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Concept:

For a set x₁, x₂, ..., x_n of n observations:

• Mean: \(\rm \bar{x} =\dfrac{x_{1}+x_{2}+\ ...\ +x_{n}}{n}\).
• Median: is the value at the central position when the data is arranged in ascending or descending order.

For odd n, median is the value at the \(\rm \left(\dfrac{n+1}{2}\right)\) position.

For even n, median is the mean of values at \(\rm \left(\dfrac{n}{2}\right)\) and \(\rm \left(\dfrac{n}{2}+1\right)\) position.

• Mode: is the value which occurs the most number of times, i.e. the value with the highest frequency.

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Calculation:

There are 7 observations in each set, so the Median will be the value of observation at the \(\rm \left(\dfrac{7+1}{2}\right)\) = 8^th position, when they are arranged in ascending/descending order.

The values of Mean, Median and Mode for the given sets is summarized below:

 

'Mean, Median and Mode of A are same.' is correct.

Concept:

For a set x₁, x₂, ..., x_n of n observations:

• Mean: \(\rm \bar{x} =\dfrac{x_{1}+x_{2}+\ ...\ +x_{n}}{n}\).
• Variance: \(\rm \sigma^2 =\dfrac{(x_{1}-\bar{x})^2+(x_{2}-\bar{x})^2+...+(x_{n}-\bar{x})^2}{n}=\dfrac{\sum (x_i-\bar x)^2}{n}=\dfrac{\sum x_i^2}{n}-\bar x^2\).
• Standard Deviation: \(\rm \sigma=\sqrt{\sigma^2}=\sqrt{Variance}\).

​

Calculation:

Total number of items in the distribution = Σ f_i = 3 + 6 + 9 + 13 + 8 + 5 + 4 = 48.

The Mean (x̅) of the given set = \(\rm \dfrac{\sum f_i x_i}{\sum f_i}\).

⇒ x̅ = \(\rm \dfrac{6\times3+7\times6+8\times9+9\times13+10\times8+11\times5+12\times4}{48}=\dfrac{432}{48}\) = 9.

Let's calculate the variance using the formula: \(\rm \sigma^2 =\dfrac{\sum x_i^2}{n}-\bar x^2\).

\(\rm \dfrac{\sum {x_i}^2}{n}=\dfrac{6^2\times3+7^2\times6+8^2\times9+9^2\times13+10^2\times8+11^2\times5+12^2\times4}{48}=\dfrac{4012}{48}\) = 83.58.

∴ σ² = 83.58 - 9² = 83.58 - 81 = 2.58.

And, Standard Deviation (σ) = \(\rm \sqrt{\sigma^2}=\sqrt{Variance}=\sqrt{2.58}\) ≈ 1.607.

Concept:

Matrix Multiplication:

• Matrices are multiplied by multiplying each element of a row of the first m×n matrix with the corresponding elements of all the columns of the second n×p matrix to obtain the first row of the product matrix with p columns, and so on for all the m rows of the first matrix.
• Multiplication is only possible when the number of columns of the first matrix is equal to the number of rows of the second matrix.
• A m×n matrix multiplied by a n×p matrix results in a m×p matrix.

 

Trigonometric Identities:

• cos (α ± β) = cos α cos β ∓ sin α sin β.
• sin (α ± β) = sin α cos β ± sin β cos α.
• cos 2θ = cos² θ - sin² θ.
• sin 2θ = 2 sin θ cos θ.

 

Calculations:

Let \(\rm M= \begin{bmatrix}\ \ \ \rm \cos \alpha & \rm \sin \alpha \ \rm -\sin \alpha & \rm \cos \alpha \end{bmatrix}\) and \(\rm N= \begin{bmatrix}\ \ \ \rm \cos \beta & \rm \sin \beta \ \rm -\sin \beta & \rm \cos \beta \end{bmatrix}\).

⇒ \(\rm M\times N= \begin{bmatrix}\ \ \ \cos\alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}\times \begin{bmatrix}\ \ \ \cos \beta & \sin \beta \ -\sin \beta & \cos \beta \end{bmatrix}\)

⇒ \(\rm M\times N = \begin{bmatrix}\cos \alpha \cos \beta - \sin \alpha \sin \beta & \cos \alpha \sin \beta+\sin \alpha \cos \beta \ -\sin \alpha \cos \beta - \cos \alpha \sin \beta & -\sin \alpha \sin \beta + \cos \alpha \cos \beta \end{bmatrix}\)

⇒ \(\rm M\times N= \begin{bmatrix}\cos (\alpha +\beta) & \sin (\alpha +\beta) \ -\sin (\alpha +\beta) & \cos (\alpha +\beta) \end{bmatrix}\).

∴ \(\rm A\times A= A^2=\begin{bmatrix}\cos (\alpha +\alpha) & \sin (\alpha +\alpha) \ -\sin (\alpha +\alpha) & \cos (\alpha +\alpha) \end{bmatrix}=\begin{bmatrix}\ \ \ \rm \cos 2\alpha & \rm \sin 2\alpha \ \rm -\sin 2\alpha & \rm \cos 2\alpha \end{bmatrix}\)

⇒ \(\rm A^2\times A= A^3=\begin{bmatrix}\cos (2\alpha +\alpha) & \sin (2\alpha +\alpha) \ -\sin (2\alpha +\alpha) & \cos (2\alpha +\alpha) \end{bmatrix}=\begin{bmatrix}\ \ \ \rm \cos 3\alpha & \rm \sin 3\alpha \ \rm -\sin 3\alpha & \rm \cos 3\alpha \end{bmatrix}\)

And so on.

Hence, Aⁿ = \(\rm \begin{bmatrix} \rm \cos n\alpha & \ \ \ \rm \sin n\alpha \ \rm -\sin n\alpha & \rm \cos n\alpha \end{bmatrix}\).

Concept:

If α and β are the two roots of the quadratic equation Ax² + Bx + C = 0, then α + β = \(\rm -\frac{B}{A}\) and αβ = \(\rm \frac{C}{A}\).

​

Calculation:

Given: ax2 - 2bx + c = 0

Using the formula for the sum and the product of the roots we get:

m + n = \(\rm \frac{2b}{a}\)                ...(1)

mn = \(\rm \frac{c}{a}\)                ...(2)

Now, the given expression \(\rm \frac{b}{an^2+c}+\frac{b}{am^2+c}\):

= \(\rm \frac{\tfrac{2b}{a}}{2\left (n^2+\tfrac{c}{a} \right )}+\frac{\tfrac{2b}{a}}{2\left (m^2+\tfrac{c}{a} \right )}\)

= \(\rm \frac{m+n}{2(n^2+mn)}+\frac{m+n}{2(m^2+mn)}\)

= \(\rm \frac{m+n}{2n(n+m)}+\frac{m+n}{2m(m+n)}\)

= \(\rm \frac{1}{2}\left (\frac{1}{m}+\frac{1}{n}\right )\)

= \(\rm \frac{1}{2}\left (\frac{m+n}{mn}\right )\)

= \(\rm \frac{1}{2}\left (\frac{\tfrac{2b}{a}}{\tfrac{c}{a}}\right )\)

= \(\rm \frac{b}{c}\).

Concept:

Cramer's rule for Linear Equations of Three Variables:

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

\(\rm D=\begin{vmatrix} \rm a_{1} & \rm a_{2} & \rm a_{3}\ \rm b_{1} & \rm b_{2} & \rm b_{3}\ \rm c_{1} & \rm c_{2} & \rm c_{3}\end{vmatrix}\)

\(\rm \rm D_{x}=\begin{vmatrix}\rm d_{1} & \rm d_{2} & \rm d_{3}\ \rm b_{1} & \rm b_{2} & \rm b_{3}\ \rm c_{1} & \rm c_{2} & \rm c_{3}\end{vmatrix}\quad \rm D_{y}=\begin{vmatrix}\rm a_{1} & \rm a_{2} & \rm a_{3}\ \rm d_{1} & \rm d_{2} & \rm d_{3}\ \rm c_{1} & \rm c_{2} & \rm c_{3}\end{vmatrix}\quad \rm D_{z}=\begin{vmatrix}\rm a_{1} & \rm a_{2} & \rm a_{3}\ \rm b_{1} & \rm b_{2} & \rm b_{3}\ \rm d_{1} & \rm d_{2} & \rm d_{3}\end{vmatrix}\)

If D ≠ 0: a unique solution (consistent).

The solution is: \(\rm x=\frac{D_x}{D}\quad y=\frac{D_y}{D}\quad z=\frac{D_z}{D}\).

If D = 0: either infinitely many solutions (consistent and dependent) or no solution (inconsistent). To find out if the system is dependent or inconsistent, another method, such as elimination or Rouché–Capelli theorem, will have to be used.

Calculation:

For the given set of equations:

4x + ky + z = 0

kx + 4y + z = 0

2x + 2y + z = 0

It can be observed that x = y = z = 0 is one solution of the system.

In order to have infinitely many (including non-zero) solutions, D must be zero.

⇒ \(\rm D=\begin{vmatrix} 4 & \rm k & 2 \ \rm k & 4 & 2\ 1 & 1 & 1\end{vmatrix}\) = 0

⇒ 4(4 - 2) + k(2 - k) + 2(k - 4) = 0

⇒ 8 + 2k - k² + 2k - 8 = 0

⇒ k² - 4k = 0

⇒ k(k - 4) = 0

⇒ k = 0 OR k - 4 = 0

⇒ k = 0 OR k = 4

Hence, there are 2 possible values of k but a non-zero solution is one.

Concept:

Properties of determinants:

• The determinant of a diagonal matrix is the product of the diagonal entries.
• If A and B are both n×n matrices, then det(AB) = det(A) det(B).
• For a n×n matrix A, det(kA) = kⁿ det(A).
• The determinant of a square matrix is the same as the determinant of its transpose.

 

Calculation:

From the properties of determinants stated above, it can be seen that det(cA) = c [det(A)] is not correct.

The correct statement would be det(cA) = cⁿ [det(A)].

Concept:

Straight Lines:

• The general equation of a line is y = mx + c, where m is the slope of the line.
• If the line y = mx + c makes an angle θ with the positive direction of the x-axis, then m = tan θ.

 

Tangents to an Ellipse:

• If the line y = mx + c touches the ellipse \(\rm \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\), then c² = a²m² + b².
• The straight lines \(\rm y = mx \pm \sqrt{a^2m^2 + b^2}\) represent the tangents to the ellipse.

 

Calculation:

The given tangent makes an angle of 60^∘ with the positive direction of the x-axis.

∴ m = tan 60∘ = √3.

Converting the given equation of the ellipse into the standard form \(\rm \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 \):

x2 + 16y2 = 16

⇒ \(\rm \dfrac{x^2}{4^2}+\dfrac{y^2}{1^2}=1 \)

∴ a = 4 and b = 1.

Using the formula \(\rm y = mx \pm \sqrt{a^2m^2 + b^2} \) for the tangents to the ellipse, the required equation is:

\(\rm y = \sqrt3x + \sqrt{4^2({\sqrt3})^2 + 1^2} \)

⇒ y = √3x + 7

⇒ ​**√3x - y + 7 = 0**.

Concept:

Two curves f(x, y) = 0 and g(x, y) = 0 cut/touch at a point (a, b) if f(a, b) = g(a, b) = 0.

Calculation:

Let f(x, y) = \(\rm \dfrac{x^2}{4} + \dfrac{(y-1)^2}{9}-1\) = 0 and g(x, y) = x2 + y2 - 4 = 0 intersect at a point (a, b).

∴ f(a, b) = g(a, b) = 0

⇒ \(\rm \dfrac{a^2}{4} + \dfrac{(b-1)^2}{9}-1\) = a2 + b2 - 4 = 0

⇒ \(\rm a^2 + \dfrac{4(b-1)^2}{9}-4\) = a2 + b2 - 4 = 0

⇒ \(\rm \dfrac{4(b-1)^2}{9}\) = b²

⇒ 4(b - 1)² = 9b²

⇒ (2b - 2)2 - (3b)2 = 0

⇒ (2b - 2 + 3b)(2b - 2 - 3b) = 0

⇒ (5b - 2)(-b - 2) = 0

⇒ 5b - 2 = 0 OR -b - 2 = 0

⇒ b = \(\dfrac25\) OR b = -2.

Now, using a2 + b2 - 4 = 0, we get:

\(\rm a^2+\left(\dfrac{2}{5}\right)^2-4=0\) OR a2 + (-2)2 - 4 = 0

⇒ a² = \(\rm \dfrac{96}{25}\) OR a² = 0

⇒ a = \(\rm \pm \dfrac{4\sqrt6}{5}\) OR a = 0.

∴ The points of intersection are \(\rm \left (\dfrac{4\sqrt6}{5}, \dfrac25 \right )\), \(\rm \left (-\dfrac{4\sqrt6}{5}, \dfrac25 \right )\) and (0, -2).

The given ellipse and the circle intersect at three points (intersect at two and touch at one) as shown below:

Concept:

Arithmetic Progression (AP): The series of numbers where the difference of any two consecutive terms is the same, is called an Arithmetic Progression.

• An Arithmetic Progression of n terms, with first term a and common difference d, is represented as:

a, a + d, a + 2d, a + 3d, ..., a + (n - 2)d, a + (n - 1)d.

• The sum of n terms of the above series is given by:

S_n = \(\rm \dfrac{n}{2}[a+{a+(n-1)d}]=\left (\dfrac{First\ Term+Last\ Term}{2} \right )\times n\).

 

Calculation:

It is given that first term a = 3. Let the common difference be d.

Also, the sum of the first 8 terms is twice the sum of the first 5 terms.

⇒ S₈ = 2S₅

⇒ \(\rm \dfrac{8}{2}[3+{3+(8-1)d}]=2\times\dfrac{5}{2}[3+{3+(5-1)d}]\)

⇒ 4(6 + 7d) = 5(6 + 4d)

⇒ 24 + 28d = 30 + 20d

⇒ 8d = 6

⇒ d = \(\dfrac 34\).

∴ The common difference of the given arithmetic progression is \(\dfrac 34\).

Concept:

• (a + b + c)(a² + b² + c² - ab - bc - ca) = a³ + b³ + c³ - 3abc.
• If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Calculation:

Since, a + b + c = 0, we have a³ + b³ + c³ = 3abc.

∴ The given expression is:

\(\rm \dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab}\)

= \(\rm \dfrac{a^3}{abc} + \dfrac{b^3}{bca} + \dfrac{c^3}{cab}\)

= \(\rm \dfrac{3abc}{abc}\) = 3.