfrac{PT}{PR} = \dfrac{ST}{QR}. \]
But \(\dfrac{PS}{PQ} = \dfrac{k}{k+1}\), so \(\dfrac{PT}{PR} = \dfrac{k}{k+1}\).
But given \(\dfrac{PT}{TR} = k\), \(PR = PT + TR = k TR + TR = (k+1)TR\), so \(\dfrac{PT}{PR} = \dfrac{k}{k+1}\).
Consistent.


To show isosceles: need \(PQ = PR\) or angles equal.


Since ratios equal and \(\triangle PST \sim \triangle PQR\), and \(S, T\) divide in same ratio, by converse of basic proportionality, \(ST \parallel QR\).


Then \(\angle PST = \angle PQR\) (corresponding), but given \(\angle PST = \angle PRQ\).


So \(\angle PQR = \angle PRQ\) → \(\triangle PQR\) isosceles with \(PQ = PR\).


Explanation: Parallel lines and equal angles imply base angles equal.
Quick Tip: Draw line \(ST \parallel QR\) using ratio, then use corresponding angles.

Concept: Assume rational, lead to contradiction.

Calculation:
Assume \(5 - \sqrt{3} = \dfrac{p}{q}\), \(p, q\) coprime, \(q \neq 0\). \[ \sqrt{3} = 5 - \dfrac{p}{q} = \dfrac{5q - p}{q}. \]
Square: \[ 3 = \dfrac{(5q - p)^2}{q^2} \quad \Rightarrow \quad 3q^2 = (5q - p)^2. \] \(3q^2\) integer, so \(5q - p\) must be multiple of \(q\), but leads to \(\sqrt{3}\) rational, contradiction since \(\sqrt{3}\) irrational.

Explanation: Contradiction with irrationality of \(\sqrt{3}\).
Quick Tip: Isolate the square root and square both sides.

Concept: Area of triangle = 0 for collinear points.

Calculation:
Using formula: \[ \dfrac{1}{2} |1(k - 4) + 3(4 - 1) + (-1)(1 - k)| = 0 \] \[ |k - 4 + 9 - 1 - 1 + k| = 0 \quad \Rightarrow \quad |2k + 4| = 0 \quad \Rightarrow \quad k = -2. \]

Explanation: \(k = -2\) makes area zero, hence collinear.
Quick Tip: Slope method: \(\dfrac{k-1}{3-1} = \dfrac{4-1}{-1-1}\).

Concept: Point \((0, y)\) equidistant from \(A(6,5)\), \(B(-4,3)\).

Calculation: \[ \sqrt{(0-6)^2 + (y-5)^2} = \sqrt{(0+4)^2 + (y-3)^2} \]
Square: \[ 36 + (y-5)^2 = 16 + (y-3)^2 \] \[ 36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9 \] \[ 61 - 10y = 25 - 6y \quad \Rightarrow \quad 36 = 4y \quad \Rightarrow \quad y = 9. \]
Wait: Recheck.
Left: \(36 + (y-5)^2\), right: \(16 + (y-3)^2\).
Expand:
Left: \(36 + y^2 - 10y + 25 = y^2 - 10y + 61\),
Right: \(16 + y^2 - 6y + 9 = y^2 - 6y + 25\). \[ y^2 - 10y + 61 = y^2 - 6y + 25 \quad \Rightarrow \quad -4y = -36 \quad \Rightarrow \quad y = 9. \]
Point: \((0, 9)\).
Verify:
Distance to (6,5): \(\sqrt{36 + 16} = \sqrt{52}\),
To (-4,3): \(\sqrt{16 + 36} = \sqrt{52}\). Yes.

Explanation: Point \((0, 9)\).
Quick Tip: Set distance equations and square to eliminate roots.

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 \(f_i x_i\):
\begin{tabular{c|c|c
\hline
Class & \(f_i\) & \(x_i\) & \(f_i x_i\)

\hline
11--13 & 7 & 12 & 84

13--15 & 6 & 14 & 84

15--17 & 9 & 16 & 144

17--19 & 13 & 18 & 234

19--21 & 20 & 20 & 400

21--23 & 5 & 22 & 110

23--25 & 4 & 24 & 96

\hline
& \(\sum f_i = 64\) & & \(\sum f_i x_i = 1152\)

\hline
\end{tabular
\[ \bar{x} = \dfrac{1152}{64} = 18 \]
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.