ISC Class 12 Mathematics Board Exam Question Paper 2017 with Solutions

ISC Class 12 Mathematics Board Exam Question Paper with Solutions

 

SECTION A (80 Marks)

 

Question 1. [10 × 3]
(i) If the matrix \(\begin{pmatrix} 6 & -x^2 \\ 2x - 15 & 10 \end{pmatrix}\) is symmetric, find the value of x. [3 Marks]

Answer:
Let \(A = \begin{pmatrix} 6 & -x^2 \\ 2x - 15 & 10 \end{pmatrix}\).
For a symmetric matrix, \(A = A'\).
\(A' = \begin{pmatrix} 6 & 2x - 15 \\ -x^2 & 10 \end{pmatrix}\).
Equating corresponding elements: \(-x^2 = 2x - 15 \implies x^2 + 2x - 15 = 0 \implies (x + 5)(x - 3) = 0\).
Therefore, \(x = -5\) or \(x = 3\).

Teacher's Note:
a) A matrix is symmetric if and only if \(a_{ij} = a_{ji}\) for all \(i\) and \(j\).
b) Always solve the resulting quadratic equation completely to find all possible values of \(x\).

 

(ii) If \(y - 2x - k = 0\) touches the conic \(3x^2 - 5y^2 = 15\), find the value of k. [3 Marks]

Answer:
The given line is \(y = 2x + k\), so \(m = 2\), \(c = k\).
The given conic is \(3x^2 - 5y^2 = 15\), which can be written as \(\frac{x^2}{5} - \frac{y^2}{3} = 1\) (Hyperbola). Here \(a^2 = 5\) and \(b^2 = 3\).
The condition for the line \(y = mx + c\) to touch the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) is \(c^2 = a^2m^2 - b^2\).
\(k^2 = 5(2)^2 - 3 = 5(4) - 3 = 20 - 3 = 17\).
Therefore, \(k = \pm \sqrt{17}\).

Teacher's Note:
a) Reduce the given conic equation to standard form before identifying \(a^2\) and \(b^2\).
b) Remember that the condition of tangency for a hyperbola has a minus sign (\(c^2 = a^2m^2 - b^2\)), unlike the ellipse which has a plus sign.

 

(iii) Prove that \(\frac{1}{2} \cos^{-1}\left(\frac{1 - x}{1 + x}\right) = \tan^{-1}\sqrt{x}\) [3 Marks]

Answer:
Let \(x = \tan^2\theta\), so \(\theta = \tan^{-1}\sqrt{x}\).
L.H.S. = \(\frac{1}{2} \cos^{-1}\left(\frac{1 - \tan^2\theta}{1 + \tan^2\theta}\right)\)
\(= \frac{1}{2} \cos^{-1}(\cos 2\theta) = \frac{1}{2} \times 2\theta = \theta = \tan^{-1}\sqrt{x} = \text{R.H.S.}\)

Teacher's Note:
a) Use the standard substitution \(x = \tan^2\theta\) to simplify the inverse trigonometric function.
b) Recall the double angle formula \(\frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta\).

 

(iv) Using L 'Hospital's Rule, evaluate: \(\lim_{x \to \pi/2} \left[ x \tan x - \frac{\pi}{4} \sec x \right]\) [3 Marks]

Answer:
\(\lim_{x \to \pi/2} \left[ \frac{x \sin x - \frac{\pi}{4}}{\cos x} \right]\) (which is of the form \(\frac{0}{0}\))
Using L 'Hospital's Rule by differentiating numerator and denominator with respect to \(x\):
\(= \lim_{x \to \pi/2} \frac{x \cos x + \sin x \cdot 1 - 0}{-\sin x} = \frac{\frac{\pi}{2}(0) + 1}{-1} = \frac{1}{-1} = -1\).

Teacher's Note:
a) Combine the terms into a single fraction to obtain the \(\frac{0}{0}\) indeterminate form before applying L 'Hospital's Rule.
b) Apply the product rule correctly when differentiating the numerator.

 

(v) Evaluate: \(\int \frac{1}{x^2} \sin^2\left(\frac{1}{x}\right) dx\) [3 Marks]

Answer:
Let \(t = \frac{1}{x}\), then \(dt = -\frac{1}{x^2} dx \implies -\frac{1}{x^2} dx = dt\).
Integral becomes \(\int -\sin^2 t \, dt = -\int \frac{1 - \cos 2t}{2} dt\)
\(= -\frac{1}{2} \left( t - \frac{\sin 2t}{2} \right) + C = -\frac{1}{2x} + \frac{1}{4} \sin\left(\frac{2}{x}\right) + C\).

Teacher's Note:
a) Substitution method simplifies integrals of composite functions involving reciprocal powers.
b) Use the trigonometric identity \(\sin^2 t = \frac{1 - \cos 2t}{2}\) to integrate easily.

 

(vi) Evaluate: \(\int_{0}^{\pi/4} \log(1 + \tan\theta) d\theta\) [3 Marks]

Answer:
Let \(I = \int_{0}^{\pi/4} \log(1 + \tan\theta) d\theta\) ... (1)
Using property \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx\):
\(I = \int_{0}^{\pi/4} \log\left(1 + \tan\left(\frac{\pi}{4} - \theta\right)\right) d\theta = \int_{0}^{\pi/4} \log\left(1 + \frac{1 - \tan\theta}{1 + \tan\theta}\right) d\theta\)
\(= \int_{0}^{\pi/4} \log\left(\frac{2}{1 + \tan\theta}\right) d\theta = \int_{0}^{\pi/4} [\log 2 - \log(1 + \tan\theta)] d\theta = \int_{0}^{\pi/4} \log 2 \, d\theta - I\)
\(2I = [\theta \log 2]_{0}^{\pi/4} = \frac{\pi}{4} \log 2 \implies I = \frac{\pi}{8} \log 2\).

Teacher's Note:
a) This is a standard definite integral solved using the property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx\).
b) Add the original integral and the transformed integral to eliminate the logarithmic term.

 

(vii) By using the data \(\bar{x} = 25\), \(\bar{y} = 30\), \(b_{yx} = 1.6\) and \(b_{xy} = 0.4\), find:
(a) The regression equation y on x.
(b) What is the most likely value of y when x = 60?
(c) What is the coefficient of correlation between x and y? [3 Marks]

Answer:
(a) Regression equation of \(y\) on \(x\) is given by \(y - \bar{y} = b_{yx}(x - \bar{x})\)
\(y - 30 = 1.6(x - 25) \implies y - 30 = 1.6x - 40 \implies 1.6x - y - 10 = 0\) (or \(8x - 5y - 50 = 0\)).
(b) When \(x = 60\), \(8(60) - 5y - 50 = 0 \implies 480 - 50 = 5y \implies 5y = 430 \implies y = 86$.
(c) Coefficient of correlation \(r = \pm\sqrt{b_{yx} \cdot b_{xy}} = \sqrt{1.6 \times 0.4} = \sqrt{0.64} = 0.8\).

Teacher's Note:
a) Both regression coefficients must have the same sign as the correlation coefficient \(r\).
b) Ensure proper substitution of mean values and regression coefficients in the standard linear equations.

 

(viii) A problem is given to three students whose chances of solving it are \(\frac{1}{4}\), \(\frac{1}{5}\) and \(\frac{1}{3}\) respectively. Find the probability that the problem is solved. [3 Marks]

Answer:
Let \(P(A) = \frac{1}{4}\), \(P(B) = \frac{1}{5}\), \(P(C) = \frac{1}{3}\).
Probability of not solving: \(P(\bar{A}) = 1 - \frac{1}{4} = \frac{3}{4}\), \(P(\bar{B}) = 1 - \frac{1}{5} = \frac{4}{5}\), \(P(\bar{C}) = 1 - \frac{1}{3} = \frac{2}{3}\).
Probability that the problem is solved = \(1 - P(\text{none solves}) = 1 - [P(\bar{A}) \cdot P(\bar{B}) \cdot P(\bar{C})]\)
\(= 1 - \left(\frac{3}{4} \times \frac{4}{5} \times \frac{2}{3}\right) = 1 - \frac{2}{5} = \frac{3}{5}\).

Teacher's Note:
a) It is much easier to find the probability that the problem is not solved by any student first.
b) Subtract the joint probability of failure from 1 to get the probability of at least one success.

 

(ix) If \(a + ib = \frac{x + iy}{x - iy}\) prove that \(a^2 + b^2 = 1\) and \(\frac{b}{a} = \frac{2xy}{x^2 - y^2}\) [3 Marks]

Answer:
Given \(a + ib = \frac{x + iy}{x - iy}\) ... (1)
Taking conjugate on both sides, \(a - ib = \frac{x - iy}{x + iy}\) ... (2)
Multiplying (1) and (2): \((a + ib)(a - ib) = \left(\frac{x + iy}{x - iy}\right) \left(\frac{x - iy}{x + iy}\right) \implies a^2 + b^2 = 1\).
Also, \(a + ib = \frac{(x + iy)(x + iy)}{(x - iy)(x + iy)} = \frac{x^2 - y^2 + 2xyi}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2} + i\frac{2xy}{x^2 + y^2}\).
Equating real and imaginary parts: \(a = \frac{x^2 - y^2}{x^2 + y^2}\) and \(b = \frac{2xy}{x^2 + y^2}\).
Therefore, \(\frac{b}{a} = \frac{2xy}{x^2 - y^2}\).

Teacher's Note:
a) Using properties of complex conjugates makes proving \(a^2 + b^2 = 1\) very direct.
b) Rationalise the denominator to separate real and imaginary parts for finding \(a\) and \(b\).

 

(x) Solve: \(\frac{dy}{dx} = 1 - xy + y - x\) [3 Marks]

Answer:
\(\frac{dy}{dx} = (1 + y) - x(1 + y) = (1 + y)(1 - x)\)
Separating variables: \(\frac{dy}{1 + y} = (1 - x) dx\)
Integrating both sides: \(\int \frac{dy}{1 + y} = \int (1 - x) dx\)
\(\log|1 + y| = x - \frac{x^2}{2} + C\).

Teacher's Note:
a) Factorise the right-hand side expression by grouping terms to separate variables.\(
b) Integrate both sides independently and include the constant of integration \(C\).

 

Question 2.
(a) Using properties of determinants, prove that: \(\begin{vmatrix} a & b & b+c \\ c & a & c+a \\ b & c & a+b \end{vmatrix} = (a + b + c)(a - c)^2\) [4 Marks]

Answer:
Let \(\Delta = \begin{vmatrix} a & b & b+c \\ c & a & c+a \\ b & c & a+b \end{vmatrix}\)
Applying \(C_3 \to C_3 - C_2\):
\(\Delta = \begin{vmatrix} a & b & c \\ c & a & c \\ b & c & a\end{vmatrix}\)
Applying \(R_1 \to R_1 + R_2 + R_3\):
\(\Delta = \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ c & a & c \\ b & c & a \end{vmatrix}\)
Taking \((a + b + c)\) common from \(R_1\):
\(\Delta = (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ c & a & c \\ b & c & a \end{vmatrix}\)
Applying \(C_1 \to C_1 - C_3\), \(C_2 \to C_2 - C_3\):
\(\Delta = (a+b+c) \begin{vmatrix} 0 & 0 & 1 \\ 0 & a-c & c \\ c-a & 2c-a-b & a+b-c \end{vmatrix}\)
Expanding along \(R_1\):
\(\Delta = (a + b + c) [1 \cdot \{0 - (a-c)(c-a)\}] = (a + b + c)(a - c)^2\).

Teacher's Note:
a) Strategic row and column operations simplify determinants before expanding.
b) Factor out common terms as early as possible to simplify calculations.

 

(b) Given that: \(A = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix}\) and \(B = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix}\), find \(AB\). Using this result, solve the following system of equation: \(x - y = 3\), \(2x + 3y + 4z = 17\) and \(y + 2z = 7\) [6 Marks]

Answer:
\(AB = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} = \begin{pmatrix} 2+4+0 & 2-2-0 & -4+4+0 \\ 4-12+8 & 4+6-4 & -8-12+20 \\ 0-4+4 & 0+2-2 & 0-4+10 \end{pmatrix} = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} = 6I_3\).
Therefore, \(A^{-1} = \frac{1}{6} B\).
The matrix form of the given system of equations is:\(
\begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 17 \\ 7 \end{pmatrix} \implies AX = C\).
\(X = A^{-1}C = \frac{1}{6} \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} \begin{pmatrix} 3 \\ 17 \\ 7 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 6 + 34 - 28 \\ -12 + 34 - 28 \\ 6 - 17 + 35 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 12 \\ -6 \\ 24 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\).
Hence, \(x = 2\), \(y = -1\), \(z = 4\).

Teacher's Note:
a) Verify matrix multiplication carefully element by element.
b) Ensure that the coefficient matrix matches matrix \(A\) so that the inverse relation \(AB = 6I\) can be directly applied.

 

Question 3.
(a) Solve the equation for x: \(\sin^{-1} x + \sin^{-1}(1 - x) = \cos^{-1} x\), \(x \neq 0\) [4 Marks]

Answer:
\(\sin^{-1} x + \sin^{-1}(1 - x) = \cos^{-1} x \implies \sin^{-1}\{x\sqrt{1 - (1-x)^2} + (1-x)\sqrt{1 - x^2}\} = \sin^{-1}\sqrt{1 - x^2}\)
\(x\sqrt{2x - x^2} + (1 - x)\sqrt{1 - x^2} = \sqrt{1 - x^2} \implies x\sqrt{2x - x^2} = x\sqrt{1 - x^2}\).
This gives \(x = 0\) or \(\sqrt{2x - x^2} = \sqrt{1 - x^2}\).
Squaring both sides: \(2x - x^2 = 1 - x^2 \implies 2x = 1 \implies x = \frac{1}{2}\).
Since \(x \neq 0\), the valid solution is \(x = \frac{1}{2}\).

Answer: (Note: \(x = 0$ is rejected as it does not satisfy the given equation or domain constraints, thus \(x = 1/2\)).

Teacher's Note:
a) Convert all inverse trigonometric functions to a common function (like sine) using standard identities.
b) Always check for extraneous roots introduced by squaring equations.

 

(b) If A, B and C are the elements of Boolean algebra, simplify the expression \((A' + B')(A + C') + B'(B + C)\). Draw the simplified circuit. [6 Marks]

Answer:
Expression = \((A' + B')(A + C') + B'(B + C)\)
\(= A'A + A'C' + B'A + B'C' + B'B + B'C\)
\(= 0 + A'C' + B'A + B'C' + 0 + B'C\) (since \(A'A = 0\), \(B'B = 0\))
\(= A'C' + B'A + B'(C' + C)\)
\(= A'C' + B'A + B'(1)\) (since \(C' + C = 1\))
\(= A'C' + B'A + B'\)
\(= A'C' + B'(A + 1)\)
\(= A'C' + B'(1) = A'C' + B'\).
Simplified circuit: Two inputs \(A'\) and \(C'\) connected via an AND gate, whose output is combined with input \(B'\) via an OR gate.

Teacher's Note:
a) Apply basic Boolean laws such as complementarity (\(A \cdot A' = 0$, \(A + A' = 1\)) and absorption step by step.
b) Clearly draw logic gates representing the final reduced expression.

 

Question 4.
(a) Verify Lagrange's mean value theorem for the function: \(f(x) = x(1 - \log x)\) and find the value of 'c' in the interval \([1, 2]\) [5 Marks]

Answer:
The function \(f(x) = x - x\log x\) is continuous on \([1, 2]\) and differentiable on \((1, 2)\).
\(f'(x) = 1 - \left(x \cdot \frac{1}{x} + \log x \cdot 1\right) = 1 - 1 - \log x = -\log x\).
According to Lagrange's Mean Value Theorem, there exists \(c \in (1, 2)\) such that:
\(f'(c) = \frac{f(2) - f(1)}{2 - 1}\)
\(f(1) = 1(1 - \log 1) = 1\)
\(f(2) = 2(1 - \log 2) = 2 - 2\log 2 = \log\left(\frac{e^2}{4}\right)\) or simply \(2 - 2\log 2\).
\(\frac{f(2) - f(1)}{2 - 1} = (2 - 2\log 2) - 1 = 1 - 2\log 2 = 1 - \log 4 = \log e - \log 4 = \log\left(\frac{e}{4}\right)\).
So, \(-\log c = \log\left(\frac{e}{4}\right) \implies \log\left(\frac{1}{c}\right) = \log\left(\frac{e}{4}\right) \implies c = \frac{4}{e}\).
Since \(2 < e < 3\), \(\frac{4}{3} < c < 2\), hence \(c = \frac{4}{e} \in (1, 2)\).

Teacher's Note:
a) Check continuity and differentiability conditions before applying LMVT.
b) Verify that the obtained value of \(c\) strictly lies within the given open interval \((1, 2)\).

 

(b) Find the coordinates of the centre, foci and equation of directrix of the hyperbola \(x^2 - 3y^2 - 4x = 8\). [5 Marks]

Answer:
Given equation: \(x^2 - 4x - 3y^2 = 8 \implies (x - 2)^2 - 4 - 3y^2 = 8 \implies (x - 2)^2 - 3y^2 = 12\).
Dividing by 12: \(\frac{(x - 2)^2}{12} - \frac{y^2}{4} = 1\).
Here \(a^2 = 12\) (\(a = 2\sqrt{3}\)) and \(b^2 = 4\) (\(b = 2\)).
Let \(X = x - 2\) and \(Y = y\).
Eccentricity \(e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{12}} = \sqrt{\frac{16}{12}} = \frac{2}{\sqrt{3}}\).
1. Centre: \(X = 0, Y = 0 \implies x - 2 = 0, y = 0 \implies\) Centre is \((2, 0)\).
2. Foci: \(X = \pm ae, Y = 0 \implies x - 2 = \pm 2\sqrt{3} \cdot \frac{2}{\sqrt{3}} = \pm 4 \implies x = 2 \pm 4 \implies x = 6, -2\). Foci are \((6, 0)\) and \((-2, 0)\).
3. Directrices: \(X = \pm \frac{a}{e} \implies x - 2 = \pm \frac{2\sqrt{3}}{2/\sqrt{3}} = \pm 3 \implies x = 2 \pm 3 \implies x = 5, -1\).

Teacher's Note:
a) Complete the square for the \(x\)-terms to transform the hyperbola equation into standard shifted form.
b) Use shifted coordinate relations \(X = x - 2\) and \(Y = y\) to find foci and directrices accurately.

 

Question 5.
(a) If \(y = \cos(\sin x)\), show that: \(\frac{d^2y}{dx^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0\) [5 Marks]

Answer:
Given \(y = \cos(\sin x)\) ... (1)
\(\frac{dy}{dx} = -\sin(\sin x) \cdot \cos x\) ... (2)
\(\frac{d^2y}{dx^2} = -\left[ \sin(\sin x)(-\sin x) + \cos(\sin x)\cos x \cdot \cos x \right] = \sin(\sin x)\sin x \cdot \cos x - \cos(\sin x)\cos^2 x\)
From (2), \(\sin(\sin x) \cos x = -\frac{dy}{dx}\), so \(\sin(\sin x) = -\frac{1}{\cos x}\frac{dy}{dx}\).
Substituting this and \(y = \cos(\sin x)\):
\(\frac{d^2y}{dx^2} = \left(-\frac{1}{\cos x}\frac{dy}{dx}\right)\sin x - y\cos^2 x = -\tan x \frac{dy}{dx} - y\cos^2 x\)
Therefore, \(\frac{d^2y}{dx^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0\).

Teacher's Note:
a) Apply the chain rule carefully for successive differentiation of composite trigonometric functions.
b) Substitute first derivative expressions back into the second derivative equation to prove the required relation.

 

(b) Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube. [5 Marks]

Answer:
Let the side of the square base be \(x\) and the height of the cuboid be \(y\).
Volume \(V = x^2y\) (constant) \(\implies y = \frac{V}{x^2}\).
Total surface area \(S = 2x^2 + 4xy = 2x^2 + 4x\left(\frac{V}{x^2}\right) = 2x^2 + \frac{4V}{x}\).
Differentiating with respect to \(x\):
\(\frac{dS}{dx} = 4x - \frac{4V}{x^2}\).
For critical points, \(\frac{dS}{dx} = 0 \implies 4x = \frac{4V}{x^2} \implies V = x^3 \implies y = \frac{x^3}{x^2} = x\).
Since \(x = y\), the cuboid is a cube.
Second derivative: \(\frac{d^2S}{dx^2} = 4 + \frac{8V}{x^3}\).
When \(V = x^3\), \(\frac{d^2S}{dx^2} = 4 + 8 = 12 > 0\).
Thus, the surface area is minimum when the cuboid is a cube.

Teacher's Note:
a) Express the surface area as a function of a single variable using the given volume constraint.
b) Use the second derivative test to confirm that the critical point yields a minimum.

 

Question 6.
(a) Evaluate: \(\int \frac{\sin 2x}{(1 + \sin x)(2 + \sin x)} dx\) [5 Marks]

Answer:
Let \(I = \int \frac{2\sin x \cos x}{(1 + \sin x)(2 + \sin x)} dx\).
Put \(\sin x = t \implies \cos x dx = dt\).
\(I = \int \frac{2t}{(1 + t)(2 + t)} dt\).
Using partial fractions: \(\frac{2t}{(1 + t)(2 + t)} = \frac{A}{1 + t} + \frac{B}{2 + t} \implies 2t = A(2 + t) + B(1 + t)\).
Putting \(t = -2 \implies -4 = B(-1) \implies B = 4\).
Putting \(t = -1 \implies -2 = A(1) \implies A = -2\).
\(I = \int \left( \frac{-2}{1 + t} + \frac{4}{2 + t} \right) dt = -2\log|1 + t| + 4\log|2 + t| + C\)
\(= -2\log(1 + \sin x) + 4\log(2 + \sin x) + C\).

Teacher's Note:
a) Use trigonometric identity \(\sin 2x = 2\sin x \cos x\) before substitution.
b) Decompose the rational integrand into partial fractions for straightforward integration.

 

(b) Draw a rough sketch of the curve \(y^2 = 4x\) and find the area of region enclosed by the curve and the line \(y = x\). [5 Marks]

Answer:

[Figure: Right-handed parabola \(y^2 = 4x\) with vertex at origin and line \(y = x\) passing through origin intersecting at \((0, 0)\) and \((4, 4)\), enclosing a region between them in the first quadrant]

Solving \(y^2 = 4x\) and \(y = x\): \(x^2 = 4x \implies x(x - 4) = 0 \implies x = 0, 4\).
Points of intersection are \((0, 0)\) and \((4, 4)\).
Required Area = \(\int_{0}^{4} (2\sqrt{x} - x) dx = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{4}\)
\(= \left( \frac{4}{3}(8) - \frac{16}{2} \right) - 0 = \frac{32}{3} - 8 = \frac{32 - 24}{3} = \frac{8}{3}\) sq. units.

Teacher's Note:
a) Find the points of intersection correctly to set up the proper limits of integration.
b) Integrate the upper curve minus the lower curve with respect to \(x\).

 

Question 7.
(a) Calculate the Spearman's rank correlation coefficient for the following data and interpret the result: [5 Marks]

X	35	54	80	95	73	73	35	91	83	81
Y	40	60	75	90	70	75	38	95	75	70

Answer:
Arranging ranks and calculating differences: \(N = 10\).
Ties in \(X\): value 35 repeats 2 times (ranks 9.5), value 73 repeats 2 times (ranks 6.5).
Ties in \(Y\): value 75 repeats 3 times (ranks 4), value 70 repeats 2 times (ranks 6.5).
Correction factor for \(X\): \(\sum \frac{t^3 - t}{12} = \frac{2^3-2}{12} + \frac{2^3-2}{12} = 0.5 + 0.5 = 1\).
Correction factor for \(Y\): \(\frac{3^3-3}{12} + \frac{2^3-2}{12} = 2 + 0.5 = 2.5\).
Total correction factor \(\sum \frac{t^3 - t}{12} = 1 + 2.5 = 3.5\).
Sum of squared differences \(\sum d^2 = 17\).
\(r = 1 - \frac{6 \left[ \sum d^2 + \sum \frac{t^3-t}{12} \right]}{N(N^2 - 1)} = 1 - \frac{6[17 + 3.5]}{10(100 - 1)} = 1 - \frac{6(20.5)}{990} = 1 - \frac{123}{990} = 1 - 0.1242 = 0.876\).
Interpretation: There is a strong positive correlation between X and Y.

Teacher's Note:
a) Adjust ranks appropriately when duplicate values occur in the data series.
b) Add the tie correction factors to \(\sum d^2\) in Spearman's formula.

 

(b) Find the line of best fit for the following data, treating x as the dependent variable (Regression equation x on y): [5 Marks]
Hence, estimate the value of x when y = 16.

X	14	12	13	14	16	10	13	12
Y	14	23	17	24	18	25	23	24

Answer:
\(N = 8\), \(\sum X = 104 \implies \bar{X} = 13\), \(\sum Y = 168 \implies \bar{Y} = 21\).
\(\sum x^2 = 22\), \(\sum y^2 = 116\), \(\sum xy = -30\) (where \(x = X - \bar{X}\), \(y = Y - \bar{Y}\)).
Regression coefficient \(b_{xy} = \frac{\sum xy}{\sum y^2} = \frac{-30}{116} = -0.259\).
Regression equation of \(X\) on \(Y\): \(X - \bar{X} = b_{xy}(Y - \bar{Y})\)
\(X - 13 = -0.259(Y - 21) \implies X + 0.259Y = 13 + 5.439 = 18.439\) (or \(58X + 15Y = 1069\)).
When \(Y = 16\), \(58X + 15(16) = 1069 \implies 58X + 240 = 1069 \implies 58X = 829 \implies X = 14.29\).

Teacher's Note:
a) Ensure you calculate the correct regression coefficient \(b_{xy}\) when \(X\) is treated as dependent on \(Y\).
b) Substitute \(Y = 16\) into the derived regression line equation to estimate \(X\).

 

Question 8.
(a) In a class of 60 students, 30 opted for Mathematics, 32 opted for Biology and 24 opted for both Mathematics and Biology. If one of these students is selected at random, find the probability that: [5 Marks]
(i) The student opted for Mathematics or Biology.
(ii) The student has opted neither Mathematics nor Biology.
(iii) The student has opted Mathematics but not Biology.

Answer:
Total students \(n(U) = 60\).
\(n(M) = 30\), \(n(B) = 32\), \(n(M \cap B) = 24\).
(i) \(P(M \cup B) = \frac{n(M) + n(B) - n(M \cap B)}{n(U)} = \frac{30 + 32 - 24}{60} = \frac{38}{60} = \frac{19}{30}\).
(ii) \(P(\text{neither}) = \frac{60 - 38}{60} = \frac{22}{60} = \frac{11}{30}\).
(iii) \(P(\text{Mathematics only}) = \frac{n(M) - n(M \cap B)}{n(U)} = \frac{30 - 24}{60} = \frac{6}{60} = \frac{1}{10}\).

Teacher's Note:
a) Apply standard set theory formulas for union and difference of sets.
b) Divide the required number of favorable students by the total number of students in the class.

 

(b) Bag A contains 1 white, 2 blue and 3 red balls. Bag B contains 3 white, 3 blue and 2 red balls. Bag C contains 2 white, 3 blue and 4 red balls. One bag is selected at random and then two balls are drawn from the selected bag. Find the probability that the balls drawn are white and red. [5 Marks]

Answer:
Let \(B_1, B_2, B_3\) be the events of selecting Bag A, Bag B, and Bag C respectively. \(P(B_1) = P(B_2) = P(B_3) = \frac{1}{3}\).
Let \(E\) be the event of drawing 1 white and 1 red ball.
Bag A (Total 6): \(P(E/B_1) = \frac{{}^1C_1 \times {}^3C_1}{{}^6C_2} = \frac{1 \times 3}{15} = \frac{3}{15} = \frac{1}{5}\).
Bag B (Total 8): \(P(E/B_2) = \frac{{}^3C_1 \times {}^2C_1}{{}^8C_2} = \frac{3 \times 2}{28} = \frac{6}{28} = \frac{3}{14}\).
Bag C (Total 9): \(P(E/B_3) = \frac{{}^2C_1 \times {}^4C_1}{{}^9C_2} = \frac{2 \times 4}{36} = \frac{8}{36} = \frac{2}{9}\).
Total Probability \(P(E) = \sum P(B_i)P(E/B_i) = \frac{1}{3}\left(\frac{1}{5} + \frac{3}{14} + \frac{2}{9}\right)\)
\(= \frac{1}{3}\left(\frac{126 + 135 + 140}{630}\right) = \frac{1}{3}\left(\frac{401}{630}\right) = \frac{401}{1890}\).

Teacher's Note:
a) Use the Law of Total Probability for multi-stage selection problems.
b) Compute combinations correctly using \({}^nC_r = \frac{n!}{r!(n-r)!}\).

 

Question 9.
(a) Prove that locus of z is circle and find its centre and radius if \(\frac{z - i}{z - 1}\) is purely imaginary. [5 Marks]

Answer:
Let \(z = x + iy\).
\(\frac{z - i}{z - 1} = \frac{x + i(y - 1)}{(x - 1) + iy} = \frac{[x + i(y - 1)][(x - 1) - iy]}{(x - 1)^2 + y^2}\)
Real part of the numerator = \(x(x - 1) + y(y - 1)\).
Since the complex number is purely imaginary, its real part must be zero:
\(x(x - 1) + y(y - 1) = 0 \implies x^2 + y^2 - x - y = 0\).
This represents a circle.
Centre = \(\left(\frac{1}{2}, \frac{1}{2}\right)\).
Radius \(r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - 0} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}}\).

Teacher's Note:
a) A complex number is purely imaginary when its real part equals zero.
b) Complete the coefficient terms to identify the standard centre and radius of the circle equation.

 

(b) Solve: \((x^2 - yx^2) dy + (y^2 + xy^2) dx = 0\) [5 Marks]

Answer:
\((1 - y)x^2 dy + (1 + x)y^2 dx = 0\)
Separating variables: \(\frac{1 - y}{y^2} dy + \frac{1 + x}{x^2} dx = 0 \implies \left(\frac{1}{y^2} - \frac{1}{y}\right) dy + \left(\frac{1}{x^2} + \frac{1}{x}\right) dx = 0\).
Integrating both terms:
\(\int y^{-2} dy - \int \frac{1}{y} dy + \int x^{-2} dx + \int \frac{1}{x} dx = C\)
\(-\frac{1}{y} - \log y - \frac{1}{x} + \log x = C \implies \log\left(\frac{x}{y}\right) - \frac{1}{x} - \frac{1}{y} = C\).

Teacher's Note:
a) Factorize terms to separate variables \(x\) and \(y\) completely.
b) Integrate term by term and combine logarithmic expressions using log properties.

 

SECTION B (20 Marks)

 

Question 10.
(a) If \(\vec{a}, \vec{b}, \vec{c}\) are three mutually perpendicular vectors of equal magnitude, prove that \((\vec{a} + \vec{b} + \vec{c})\) is equally inclined with vectors \(\vec{a}, \vec{b}\) and \(\vec{c}\). [5 Marks]

Answer:
Given \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0\) and \(|\vec{a}| = |\vec{b}| = |\vec{c}| = k\).
Let \(\theta_1\) be the angle between \((\vec{a} + \vec{b} + \vec{c})\) and \(\vec{a}\):
\(\cos\theta_1 = \frac{\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c})}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{|\vec{a}|^2 + 0 + 0}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{|\vec{a}|}{|\vec{a} + \vec{b} + \vec{c}|}\).
Similarly, cosine of angles with \(\vec{b}\) and \(\vec{c}\) are \(\frac{|\vec{b}|}{|\vec{a} + \vec{b} + \vec{c}|}\) and \(\frac{|\vec{c}|}{|\vec{a} + \vec{b} + \vec{c}|}\) respectively.
Since magnitudes are equal, \(\cos\theta_1 = \cos\theta_2 = \cos\theta_3\), meaning the vector is equally inclined.

Teacher's Note:
a) Use the definition of dot product to find angles between vectors.
b) Mutual orthogonality simplifies cross-term dot products to zero.

 

(b) Find the value of \(\lambda\) for which the four points with position vectors \(6\hat{i} - 7\hat{j}\), \(16\hat{i} - 19\hat{j} - 4\hat{k}\), \(\lambda\hat{i} - 6\hat{k}\) and \(2\hat{i} - 5\hat{j} + 10\hat{k}\) are coplanar. [5 Marks]

Answer:
Let the points be A, B, C, D.
\(\vec{AB} = (16 - 6)\hat{i} + (-19 - (-7))\hat{j} + (-4 - 0)\hat{k} = 10\hat{i} - 12\hat{j} - 4\hat{k}\).
\(\vec{AC} = (\lambda - 6)\hat{i} + (0 - (-7))\hat{j} + (-6 - 0)\hat{k} = (\lambda - 6)\hat{i} + 7\hat{j} - 6\hat{k}\).
\(\vec{AD} = (2 - 6)\hat{i} + (-5 - (-7))\hat{j} + (10 - 0)\hat{k} = -4\hat{i} + 2\hat{j} + 10\hat{k}\).
For coplanarity, scalar triple product \([\vec{AB}, \vec{AC}, \vec{AD}] = 0\):
\(\begin{vmatrix} 10 & -12 & -4 \\ \lambda-6 & 7 & -6 \\ -4 & 2 & 10 \end{vmatrix} = 0\)
\(10(70 - (-12)) + 12(10(\lambda-6) - 24) - 4(2(\lambda-6) - (-28)) = 0\)
Solving this gives \(100\lambda + 820 - 1008 - 64 - 16\lambda = 0 \implies 84\lambda = 252 \implies \lambda = 3\).

Teacher's Note:
a) Four points are coplanar if the scalar triple product of three vectors formed by them is zero.
b) Form vectors taking a common initial point like A to set up the determinant.

 

Question 11.
(a) Show that the lines \(\frac{x - 4}{1} = \frac{y + 3}{-4} = \frac{z + 1}{7}\) and \(\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8}\) intersect. Find the coordinates of their point of intersection. [5 Marks]

Answer:
Any general point on line 1 is \((\lambda + 4, -4\lambda - 3, 7\lambda - 1)\).
Any general point on line 2 is \((2\mu + 1, -3\mu - 1, 8\mu - 10)\).
Equating coordinates for intersection:
\(\lambda + 4 = 2\mu + 1 \implies \lambda - 2\mu = -3\) ... (1)
\(-4\lambda - 3 = -3\mu - 1 \implies 4\lambda - 3\mu = -2\) ... (2)
Solving (1) and (2) gives \(\lambda = 1\), \(\mu = 2\).
Checking with third equation: \(7(1) - 1 = 6\) and \(8(2) - 10 = 6\) (satisfied).
Point of intersection: substituting \(\lambda = 1\) gives \((5, -7, 6)\).

Teacher's Note:
a) Equating general points from two lines gives a system of equations to find parameters.\(
b) Verify consistency by substituting the parameters into the third coordinate equation.

 

(b) Find the equation of the plane passing through the point \((1, -2, 1)\) and perpendicular to the line joining the points \(A(3, 2, 1)\) and \(B(1, 4, 2)\). [5 Marks]

Answer:
Direction ratios of line AB are \(\langle 1 - 3, 4 - 2, 2 - 1 \rangle = \langle -2, 2, 1 \rangle\).
Since the plane is perpendicular to AB, these direction ratios serve as normal vector components for the plane.
Equation of plane passing through \((x_1, y_1, z_1) = (1, -2, 1)\):
\(A(x - x_1) + B(y - y_1) + C(z - z_1) = 0\)
\(-2(x - 1) + 2(y + 2) + 1(z - 1) = 0\)
\(-2x + 2 + 2y + 4 + z - 1 = 0 \implies -2x + 2y + z + 5 = 0\) (or \(2x - 2y - z - 5 = 0\)).

Teacher's Note:
a) The direction ratios of the line joining two points act as the normal vector coefficients for a perpendicular plane.
b) Use the point-normal form to write the equation of the plane directly.

 

Question 12.
(a) A fair die is rolled. If face 1 turns up, a ball is drawn from Bag A. If face 2 or 3 turns up, a ball is drawn from Bag B. If face 4 or 5 or 6 turns up, a ball is drawn from Bag C. Bag A contains 3 red and 2 white balls, Bag B contains 3 red and 4 white balls and Bag C contains 4 red and 5 white balls. The die is rolled, a Bag is picked up and a ball is drawn. If the drawn ball is red; what is the probability that it is drawn from Bag B? [5 Marks]

Answer:
Let \(E_1, E_2, E_3\) be the events of choosing Bag A, Bag B, Bag C.
\(P(E_1) = \frac{1}{6}\), \(P(E_2) = \frac{2}{6} = \frac{1}{3}\), \(P(E_3) = \frac{3}{6} = \frac{1}{2}\).
Let \(R\) be the event of drawing a red ball.
\(P(R/E_1) = \frac{3}{5}\), \(P(R/E_2) = \frac{3}{7}\), \(P(R/E_3) = \frac{4}{9}\).
Using Bayes' Theorem, probability that it is from Bag B given it is red:\(
\(P(E_2/R) = \frac{P(E_2)P(R/E_2)}{P(E_1)P(R/E_1) + P(E_2)P(R/E_2) + P(E_3)P(R/E_3)}\)
\(= \frac{\frac{1}{3} \times \frac{3}{7}}{\frac{1}{6} \times \frac{3}{5} + \frac{1}{3} \times \frac{3}{7} + \frac{1}{2} \times \frac{4}{9}} = \frac{\frac{1}{7}}{\frac{1}{10} + \frac{1}{7} + \frac{2}{9}} = \frac{\frac{1}{7}}{\frac{63 + 90 + 140}{630}} = \frac{\frac{1}{7}}{\frac{293}{630}} = \frac{90}{293}\).

Teacher's Note:
a) Identify probabilities of selecting each bag based on die outcomes.
b) Apply Bayes' theorem formula accurately for conditional probability.

 

(b) An urn contains 25 balls of which 10 balls are red and the remaining green. A ball is drawn at random from the urn, the colour is noted and the ball is replaced. If 6 balls are drawn in this way, find the probability that: [5 Marks]
(i) All the balls are red.
(ii) Not more than 2 balls are green.
(iii) The number of red balls and green balls is equal.

Answer:
Total balls \(N = 25\), Red \(p = \frac{10}{25} = \frac{2}{5}\), Green \(q = \frac{15}{25} = \frac{3}{5}\), \(n = 6\).
(i) P(all are red) = \({}^6C_6 p^6 = \left(\frac{2}{5}\right)^6 = \frac{64}{15625}\).
(ii) P(not more than 2 are green) = \(P(0 \text{ green}) + P(1 \text{ green}) + P(2 \text{ green})\)
\(= {}^6C_0 p^6 q^0 + {}^6C_1 p^5 q^1 + {}^6C_2 p^4 q^2 = \left(\frac{2}{5}\right)^6 + 6\left(\frac{2}{5}\right)^5\left(\frac{3}{5}\right) + 15\left(\frac{2}{5}\right)^4\left(\frac{3}{5}\right)^2 = \frac{64 + 576 + 2160}{15625} = \frac{2800}{15625} = \frac{112}{625}\).
(iii) P(equal number of red and green, i.e., 3 red and 3 green) = \({}^6C_3 p^3 q^3 = 20 \left(\frac{2}{5}\right)^3 \left(\frac{3}{5}\right)^3 = 20 \times \frac{8}{125} \times \frac{27}{125} = \frac{864}{3125}\).

Teacher's Note:
a) Use binomial distribution formula \(P(X = k) = {}^nC_k p^k q^{n-k}\) with replacement.
b) Sum individual binomial probabilities for cumulative conditions like "not more than".

 

SECTION C (20 Marks)

 

Question 13.
(a) A machine costs Rs. 60000 and its effective life is estimated to be 25 years. A sinking fund is to be created for replacing the machine at the end of its life when its scrap value is estimated as Rs. 5000. The price of the new machine is estimated to be 100% more than the price of the present one. Find the amount that should be set aside at the end of each year, out of the profits, for the sinking fund it accumulates at an interest of 6% per annum compounded annually. [5 Marks]

Answer:
Cost of new machine = 100% more than Rs. 60000 = Rs. 60000 + Rs. 60000 = Rs. 1,20,000.
Net replacement cost \(A = \text{Cost of new machine} - \text{Scrap value} = 1,20,000 - 5000 = \text{Rs. } 1,15,000\).
Sinking fund formula: \(A = \frac{P}{i} [(1 + i)^n - 1]\), where \(n = 25\), \(i = 0.06\).
\(1,15,000 = \frac{P}{0.06} [(1 + 0.06)^{25} - 1] \implies 6900 = P[(1.06)^{25} - 1]\).
Using \((1.06)^{25} = 4.29\):
\(6900 = P[4.29 - 1] = 3.29P \implies P = \frac{6900}{3.29} = \text{Rs. } 2097.26\).

Teacher's Note:
a) Net replacement cost equals new machine price minus scrap value realization.
b) Apply the standard sinking fund annuity formula to compute annual installment \(P\).

 

(b) A farmer has a supply of chemical fertilizer of type A which contains 10% nitrogen and 6% phosphoric acid and of type B which contains 5% nitrogen and 10% phosphoric acid. After the soil test, it is found that at least 7 kg of nitrogen and the same quantity of phosphoric acid is required for a good crop. The fertilizer of type A costs Rs. 5.00 per kg and the type B costs Rs. 8.00 per kg. Using Linear programming, find how many kilograms of each type of fertilizer should be bought to meet the requirement and for the cost to be minimum. Find the feasible region in the graph. [5 Marks]

Answer:
Let \(x\) kg of type A and \(y\) kg of type B be bought.
Objective function: Minimize \(Z = 5x + 8y\).
Constraints:
\(0.10x + 0.05y \geq 7 \implies 2x + y \geq 140\)
\(0.06x + 0.10y \geq 7 \implies 3x + 5y \geq 350\)
\(x \geq 0, y \geq 0\).
Corner points of feasible region: \((0, 140)\), \((50, 40)\), and \((70, 0)\).
Evaluating \(Z\) at corner points:
At \((0, 140)\): \(Z = 0 + 8(140) = 1120\).
At \((50, 40)\): \(Z = 5(50) + 8(40) = 250 + 320 = 570\) (Minimum).
At \((70, 0)\): \(Z = 5(70) + 0 = 350\) (Wait, boundary check: intersection of \(2x+y=140\) and \(3x+5y=350\): multiplying first by 5 gives \(10x+5y=700\), subtracting gives \(7x=350 \implies x=50, y=40\). At \((70,0)\), \(3(70)+0 = 210 < 350\), not feasible).
Checking corner points of unbounded region: \((0, 140)\), \((50, 40)\), \((350/3, 0)\).
At \((350/3, 0)\): \(Z = 5(350/3) = 583.33\).
Minimum cost is Rs. 570 at \((50, 40)\).

Teacher's Note:
a) Formulate constraints carefully matching percentage nutrient requirements.
b) Test all feasible corner points to determine the optimal minimum cost.

 

Question 14.
(a) The demand for a certain product is represented by the equation \(p = 500 + 25x - \frac{x^2}{3}\) in rupees, where x is the number of units and p is the price per unit. Find:
(i) Marginal revenue function.
(ii) The marginal revenue when 10 units are sold. [5 Marks]

Answer:
(i) Total Revenue \(R = p \cdot x = 500x + 25x^2 - \frac{x^3}{3}\).
Marginal Revenue \(MR = \frac{dR}{dx} = 500 + 50x - x^2\).
(ii) When \(x = 10\):
\((MR)_{10} = 500 + 50(10) - (10)^2 = 500 + 500 - 100 = 900\).

Teacher's Note:
a) Total revenue is obtained by multiplying price function by quantity \(x\).
b) Marginal revenue is the first derivative of total revenue with respect to \(x\).

 

(b) A bill of Rs. 60000 payable 10 months after the date was discounted for Rs. 57300 on 30th June 2007. If the rate of interest was \(11\frac{1}{4}\%\) per annum, on what date was the bill drawn? [5 Marks]

Answer:
Face Value = Rs. 60,000, Amount Received = Rs. 57,300.
Banker's Discount (BD) = Face Value - Amount Received = \(60000 - 57300 = \text{Rs. } 2700\).
Rate \(r = 11.25\% = \frac{45}{400}\).
\(BD = \frac{F \cdot n \cdot r}{100} \implies 2700 = \frac{60000 \cdot n \cdot 11.25}{100} \implies 2700 = 6750 n \implies n = \frac{2700}{6750} = 0.4\text{ years} = 0.4 \times 365 = 146\text{ days}\).
Legally due date for a bill drawn on 30 June 2007 for 10 months is 3 April 2008.
Counting 146 days backwards from 3 April 2008: April (3), March (31), Feb (28), Jan (31), Dec (31), Nov (19).
Thus, the bill was discounted on 11 November 2007.
Since it was discounted 10 months before maturity (or factoring backward from discounting date), bill date is calculated as 11 Nov 2007 minus period.

Teacher's Note:
a) Banker's discount equals Face Value minus Present Value (Amount received).
b) Calculate the unexpired legal period in days using standard calendar days.

 

Question 15.
(a) The price relatives and weights of a set of commodities are given below: [5 Marks]

Commodity	A	B	C	D
Price relatives	125	120	127	119
Weights	\(x\)	\(2x\)	\(y\)	\(y + 3\)

If the sum of the weights is 40 and the weighted average of price relatives index number is 122, find the numerical values of x and y.

Answer:
Sum of weights: \(x + 2x + y + (y + 3) = 40 \implies 3x + 2y = 37\) ... (1)
Weighted average index \(I = \frac{\sum IW}{\sum W} = 122\).
\(\frac{125(x) + 120(2x) + 127(y) + 119(y + 3)}{40} = 122\)
\(125x + 240x + 127y + 119y + 357 = 4880 \implies 365x + 246y = 4523\) ... (2)
Solving (1) and (2): from (1), \(2y = 37 - 3x\).
Substituting and solving gives \(x = 7\) and \(y = 8\).

Teacher's Note:
a) Use the weighted index formula \(I = \frac{\sum IW}{\sum W}\) to set up simultaneous equations.
b) Solve linear equations in \(x\) and \(y\) to find their numerical values.

 

(b) Construct 3 yearly moving averages from the following data and show on a graph against the original data: [5 Marks]

Year	2000	2001	2002	2003	2004	2005	2006	2007	2008	2009
Annual sale (in lakhs)	18	22	20	26	30	22	24	28	32	35

Answer:
3-yearly moving totals and averages:
2000: - , -
2001: 18 + 22 + 20 = 60, Average = 20
2002: 22 + 20 + 26 = 68, Average = 22.67
2003: 20 + 26 + 30 = 76, Average = 25.33
2004: 26 + 30 + 22 = 78, Average = 26
2005: 30 + 22 + 24 = 76, Average = 25.33
2006: 22 + 24 + 28 = 74, Average = 24.67
2007: 24 + 28 + 32 = 84, Average = 28
2008: 28 + 32 + 35 = 95, Average = 31.67
2009: - , -

[Figure: Line graph showing original annual sales and 3-yearly moving average trend line plotted against years 2000 to 2009 on grid paper]

Teacher's Note:
a) Compute 3-yearly moving totals by summing values in groups of three consecutive years.
b) Center each moving average against the middle year of the corresponding group.