Smallest 2-digit number = 10

Smallest composite number = 4

We need to find the HCF of 10 and 4.

Prime factorization of 10: \( 10 = 2 \times 5 \)

Prime factorization of 4: \( 4 = 2 \times 2 \)

The common factor is 2, so the HCF is 2.

For infinitely many solutions, the ratios of coefficients must be equal:

\( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \)

Here, \( a_1 = 1, b_1 = 1, c_1 = -4 \)

\( a_2 = 2, b_2 = k, c_2 = -8 \)

So, \( \dfrac{1}{2} = \dfrac{1}{k} = \dfrac{-4}{-8} \)

From \( \dfrac{1}{2} = \dfrac{1}{k} \), we get \( k = 2 \)

Also, \( \dfrac{-4}{-8} = \dfrac{1}{2} \), which is consistent.

Therefore, \( k = 2 \)

Substitute \( x = 2 \) in each equation:

A: \( (2)^2 - 4(2) + 5 \) \(= 4 - 8 + 5 \) \(= 1 \ne 0 \)

B: \( (2)^2 + 3(2) - 12 \) \(= 4 + 6 - 12 \) \(= -2 \ne 0 \)

C: \( 2(2)^2 - 7(2) + 6 \) \(= 8 - 14 + 6 \) \(= 0 \) ✓

D: \( 3(2)^2 - 6(2) - 2 \) \(= 12 - 12 - 2 \) \(= -2 \ne 0 \)

Only equation C has 2 as a root.

The nth term of an A.P. is given by \( a_n = a + (n-1)d \)

Given: \( a_7 = 4 \) and \( d = -4 \)

So, \( a + (7-1)(-4) = 4 \)

\( a + 6(-4) = 4 \)

\( a - 24 = 4 \)

\( a = 4 + 24 = 28 \)

Therefore, the first term is 28.

The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

\( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Here, the points are (5, 4) and (0, 0)

Distance = \( \sqrt{(5-0)^2 + (4-0)^2} \) \( = \sqrt{25 + 16} \) \(= \sqrt{41} \)

Therefore, the distance is \( \sqrt{41} \).

Given: \( \sin A = \dfrac{3}{5} \)

In a right triangle, \( \sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} \)

So, opposite side = 3, hypotenuse = 5

Using Pythagoras theorem: adjacent side = \( \sqrt{5^2 - 3^2} \) \(= \sqrt{25 - 9} \) \(= \sqrt{16} = 4 \)

\( \cot A = \dfrac{\text{adjacent}}{\text{opposite}} \) \( = \dfrac{4}{3} \)

Therefore, \( \cot A = \dfrac{4}{3} \).

We know that:

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

\( 1 + \cot^2 A = \csc^2 A \)

So, \( \dfrac{1 + \tan^2 A}{1 + \cot^2 A} \) \( = \dfrac{\sec^2 A}{\csc^2 A} \) \( = \dfrac{\dfrac{1}{\cos^2 A}}{\dfrac{1}{\sin^2 A}} \) \( = \dfrac{\sin^2 A}{\cos^2 A} \) \( = \tan^2 A \)

Therefore, the expression equals \( \tan^2 A \).

We know that \( \tan 30^\circ = \dfrac{1}{\sqrt{3}} \)

So, \( \tan^2 30^\circ = \left( \dfrac{1}{\sqrt{3}} \right)^2 = \dfrac{1}{3} \)

Now, \( \dfrac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} \) \( = \dfrac{2 \times \dfrac{1}{\sqrt{3}}}{1 - \dfrac{1}{3}} \) \( = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} \) \( = \dfrac{2}{\sqrt{3}} \times \dfrac{3}{2} \) \( = \dfrac{3}{\sqrt{3}} \) \( = \sqrt{3} \)

And \( \tan 60^\circ = \sqrt{3} \)

Therefore, the expression equals \( \tan 60^\circ \).

A quadratic polynomial with zeroes \( \alpha \) and \( \beta \) is given by:

\( x^2 - (\alpha + \beta)x + \alpha\beta \)

Given: \( \alpha + \beta = -5 \) and \( \alpha\beta = 6 \)

So, the polynomial is \( x^2 - (-5)x + 6 = x^2 + 5x + 6 \)

Therefore, the quadratic polynomial is \( x^2 + 5x + 6 \).

To find the zeroes of \( 3x^2 + 11x - 4 = 0 \), we can use the quadratic formula:

\( x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3, b = 11, c = -4 \)

Discriminant = \( b^2 - 4ac = 121 - 4(3)(-4) \) \(= 121 + 48 = 169 \)

\( x = \dfrac{-11 \pm \sqrt{169}}{2 \times 3} = \dfrac{-11 \pm 13}{6} \)

So, \( x = \dfrac{-11 + 13}{6} \) \( = \dfrac{2}{6} = \dfrac{1}{3} \) or \( x = \dfrac{-11 - 13}{6} \) \(= \dfrac{-24}{6} = -4 \)

Therefore, the zeroes are \( \dfrac{1}{3} \) and \(-4\).

Modal class is the class with the highest frequency. Here, frequency 22 is highest for class 0-10.

So, modal class is 0-10, and its upper limit is 10.

For median class, we need to find the class containing the \( \dfrac{n}{2} \)th term, where n = 66.

\( \dfrac{n}{2} = \dfrac{66}{2} = 33 \)

Cumulative frequencies: 0-10: 22, 10-20: 22+10=32, 20-30: 32+8=40

The 33rd term lies in the class 20-30 (since cumulative frequency reaches 40 at this class).

So, median class is 20-30, and its upper limit is 30.

Difference = 30 - 10 = 20

Therefore, the difference is 20.

By definition, probability of any event A always lies between 0 and 1, inclusive.

P(A) = 0 means the event is impossible.

P(A) = 1 means the event is certain.

For any event A, \( 0 \leq P(A) \leq 1 \).

Therefore, the correct option is \( 0 \leq P(A) \leq 1 \).

Total surface area of a solid hemisphere = Curved surface area + Area of base

Curved surface area = \( 2\pi r^2 \)

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

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

Given r = 7 cm

So, total surface area = \( 3\pi (7)^2 = 3\pi \times 49 = 147\pi \, \text{cm}^2 \)

Therefore, the total surface area is \( 147\pi \, \text{cm}^2 \).

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

For minor sector, \(θ = 120°\), \(r = 21 \text {cm}\)

Area of minor sector = \( \dfrac{120}{360} \times \pi \times 21^2 \) \(= \dfrac{1}{3} \times \pi \times 441 \) \(= 147\pi \, \text{cm}^2 \)

For major sector, \(θ = 360° - 120° \) \(= 240°\)

Area of major sector = \( \dfrac{240}{360} \times \pi \times 21^2 \) \(= \dfrac{2}{3} \times \pi \times 441 \) \(= 294\pi \, \text{cm}^2 \)

Difference = Area of major sector - Area of minor sector = \( 294\pi - 147\pi = 147\pi \, \text{cm}^2 \)

Using \( \pi = \dfrac{22}{7} \), difference = \( 147 \times \dfrac{22}{7} \) \(= 21 \times 22 = 462 \, \text{cm}^2 \)

Therefore, the difference is 462 cm².

For two lines to be parallel, their slopes must be equal but their intercepts must be different.

The slope of the line \( a_1x + b_1y = c_1 \) is \( -\dfrac{a_1}{b_1} \)

The slope of the line \( a_2x + b_2y = c_2 \) is \( -\dfrac{a_2}{b_2} \)

For parallel lines: \( -\dfrac{a_1}{b_1} = -\dfrac{a_2}{b_2} \) ⇒ \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \)

But the intercepts must be different: \( \dfrac{c_1}{b_1} \ne \dfrac{c_2}{b_2} \), which implies \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2} \)

Therefore, the correct condition is \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2} \).

Given: ∠APB = 80°

Since PA and PB are tangents from an external point P, PA = PB.

So, triangle APB is isosceles with ∠PAB = ∠PBA.

In triangle APB: ∠PAB + ∠PBA + ∠APB = 180°

Let ∠PAB = ∠PBA = x

Then, x + x + 80° = 180° ⇒ 2x = 100° ⇒ x = 50°

So, ∠PAB = 50°

Now, OA is perpendicular to PA (radius to tangent)

So, ∠OAP = 90°

Therefore, ∠OAB = ∠OAP - ∠PAB = 90° - 50° = 40°

In triangle OAB, OA = OB (radii), so triangle OAB is isosceles.

Thus, ∠OBA = ∠OAB = 40°

Therefore, ∠ABO = 40°.

In geometry:

- A secant is a line that intersects a circle at two distinct points.

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

- A diameter is a chord that passes through the center of the circle.

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

Therefore, a line intersecting a circle in two distinct points is called a secant.

Let θ be the angle of elevation of the sun.

In the right triangle formed by the pole, its shadow, and the sun's rays:

Height of pole = 6 m (opposite side)

Length of shadow = \( 2\sqrt{3} \) m (adjacent side)

So, \( \tan \theta \) \(= \dfrac{\text{opposite}}{\text{adjacent}} \) \(= \dfrac{6}{2\sqrt{3}} \) \(= \dfrac{3}{\sqrt{3}} \) \(= \sqrt{3} \)

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

Therefore, \(θ = 60°\)

The sun's elevation is \(60°\).

Assertion (A) is true - This is the Basic Proportionality Theorem (Thales' Theorem), which states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

Reason (R) is false - We can indeed draw lines parallel to any side of a triangle. In fact, the Basic Proportionality Theorem requires drawing such parallel lines.

Therefore, Assertion (A) is true but Reason (R) is false.

The correct option is C.

Assertion (A) is true - The point (0, 4) has x-coordinate 0, so it lies on the y-axis.

Reason (R) is true - By definition, any point on the y-axis has x-coordinate equal to 0.

Reason (R) correctly explains why Assertion (A) is true.

Therefore, both Assertion (A) and Reason (R) are correct and Reason (R) is the correct explanation of Assertion (A).

The correct option is A.