We know that for a quadratic equation \(ax^2 + bx + c\), the product of the zeroes (roots) is given by: \(\dfrac{c}{a}\).

In our case, the polynomial is \(kx^2 - 4x - 7\), where: \(a = k\), \(b = -4\), \(c = -7\).

We are given that the product of the zeroes is \(2\). Therefore, we can set up the equation: \(\dfrac{c}{a} = 2\).

Substituting the values of \(c\) and \(a\): \(\dfrac{-7}{k} = 2\).

Now, solve for \(k\): \(-7 = 2k \;\Rightarrow\; k = -\dfrac{7}{2}\).

Thus, the correct answer is: (b) \(-\dfrac{7}{2}\).

Quick Tip: For any quadratic equation \(ax^2 + bx + c\), the product of the zeroes is \(\dfrac{c}{a}\).

In an arithmetic progression (A.P.), the \(n\)-th term is given by the formula:

\(a_n = a + (n-1)d\)

Where:
- \(a_n\) is the \(n\)-th term,
- \(a\) is the first term,
- \(d\) is the common difference.

We are given:
- \(a = 8\) (the first term),
- \(a_{10} = -19\) (the 10th term),
- We need to find \(d\) (the common difference).

Substitute the known values into the formula for the 10th term:

\(a_{10} = a + (10-1)d \) \(\;\Rightarrow\; -19 = 8 + 9d\)

Now, solve for \(d\):

\(-19 - 8 = 9d \) \(\;\Rightarrow\; -27 = 9d \) \(\;\Rightarrow\; d = -3\)

Thus, the correct answer is: (d) -3

Quick Tip: To find the common difference \(d\) in an A.P., use the formula \(a_n = a + (n-1)d\).

The formula for the mid-point of a line segment joining the points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: \( \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\)

Substitute the coordinates of the points \((-1, 3)\) and \(\left(8, \dfrac{3}{2}\right)\):

\( \left(\dfrac{-1 + 8}{2}, \dfrac{3 + \dfrac{3}{2}}{2}\right) \) \(= \left(\dfrac{7}{2}, \dfrac{\dfrac{6}{2} + \dfrac{3}{2}}{2}\right) \) \(= \left(\dfrac{7}{2}, \dfrac{9}{4}\right) \)

Thus, the mid-point of the line segment is \(\left(\dfrac{7}{2}, \dfrac{9}{4}\right)\)

QuickTip : To find the mid-point of a line segment, simply average the x-coordinates and y coordinates of the two points.

Given \(\sin \theta = \dfrac{1}{3}\)

Using the Pythagorean identity,

\( \sin^2 \theta + \cos^2 \theta = 1 \)

\( \Rightarrow \left(\dfrac{1}{3}\right)^2 + \cos^2 \theta = 1 \)

\( \Rightarrow \dfrac{1}{9} + \cos^2 \theta = 1 \)

\( \Rightarrow \cos^2 \theta = 1 - \dfrac{1}{9} = \dfrac{8}{9} \)

\( \Rightarrow \cos \theta = \sqrt{\dfrac{8}{9}} = \dfrac{2\sqrt{2}}{3} \)

\(\therefore \sec \theta = \dfrac{1}{\cos \theta} = \dfrac{1}{\dfrac{2\sqrt{2}}{3}} = \dfrac{3}{2\sqrt{2}} \)

Quick Tip: To find \(\sec \theta\), use the identity \(\sec \theta = \dfrac{1}{\cos \theta}\) and apply the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) when \(\sin \theta\) is given.

To find the Highest Common Factor (HCF) of 132 and 77, use the Euclidean algorithm:

- Divide 132 by 77: \(132 = 1 \times 77 + 55\)

- Divide 77 by 55: \(77 = 1 \times 55 + 22\)

- Divide 55 by 22: \(55 = 2 \times 22 + 11\)

- Divide 22 by 11: \(22 = 2 \times 11 + 0\)

The remainder is now 0, so the HCF is the last non-zero remainder = 11.

Quick Tip: Use the Euclidean algorithm to find HCF by successive division and remainders.

For the quadratic equation \(ax^2 + bx + c = 0\), the condition for real and equal roots is:

\(\Delta = b^2 - 4ac = 0\)

Here, \(a = 4\), \(b = -5\), and \(c = k\)

Using the discriminant condition:

\(\Delta = (-5)^2 - 4 \times 4 \times k = 0\)

\( \Rightarrow 25 - 16k = 0 \)

\(\Rightarrow 16k = 25 \)

\( \Rightarrow k = \dfrac{25}{16} \)

Quick Tip: To find the value of \(k\) for real and equal roots, set the discriminant equal to zero and solve for \(k\).

The total probability of all possible outcomes must sum to 1.

If the probability of winning a game is \(p\), then the probability of losing is : \(1 - p\)

Quick Tip: The sum of the probabilities of all possible outcomes of an event is always \(1\).

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

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

Substituting the points \((x_1, y_1) =(-2, 3)\) and \( (x_2, y_2)=(2, -3)\):

\( d = \sqrt{(2 - (-2))^2 + (-3 - 3)^2} \) \(= \sqrt{(4)^2 + (-6)^2} \) \(= \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \)

\(\therefore\) The distance is \(2\sqrt{13}\) units

Quick Tip: Use the distance formula to find the length of the line segment between two points in coordinate geometry.

Given the expression: \( \sin^2 \theta + \sin \theta + \cos^2 \theta = 2 \)

We know the identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)

So substitute in the given expression:

\( \sin^2 \theta + \sin \theta + \cos^2 \theta = 2 \)

\( \Rightarrow (\sin^2 \theta + \cos^2 \theta) + \sin \theta =2\)

\(\Rightarrow 1 + \sin \theta = 2 \)

\( \Rightarrow \sin \theta = 1 \)

The \(\sin \theta\) is 1 only when: \( \theta = 90^\circ \)

Therefore, the value of \(\theta\) is \(90^\circ\).

Quick Tip: Use the identity \(\sin^2 \theta + \cos^2 \theta = 1\) to simplify expressions, and then solve for \(\theta\) using basic trigonometric values.

There are 2 red queens in a standard deck of 52 cards: one queen of hearts and one queen of diamonds.

Probability of drawing a red queen = \(\dfrac{2}{52} = \dfrac{1}{26}\)

Quick Tip: In a deck of 52 cards, calculate probability by dividing the number of favorable outcomes by the total number of cards.

The median of a data set is the value that divides the data into two equal halves when the data is arranged in order. It represents the middle value, so it splits the data into two equal parts.

Quick Tip: When data is arranged in ascending or descending order, the median is the middle value that divides the dataset into two equal halves.

The volume \(V\) of a sphere of radius \(r\) is given by:

\( V = \dfrac{4}{3} \pi r^3 \)

Given radius \(r = \dfrac{7}{2}\) cm, substitute:

\( V = \dfrac{4}{3} \pi \left(\dfrac{7}{2}\right)^3 \) \( = \dfrac{4}{3} \pi \times \dfrac{343}{8} \) \(= \dfrac{343}{6} \pi \)

Using \(\pi = \dfrac{22}{7}\)

\( V = \dfrac{343}{6} \times \dfrac{22}{7} \) \(= \dfrac{343 \times 22}{42} \) \(= \dfrac{539}{3} \text{ cu cm} \)

Quick Tip: The volume of a sphere is calculated using \(\dfrac{4}{3}\pi r^3\)

In statistics, the relationship between the mean, median, and mode is given by:

\( \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \)

Substitute the values: \( 3 \times 23 - 2 \times 21 \) \(= 69 - 42 = 27 \)

Quick Tip: When mean and median are given, find mode using the formula \(\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}\)

The slant height \(l\) of a right circular cone is given by the Pythagoras theorem:

\( l = \sqrt{r^2 + h^2} \)

Where \(r = 7\) cm (radius) and \(h = 24\) cm (height).

Calculate: \( l = \sqrt{7^2 + 24^2} \) \(= \sqrt{49 + 576} \) \(= \sqrt{625} \) \(= 25 \text{ cm} \)

Quick Tip: Use Pythagoras theorem \(l = \sqrt{r^2 + h^2}\) to find the slant height of a cone when the height and radius are known.

Given quadratic polynomial: \((\alpha -1)x^2 + \alpha x + 1 = 0\)

Since \(-3\) is a zero, substitute \(x = -3\) in the polynomial:

\( (\alpha -1)(-3)^2 + \alpha(-3) + 1 = 0 \)

\(\Rightarrow (\alpha -1) \times 9 - 3\alpha + 1 = 0 \)

\(\Rightarrow 9\alpha - 9 - 3\alpha + 1 = 0 \)

\(\Rightarrow 6\alpha - 8 = 0 \)

\(\Rightarrow 6\alpha = 8 \)

\(\Rightarrow \alpha = \dfrac{8}{6} = \dfrac{4}{3} \)

Quick Tip: To find the value of a constant in a polynomial when a root is given, substitute the root into the polynomial and solve for the constant.

The length of the diameter is 6 cm, so the distance between the two endpoints on the x-axis is 6 units.

Given one endpoint is \((-4, 0)\), let the other endpoint be \((x, 0)\)

Distance between the points:

\( |x - (-4)| = 6 \) \( \Rightarrow |x + 4| = 6 \)

So, \( x + 4 = 6 \Rightarrow x = 2 \) or \( x + 4 = -6 \Rightarrow x = -10 \)

But since it's on the x-axis and consistent with a diameter within the circle, the other endpoint is (2, 0).

Quick Tip: For points on the x-axis with a known segment length, use the distance formula to find the second point.

The condition for no solution of two linear equations \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) is:

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

Given equations:

\( 5x + 2y - 7 = 0\) \( \Rightarrow a_1=5, b_1=2, c_1 = -7 \)

\( 2x + ky + 1 = 0 \) \( \Rightarrow a_2=2, b_2=k, c_2 = 1 \)

Check ratio:

\( \dfrac{5}{2} = \dfrac{2}{k} \) \(\Rightarrow 5k = 4\) \( \Rightarrow k = \dfrac{4}{5} \)

\(\because \dfrac{5}{2} = \dfrac{2}{k} \neq \dfrac{-7}{1} \) the condition of no solution holds for \(k = \dfrac{4}{5}\).

Quick Tip: For no solution, set the ratios of coefficients of \(x\) and \(y\) equal, but not equal to the ratio of constants.

A doublet occurs when both dice show the same number. There are 6 possible doublets:

\( (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \)

Total outcomes when two dice are rolled = \(6 \times 6 = 36\).

Therefore, probability of a doublet = \( \dfrac{6}{36} = \dfrac{1}{6} \)

Quick Tip: When rolling dice, total possible outcomes equal \(6^2 = 36\). Doublets are favorable outcomes where both dice show the same face.

Assertion A: The quadrilateral OAPB is indeed cyclic. This is a property of tangents drawn from an external point to a circle. The angle between the tangents is supplementary to the angle subtended by the chord at the center, hence forming a cyclic quadrilateral.

Reason R: The reason states that in a cyclic quadrilateral, opposite angles are equal. However, this is not necessarily true for the given scenario. The angles in the cyclic quadrilateral formed by tangents from an external point may not be equal, as the quadrilateral is not necessarily inscribed in a circle in the general sense. Therefore, this statement is false

Quick Tip: Remember, a quadrilateral is cyclic if and only if the sum of the opposite angles is 180°. In this case, although the quadrilateral is cyclic, the statement about opposite angles being equal is not universally applicable

Factor the polynomial:

\( x^2 - 2x - 3 = = \)

\(\Rightarrow (x - 3)(x + 1) = 0 \)

So, zeroes are \(x = 3\) and \(x = -1\).

The graph of the polynomial intersects the x-axis where the polynomial is zero, hence at points \((-1, 0)\) and \((3, 0)\).

Both the assertion and reason are true, and the reason explains the assertion.

Quick Tip: The zeroes of a quadratic polynomial are the points where the graph intersects the x-axis.