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: Monic quadratic with sum \(S\) and product \(P\) is \(x^{2}-Sx+P\).


Calculation:
\(S=3,\ P=-2 \Rightarrow x^{2}-3x-2\).


Explanation: Direct application of coefficient–root relations.
Quick Tip: Use \(x^{2}-(sum)x+(product)\) to build the quadratic quickly.

Concept: Degree of quotient = degree(dividend) − degree(divisor).


Calculation:

deg\,\(p(x)=4\), deg\,\(q(x)=1\), so quotient degree \(=4-1=3\).


Explanation: Division by a linear polynomial reduces degree by 1.
Quick Tip: For division by linear factor, quotient degree = original degree − 1.

Concept: If one equation is a scalar multiple of the other, they represent the same line → infinitely many common points.


Calculation:

Multiply first equation by \(3\): \(3x+6y+9=0\), which is exactly the second equation.


Explanation: Both equations describe the same line, so every point on that line is a solution.
Quick Tip: If ratios of coefficients (including constants) are equal, the lines coincide → infinitely many solutions.

Concept: Parallel distinct lines have same slope but different intercepts → they never meet.


Explanation: Since the lines do not intersect, the system has no solution (inconsistent).
Quick Tip: Equal slopes with different constants → parallel distinct lines → no solution.

Concept: Solve to check whether a unique solution exists (consistent & independent), no solution (inconsistent), or infinitely many (dependent).


Calculation (elimination):

Write in standard form: \(5x-4y=-8\) and \(7x+6y=9\). Multiply first by \(3\): \(15x-12y=-24\). Multiply second by \(2\): \(14x+12y=18\). Add: \(29x=-6 \Rightarrow x=-\dfrac{6}{29}\). Substitute to find \(y\) → unique solution.


Explanation: A unique solution exists so the system is consistent (and independent).
Quick Tip: If elimination/substitution yields a unique pair, the system is consistent & independent.

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


Calculation:

For \(3x^{2}-5x+2=0\), \(\alpha+\beta=\dfrac{5}{3}\), \(\alpha\beta=\dfrac{2}{3}\).

So \(\alpha^{2}+\beta^{2}=\left(\dfrac{5}{3}\right)^{2}-2\cdot\dfrac{2}{3}=\dfrac{25}{9}-\dfrac{4}{3}=\dfrac{25-12}{9}=\dfrac{13}{9}\).
Quick Tip: Use \( \alpha+\beta=-\dfrac{b}{a}\) and \(\alpha\beta=\dfrac{c}{a}\) then build required symmetric sums.

Concept: Substitute known root into equation to find unknown parameter.


Calculation:

Put \(x=2\): \(2(2)^2 -7(2) - p =0 \Rightarrow 8-14-p=0 \Rightarrow -6-p=0 \Rightarrow p=-6\).
Quick Tip: Direct substitution of a known root is the quickest method to determine unknown coefficients.

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


Calculation:

For \(2x^{2}-x-6=0\), sum of roots \(=\dfrac{1}{2}\). Given one root \(=-\dfrac{3}{2}\), so other root \(= \dfrac{1}{2} - \left(-\dfrac{3}{2}\right)=\dfrac{1}{2}+\dfrac{3}{2}=2\).
Quick Tip: Use sum/product relations to find the unknown root quickly.

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


Calculation:
\(a=2,\ b=-6,\ c=3 \Rightarrow D=(-6)^{2}-4\cdot2\cdot3=36-24=12>0\).


Explanation: \(D>0\) implies two distinct real roots (real and unequal).
Quick Tip: Compute \(D=b^{2}-4ac\) first: \(D>0\) → two distinct real roots.

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


Calculation:
\(a_5 = a + 4d = 11\), \(d=2\) \(\Rightarrow a + 8 = 11 \Rightarrow a = 3\).
Quick Tip: Apply \(a_n = a + (n-1)d\) and solve for \(a\) when \(a_n\) and \(d\) are given.

Concept: \(t_n = S_n - S_{n-1}\) where \(S_n\) is sum of first \(n\) terms.


Calculation:
\(S_n = n^{2}+2n+1=(n+1)^{2}\).
\(S_6=(6+1)^{2}=49,\quad S_5=(5+1)^{2}=36\).
\(t_6=S_6-S_5=49-36=13\).


Explanation: 13 is not among options A–C, so (D) is correct.
Quick Tip: Find \(t_n\) by \(S_n-S_{n-1}\) when given the sum formula.

Concept: Sequence is A.P. if consecutive differences are constant.


Calculation:

Differences in (C): \(2x-x=x,\ 3x-2x=x,\ \ldots\) constant \(=x\). Others do not have constant differences.
Quick Tip: Check consecutive differences; constant difference ⇒ A.P.

Concept: A.P. requires constant difference; sequence of even squares has increasing differences.


Calculation:

Sequence (D): \(4,16,36,64,\ldots\) differences \(12,20,28,\ldots\) not constant → not A.P.
Quick Tip: Sequences with quadratic growth (squares) are not arithmetic—check differences.

Concept: \(S_n=\dfrac{n}{2}[2a+(n-1)d]\).


Calculation:
\(a=1,\ d=3,\ n=20\).
\(S_{20}=\dfrac{20}{2}[2\cdot1+19\cdot3]=10[2+57]=10\cdot59=590\).
Quick Tip: Always identify \(a,d,n\) precisely and substitute into \(S_n=\dfrac{n}{2}[2a+(n-1)d]\).

Evaluate each option step by step:

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

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

- (C):
\[ \sin^2 60^\circ = \dfrac{3}{4} \neq 1 \]

- (D):
\[ \sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt{2}} \]
\[ \dfrac{1}{\sqrt{2}} \times \dfrac{1}{\frac{1}{\sqrt{2}}} = \dfrac{1}{\sqrt{2}} \times \sqrt{2} = 1 \]
(Alternatively: \(\dfrac{\sin 45^\circ}{\cos 45^\circ} = \tan 45^\circ = 1\))

Only (D) equals 1.

Explanation: Option (D) simplifies to \(\tan 45^\circ = 1\).
Quick Tip: \(\dfrac{\sin \theta}{\cos \theta} = \tan \theta\); use standard values.

Use the identity: \[ 1 + \tan^2 A = \sec^2 A \]
Substitute: \[ \cos^2 A \cdot (1 + \tan^2 A) = \cos^2 A \cdot \sec^2 A \] \[ \sec A = \dfrac{1}{\cos A} \quad \Rightarrow \quad \sec^2 A = \dfrac{1}{\cos^2 A} \] \[ \cos^2 A \cdot \dfrac{1}{\cos^2 A} = 1 \]

Alternative derivation: \[ 1 + \tan^2 A = 1 + \dfrac{\sin^2 A}{\cos^2 A} = \dfrac{\cos^2 A + \sin^2 A}{\cos^2 A} = \dfrac{1}{\cos^2 A} = \sec^2 A \]
Then: \[ \cos^2 A \cdot \sec^2 A = 1 \]

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

Using standard 30-60-90 triangle:
- Opposite to 30°: 1
- Adjacent: \(\sqrt{3}\)
- Hypotenuse: 2
\[ \tan 30^\circ = \dfrac{opposite}{adjacent} = \dfrac{1}{\sqrt{3}} \]
Rationalized form: \[ \dfrac{1}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} \]
Both forms are correct, but the option matches \(\dfrac{1}{\sqrt{3}}\).

Alternative: \[ \tan 30^\circ = \dfrac{\sin 30^\circ}{\cos 30^\circ} = \dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \dfrac{1}{\sqrt{3}} \]

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

In a 30-60-90 triangle:
- Side opposite 30°: 1
- Side opposite 60°: \(\sqrt{3}\)
- Hypotenuse: 2
\[ \cos 60^\circ = \dfrac{adjacent}{hypotenuse} = \dfrac{1}{2} \]

Alternative: \[ \cos 60^\circ = \dfrac{adjacent}{hypotenuse} = \dfrac{1}{2} \]
(Note: \(\dfrac{\sqrt{3}}{2}\) is \(\cos 30^\circ\), not 60°.)

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

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

Verification with values:
- \(\sin 90^\circ = 1\) (top of unit circle)
- \(\tan 45^\circ = 1\) (slope of line at 45°)

Explanation: Both terms are exactly 1, so difference is zero.
Quick Tip: \(\sin 90^\circ = 1\), \(\tan 45^\circ = 1\) → squares are 1.

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

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

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

Verification:
- Horizontal: \(4\sqrt{3} \approx 6.928\)
- Vertical: 4
- \(\sqrt{6.928^2 + 4^2} \approx \sqrt{48 + 16} = 8\)

Explanation: Points are on axes; simplifies to \(\sqrt{(4\sqrt{3})^2 + 4^2} = 8\).
Quick Tip: Substitute \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\cos 60^\circ = \frac{1}{2}\) early.

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

This is the standard distance formula from the origin.

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

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

The distance from a point \((x, y)\) to the y-axis (where \(x = 0\)) is the absolute value of the x-coordinate.

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

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

Explanation: Only the x-coordinate determines distance to the y-axis.
Quick Tip: Distance to y-axis = \(|x|\).

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

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

Explanation: Ordinate refers to the y-value in the coordinate pair.
Quick Tip: Ordinate = y-coordinate.

Quadrant determination by signs of coordinates:
- I: \(x > 0\), \(y > 0\)
- II: \(x < 0\), \(y > 0\)
- III: \(x < 0\), \(y < 0\)
- IV: \(x > 0\), \(y < 0\)

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

→ Fourth quadrant

Explanation: Positive x and negative y places the point in quadrant IV.
Quick Tip: Sign pattern \((+, -)\) → Fourth quadrant.

Quadrants are determined by the signs of coordinates:
- I: \(x > 0\), \(y > 0\)

- II: \(x < 0\), \(y > 0\)

- III: \(x < 0\), \(y < 0\)

- IV: \(x > 0\), \(y < 0\)

Check each:
- (A) (3,2): \(x = 3 > 0\), \(y = 2 > 0\) → First quadrant

- (B) (-3,2): \(x = -3 < 0\), \(y = 2 > 0\) → Second quadrant

- (C) (3,-2): \(x = 3 > 0\), \(y = -2 < 0\) → Fourth quadrant

- (D) (-3,-2): \(x = -3 < 0\), \(y = -2 < 0\) → Third quadrant


Only (B) satisfies \(x < 0\), \(y > 0\).


Explanation: Second quadrant is left side, above x-axis.
Quick Tip: II: Negative x, positive y.

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

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

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

The points are symmetric about the origin, so midpoint is origin.

Explanation: Average of coordinates gives midpoint.
Quick Tip: Mid-point = \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\).

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

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

Verification:
Mid-point of \(A(5,7)\) and \(B(-1,1)\): \[ x = \frac{5 + (-1)}{2} = \frac{4}{2} = 2, \quad y = \frac{7 + 1}{2} = \frac{8}{2} = 4 \quad \]

Explanation: Set up equations using mid-point formula.
Quick Tip: Use: \(x = 2 \times mid-x - x_A\), same for y.

The centre is the mid-point of the diameter.

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

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

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

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

Vertices:
- \(A(4,6)\)
- \(B(0,4)\)
- \(C(5,5)\)
\[ x = \frac{4 + 0 + 5}{3} = \frac{9}{3} = 3 \] \[ y = \frac{6 + 4 + 5}{3} = \frac{15}{3} = 5 \] \[ G = (3, 5) \]

Verification:
Sum of x: \(4 + 0 + 5 = 9\), average = 3
Sum of y: \(6 + 4 + 5 = 15\), average = 5

Explanation: Centroid is average of vertices' coordinates.
Quick Tip: Centroid = \(\left( \frac{\sum x}{3}, \frac{\sum y}{3} \right)\).

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

Check each:
- (A): \(\dfrac{14}{2^0 \times 3^2} = \dfrac{14}{9}\) → denominator has 3 → non-terminating
- (B): \(\dfrac{9}{5 \times 49} = \dfrac{9}{245}\) → denominator has 7 → non-terminating
- (C): \(\dfrac{8}{4 \times 9} = \dfrac{8}{36} = \dfrac{2}{9}\) → denominator has 3 → non-terminating
- (D): \(\dfrac{15}{4 \times 125} = \dfrac{15}{500} = \dfrac{3}{100}\) → denominator = \(2^2 \times 5^3\) → terminating
\[ \dfrac{3}{100} = 0.03 \quad (terminates) \]

Explanation: Only (D) has denominator of form \(2^n \times 5^m\).
Quick Tip: Terminating iff denominator (after simplifying) = \(2^a \times 5^b\).

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

Simplify the fraction: \[ \dfrac{505}{1000} = \dfrac{505 \div 5}{1000 \div 5} = \dfrac{101}{200} \]

Factorize the denominator: \[ 200 = 2^3 \times 5^2 \quad (8 \times 25 = 200) \]

Thus: \[ \dfrac{101}{200} = \dfrac{101}{2^3 \times 5^2} \]

This matches option (D).

Verification by calculating each option:

- (A): \(\dfrac{101}{2 \times 25} = \dfrac{101}{50} = 2.02\)

- (B): \(\dfrac{101}{2 \times 125} = \dfrac{101}{250} = 0.404\)

- (C): \(\dfrac{101}{4 \times 25} = \dfrac{101}{100} = 1.01\)

- (D): \(\dfrac{101}{8 \times 25} = \dfrac{101}{200} = 0.505\)


Explanation: Convert decimal to fraction, simplify, and express denominator as \(2^n \times 5^m\).
Quick Tip: Multiply numerator and denominator by \(10^k\) (here \(k=3\)), then simplify.

Division algorithm: \[ a = b \cdot q + r, \quad 0 \leq r < b \]
Given: \(b = 4\), \(q = 5\), \(r = 1\) \[ a = 4 \cdot 5 + 1 = 20 + 1 = 21 \]
Check: \(0 \leq 1 < 4\) → valid.

Verification: \[ 21 \div 4 = 5 quotient, remainder 1 \quad \]

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

Solve \(2x^2 - 4x - 6 = 0\):
Divide by 2: \[ x^2 - 2x - 3 = 0 \]
Factorize: \[ (x - 3)(x + 1) = x^2 + x - 3x - 3 = x^2 - 2x - 3 \] \[ (x - 3)(x + 1) = 0 \] \[ x = 3 \quad or \quad x = -1 \]

Verification:
- At \(x = 3\): \(2(9) - 4(3) - 6 = 18 - 12 - 6 = 0\)
- At \(x = -1\): \(2(1) + 4 - 6 = 2 + 4 - 6 = 0\)

Explanation: Factorize or use quadratic formula.
Quick Tip: Try integer factors of constant term.

Degree of product = sum of degrees of factors.

- First polynomial: \(x^3 + x^2 + 2x + 1\) → degree 3
- Second: \(x^2 + 2x + 1\) → degree 2
\[ \deg(product) = 3 + 2 = 5 \]

Leading term: \[ x^3 \cdot x^2 = x^5 \]
So highest power is 5.

Explanation: Degree adds on multiplication.
Quick Tip: deg(f × g) = deg(f) + deg(g).

A polynomial has non-negative integer exponents.

- (A): \(x^2 - 7\) → exponents 2, 0 → polynomial

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

- (C): \(\frac{1}{2}x^2 + \frac{1}{2}x + 4\) → exponents 2, 1, 0 → polynomial

- (D): \(x + \frac{4}{x} = x^1 + 4x^{-1}\) → exponent -1 → not a polynomial


Explanation: Negative exponent violates polynomial definition.
Quick Tip: No negative or fractional exponents.

For zeroes \(\alpha = 2\), \(\beta = -2\):
Sum of zeroes: \(\alpha + \beta = 2 + (-2) = 0\)
Product: \(\alpha \beta = 2 \times (-2) = -4\)

Standard form: \[ x^2 - (sum)x + product = x^2 - 0 \cdot x + (-4) = x^2 - 4 \]

Factorize: \[ x^2 - 4 = (x - 2)(x + 2) = 0 \quad \Rightarrow \quad x = 2, -2 \]

Check options:
- (A): \(x^2 + 4 = 0 \Rightarrow x = \pm 2i\) (not real)
- (B): \(x^2 - 4 = 0 \Rightarrow x = \pm 2\) \(\checkmark\)
- (C): \(x^2 - 2x + 4 = 0 \Rightarrow D = 4 - 16 = -12\) (no real roots)
- (D): \(x^2 + \sqrt{8} = 0 \Rightarrow x = \pm i\sqrt{8}\) (not real)

Explanation: Use sum = 0, product = -4.
Quick Tip: Zeros \(r, -r\) \(\Rightarrow\) \(x^2 - r^2\).

For \(at^2 + bt + c = 0\):
Sum of zeroes = \(-\frac{b}{a}\)

Here: \(a = 1\), \(b = 7\), \(c = 10\) \[ \alpha + \beta = -\frac{7}{1} = -7 \]

Verification:
Factorize: \(t^2 + 7t + 10 = (t + 2)(t + 5) = 0\)
Zeros: \(t = -2, -5\)
Sum: \(-2 + (-5) = -7\)

Explanation: Use sum of roots formula.
Quick Tip: Sum = \(-\frac{b}{a}\).

Standard values: \[ \sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2} \] \[ \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2} \]

First part: \[ \sin 30^\circ + \cos 30^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \]

Second part: \[ \sin 60^\circ + \cos 60^\circ = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \]

Subtract: \[ \frac{1 + \sqrt{3}}{2} - \frac{\sqrt{3} + 1}{2} = 0 \]

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

Since \(x = 4\) is a root of the polynomial \((k-1)x^2 + kx + 1 = 0\), substitute \(x = 4\): \[ (k-1)(4)^2 + k(4) + 1 = 0 \] \[ (k-1)(16) + 4k + 1 = 0 \] \[ 16k - 16 + 4k + 1 = 0 \] \[ 20k - 15 = 0 \] \[ 20k = 15 \quad \Rightarrow \quad k = \dfrac{15}{20} = \dfrac{3}{4} \]

Correct value: \(k = \dfrac{3}{4}\)

Now check the given options:
- (A) \(-\dfrac{5}{4}\)
- (B) \(\dfrac{5}{4}\)
- (C) \(-\dfrac{4}{3}\)
- (D) \(\dfrac{4}{3}\)

None match \(\dfrac{3}{4}\).

Verification with \(k = \dfrac{3}{4}\): \[ a = k - 1 = \dfrac{3}{4} - 1 = -\dfrac{1}{4} \]
Polynomial: \[ -\dfrac{1}{4}x^2 + \dfrac{3}{4}x + 1 = 0 \]
Multiply by \(-4\): \[ x^2 - 3x - 4 = 0 \]
Factorize: \[ (x - 4)(x + 1) = 0 \quad \Rightarrow \quad x = 4, -1 \]
Root 4 is confirmed.

Conclusion: The correct value of \(k\) is \(\dfrac{3}{4}\), which is not listed in the options. There may be a printing error in the question or options. Quick Tip: Substitute the root into the polynomial and solve for \(k\).

Theorem: The lengths of tangents drawn from an external point to a circle are equal.

That is, if PA and PB are tangents from point P to the circle (touching at A and B respectively), then: \[ PA = PB \]

Given: \[ PA = 8 cm \]

Therefore: \[ PB = 8 cm \]

Verification:
- Both tangents share the same external point P.
- The radii OA and OB are perpendicular to the tangents at points of contact.
- Triangles OAP and OBP are congruent (RHS: OA = OB, OP common, right angles).
- Hence, PA = PB.

Explanation: This is a standard property of tangents from a common external point.
Quick Tip: Tangents from a point outside the circle are equal in length.

Properties:
- OA \(\perp\) PA, OB \(\perp\) PB (radius \(\perp\) tangent)
- OA = OB (radii)
- PA = PB (tangents from external point)
\(\triangle OAP\) and \(\triangle OBP\) are congruent (RHS).

Also, quadrilateral OAPB has:
- \(\angle OAP = \angle OBP = 90^\circ\)
- \(\angle APB = 80^\circ\)

Sum of angles in quadrilateral = 360\(^\)\circ
\[ \angle AOB + 90^\circ + 90^\circ + 80^\circ = 360^\circ \] \[ \angle AOB + 260^\circ = 360^\circ \quad \Rightarrow \quad \angle AOB = 100^\circ \]

Now, \(\triangle AOB\) is isosceles with OA = OB.

Base angles equal: \[ \angle OAP = \angle OBP = 90^\circ \quad (already) \]
Wait, no, the base angles for \(\triangle AOB\) are \(\angle OAP\), no.

Wait, in \(\triangle AOP\): \(\angle OAP = 90^\circ\), \(\angle APO = ?\)

The angle \(\angle APB = 80^\circ\) is at P.

The line OP bisects \(\angle APB\) because the two tangents are equal, so \(\triangle OAP \cong \triangle OBP\), so OP is angle bisector.

So \(\angle APO = \angle BPO = 40^\circ\)

Now in \(\triangle AOP\):

- \(\angle OAP = 90^\circ\)

- \(\angle APO = 40^\circ\)

- \(\angle AOP = 180^\circ - 90^\circ - 40^\circ = 50^\circ\)

So \(\angle POA = 50^\circ\)

Since \(\angle AOB = 2 \times \angle AOP = 100^\circ\), but the question asks \(\angle POA\), which is \(\angle AOP = 50^\circ\).

Explanation: OP bisects \(\angle APB\), and right triangle gives 50\(^\)\circ.
Quick Tip: OP bisects \(\angle APB\); use right triangle.

Theorem: The radius to the point of contact is perpendicular to the tangent.

So angle between radius and tangent = 90\(^\)\circ.

Proof sketch:
- Let O be centre, A point of contact.
- OA is radius.
- Tangent at A.
- The tangent is perpendicular to radius (standard theorem).

Explanation: Fundamental property of circle and tangent.
Quick Tip: Radius \(\perp\) tangent at contact point.

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

Ratio of radii \(r_1 : r_2 = 3 : 4\)

Ratio of areas = \(\pi r_1^2 : \pi r_2^2 = r_1^2 : r_2^2 = 3^2 : 4^2 = 9 : 16\)

Explanation: Area proportional to square of radius.
Quick Tip: Ratio of areas = (ratio of radii)^2.

Area of sector = \(\dfrac{\theta}{360^\circ} \times \pi r^2\)

Given: \(\theta = 30^\circ\), \(r = 42\) cm
\[ Area = \dfrac{30}{360} \times \pi \times 42^2 = \dfrac{1}{12} \times \pi \times 1764 \] \[ = \dfrac{1764}{12} \pi = 147 \pi \]
Using \(\pi \approx 3.14\): \[ 147 \times 3.14 = 147 \times 3 + 147 \times 0.14 = 441 + 20.58 = 461.58 \approx 462 \]
Or exact: \(147 \times \frac{22}{7} = 21 \times 22 = 462\)

Explanation: Fraction of full circle area.
Quick Tip: Sector area = \(\frac{\theta}{360} \pi r^2\).

Circumference = \(2\pi r\)

Ratio of circumferences = \(2\pi r_1 : 2\pi r_2 = r_1 : r_2\)

Given 5:7, so ratio of radii = 5:7

Explanation: Circumference proportional to radius.
Quick Tip: Ratio of circumferences = ratio of radii.

Use identity: \[ \sec^2 A - \tan^2 A = 1 \]
Factor out 7: \[ 7 (\sec^2 A - \tan^2 A) = 7 \times 1 = 7 \]

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

Substitute the given expressions: \[ x = a \cos \theta \quad \Rightarrow \quad x^2 = a^2 \cos^2 \theta \] \[ y = b \sin \theta \quad \Rightarrow \quad y^2 = b^2 \sin^2 \theta \]

Now compute: \[ b^2 x^2 = b^2 \cdot a^2 \cos^2 \theta = a^2 b^2 \cos^2 \theta \] \[ a^2 y^2 = a^2 \cdot b^2 \sin^2 \theta = a^2 b^2 \sin^2 \theta \]

Add them: \[ b^2 x^2 + a^2 y^2 = a^2 b^2 \cos^2 \theta + a^2 b^2 \sin^2 \theta = a^2 b^2 (\cos^2 \theta + \sin^2 \theta) \]
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\): \[ a^2 b^2 (1) = a^2 b^2 \]

Verification:
- For \(\theta = 0\): \(x = a\), \(y = 0\) → \(b^2 a^2 + a^2 (0) = a^2 b^2\)
- For \(\theta = 90^\circ\): \(x = 0\), \(y = b\) → \(b^2 (0) + a^2 b^2 = a^2 b^2\)

Explanation: This is the standard equation of an ellipse in parametric form.
Quick Tip: Factor out \(a^2 b^2\) and use \(\cos^2 \theta + \sin^2 \theta = 1\).

The question states:
- The tower is Solution10 m highSolution.
- The angle of elevation of the sun Solutionfrom its baseSolution is \(60^\circ\).

The height of the tower is explicitly given as 10 m. The angle of elevation refers to the sun's position as seen from the base of the tower, but it does SolutionnotSolution ask to calculate the height — it asks for the height, which is already provided.

Thus, the height of the tower is Solution10 mSolution.

(Note: If the question intended to ask for the length of the shadow cast by the tower, then: \[ \tan 60^\circ = \frac{height}{shadow} \quad \Rightarrow \quad \sqrt{3} = \frac{10}{shadow} \quad \Rightarrow \quad shadow = \frac{10}{\sqrt{3}} \approx 5.77 \, m \]
But the question clearly asks for the Solutionheight of the towerSolution, not the shadow.)

Explanation: Read the question carefully — the height is given directly.
Quick Tip: When height is stated, no calculation is needed unless asked for shadow or other.

The kite forms a right triangle with:
- SolutionOpposite sideSolution to the angle = height = 30 m
- SolutionAngle with groundSolution = \(60^\circ\)
- SolutionHypotenuseSolution = length of string (let’s call it \(l\))

Using \(\sin \theta = \frac{opposite}{hypotenuse}\): \[ \sin 60^\circ = \frac{30}{l} \] \[ \frac{\sqrt{3}}{2} = \frac{30}{l} \quad \Rightarrow \quad l = \frac{30 \times 2}{\sqrt{3}} = \frac{60}{\sqrt{3}} = 20\sqrt{3} \approx 34.64 \, m \]

Wait: \[ \frac{60}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{60\sqrt{3}}{3} = 20\sqrt{3} \, m \]

So the length of the string is Solution\(20\sqrt{3} \, m\)Solution, which Solutionis option (C)Solution.

SolutionCorrection: The answer is (C)Solution

Verification: \[ \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866, \quad 30 \div 0.866 \approx 34.64, \quad 20 \times 1.732 = 34.64 \quad \]

Explanation: Height is opposite to the angle; string is hypotenuse.
Quick Tip: Use \(\sin \theta = \frac{height}{string length}\).

Use the shoelace formula for area of triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\): \[ Area = \dfrac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

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

Alternative method (base-height):
- Base AB (along y-axis): \(|5 - 1| = 4\)
- Height = x-coordinate of C = 3 \[ Area = \dfrac{1}{2} \times base \times height = \dfrac{1}{2} \times 4 \times 3 = 6 \]

Verification: Both methods give 6 square units.

Explanation: Shoelace or base-height works; points A and B on y-axis simplify calculation.
Quick Tip: If two points on same axis, base = distance between them, height = perpendicular distance of third point.

Use identity: \(\tan(90^\circ - \theta) = \cot \theta\) \[ \tan 80^\circ = \tan(90^\circ - 10^\circ) = \cot 10^\circ, \quad \tan 67^\circ = \tan(90^\circ - 23^\circ) = \cot 23^\circ \]

So: \[ \tan 10^\circ \cdot \tan 23^\circ \cdot \cot 10^\circ \cdot \cot 23^\circ \] \[ = (\tan 10^\circ \cdot \cot 10^\circ) \cdot (\tan 23^\circ \cdot \cot 23^\circ) = 1 \cdot 1 = 1 \]

Since \(\tan \theta \cdot \cot \theta = \tan \theta \cdot \dfrac{1}{\tan \theta} = 1\).

Explanation: Pair complementary angles using co-tangent identity.
Quick Tip: \(\tan(90^\circ - \theta) = \cot \theta\) → product with \(\tan \theta\) = 1.

For similar triangles: \[ \dfrac{Area_1}{Area_2} = \left( \dfrac{side_1}{side_2} \right)^2 \]

Given: \[ \dfrac{Area_1}{Area_2} = \dfrac{100}{144} = \dfrac{10^2}{12^2} \] \[ \dfrac{side_1}{side_2} = \sqrt{\dfrac{100}{144}} = \dfrac{10}{12} = 10:12 \]
(Simplify by dividing by 2: \(5:6\))

Verification: \[ \left(\dfrac{10}{12}\right)^2 = \dfrac{100}{144} \quad \checkmark \]

Explanation: Ratio of sides = square root of ratio of areas.
Quick Tip: Side ratio = \(\sqrt{area ratio}\).

- Chord: Line segment joining two points on the circle.
- Secant: Line that intersects the circle at two distinct points.
- Tangent: Line that touches the circle at exactly one point.

The line passes through the circle, cutting it at two points → secant.

Explanation: Secant extends beyond the circle; chord is the segment between intersection points.
Quick Tip: Secant: cuts at 2 points; Tangent: touches at 1.

For similar triangles: \[ \dfrac{Area_1}{Area_2} = \left( \dfrac{side_1}{side_2} \right)^2 \]

Given side ratio = 4:9 \[ Area ratio = (4:9)^2 = 16:81 \]

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

Explanation: Area ratio is square of side ratio.
Quick Tip: Areas ∝ (sides)\(^2\).

Since \(\triangle ABC \sim \triangle DEF\), corresponding sides: \[ BC \leftrightarrow EF \quad \Rightarrow \quad \dfrac{BC}{EF} = \dfrac{3}{4} \]

Ratio of areas: \[ \dfrac{Area_{ABC}}{Area_{DEF}} = \left( \dfrac{BC}{EF} \right)^2 = \left( \dfrac{3}{4} \right)^2 = \dfrac{9}{16} \]

Given: \(Area_{ABC} = 54\) \[ \dfrac{54}{Area_{DEF}} = \dfrac{9}{16} \quad \Rightarrow \quad Area_{DEF} = 54 \times \dfrac{16}{9} = 6 \times 16 = 96 \]

Verification: \[ \dfrac{54}{96} = \dfrac{9}{16} \quad \checkmark \]

Explanation: Area scales with square of side ratio.
Quick Tip: Area ratio = (side ratio)\(^2\).

This is a right-angled triangle at A.
- Hypotenuse = BC = 13 cm
- One leg = AB = 12 cm
- Other leg = AC = ?

By SolutionPythagoras theoremSolution: \[ AB^2 + AC^2 = BC^2 \] \[ 12^2 + AC^2 = 13^2 \] \[ 144 + AC^2 = 169 \] \[ AC^2 = 25 \quad \Rightarrow \quad AC = 5 cm \]

Verification: \(3^2 + 4^2 + 5^2 = 9 + 16 + 25 = 50\), but here \(5^2 + 12^2 = 25 + 144 = 169 = 13^2\) → 5-12-13 triangle.

Explanation: Use Pythagoras: legs square sum to hypotenuse square.
Quick Tip: Right triangle: \(a^2 + b^2 = c^2\) (c = hypotenuse).

In any triangle, sum of angles = \(180^\circ\).

Given: \[ \angle D = \angle Q, \quad \angle R = \angle E \]

For \(\triangle DEF\): \[ \angle D + \angle E + \angle F = 180^\circ \]

For \(\triangle PQR\): \[ \angle P + \angle Q + \angle R = 180^\circ \]

Substitute: \[ \angle D + \angle R + \angle F = 180^\circ \quad (since \angle E = \angle R) \] \[ \angle P + \angle D + \angle E = 180^\circ \quad (since \angle Q = \angle D, \angle R = \angle E) \]

But both equal \(180^\circ\), so: \[ \angle F = \angle P \]

Explanation: Third angles are equal by angle sum property.
Quick Tip: If two angles equal, third must be equal (\(\sum = 180^\circ\)).

All corresponding sides proportional: \[ \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{DF} \]
→ \(\triangle ABC \sim \triangle DEF\) by SolutionSSS similaritySolution.

Corresponding angles equal:
- \(\angle A \leftrightarrow \angle D\)
- \(\angle B \leftrightarrow \angle E\)
- \(\angle C \leftrightarrow \angle F\)

Given: \[ \angle A = 40^\circ, \quad \angle B = 80^\circ \] \[ \angle C = 180^\circ - 40^\circ - 80^\circ = 60^\circ \] \[ \angle F = \angle C = 60^\circ \]

Explanation: SSS similarity → corresponding angles equal.
Quick Tip: SSS → similar → angles correspond by side order.

For Solutiontwo intersecting circlesSolution (intersect at two points):
- Solution2 common external tangentsSolution
- SolutionNo common internal tangentsSolution (they cross between circles)

Total common tangents = Solution2Solution.

(For separate circles: 4; touching externally: 3; one inside other without touching: 2; intersecting: 2.)

Explanation: Intersecting circles have only two external common tangents.
Quick Tip: Intersecting circles → 2 common tangents.

Class interval width = upper limit − lower limit.

- 2–5 → \(5 - 2 = 3\)
- 5–8 → \(8 - 5 = 3\)
- 8–11 → \(11 - 8 = 3\)

All intervals have width Solution3Solution.

(Note: Exclusive classes, but width is still 3.)

Explanation: Difference between consecutive boundaries.
Quick Tip: Class width = upper − lower.

Let numbers: \(x, x+2, x+4, x+6\)
Mean: \[ \dfrac{4x + 12}{4} = 6 \quad \Rightarrow \quad x + 3 = 6 \quad \Rightarrow \quad x = 3 \]
Numbers: 3, 5, 7, Solution9Solution

Largest = Solution9Solution

Verification: Sum = 24, mean = 6.

Explanation: Mean is average of middle two terms.
Quick Tip: Four consecutive odds → mean = average of 2nd and 3rd.

Numbers: 2, 4, 6, 8, 10, 12
Sum = 42
Mean = \(42 \div 6 = 7\)

Or: Mean = \(\dfrac{first + last}{2} = \dfrac{2 + 12}{2} = 7\)

Explanation: A.P. mean = average of ends.
Quick Tip: Mean of first \(n\) evens = \(n+1\).

From \(\sin^2 \theta + \cos^2 \theta = 1\), divide by \(\sin^2 \theta\): \[ 1 + \cot^2 \theta = \csc^2 \theta \]

Explanation: Standard identity.
Quick Tip: \(1 + \cot^2 \theta = \csc^2 \theta\).

Frequency:
- 5 → 3 times
- 7 → 2
- 9 → 2
- Others → 1

Mode = Solution5Solution

Explanation: Highest frequency.
Quick Tip: Mode = most frequent value.

\[ P(E') = 1 - P(E) = 1 - 0.02 = 0.98 \]

Explanation: Complement rule.
Quick Tip: \(P(not E) = 1 - P(E)\).

Total outcomes = 36
Same numbers: (1,1), (2,2), ..., (6,6) → 6 \[ P = \dfrac{6}{36} = \dfrac{1}{6} \]

Explanation: Equal faces → 6 cases.
Quick Tip: Same on both dice → 6 out of 36.

Outcomes: HH, HT, TH, TT → 4
Favorable: HH → 1 \[ P = \dfrac{1}{4} \]

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

Total months = 12
Favorable = 2 \[ P = \dfrac{2}{12} = \dfrac{1}{6} \]

Explanation: Two specific months.
Quick Tip: Equal chance per month.

Favorable: 4, 5 → 2
Total: 6 \[ P = \dfrac{2}{6} = \dfrac{1}{3} \]

Explanation: Two outcomes out of six.
Quick Tip: \(P(or) = P(A) + P(B)\).

Volume \(V = \dfrac{4}{3} \pi r^3\) → \(V \propto r^3\)
Surface area \(S = 4 \pi r^2\) → \(S \propto r^2\)

Given \(V_1 : V_2 = 64 : 125\) \[ \left(\dfrac{r_1}{r_2}\right)^3 = \dfrac{64}{125} \quad \Rightarrow \quad \dfrac{r_1}{r_2} = \dfrac{4}{5} \] \[ S_1 : S_2 = r_1^2 : r_2^2 = 16 : 25 \]

Explanation: Ratio of areas = (ratio of radii)\(^2\).
Quick Tip: Volume ratio → cube root → square for area.

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

Ratio: \[ \dfrac{V_1}{V_2} = \dfrac{r_1^2 h_1}{r_2^2 h_2} = \left(\dfrac{r_1}{r_2}\right)^2 \times \dfrac{h_1}{h_2} = (4:5)^2 \times (6:7) \] \[ = 16:25 \times 6:7 = 96:175 ? \quad Wait \] \[ 16 \times 6 : 25 \times 7 = 96 : 175 \]
But option (A) is 96:125?

Wait: \[ (4)^2 \times 6 : (5)^2 \times 7 = 16 \times 6 : 25 \times 7 = 96 : 175 \]
So Solution96:175Solution → option (B)

SolutionCorrect Answer: (B) 96:175Solution

Wait: But let's confirm: \[ \dfrac{96}{175} = \dfrac{96 \div 1}{175 \div 1} = 96:175 \]
Yes.

Explanation: Volume ∝ \(r^2 h\).
Quick Tip: Multiply square of radius ratio by height ratio.

Total surface area of hemisphere = Curved surface area + Base area
- Curved surface area = \(2\pi R^2\)
- Base area (circle) = \(\pi R^2\)
\[ Total = 2\pi R^2 + \pi R^2 = 3\pi R^2 \]

Verification:
- Full sphere: \(4\pi R^2\)
- Hemisphere: half curved (\(2\pi R^2\)) + base (\(\pi R^2\)) = \(3\pi R^2\)

Explanation: Includes lateral (curved) and flat circular base.
Quick Tip: Hemisphere TSA = \(3\pi R^2\).

Curved surface area of cone: \[ \pi r l = 880 \]
Given: \(r = 14\), \(\pi = \dfrac{22}{7}\) \[ \dfrac{22}{7} \times 14 \times l = 880 \] \[ 22 \times 2 \times l = 880 \quad \Rightarrow \quad 44l = 880 \quad \Rightarrow \quad l = \dfrac{880}{44} = 20 \]

Verification: \[ \pi \times 14 \times 20 = \dfrac{22}{7} \times 280 = 22 \times 40 = 880 \quad \]

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

Space diagonal of cube with edge \(a\): \[ d = a\sqrt{3} \]
Given: \(d = 2\sqrt{3}\) \[ a\sqrt{3} = 2\sqrt{3} \quad \Rightarrow \quad a = 2 \]

Verification:
- Edge 2 cm → diagonal = \(2\sqrt{3} \approx 3.46\) cm
- Matches given.

Explanation: Diagonal passes through three dimensions.
Quick Tip: Cube diagonal = \(a\sqrt{3}\).

Original TSA = \(6a^2\)
New edge = \(2a\)
New TSA = \(6(2a)^2 = 6 \times 4a^2 = 24a^2\) \[ \dfrac{New}{Old} = \dfrac{24a^2}{6a^2} = 4 \]

Explanation: TSA \(\propto a^2\), so doubles edge → \(2^2 = 4\) times.
Quick Tip: Scale factor \(k\) → area scales by \(k^2\).

Sphere TSA = \(4\pi R^2\)
Hemisphere TSA = \(3\pi R^2\) \[ \dfrac{Sphere}{Hemisphere} = \dfrac{4\pi R^2}{3\pi R^2} = \dfrac{4}{3} \]

Explanation: Sphere has full surface; hemisphere has curved + base.
Quick Tip: Sphere: \(4\pi R^2\); Hemisphere: \(3\pi R^2\).

Curved surface area of hemisphere: \[ 2\pi R^2 = 1232 \] \[ R^2 = \dfrac{1232}{2\pi} = \dfrac{616}{\pi} \]
Using \(\pi = \dfrac{22}{7}\): \[ R^2 = 616 \times \dfrac{7}{22} = 28 \times 7 = 196 \quad \Rightarrow \quad R = 14 \]

Verification: \[ 2 \times \dfrac{22}{7} \times 196 = 2 \times 22 \times 28 = 1232 \quad \]

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

Given: \[ \cos^2 \theta + \cos \theta = 1 \]
Let \(u = \cos \theta\): \[ u^2 + u - 1 = 0 \] \[ u = \dfrac{-1 \pm \sqrt{5}}{2} \]
Take \(u = \dfrac{-1 + \sqrt{5}}{2} \approx 0.618\) (since \(|\cos \theta| \leq 1\))

Then: \[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - u^2 \]
From equation: \(u^2 = 1 - u\) \[ \sin^2 \theta = 1 - (1 - u) = u \] \[ \sin^4 \theta = (\sin^2 \theta)^2 = u^2 = 1 - u \] \[ \sin^2 \theta + \sin^4 \theta = u + (1 - u) = 1 \]

Explanation: Algebraic manipulation using given equation.
Quick Tip: Substitute \(\cos^2 \theta = 1 - \cos \theta\).

Numerator: \(1 + \tan^2 A = \sec^2 A\)
Denominator: \(1 + \cot^2 A = \csc^2 A\) \[ \dfrac{\sec^2 A}{\csc^2 A} = \dfrac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A \]

Explanation: Use reciprocal identities.
Quick Tip: \(\dfrac{\sec^2}{\csc^2} = \tan^2\).

For equal roots, discriminant \(D = 0\): \[ D = b^2 - 4ac = (-6)^2 - 4(k)(1) = 36 - 4k = 0 \] \[ 36 = 4k \quad \Rightarrow \quad k = 9 \]

Verification:
- \(k = 9\): \(9x^2 - 6x + 1 = 0\) → \(D = 36 - 36 = 0\) → equal roots.

Explanation: \(D = 0\) for repeated root.
Quick Tip: Set \(b^2 = 4ac\).

By Factor Theorem: If \(p(a) = 0\), then \((x - a)\) is a factor.
Given zero = 2 → \(p(2) = 0\) → \((x - 2)\) is a factor.

Explanation: Root determines linear factor.
Quick Tip: Root \(r\) → factor \((x - r)\).

For quadratic \(cx^2 + ax + b = 0\),
Product of roots: \[ \alpha \beta = \dfrac{c}{c} = \dfrac{b}{c} \]
(Standard: \(\alpha \beta = \dfrac{constant term}{leading coefficient}\))

Verification:
- \(2x^2 + 5x + 3 = 0\) → roots \(-\frac{3}{2}, -1\) → product = \(\frac{3}{2}\) = \(\frac{b}{c}\).

Explanation: From Vieta's formulas.
Quick Tip: Product of roots = \(\dfrac{constant}{leading}\).

A quadratic equation has highest power 2.

- (A): RHS has \(x^3\) → cubic → not quadratic


- (B): Expand LHS: \(x^2 + 6x + 9\)
RHS: \(4x + 16\)
\(\Rightarrow x^2 + 6x + 9 - 4x - 16 = 0 \Rightarrow x^2 + 2x - 7 = 0\) → quadratic


- (C): LHS: \(4x^2 - 8x + 4\)
RHS: \(4x^2 + 7\)
\(\Rightarrow -8x - 3 = 0\) → linear


- (D): Multiply by \(4x\): \(16x^2 + 1 = 16x^2\) → \(1 = 0\) → contradiction, not quadratic


Explanation: Simplify to standard form \(ax^2 + bx + c = 0\).
Quick Tip: Bring to one side → check highest power.

A quadratic equation must have the highest power of \(x\) equal to 2 after simplification.

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

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

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

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

Only (D) simplifies to a linear equation.

Explanation: Expand and bring all terms to one side; check the highest degree.
Quick Tip: Simplify to \(ax^2 + bx + c = 0\) → if \(a = 0\), not quadratic.

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

Verification:
- Roots: \(\dfrac{7 \pm 1}{4}\) → \(\dfrac{8}{4} = 2\), \(\dfrac{6}{4} = 1.5\)

Explanation: \(D > 0\) → two real roots.
Quick Tip: \(D = b^2 - 4ac\).

Equation \(x = 2\) is a vertical line at \(x = 2\).
Any point with \(x\)-coordinate = 2 lies on it.
- (2,0): \(x = 2\) → yes
- (2,1): \(x = 2\) → yes
- (2,2): \(x = 2\) → yes

→ All points lie on the line.

Explanation: Vertical line: all points with same x.
Quick Tip: \(x = k\) → all \((k, y)\).

Three terms are in Arithmetic Progression (A.P.) if the difference between consecutive terms is constant.

Let the terms be:
- First: \(P + 1\)
- Second: \(2P + 1\)
- Third: \(4P - 1\)

For A.P.: \[ Second - First = Third - Second \] \[ (2P + 1) - (P + 1) = (4P - 1) - (2P + 1) \]

Simplify:
Left: \[ 2P + 1 - P - 1 = P \]

Right: \[ 4P - 1 - 2P - 1 = 2P - 2 \]

So: \[ P = 2P - 2 \] \[ P - 2P = -2 \quad \Rightarrow \quad -P = -2 \quad \Rightarrow \quad P = 2 \]

Verification:
Substitute \(P = 2\):
- \(P + 1 = 3\)
- \(2P + 1 = 5\)
- \(4P - 1 = 7\)

Terms: 3, 5, 7
Common difference: \(5 - 3 = 2\), \(7 - 5 = 2\) → A.P.

Check other options:
- \(P = 1\): 2, 3, 3 → not A.P.
- \(P = 3\): 4, 7, 11 → differences 3, 4 → no
- \(P = 4\): 5, 9, 15 → differences 4, 6 → no

Only \(P = 2\) works.

Explanation: Equal common differences ensure A.P.
Quick Tip: For three terms in A.P.: \(2 \times middle = first + third\).

A.P.: 1, 5, 9, ...
Common difference \(d = a_2 - a_1 = 5 - 1 = 4\)
Also: 9 - 5 = 4

Explanation: Subtract consecutive terms.
Quick Tip: \(d = a_{n+1} - a_n\).

\(a = 5\), \(d = 3\) \(a_n = a + (n-1)d = 38\) \[ 5 + (n-1)3 = 38 \] \[ (n-1)3 = 33 \quad \Rightarrow \quad n-1 = 11 \quad \Rightarrow \quad n = 12 \]
Wait: 11 × 3 = 33, yes → n = 12

Wait: n-1 = 11 → n = 12

But options say 11th?

Wait: Let's check:
11th term: 5 + 10×3 = 5 + 30 = 35
12th: 5 + 11×3 = 5 + 33 = 38 → 12th

Correct Answer: (C) 12th

Verification:
- 1st: 5
- 2nd: 8
- ...
- 12th: 5 + 11×3 = 38

Explanation: \(n = \dfrac{a_n - a}{d} + 1\).
Quick Tip: \(n = \dfrac{term - first}{d} + 1\).

Co-function identity: \[ \sin(90^\circ - \theta) = \cos \theta \]

Explanation: Complementary angles.
Quick Tip: \(\sin(90^\circ - \theta) = \cos \theta\).

\[ \cos(\alpha - \beta) = \cos(60^\circ - 60^\circ) = \cos 0^\circ = 1 \]

Explanation: Difference is zero.
Quick Tip: \(\cos 0^\circ = 1\).

Given \(\theta = 45^\circ\): \[ \sin 45^\circ = \dfrac{1}{\sqrt{2}}, \quad \cos 45^\circ = \dfrac{1}{\sqrt{2}} \] \[ \sin \theta + \cos \theta = \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \sqrt{2} \]

Alternative: \[ \sin \theta + \cos \theta = \sqrt{2} \left( \dfrac{1}{\sqrt{2}} \sin \theta + \dfrac{1}{\sqrt{2}} \cos \theta \right) = \sqrt{2} \sin(\theta + 45^\circ) \]
At \(\theta = 45^\circ\): \(\sin 90^\circ = 1\) → \(\sqrt{2} \times 1 = \sqrt{2}\).

Explanation: Both \(\sin\) and \(\cos\) are equal at \(45^\circ\).
Quick Tip: At \(45^\circ\), \(\sin = \cos = \dfrac{1}{\sqrt{2}}\).

The expression is the double-angle formula: \[ \tan 2A = \dfrac{2 \tan A}{1 - \tan^2 A} \]

Given \(A = 30^\circ\): \[ \tan 2A = \tan 60^\circ = \sqrt{3} \]

Compute directly: \[ \tan 30^\circ = \dfrac{1}{\sqrt{3}} \] \[ \tan^2 30^\circ = \dfrac{1}{3} \] \[ \dfrac{2 \cdot \dfrac{1}{\sqrt{3}}}{1 - \dfrac{1}{3}} = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} = \dfrac{2}{\sqrt{3}} \cdot \dfrac{3}{2} = \sqrt{3} = \tan 60^\circ \]

Explanation: Identity for \(\tan 2A\).
Quick Tip: \(\dfrac{2 \tan A}{1 - \tan^2 A} = \tan 2A\).

\[ \tan \theta = \dfrac{opposite}{adjacent} = \dfrac{12}{5} \]
Let opposite = 12, adjacent = 5.
Hypotenuse: \[ \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \] \[ \sin \theta = \dfrac{opposite}{hypotenuse} = \dfrac{12}{13} \]

Verification: \[ \cos \theta = \dfrac{5}{13}, \quad \tan \theta = \dfrac{12/13}{5/13} = \dfrac{12}{5} \quad \]

Explanation: Use Pythagorean triple 5-12-13.
Quick Tip: \(\sin \theta = \dfrac{\tan \theta}{\sqrt{1 + \tan^2 \theta}}\).

Use co-function identities: \[ \cos 59^\circ = \cos(90^\circ - 31^\circ) = \sin 31^\circ \] \[ \tan 80^\circ = \tan(90^\circ - 10^\circ) = \cot 10^\circ \]

So: \[ \dfrac{\cos 59^\circ}{\sin 31^\circ} = \dfrac{\sin 31^\circ}{\sin 31^\circ} = 1 \] \[ \dfrac{\tan 80^\circ}{\cot 10^\circ} = \dfrac{\cot 10^\circ}{\cot 10^\circ} = 1 \] \[ 1 \times 1 = 1 \]

Explanation: Complementary angles make ratios 1.
Quick Tip: \(\cos(90^\circ - \theta) = \sin \theta\), \(\tan(90^\circ - \theta) = \cot \theta\).

\[ \tan 65^\circ = \tan(90^\circ - 25^\circ) = \cot 25^\circ = \dfrac{1}{\tan 25^\circ} \] \[ \tan 25^\circ \times \tan 65^\circ = \tan 25^\circ \times \dfrac{1}{\tan 25^\circ} = 1 \] \[ \sin A = 1 \quad \Rightarrow \quad A = 90^\circ \]

Explanation: Product of \(\tan \theta\) and \(\tan(90^\circ - \theta)\) = 1.
Quick Tip: \(\tan \theta \cdot \tan(90^\circ - \theta) = 1\).

\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - x^2 \] \[ \sin \theta = \sqrt{1 - x^2} \quad (\sin \theta > 0 in acute angle) \] \[ \tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{\sqrt{1 - x^2}}{x} \]

Explanation: Use identity \(\sin^2 \theta + \cos^2 \theta = 1\).
Quick Tip: \(\tan \theta = \dfrac{\sqrt{1 - \cos^2 \theta}}{\cos \theta}\).

\[ 1 - \cos^4 \theta = (1 - \cos^2 \theta)(1 + \cos^2 \theta) \quad (difference of squares) \] \[ 1 - \cos^2 \theta = \sin^2 \theta \] \[ \Rightarrow 1 - \cos^4 \theta = \sin^2 \theta (1 + \cos^2 \theta) \]

Verification:
Let \(\cos^2 \theta = 0.36\):
LHS: \(1 - (0.36)^2 = 1 - 0.1296 = 0.8704\)
RHS: \(\sin^2 \theta = 0.64\), \(1 + 0.36 = 1.36\) → \(0.64 \times 1.36 = 0.8704\) \(\checkmark\)

Explanation: Factorize using \(a^2 - b^2 = (a-b)(a+b)\).
Quick Tip: \(1 - \cos^4 \theta = (1 - \cos^2 \theta)(1 + \cos^2 \theta)\).

The y-axis is the line where \(x = 0\).
Any point on y-axis: \((0, y)\), where \(y\) can vary.

- (A) \((y, 0)\): on x-axis
- (B) \((2, y)\): parallel to y-axis at \(x = 2\)
- (C) \((0, x)\): notation incorrect — should be \((0, y)\)

None of the options are correct.
Correct form: \((0, y)\)

Explanation: x-coordinate is zero on y-axis.
Quick Tip: Y-axis: \(x = 0\) → \((0, y)\).

The ladder forms a right triangle with the wall and the ground.

Hypotenuse (ladder) = 7 m, angle with wall = \(30^\circ\).

The height \(h\) on the wall is the side adjacent to the \(30^\circ\) angle.
\[ \cos 30^\circ = \dfrac{h}{7} \]
\[ \cos 30^\circ = \dfrac{\sqrt{3}}{2} \quad \Rightarrow \quad h = 7 \times \dfrac{\sqrt{3}}{2} = \dfrac{7\sqrt{3}}{2} \, m \]

Alternatively, the angle with the ground is \(60^\circ\), so
\[ \sin 60^\circ = \dfrac{h}{7} \quad \Rightarrow \quad h = 7 \times \dfrac{\sqrt{3}}{2} = \dfrac{7\sqrt{3}}{2} \, m \]
Quick Tip: When the angle is given with the wall, use \(\cos\) for height.

Given: ABCD is a parallelogram \(\Rightarrow\) AB \(\parallel\) CD, AD \(\parallel\) BC.

E lies on AD extended, and BE intersects CD at F.

To prove: \(\triangle ABE \sim \triangle CFB\).



Step 1: \(\angle BAE = \angle FCB\)

Since AB \(\parallel\) CD and BE is a transversal,
\(\angle BAE\) and \(\angle FCB\) are alternate interior angles \(\Rightarrow\) equal.



Step 2: \(\angle ABE = \angle CBF\)

These are vertically opposite angles at B \(\Rightarrow\) equal.



Step 3: Since two pairs of angles are equal, the third pair is also equal:
\(\angle AEB = \angle FBC\) (by angle sum in a triangle).



Thus, \(\triangle ABE \sim \triangle CFB\) by AAA similarity.
Quick Tip: Use parallel lines to get alternate interior angles and vertically opposite angles.

Given: \(\triangle ABC\) with \(\angle C = 90^\circ\) and AB = AC (isosceles).

Let AC = BC = \(x\).

By Pythagoras theorem:
\[ AB^2 = AC^2 + BC^2 = x^2 + x^2 = 2x^2 \]
\[ AB^2 = 2AC^2 \]

Hence proved.
Quick Tip: In an isosceles right triangle, the hypotenuse is \(x\sqrt{2}\).

Given: AB = AC \(\Rightarrow\) isosceles, AD \(\perp\) BC, E on CB extended, EF \(\perp\) AC.



1. \(\angle ADB = 90^\circ\) (given), \(\angle EFC = 90^\circ\) (given).

2. Since AB = AC, \(\angle ABC = \angle ACB\).

E is on CB extended \(\Rightarrow\) BE is a transversal to BC and AC.

\(\angle FCE = \angle ACB\) (corresponding angles).

Thus, \(\angle ABD = \angle FCE\).



Both triangles have one right angle and one equal angle \(\Rightarrow\) third angles equal.

Hence, \(\triangle ABD \sim \triangle ECF\) by AA similarity.
Quick Tip: Use right angles and base angles of isosceles triangle.

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

Construct \(\triangle PQM\) such that PQ = AB, PR = BC, PM = AD.

Then \(\triangle PQM \cong \triangle ABC\) by SSS.

M is the midpoint of QR (since PM is median and AD is median).

Thus, \(\triangle PQM \cong \triangle ABC\) implies corresponding angles equal.

Hence, \(\triangle PQR \sim \triangle ABC\) by SAS similarity with included median.
Quick Tip: Construct congruent triangle using two sides and median.

Area ratio: \(\dfrac{9}{64}\).

Side ratio: \(\sqrt{\dfrac{9}{64}} = \dfrac{3}{8}\).
\[ \dfrac{AB}{DE} = \dfrac{3}{8} \quad \Rightarrow \quad AB = 5.1 \times \dfrac{3}{8} \]
\[ 5.1 = \dfrac{51}{10} \quad \Rightarrow \quad AB = \dfrac{51}{10} \times \dfrac{3}{8} = \dfrac{153}{80} = 1.9125 \, cm \]
Quick Tip: Side ratio = square root of area ratio.

Using polynomial division:
\[ x^3 + 1 = (x + 1)(x^2 - x + 1) \]

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

Quotient = \(x^2 - x + 1\), Remainder = 0.
Quick Tip: Use identity \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\).

\[ 1188 = 504 \times 2 + 180 \]
\[ 504 = 180 \times 2 + 144 \]
\[ 180 = 144 \times 1 + 36 \]
\[ 144 = 36 \times 4 + 0 \]

H.C.F. = 36.
Quick Tip: Continue until remainder is 0.

\(a = 2\), \(b = 5\), \(c = -3\).
\[ D = b^2 - 4ac = 25 - 4(2)(-3) = 25 + 24 = 49 \]

Since \(D > 0\), there are two distinct real roots.
Quick Tip: \(D > 0\) \(\Rightarrow\) two real distinct roots.

Section formula (internal):
\[ x = \dfrac{mx_2 + nx_1}{m+n}, \quad y = \dfrac{my_2 + ny_1}{m+n} \]

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

Point: \(\left(1, \dfrac{27}{5}\right)\).
Quick Tip: Weighted average of coordinates.

Shoelace formula:
\[ Area = \dfrac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
\[ = \dfrac{1}{2} \left| -5(-5 - 2) + 3(2 - (-1)) + 5(-1 - (-5)) \right| \]
\[ = \dfrac{1}{2} \left| 35 + 9 + 20 \right| = \dfrac{64}{2} = 32 \]
Quick Tip: List vertices in order and repeat first at end.

Space diagonal of cube: \(a\sqrt{3} = 9\sqrt{3}\).
\[ a = 9 \, cm \]

Total surface area = \(6a^2 = 6 \times 81 = 486 \, cm^2\).
Quick Tip: Cube diagonal = \(a\sqrt{3}\).

Assume \(5 - \sqrt{3} = r\) (rational).
\[ \sqrt{3} = 5 - r \]

Square both sides:
\[ 3 = (5 - r)^2 = 25 - 10r + r^2 \]
\[ r^2 - 10r + 22 = 0 \]

Discriminant = \(100 - 88 = 12\) (not a perfect square) \(\Rightarrow\) \(r\) irrational.

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

Points are collinear if area of triangle is zero:
\[ \dfrac{1}{2} \left| 1(k - 4) + 3(4 - 1) + (-1)(1 - k) \right| = 0 \]
\[ | k - 4 + 9 - 1 + k | = 0 \]
\[ | 2k + 4 | = 0 \quad \Rightarrow \quad k = -2 \]

Alternatively, 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: Use area = 0 or equal slopes.

Let point be \((0, y)\).

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:
\[ 36 + (y-5)^2 = 16 + (y-3)^2 \]
\[ 36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9 \]
\[ 61 - 10y = 25 - 6y \]
\[ 36 = 4y \quad \Rightarrow \quad y = 9 \]

Point: \((0, 9)\).
Quick Tip: Set distance formulas equal.

Given: \(\tan \theta = \dfrac{5}{12}\).

Consider a right triangle where opposite = 5, adjacent = 12.

Hypotenuse = \(\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\).
\[ \sin \theta = \dfrac{opposite}{hypotenuse} = \dfrac{5}{13} \]
\[ \cos \theta = \dfrac{adjacent}{hypotenuse} = \dfrac{12}{13} \]
\[ \sin \theta + \cos \theta = \dfrac{5}{13} + \dfrac{12}{13} = \dfrac{17}{13} \]
Quick Tip: Use the 5-12-13 Pythagorean triple.

Given: \(\sin 3A = \cos(A - 26^\circ)\).

We know \(\sin x = \cos(90^\circ - x)\).
\[ \sin 3A = \cos(90^\circ - 3A) \]

So: \[ 90^\circ - 3A = A - 26^\circ \]
\[ 90^\circ + 26^\circ = 3A + A \]
\[ 116^\circ = 4A \quad \Rightarrow \quad A = 29^\circ \]

Check: \(3A = 87^\circ\) (acute), \(\sin 87^\circ = \cos(90^\circ - 87^\circ) = \cos 3^\circ\).

Right side: \(\cos(29^\circ - 26^\circ) = \cos 3^\circ\).

Equal.
Quick Tip: Use \(\sin x = \cos(90^\circ - x)\).

Let smaller number = \(x\).

Larger number = \(\dfrac{7}{3}x\).
\[ x + \dfrac{7}{3}x = 50 \]
\[ \dfrac{3x + 7x}{3} = 50 \quad \Rightarrow \quad 10x = 150 \quad \Rightarrow \quad x = 15 \]

Larger = \(\dfrac{7}{3} \times 15 = 35\).

Numbers: 15 and 35.
Quick Tip: Let one variable, express other in terms of it.

Curved surface area = \(\pi r l\).

First, find slant height \(l\): \[ l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \]
\[ CSA = \pi \times 7 \times 25 = 175\pi \, cm^2 \]
Quick Tip: \(l = \sqrt{r^2 + h^2}\), then \(\pi r l\).

The minute hand completes one full circle (360°) in 60 minutes.

In 40 minutes, it sweeps:
\[ \dfrac{40}{60} \times 360^\circ = 240^\circ \]

Area of sector with radius \(r = 7\) cm and angle \(240^\circ\):
\[ Area = \dfrac{\theta}{360^\circ} \times \pi r^2 = \dfrac{240}{360} \times \pi \times 7^2 = \dfrac{2}{3} \times \pi \times 49 = \dfrac{98}{3} \pi \]

Wait: \(\dfrac{2}{3} \times 49 = \dfrac{98}{3}\), yes.

But let's recompute:
\[ \dfrac{240}{360} = \dfrac{2}{3}, \quad 49 \times \dfrac{2}{3} = \dfrac{98}{3} \]

No: \(\dfrac{98}{3} \pi \approx 102.6\), but full circle is \(49\pi \approx 154\), \(\dfrac{2}{3}\) of it is about 102.6, correct.

But earlier mistake: I said \(\dfrac{308}{3}\), wrong.

Correct: \(\dfrac{98}{3} \pi\).

Wait: 7² = 49, \(\dfrac{240}{360} \times 49\pi = \dfrac{2}{3} \times 49\pi = \dfrac{98}{3}\pi\).

Yes.

Correct Answer: \(\dfrac{98}{3} \pi\) cm²
Quick Tip: Fraction of circle = \(\dfrac{time in minutes}{60}\).

Given: \(\tan 7^\circ \cdot \tan 60^\circ \cdot \tan 83^\circ\).

We know \(\tan 60^\circ = \sqrt{3}\).

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

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

Let integers be \(n\) and \(n+1\).
\[ n^2 + (n+1)^2 = 365 \]
\[ n^2 + n^2 + 2n + 1 = 365 \]
\[ 2n^2 + 2n + 1 = 365 \]
\[ 2n^2 + 2n - 364 = 0 \]
\[ n^2 + n - 182 = 0 \]

Discriminant = \(1 + 728 = 729 = 27^2\).
\[ n = \dfrac{-1 \pm 27}{2} \]
\(n = 13\) (positive).

Numbers: 13 and 14.
Quick Tip: Let \(n\) and \(n+1\), solve quadratic.

Let the larger number be \(x\).

Let the smaller number be \(y\).

Given:
1. Difference of squares: \(x^2 - y^2 = 180\)
2. Square of smaller = 8 times larger: \(y^2 = 8x\)
Substitute (2) into (1): \[ x^2 - 8x = 180 \]
\[ x^2 - 8x - 180 = 0 \]

This is the required quadratic equation in terms of the larger number.
Quick Tip: Define larger and smaller clearly before forming equations.

Given: \(\dfrac{PS}{SQ} = \dfrac{PT}{TR} = k\) (say), and \(\angle PST = \angle PRQ\).

Let \(\angle PRQ = \gamma\).

Then \(\angle PST = \gamma\).

In \(\triangle PST\) and \(\triangle PQR\):

- \(\angle PST = \angle PRQ = \gamma\)

- \(\dfrac{PS}{PQ} = \dfrac{PS}{PS + SQ} = \dfrac{k}{k+1}\), similarly for PT.

By basic proportionality theorem (Thales), ST \(\parallel\) QR.

Since ST \(\parallel\) QR, \(\angle PST = \angle PQR\) (corresponding angles).

But \(\angle PST = \angle PRQ = \gamma\).
\(\angle PQR = \gamma = \angle PRQ\).

Thus, PQ = PR (base angles equal).

Hence, \(\triangle PQR\) is isosceles with PQ = PR.
Quick Tip: Use Thales' theorem and corresponding angles.

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

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

Roots: \[ x = \dfrac{-b \pm \sqrt{D}}{2a} = \dfrac{2\sqrt{2} \pm 0}{4} = \dfrac{2\sqrt{2}}{4} = \dfrac{\sqrt{2}}{2} \]

Repeated root: \(\dfrac{\sqrt{2}}{2}\).
Quick Tip: \(D = 0\) \(\Rightarrow\) equal roots.

This is an A.P. with:
- First term \(a = 3\)
- Common difference \(d = 11 - 3 = 8\)
- Last term \(l = 67\)
Find number of terms \(n\): \[ l = a + (n-1)d \]
\[ 67 = 3 + (n-1) \times 8 \]
\[ 64 = (n-1) \times 8 \quad \Rightarrow \quad n-1 = 8 \quad \Rightarrow \quad n = 9 \]

Sum of A.P.: \[ S_n = \dfrac{n}{2} (a + l) = \dfrac{9}{2} (3 + 67) = \dfrac{9}{2} \times 70 = 9 \times 35 = 315 \]

Verification by listing:
3, 11, 19, 27, 35, 43, 51, 59, 67 → 9 terms.
Sum = 315.
Quick Tip: Use \(S_n = \dfrac{n}{2} (first + last)\).

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

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

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

Subtract first from second: \[ (a + 8d) - (a + 4d) = 79 - 43 \]
\[ 4d = 36 \quad \Rightarrow \quad d = 9 \]

Substitute into first equation: \[ a + 4(9) = 43 \quad \Rightarrow \quad a + 36 = 43 \quad \Rightarrow \quad a = 7 \]

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

Verification:
- 5th term: 7 + 4×9 = 43

- 9th term: 7 + 8×9 = 79
Quick Tip: Subtract term equations to eliminate \(a\).

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

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

Assuming \(\sin \theta > 0\), \[ = \dfrac{1 + \cos \theta}{\sin \theta} = RHS \]

Hence proved.
Quick Tip: Rationalize inside square root.

RHS: \(\cot 63^\circ \cdot \cot 81^\circ\).
\(\cot 63^\circ = \cot(90^\circ - 27^\circ) = \tan 27^\circ\).
\(\cot 81^\circ = \cot(90^\circ - 9^\circ) = \tan 9^\circ\).
\[ \cot 63^\circ \cdot \cot 81^\circ = \tan 27^\circ \cdot \tan 9^\circ = LHS \]

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

Given: \(\cos A = \dfrac{4}{5}\).
\[ \sin^2 A = 1 - \cos^2 A = 1 - \dfrac{16}{25} = \dfrac{9}{25} \quad \Rightarrow \quad \sin A = \dfrac{3}{5} \quad (acute angle) \]
\[ \cot A = \dfrac{\cos A}{\sin A} = \dfrac{4/5}{3/5} = \dfrac{4}{3} \]
\[ \cosec A = \dfrac{1}{\sin A} = \dfrac{5}{3} \]
Quick Tip: Use \(\sin^2 A + \cos^2 A = 1\).

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.