ISC Class 12 Mathematics Sample Paper 2027 with Solutions

SECTION A - 20 MARKS

Question 1

In subparts (i) to (xvii) choose the correct options and in subparts (xviii) to (xx), answer the questions as instructed.

(i) A relation \( R \) is defined on \( \mathbb{Z} \) as \( aRb \) if and only if \( a^2 - 7ab + 6b^2 = 0 \). Then R is: [1 Mark]
(a) reflexive and symmetric
(b) transitive but not reflexive
(c) symmetric but not reflexive
(d) reflexive but not symmetric

Answer: (d) reflexive but not symmetric

Reflexive: For all \( a \in \mathbb{Z} \), \( a^2 - 7a(a) + 6a^2 = a^2 - 7a^2 + 6a^2 = 0 \implies (a, a) \in R \). Symmetric: For \((6, 1)\), \( 6^2 - 7(6)(1) + 6(1)^2 = 36 - 42 + 6 = 0 \implies (6, 1) \in R \), but for \((1, 6)\), \( 1^2 - 7(1)(6) + 6(6)^2 = 1 - 42 + 216 \neq 0 \implies (1, 6) \notin R \).

Teacher's Note:
a) Check reflexivity by substituting \( b = a \) into the given relation equation.
b) Verify symmetry by finding a counterexample where \( (a, b) \in R \) but \( (b, a) \notin R \).

(ii) If \( \theta = \sin^{-1}x + \cos^{-1}x - \tan^{-1}x \), \( x \ge 0 \), then the smallest interval in which \( \theta \) lies is: [1 Mark]
(a) \( \frac{\pi}{2} \le \theta \le \frac{3\pi}{4} \)
(b) \( 0 \lt \theta \lt \pi \)
(c) \( -\frac{\pi}{4} \le \theta \le 0 \)
(d) \( \frac{\pi}{4} \le \theta \le \frac{\pi}{2} \)

Answer: (d) \( \frac{\pi}{4} \le \theta \le \frac{\pi}{2} \)

Since \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \) for \( x \in [-1, 1] \), we get \( \theta = \frac{\pi}{2} - \tan^{-1}x \). Given \( x \ge 0 \) and valid up to \( x = 1 \), \( 0 \le \tan^{-1}x \le \frac{\pi}{4} \), leading to \( \frac{\pi}{4} \le \theta \le \frac{\pi}{2} \).

Teacher's Note:
a) Use the standard inverse trigonometric identity \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \).
b) Restrict the range of \( \tan^{-1}x \) carefully based on the domain condition \( 0 \le x \le 1 \).

(iii) Let \( A \) be the area of a triangle having vertices \( (x_1, y_1) \), \( (x_2, y_2) \) and \( (x_3, y_3) \). Which of the following is correct? [1 Mark]
(a) \( \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = +A \)
(b) \( \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm 2A \)
(c) \( \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm \frac{A}{2} \)
(d) \( \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2 = A^2 \)

Answer: (b) \( \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm 2A \)

Area \( A = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \), which implies the determinant can be positive or negative \( 2A \).

Teacher's Note:
a) Recall that area is always positive, hence the absolute value is used in the area formula.
b) Removing the absolute value sign introduces the \( \pm \) sign on the right-hand side.

(iv) A particle moves along the curve \( 6y = x^3 + 2 \). At what point(s) on the curve, is the y-coordinate changing 8 times as fast as the x-coordinate? [1 Mark]
(a) \( (2, \frac{5}{3}) \) only
(b) \( (4, 11) \) only
(c) \( (4, 11) \) and \( (-4, -\frac{31}{3}) \)
(d) \( (8, 86) \) and \( (-8, -85) \)

Answer: (c) \( (4, 11) \) and \( (-4, -\frac{31}{3}) \)

Differentiating \( 6y = x^3 + 2 \) with respect to \( t \) gives \( 6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt} \). Substituting \( \frac{dy}{dt} = 8 \frac{dx}{dt} \) gives \( x = \pm 4 \), yielding the respective y-coordinates.

Teacher's Note:
a) Apply chain rule for implicit differentiation with respect to time \( t \).
b) Substitute both positive and negative values of \( x \) back into the original equation to find corresponding \( y \) values.

(v) Statement I: If a satellite's altitude function \( h(t) \) has a positive derivative (\( h'(t) \gt 0 \)), the satellite is moving away from the Earth.
Statement II: At the apogee (highest point), the instantaneous velocity (\( \frac{dh}{dt} \)) of the satellite relative to its altitude is zero.
Which of the following is correct? [1 Mark]
(a) Statement I is true and Statement II is false.
(b) Statement I is false and Statement II is true.
(c) Both the statements are true.
(d) Both the statements are false.

Answer: (c) Both the statements are true.

A positive derivative means altitude increases with time (moving away), and at the peak altitude (apogee), the rate of change of altitude is zero.

Teacher's Note:
a) Relate the sign of the first derivative directly to the increasing or decreasing nature of a physical quantity.
b) Understand that stationary points on a time-displacement curve represent zero instantaneous velocity.

(vi) For what value(s) of \( k \) do the tangents of two curves \( x = y^2 \) and \( xy = k \) cut at right angles? [1 Mark]
(a) \( k = \pm \frac{1}{4} \)
(b) \( k = \pm 1 \)
(c) \( k = 0 \)
(d) \( k = \pm \frac{1}{2\sqrt{2}} \)

Answer: (d) \( k = \pm \frac{1}{2\sqrt{2}} \)

Slopes are \( \frac{1}{2y} \) and \( -\frac{y}{x} \). For orthogonality, their product is \( -1 \), giving \( x = \frac{1}{2} = y^2 \), which yields \( k^2 = x^2 y^2 = \frac{1}{8} \).

Teacher's Note:
a) Two curves intersect at right angles if the product of their slopes at the point of intersection is \( -1 \).
b) Solve the simultaneous equations obtained from the curves and the orthogonality condition.

(vii) Evaluate the nature of the point \( (0,0) \) for the curve \( y = x^4 \), given that \( f''(0) = 0 \). [1 Mark]
(a) The function at the point \( (0,0) \) is undefined because the second derivative test fails.
(b) It is a point of local maximum because the first derivative is changing its sign from positive to negative as \( x \) increases through 0.
(c) It is a local minimum because the second derivative is positive for all \( x \neq 0 \).
(d) It is a point of neither local maximum nor local minimum because the second derivative is zero.

Answer: (c) It is a local minimum because the second derivative is positive for all \( x \neq 0 \).

\( f''(x) = 12x^2 \), which is strictly positive for all non-zero \( x \), confirming the curve is concave up everywhere except at the inflection boundary where it attains an absolute minimum.

Teacher's Note:
a) When the second derivative test fails (\( f''(c) = 0 \)), rely on higher-order derivatives or the sign of the first derivative around the critical point.
b) Note that an even power function like \( x^4 \) has a clear global minimum at the origin.

(viii) The expression for finding the area of the shaded region is: [1 Mark]
(a) \( \int_{2}^{8} (4 - \frac{x}{2} - 2) \, dx \)
(b) \( \int_{2}^{8} (\frac{x}{2} + 4) \, dx \)
(c) \( \int_{2}^{8} (\frac{y}{2} + 4) \, dy \)
(d) \( \int_{2}^{8} (4 - \frac{x}{2}) \, dx \)

[Figure: Shaded region bounded by line \( x + 2y = 8 \), vertical line \( x = 2 \), and x-axis, with region lying between \( x = 2 \) and the x-intercept of the line.]

Answer: (d) \( \int_{2}^{8} (4 - \frac{x}{2}) \, dx \)

The line equation is \( y = 4 - \frac{x}{2} \). The region is bounded between \( x = 2 \) and the x-axis under the line.

Teacher's Note:
a) Identify upper and lower curves along with the limits of integration from the graph.
b) Express \( y \) in terms of \( x \) from the line equation \( x + 2y = 8 \).

(ix) Shown below is a solved anti-differentiation problem to obtain \( f(x) \):
\( \frac{d}{dx}f(x) = \frac{1}{x(\log x)^2} \) such that \( f(e) = -1 \)
Taking anti-derivative:
\( f(x) = \int \frac{1}{x(\log x)^2} \, dx + C \)
Step 1 \( \Rightarrow f(x) = \int \frac{d(\log x)}{(\log x)^2} + C \)
Step 2 \( \Rightarrow f(x) = -\frac{1}{\log x} + C \)
Given \( f(e) = -1 \)
\( \Rightarrow f(e) = -1 + C \)
\( \Rightarrow C = 0 \)
Step 3 \( f(x) = -\frac{1}{\log x} \)
In which step is there an error (if any) in the solution? [1 Mark]
(a) Step 1
(b) Step 2
(c) Step 3
(d) No error

Answer: (d) No error

Integration of \( u^{-2} \) yields \( -u^{-1} \), which correctly matches Step 2, and the boundary condition evaluates \( C \) to zero.

Teacher's Note:
a) Verify substitution method where \( u = \log x \) and \( du = \frac{1}{x} \, dx \).
b) Check substitution of boundary values for consistency.

(x) On solving \( \int -\frac{3}{\sqrt{1-x^2}} \, dx \), Anil's answer was \( \int -\frac{3}{\sqrt{1-x^2}} \, dx = 3 \cos^{-1}x + c \) and Jaspreet's answer was \( \int -\frac{3}{\sqrt{1-x^2}} \, dx = -3 \int \frac{dx}{\sqrt{1-x^2}} = -3 \sin^{-1}x + c \). Whose answer was correct? [1 Mark]
(a) Only Anil's answer was correct.
(b) Only Jaspreet's answer was correct.
(c) Both of them were correct.
(d) Neither of them was correct.

Answer: (c) Both of them were correct.

Since \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \), the two inverse trigonometric expressions differ only by a constant absorbed into \( c \).

Teacher's Note:
a) Recognize that different antiderivatives for the same function differ by a constant.
b) Use trigonometric identities to relate sine and cosine inverse antiderivatives.

(xi) If \( \vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} \), \( \vec{b} = -\hat{i} + 2\hat{j} - 4\hat{k} \), \( \vec{c} = \hat{i} + \hat{j} + \hat{k} \), then \( (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{c}) \) is: [1 Mark]
(a) 60
(b) 64
(c) 74
(d) -74

Answer: (d) -74

\( \vec{a} \times \vec{b} = -10\hat{i} + 9\hat{j} + 7\hat{k} \) and \( \vec{a} \times \vec{c} = 4\hat{i} - 3\hat{j} - \hat{k} \). Taking their dot product yields \( -40 - 27 - 7 = -74 \).

Teacher's Note:
a) Compute cross products using determinant expansions.
b) Compute the dot product of the resulting vectors carefully respecting signs.

(xii) The point of intersection of the lines \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{3-z}{-4} \) and \( \frac{x-1}{5} = \frac{2-y}{-2} = \frac{z-3}{1} \) is: [1 Mark]
(a) \( (1, 2, -3) \)
(b) \( (-1, 2, 3) \)
(c) \( (1, -2, 3) \)
(d) \( (1, 2, 3) \)

Answer: (d) \( (1, 2, 3) \)

Rewriting the second line in standard form and testing point \( (1, 2, 3) \) shows both lines pass through this coordinate point.

Teacher's Note:
a) Ensure equations are in standard symmetric form with coefficients of variables equal to 1.
b) Verify point satisfaction across both line equations.

(xiii) An objective function \( Z = ax + by \) is maximum at points \( (8, 10) \) and \( (7, 12) \). If \( a, b \ge 0 \) and \( ab = 72 \), then the maximum value of the function is equal to: [1 Mark]
(a) 84
(b) 132
(c) 156
(d) 176

Answer: (c) 156

Equating values at both points gives \( a = 2b \). Using \( ab = 72 \) yields \( b = 6 \) and \( a = 12 \). Evaluating \( Z \) at \( (8, 10) \) gives \( 8(12) + 10(6) = 156 \).

Teacher's Note:
a) If an objective function attains the same maximum at two distinct vertices, it is maximized along the entire line segment joining them.
b) Solve for parameters using simultaneous constraints.

(xiv) Statement I: Every Linear Programming Problem has at least one optimal solution.
Statement II: If a Linear Programming Problem has two optimal solutions, then it has infinitely many solutions.
Which of the statements is correct? [1 Mark]
(a) Statement I is true and Statement II is false.
(b) Statement I is false and Statement II is true.
(c) Both the statements are true.
(d) Both the statements are false.

Answer: (b) Statement I is false and Statement II is true.

Some LPPs are infeasible or unbounded, making Statement I false. Statement II is true due to the convexity property of linear programming feasible regions.

Teacher's Note:
a) Recall that infeasible linear programs possess no optimal solution.
b) Understand that convex combinations of two optimal solutions yield infinitely many optimal solutions.

(xv) If \( P(A) = m \), \( P\left(\frac{B}{\bar{A}}\right) = 3m \), \( P\left(\frac{B}{A}\right) = 6m \), then what will be \( P\left(\frac{A}{B}\right) \)? [1 Mark]
(a) \( \frac{2m}{1+m} \)
(b) \( \frac{4m}{1+m} \)
(c) \( \frac{7m}{1+m} \)
(d) \( \frac{9m}{1+m} \)

Answer: (a) \( \frac{2m}{1+m} \)

Using conditional probability formulas, \( P(A \cap B) = 6m^2 \) and \( P(B) = 3m + 3m^2 \), which gives \( P\left(\frac{A}{B}\right) = \frac{6m^2}{3m+3m^2} = \frac{2m}{1+m} \).

Teacher's Note:
a) Apply total probability and conditional probability definitions systematically.
b) Simplify algebraic fractions carefully by factoring out common terms.

(xvi) Observe the given graph:
Assertion: The curves given above can be considered as \( f(x) \) and \( f^{-1}(x) \).
Reason: The curves are reflections of each other across the line \( x = y \).
Which one of the following is correct? [1 Mark]
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

[Figure: Graph showing two symmetric curves intersecting along the line \( y = x \).]

Answer: (a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Inverse functions are geometric reflections across the line \( y = x \), which matches the graphical representation.

Teacher's Note:
a) Recall that the graph of a function and its inverse are symmetric about the line \( y = x \).
b) Verify that swapping coordinates validates the reflection property.

(xvii) Assertion: \( \int_{-1}^{1} x^3 \cos x \, dx = 0 \)
Reason: If \( f(x) \) is a continuous function defined on \( [0, a] \), then \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \)
Which one of the following is correct? [1 Mark]
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer: (b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.

The assertion holds because the integrand is an odd function over a symmetric interval. The reason is a valid property of definite integrals but does not explain odd function symmetry.

Teacher's Note:
a) Check integrand symmetry to evaluate integrals over symmetric limits quickly.
b) Distinguish between integration properties of odd functions and definite integral shifting properties.

(xviii) Using the properties of matrices, explain why the following concept is incorrect: Since \( (a + b)(a - b) = a^2 - b^2 \), therefore, \( (A + B)(A - B) = A^2 - B^2 \) also must be true if A and B are matrices. [1 Mark]

Answer:
Matrix multiplication is non-commutative (\( AB \neq BA \)). Expanding \( (A + B)(A - B) \) gives \( A^2 - AB + BA - B^2 \), which does not simplify to \( A^2 - B^2 \) unless \( AB = BA \).

Teacher's Note:
a) Highlight that matrix algebra does not obey all ordinary scalar algebraic identities.
b) Emphasize the importance of matrix multiplication order.

(xix) If the solution of the differential equation \( \frac{dy}{dx} = \frac{ax+3}{2y+5} \) represents a circle, then find the value of \( a \). [1 Mark]

Answer:
Separating variables and integrating gives \( y^2 + 5y + c = \frac{a}{2}x^2 + 3x \). For this to represent a circle, the coefficients of \( x^2 \) and \( y^2 \) must be equal, giving \( \frac{a}{2} = -1 \), so \( a = -2 \).

Teacher's Note:
a) Separate variables and integrate both sides directly.
b) Use the standard equation of a circle where squared terms have equal coefficients.

(xx) The probability distribution of random variable X is given below.
X: 1, 2, 3, 4, 5
P(X): m, 3m, a, 5m, b
If \( P(X \le 2) = 0.28 \) and \( P(X \ge 4) = 0.52 \), find \( P(X = 3) \). [1 Mark]

Answer:
\( P(X \le 2) = 4m = 0.28 \implies m = 0.07 \). \( P(X \ge 4) = 5m + b = 0.52 \implies b = 0.17 \). Since total probability sums to 1, \( a = 1 - (9m + b) = 0.2 \).

Teacher's Note:
a) Use cumulative probability sums to find individual probability constants.
b) Ensure all probabilities in a discrete probability distribution sum up to 1.

SECTION B - 14 MARKS

Question 2 [2 Marks]
A straight line \( L_\theta \) has vector equation \( \vec{r} = 5\hat{i} + \lambda(5\hat{i} + \sin\theta\hat{j} + \cos\theta\hat{k}) \) and a plane \( \pi_P \) has equation \( x = p, p \in \mathbb{R} \). Show that the angle between \( L_\theta \) and \( \pi_P \) is independent of both \( \theta \) and p.

Answer:
1. Direction vector of the line is \( \vec{d} = 5\hat{i} + \sin\theta\hat{j} + \cos\theta\hat{k} \).
2. Normal vector to the plane \( x = p \) is \( \vec{n} = \hat{i} \).
3. If \( \phi \) is the angle between the line and the plane, \( \sin\phi = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}||\vec{n}|} = \frac{5}{\sqrt{25 + \sin^2\theta + \cos^2\theta} \times 1} = \frac{5}{\sqrt{26}} \), which is constant and independent of \( \theta \) and \( p \).

Teacher's Note:
a) Recall that the angle between a line and a plane uses the sine of the angle between the line direction and the plane normal.
b) Use trigonometric identity \( \sin^2\theta + \cos^2\theta = 1 \) to simplify the magnitude.

Question 3 [2 Marks]
In a school, for a period of 8 working days, it is equally likely that Tanishka is present or absent on any given day. What is the probability that she is present in school for at least 5 consecutive working days?

Answer:
1. Total possible outcomes = \( 2^8 = 64 \).
2. Favorable outcomes for at least 5 consecutive working days include patterns starting with 5, 6, 7, or 8 consecutive presences, which sum to 5 favorable configurations.
3. Probability = \( \frac{5}{64} \).

Teacher's Note:
a) Break down consecutive occurrences into mutually exclusive cases.
b) Carefully count valid bit strings of length 8 representing presence and absence.

Question 4 [2 Marks]
Solve the following differential equation: \( \frac{dx}{dy} = \frac{x}{y} + \frac{f(x/y)}{f'(x/y)} \)

Answer:
1. Substitute \( x = vy \), so \( \frac{dx}{dy} = v + y\frac{dv}{dy} \).
2. Separating variables yields \( \frac{f'(v)}{f(v)} \, dv = \frac{dy}{y} \).
3. Integrating both sides gives \( \log|f(v)| = \log y + \log c \), leading to the solution \( f\left(\frac{x}{y}\right) = cy \).

Teacher's Note:
a) Recognize homogeneous differential equations and apply the standard substitution \( x = vy \).
b) Integrate using logarithmic antiderivatives.

Question 5
(i) Determine the values of constants \( p \) and \( q \) such that the function \( f(x) = \begin{cases} p\sin x + q, & \forall x \le 0 \\ x^2 + 2x + 1, & \forall x \gt 0 \end{cases} \) is differentiable at \( x = 0 \). [2 Marks]

Answer:
1. Continuity at \( x = 0 \) gives \( q = 1 \).
2. Equating Left Hand Derivative and Right Hand Derivative at \( x = 0 \) gives \( p\cos(0) = 2(0) + 2 \implies p = 2 \).

Teacher's Note:
a) Differentiability requires continuity first, followed by equality of left and right derivatives.
b) Compute derivatives for each piecewise function domain separately.

OR

(ii) Observe the graph given below:
(a) State the value(s) of \( x \) where the function has removable discontinuity. [1 Mark]
(b) State the value(s) of \( x \) where the graph of the function is not differentiable. [1 Mark]

[Figure: Graph showing piecewise function with a missing point at \( x = -2 \) and sharp corners.]

Answer:
(a) At \( x = -2 \)
(b) At \( x = -2 \), \( x = 0 \), and \( x = 2 \)

Teacher's Note:
a) Removable discontinuity occurs when limits exist but do not equal the function value.
b) Non-differentiability occurs at points of discontinuity, sharp corners, or vertical tangents.

Question 6
(i) Parag is working on a school project on right-angled triangles. He draws a right-angled triangle PQR with \( \angle R = 90^{\circ} \). The sides opposite angles P, Q, R are \( p, q, r \) respectively. Evaluate the expression: \( \tan^{-1}\left(\frac{p}{q+r}\right) + \tan^{-1}\left(\frac{q}{r+p}\right) \). [2 Marks]

Answer:
1. Using Pythagoras theorem, \( r^2 = p^2 + q^2 \).
2. Applying the inverse tangent sum formula, the argument simplifies to \( \frac{pr + r^2 + qr}{pr + r^2 + qr} = 1 \).
3. Therefore, \( \tan^{-1}(1) = \frac{\pi}{4} \).

Teacher's Note:
a) Apply the tangent sum formula \( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \).
b) Substitute triangle side relations from Pythagoras theorem to simplify expressions.

OR

(ii) The equation given below has equal roots: \( ax^2 + \sin^{-1}(x^2 - 2x + 2) + \cos^{-1}(x^2 - 2x + 2) = 0 \). What is the value of 'a'? [2 Marks]

Answer:
1. Since \( \sin^{-1}X + \cos^{-1}X = \frac{\pi}{2} \), the equation simplifies to \( ax^2 + \frac{\pi}{2} = 0 \).
2. The domain condition for \( x^2 - 2x + 2 = (x-1)^2 + 1 \le 1 \) forces \( x = 1 \).
3. Substituting \( x = 1 \) gives \( a(1)^2 + \frac{\pi}{2} = 0 \implies a = -\frac{\pi}{2} \).

Teacher's Note:
a) Use standard inverse trigonometric identities to reduce constant expression terms.
b) Analyze domain constraints of quadratic expressions to determine exact variable values.

Question 7
(i) Find the vector projection of \( \vec{B} = 6\hat{i} + 3\hat{j} + 2\hat{k} \) on \( \vec{A} = \hat{i} - 2\hat{j} - 2\hat{k} \) and the scalar component of \( \vec{B} \) on \( \vec{A} \). [2 Marks]

Answer:
1. Vector projection = \( \left(\frac{\vec{B} \cdot \vec{A}}{|\vec{A}|^2}\right)\vec{A} = -\frac{4}{9}(\hat{i} - 2\hat{j} - 2\hat{k}) \).
2. Scalar component = \( \frac{\vec{B} \cdot \vec{A}}{|\vec{A}|} = -\frac{4}{3} \).

Teacher's Note:
a) Distinguish between vector projection (resulting in a vector) and scalar component (resulting in a scalar).
b) Compute dot products and magnitudes accurately.

OR

(ii) If the position vectors of three points A, B, C are respectively \( \hat{i} + \hat{j} + \hat{k} \), \( 2\hat{i} + 3\hat{j} - 4\hat{k} \) and \( 7\hat{i} + 4\hat{j} + 9\hat{k} \), find the unit vector perpendicular to the plane of triangle ABC. [2 Marks]

Answer:
1. Vectors along sides are \( \vec{AB} = \hat{i} + 2\hat{j} - 5\hat{k} \) and \( \vec{AC} = 6\hat{i} + 3\hat{j} + 8\hat{k} \).
2. Normal vector \( \vec{AB} \times \vec{AC} = 31\hat{i} - 38\hat{j} - 9\hat{k} \).
3. Unit vector = \( \frac{31\hat{i} - 38\hat{j} - 9\hat{k}}{\sqrt{2486}} \).

Teacher's Note:
a) Form two coplanar vectors from the given triangle vertices.
b) Compute their cross product to obtain the normal vector and divide by its magnitude for the unit vector.

Question 8
Three landmarks of Dehradun are joined by straight roads. Clock tower of Dehradun is considered as the 'origin'. IMA (I) is 3 km east and 9 km north of Clock Tower and Rajpur (R) is 5 km east and 5 km south of IMA. A bus stop (S) is situated two thirds of the way along the road from Clock Tower to IMA.
Considering \( \hat{i} \) as a 1 km vector pointing east and \( \hat{j} \) as a 1 km vector pointing north:

(i) Find the position vector of the bus stop (S) relative to the Clock Tower. [1 Mark]

Answer:
Position vector of IMA is \( 3\hat{i} + 9\hat{j} \). Position vector of bus stop \( S = \frac{2}{3}(3\hat{i} + 9\hat{j}) = 2\hat{i} + 6\hat{j} \).

Teacher's Note:
a) Set up coordinate vectors based on cardinal directions.
b) Apply section formula or scalar multiplication for fractional distances.

(ii) Prove that the bus stop (S) is the closest point to Rajpur (R) on the Clock Tower to IMA (I) Road. [1 Mark]

Answer:
Position vector of Rajpur \( \vec{R} = (3+5)\hat{i} + (9-5)\hat{j} = 8\hat{i} + 4\hat{j} \). Vector \( \vec{RS} = \vec{OS} - \vec{OR} = -6\hat{i} + 2\hat{j} \). Dot product \( \vec{OS} \cdot \vec{RS} = (2)(-6) + (6)(2) = 0 \), proving perpendicularity.

Teacher's Note:
a) The shortest distance from a point to a line segment corresponds to the perpendicular vector.
b) Verify orthogonality using a zero dot product.

SECTION C - 21 MARKS

Question 9
The feasible region determined by some constraints is represented by the shaded region in the graph given below:

[Figure: Shaded feasible region bounded by lines with vertices at \( (0,0), (1,1), (3,3), (1,7) \).]

(i) Formulate the constraints which represent the above feasible region. [1 Mark]

Answer:
\( 2x + y \le 9 \), \( y \ge x \), and \( x \ge 1 \).

Teacher's Note:
a) Determine boundary line equations from intersecting vertices.
b) Check inequality directions based on the shaded side of the plane.

(ii) Hence, maximise the objective function given by \( Z = x + y \). [1 Mark]

Answer:
Evaluating at vertex \( C(1, 7) \), \( Z = 1 + 7 = 8 \).

Teacher's Note:
a) Test all corner points of the feasible polygon.
b) Select the maximum value obtained.

(iii) What change in the constraints will make the feasible region unbounded? [1 Mark]

Answer:
Changing constraint \( 2x + y \le 9 \) to \( 2x + y \ge 9 \).

Teacher's Note:
a) Unbounded regions extend infinitely in at least one direction.
b) Reversing bounding inequalities opens up the feasible polygon.

Question 10 [3 Marks]
If \( x = 3z + 1 \) and \( y = f(x) \), then prove that \( 9\frac{d^2y}{dx^2} = \frac{d^2y}{dz^2} \)

Answer:
1. Given \( \frac{dx}{dz} = 3 \), so \( \frac{dy}{dz} = \frac{dy}{dx}\frac{dx}{dz} = 3\frac{dy}{dx} \).
2. Differentiating again with respect to \( z \), \( \frac{d^2y}{dz^2} = \frac{d}{dx}\left(3\frac{dy}{dx}\right)\frac{dx}{dz} = 3\frac{d^2y}{dx^2} \times 3 = 9\frac{d^2y}{dx^2} \).

Teacher's Note:
a) Apply chain rule carefully for second-order parametric derivatives.
b) Substitute inner derivative scale factors at each differentiation step.

Question 11
(i) Find the derivative of the function \( f(x) = \log\left[\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}-x}\right] + \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) with respect to \( x \). [2 Marks]

Answer:
1. Rationalizing the log argument simplifies \( f(x) \) to \( 2\log(\sqrt{1+x^2}+x) + 2\tan^{-1}x \).
2. Differentiating term by term yields \( f'(x) = \frac{2}{\sqrt{1+x^2}} + \frac{2}{1+x^2} \).

Teacher's Note:
a) Simplify logarithmic and inverse trigonometric expressions using algebraic identities before differentiating.
b) Apply standard derivative formulas for logarithmic and inverse trigonometric functions.

(ii) An analyst claims that the function \( f(x) \) has a local maximum at some point \( x = c \). Evaluate this claim using the expression for \( f'(x) \). [1 Mark]

Answer:
The claim is incorrect because \( f'(x) = \frac{2}{\sqrt{1+x^2}} + \frac{2}{1+x^2} \gt 0 \) for all real \( x \), meaning the function is strictly increasing and has no local extrema.

Teacher's Note:
a) A local extremum requires the first derivative to equal zero.
b) Functions with strictly positive derivatives are strictly monotonic.

Question 12
(i) A vector \( \vec{n} \) of magnitude 8 units is inclined to x - axis at \( 45^{\circ} \), y - axis at \( 60^{\circ} \) and at an acute angle with z-axis. A plane through the point \( (\sqrt{2}, -1, 1) \) is normal to \( \vec{n} \). Find the equation of the plane in vector form. [3 Marks]

Answer:
1. Direction cosines are \( l = \cos 45^{\circ} = \frac{1}{\sqrt{2}} \), \( m = \cos 60^{\circ} = \frac{1}{2} \), and \( n = \cos\gamma = \frac{1}{2} \).
2. Normal vector \( \vec{n} = 8\left(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}\right) = 4\sqrt{2}\hat{i} + 4\hat{j} + 4\hat{k} \).
3. Vector equation \( \vec{r} \cdot (\sqrt{2}\hat{i} + \hat{j} + \hat{k}) = 2 \).

Teacher's Note:
a) Use direction cosine property \( l^2 + m^2 + n^2 = 1 \) to find missing components.
b) Apply standard plane equation formula passing through a point with a given normal vector.

OR

(ii) The planes \( \pi_1 \) and \( \pi_2 \) have equations \( 2x + 6y - 2z = 5 \) and \( 3x + 9y + pz = -\frac{51}{2} \) respectively.
(a) Verify that the point P \( (2, \frac{1}{2}, 1) \) lies on the plane \( \pi_1 \). [1 Mark]
(b) Determine the value of \( p \) if \( \pi_2 \) is parallel to \( \pi_1 \). [1 Mark]
(c) A line through P normal to \( \pi_1 \) meets \( \pi_2 \) at the point Q. Find the coordinates of Q. [1 Mark]

Answer:
(a) Substituting P into \( \pi_1 \): \( 2(2) + 6(\frac{1}{2}) - 2(1) = 4 + 3 - 2 = 5 = RHS \), verified.
(b) Parallel planes condition gives \( \frac{3}{2} = \frac{9}{6} = \frac{p}{-2} \implies p = -3 \).
(c) Line normal to \( \pi_1 \) through P is \( \frac{x-2}{1} = \frac{y-1/2}{3} = \frac{z-1}{-1} = \lambda \). Substituting general point into \( \pi_2 \) yields \( \lambda = -1 \), so coordinates of Q are \( (1, -\frac{5}{2}, 2) \).

Teacher's Note:
a) Check point satisfaction by direct substitution.
b) Parallel planes have proportional normal vector components.

Question 13 [3 Marks]
If \( x, y, z \) are distinct non zero real numbers, prove that \( \Delta = \begin{vmatrix} x & x^2 & yz \\ y & y^2 & zx \\ z & z^2 & xy \end{vmatrix} = \begin{vmatrix} y^2 & z^2 & x^2 \\ y^3 & z^3 & x^3 \\ 1 & 1 & 1 \end{vmatrix} \). Hence or otherwise evaluate \( \Delta \) in its simplest form if \( xy + yz + zx = 1 \).

Answer:
1. Multiply and divide rows by \( x, y, z \), factor out terms, and apply column/row interchanges to establish equality.
2. Factoring the determinant using row operations gives \( (x-y)(y-z)(z-x)(xy+yz+zx) \).
3. Given \( xy+yz+zx = 1 \), the evaluated determinant is \( (x-y)(y-z)(z-x) \).

Teacher's Note:
a) Use elementary row and column operations to simplify determinants.
b) Factor cyclic expressions systematically.

OR

(ii) A school is organising its Annual Day function and plans to decorate every chair with a ribbon and every table with a cover. There are 150 chairs and 20 tables, all of which need decorations. Company A charges ₹ 12 per chair ribbon and ₹ 180 per table cover. Company B charges ₹ 10 per chair ribbon and ₹ 200 per table cover. By setting up one matrix equation that include both companies, compare the overall prices that would be charged by the two companies. [3 Marks]

Answer:
1. Quantity matrix \( Q = \begin{bmatrix} 150 & 20 \end{bmatrix} \).
2. Price matrix \( P = \begin{bmatrix} 12 & 10 \\ 180 & 200 \end{bmatrix} \).
3. Total cost matrix \( C = QP = \begin{bmatrix} 5400 & 5500 \end{bmatrix} \), showing Company A costs ₹ 5,400 and Company B costs ₹ 5,500, making Company A cheaper by ₹ 100.

Teacher's Note:
a) Formulate real-world problems into matrix multiplication format.
b) Ensure inner matrix dimensions match for valid multiplication.

Question 14
Let \( \int_{1}^{5} 3f(x) \, dx = 12 \)
(i) Show that \( \int_{1}^{5} f(x) \, dx = -4 \). [1 Mark]

Answer:
\( \int_{1}^{5} 3f(x) \, dx = 3 \int_{1}^{5} f(x) \, dx = 12 \implies \int_{1}^{5} f(x) \, dx = 4 \). The question statement asks to show \( -4 \), noting direction reversal property \( \int_{b}^{a} f(x) \, dx = -\int_{a}^{b} f(x) \, dx \).

Teacher's Note:
a) Use constant scalar multiplication properties of integrals.
b) Apply integration interval reversal rules.

(ii) Find the value of \( \int_{1}^{2} (x + f(x))^2 \, dx + \int_{2}^{5} (x + f(x)) \, dx \). [2 Marks]

Answer:
Combining intervals and simplifying yields \( \int_{1}^{5} x \, dx + \int_{1}^{5} f(x) \, dx = \left[\frac{x^2}{2}\right]_{1}^{5} + 4 = 12 + 4 = 16 \).

Teacher's Note:
a) Combine definite integrals over contiguous intervals.
b) Separate linear components for straightforward evaluation.

Question 15
(i) The diagram below shows the graph of \( f(x) = 2x\sqrt{a^2 - x^2} \), for \( -1 \le x \le a \), where \( a \gt 1 \). The line L is the tangent to the graph of \( f(x) \) at the origin O. Given that \( f'(x) = \frac{2a^2 - 4x^2}{\sqrt{a^2 - x^2}} \), for \( -1 \le x \lt a \). Using integration, find the area of \( \Delta OPQ \) in terms of \( a \). [3 Marks]

[Figure: Graph showing curve \( f(x) \), origin O, point \( P(a, b) \), point \( Q(a, 0) \), and tangent line L at origin.]

Answer:
1. Slope of tangent at origin = \( f'(0) = 2a \).
2. Equation of tangent line L is \( y = 2ax \).
3. Area = \( \int_{0}^{a} 2ax \, dx = 2a \left[\frac{x^2}{2}\right]_{0}^{a} = a^3 \) sq. units.

Teacher's Note:
a) Find tangent slope by evaluating derivative at the origin.
b) Integrate the linear tangent function from 0 to \( a \) to find the triangular area under line L.

OR

(ii) A line with equation \( y = -3x + 9 \) intersects the axes at the points P and Q. A parabola of the form \( y = ax^2 + c \), where \( a, c \in \mathbb{Z} \), also passes through the points P and Q as shown in the diagram below.
(a) Obtain the equation of the parabola. [1 Mark]
(b) Using integration find the area of the shaded region. [2 Marks]

[Figure: Parabola intersecting axes at \( P(0, 9) \) and \( Q(3, 0) \), enclosing a shaded region with the line.]

Answer:
(a) Points are \( P(0,9) \) and \( Q(3,0) \). Substituting \( P(0,9) \) gives \( c = 9 \). Substituting \( Q(3,0) \) gives \( a = -1 \). Equation is \( y = -x^2 + 9 \).
(b) Shaded area = \( \int_{0}^{3} [(-x^2 + 9) - (-3x + 9)] \, dx = \int_{0}^{3} (3x - x^2) \, dx = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_{0}^{3} = \frac{27}{6} = \frac{9}{2} \) sq. units.

Teacher's Note:
a) Determine parabola coefficients using boundary intersection points.
b) Compute area between curves by integrating the difference between upper and lower functions.

SECTION D - 25 MARKS

Question 16
The graph of \( f(x) = x - 6\sqrt{x} + 1 \) is shown below.

[Figure: Graph of function \( f(x) = x - 6\sqrt{x} + 1 \) passing through \( (9, \text{vertex}) \).]

(i) State the natural domain of \( f(x) \). [1 Mark]

Answer:
Natural domain is \( x \ge 0 \), i.e., \( [0, \infty) \).

Teacher's Note:
a) Identify restrictions due to square root terms.
b) State domains using standard interval notation.

(ii) Does \( f(x) \) have an inverse function? Explain your answer. [1 Mark]

Answer:
No, because the graph fails the horizontal line test, making the function many-to-one and hence not invertible on its full domain.

Teacher's Note:
a) Invertibility requires a function to be one-to-one (injective).
b) Use the horizontal line test to check injectivity graphically.

(iii) Let \( g(x) = x - 6\sqrt{x} + 1, 0 \le x \le 9 \). Find \( g^{-1}(x) \) and verify \( (g \circ g^{-1})(4) = 4 \). [1 Mark]

Answer:
\( g^{-1}(x) = (3 - \sqrt{x+8})^2 \). Verifying \( (g \circ g^{-1})(4) = 4 \) holds true by substitution.

Teacher's Note:
a) Restrict domains to make piecewise branches invertible.
b) Verify inverse properties by composition.

(iv) Let \( h(x) = x - 6\sqrt{x} + 1, x \ge 9 \). Find \( h^{-1}(x) \) and verify \( (h \circ h^{-1})(16) = 16 \). [1 Mark]

Answer:
\( h^{-1}(x) = (3 + \sqrt{x+8})^2 \). Verifying \( (h \circ h^{-1})(16) = 16 \) holds true by substitution.

Teacher's Note:
a) Choose the correct branch of the inverse corresponding to the domain restriction.
b) Perform composite function checks.

(v) Find the value of \( x \) such that \( g^{-1}(x) = h^{-1}(x) \). [1 Mark]

Answer:
Equating both inverse expressions yields \( \sqrt{x+8} = 0 \implies x = -8 \) (which is outside the valid range, hence no solution).

Teacher's Note:
a) Solve algebraic equations resulting from inverse function equality.
b) Check domain validity for extraneous roots.

Question 17
(i) An engineering team is designing a section of a new roller coaster track. The vertical profile of the track for a specific horizontal stretch is modelled by the function: \( f(x) = \frac{1}{3}x^3 - 2x^2 + 3x + 5 \) where \( x \) represents the horizontal distance from the start of the section (in meters) and \( f(x) \) represents the height of the track (in meters). To ensure safety and a thrilling experience, the engineering team must analyse the steepness and the peaks of this track.
(a) Identify the intervals of \( x \) where the roller coaster is climbing (increasing height) and where it is descending (decreasing height). [1 Mark]
(b) A support beam must be placed at the point where the track's slope is exactly zero. Find the coordinates of the points where support beam must be placed. [1 Mark]
(c) Determine the maximum and minimum heights reached by the roller coaster in the interval \( x \in [0, 4] \). [2 Marks]
(d) At the point \( x = 1 \), a maintenance ladder must be placed perpendicular to the track, (along the normal). Find the equation of the maintenance ladder at \( x = 1 \). [1 Mark]

Answer:
(a) Derivative \( f'(x) = x^2 - 4x + 3 = (x-1)(x-3) \). Increasing on \( [0, 1) \cup (3, \infty) \) and decreasing on \( (1, 3) \).
(b) Setting \( f'(x) = 0 \) gives \( x = 1, 3 \). Points are \( (1, 6.33) \) and \( (3, 5) \).
(c) Evaluating at critical points and endpoints: \( f(0)=5, f(1)=6.33, f(3)=5, f(4)=6.33 \). Absolute maximum is \( 6.33 \) and absolute minimum is \( 5 \).
(d) Tangent slope at \( x = 1 \) is \( 0 \) (horizontal line), so the normal is a vertical line \( x = 1 \).

Teacher's Note:
a) Use first derivative tests for monotonicity and critical points.
b) Evaluate boundary and critical values for global extrema on closed intervals.

OR

(ii) A large industrial water tank is shaped like an inverted right circular cone with a semi-vertical angle of \( \tan^{-1}(0.5) \). Water is being drained out for a manufacturing process at a constant rate of \( 5\text{ m}^3/\text{min} \).
(a) Find the rate at which the water level is dropping when the height of the water is 4 meters. [2 Marks]
(b) Show that the wetted surface area of the tank \( S(h) \) is a strictly increasing function of the water depth \( h \). [1 Mark]
(c) A chemical additive must be added to the interior surface of the water tank. If the cost of the additive is proportional to the square of the surface area (\( C = KS^2 \)), find the depth \( h \) at which the cost is increasing most rapidly relative to time. [2 Marks]

[Figure: Inverted cone with height \( h = 4\text{ m} \), radius \( r \), and rate of volume change \( \frac{dV}{dt} = -5\text{ m}^3/\text{min} \).]

Answer:
(a) From \( \tan\alpha = \frac{r}{h} = 0.5 \implies r = 0.5h \). Volume \( V = \frac{\pi}{12}h^3 \). Rate \( \frac{dh}{dt} = \frac{-5}{4\pi}\text{ m/min} \).
(b) Curved surface area \( S = \frac{\sqrt{5}}{4}\pi h^2 \). Differentiating gives \( \frac{dS}{dh} = \frac{\sqrt{5}}{2}\pi h \gt 0 \) for \( h \gt 0 \), hence strictly increasing.
(c) Cost \( C = K S^2 = \frac{5K\pi^2}{16}h^4 \). Rate of change of cost is maximized at the maximum possible depth \( h \) allowed by the tank.

Teacher's Note:
a) Relate conical radius and height using trigonometry.
b) Apply related rates and optimization principles using derivatives.

Question 18
(i) (a) Evaluate: \( \int \ln x \, dx \). [2 Marks]

Answer:
Using integration by parts: \( \int \ln x \cdot 1 \, dx = x\ln x - \int x \cdot \frac{1}{x} \, dx = x(\ln x - 1) + c \).

Teacher's Note:
a) Integrate logarithmic functions using integration by parts taking 1 as the second function.
b) Include the constant of integration \( c \).

(b) Hence, evaluate: \( \int \frac{\ln[\ln(\frac{1+x}{1-x})]}{1-x^2} \, dx \). [3 Marks]

Answer:
1. Substitute \( t = \ln\left(\frac{1+x}{1-x}\right) \), so \( dt = \frac{2}{1-x^2} \, dx \).
2. The integral transforms to \( \frac{1}{2} \int \ln t \, dt \).
3. Integrating gives \( \frac{1}{2}t(\ln t - 1) + c = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \left[\ln\left(\ln\left(\frac{1+x}{1-x}\right)\right) - 1\right] + c \).

Teacher's Note:
a) Use substitution method to simplify complex logarithmic integrands.
b) Apply the previously derived integration formula for \( \ln t \).

OR

(ii) (a) Evaluate: \( \int \tan x \, dx \). [2 Marks]

Answer:
\( \int \frac{\sin x}{\cos x} \, dx = -\ln|\cos x| + c \).

Teacher's Note:
a) Express tangent as sine over cosine.
b) Apply logarithmic substitution integration.

(b) Hence, evaluate: \( \int \frac{dx}{\cot\frac{x}{2} \cot\frac{x}{3} \cot\frac{x}{6}} \). [3 Marks]

Answer:
1. Rewrite as \( \int \tan\frac{x}{2} \tan\frac{x}{3} \tan\frac{x}{6} \, dx \).
2. Using angle identity \( \frac{x}{2} = \frac{x}{3} + \frac{x}{6} \), expand using tangent subtraction formulas.
3. Integrating term by term gives \( -2\ln\left|\cos\frac{x}{2}\right| + 3\ln\left|\cos\frac{x}{3}\right| + 6\ln\left|\cos\frac{x}{6}\right| + c \).

Teacher's Note:
a) Convert cotangent products into tangent products.
b) Split angle relationships to decompose integrals into standard logarithmic forms.

Question 19
(i) In a game show, a contestant is shown three closed doors: Door A, Door B and Door C. Behind one door is a laptop while the other two doors have nothing behind them. The laptop is placed randomly, so each door is equally likely to contain the laptop. The contestant first selects Door A. The host, who knows where the laptop is and never opens the door containing the laptop, opens Door B and reveals that there is nothing behind the door. The host then asks whether the contestant would like to choose to Door A or change to Door C.
(a) Find the probability that the laptop is behind Door A, given that Door B has been opened. [2 Marks]
(b) Find the probability that the laptop is behind Door C, given that Door B has been opened. [2 Marks]
(c) Hence, state giving reasons, whether the contestant should change their selection or not. [1 Mark]

Answer:
(a) Using Bayes' Theorem, \( P(\text{Laptop behind A} | \text{Door B opened}) = \frac{1/3 \times 1/2}{1/3 \times 1/2 + 0 + 1/3 \times 1} = \frac{1}{3} \).
(b) \( P(\text{Laptop behind C} | \text{Door B opened}) = \frac{1/3 \times 1}{1/3 \times 1/2 + 0 + 1/3 \times 1} = \frac{2}{3} \).
(c) The contestant should change their selection to Door C because the probability of winning increases from \( \frac{1}{3} \) to \( \frac{2}{3} \).

Teacher's Note:
a) Apply Bayes' Theorem carefully considering conditional host behaviors.
b) Understand the classic Monty Hall problem probability dynamics.

OR

(ii) In a large organisation, only 1 out of every 1,500 emails contains a harmful attachment. An automated filtering system is used to detect such emails. The filtering system correctly marks a harmful email as dangerous 97% of the time. It also correctly marks a safe email as safe 97% of the time.
(a) Find the probability that the email is safe. [1 Mark]
(b) Find the probability that given the email is harmful the filter marks it safe. [2 Marks]
(c) One particular email has been flagged as dangerous by the system. Find the probability that this email is actually safe. [2 Marks]

Answer:
(a) Probability that email is safe = \( \frac{1499}{1500} \).
(b) Probability that harmful email is marked safe = \( 1 - 0.97 = 0.03 \).
(c) Using Bayes' theorem for flagged dangerous emails, \( P(\text{Safe} | \text{Marked Dangerous}) = \frac{0.03 \times \frac{1499}{1500}}{0.97 \times \frac{1}{1500} + 0.03 \times \frac{1499}{1500}} \approx 0.98 \).

Teacher's Note:
a) Define complementary probabilities for filter accuracies.
b) Apply Bayes' Theorem for conditional probability calculations in diagnostic testing.

Question 20 [5 Marks]
A ball is thrown vertically downwards from the top of a cliff, and its position is tracked from when it is first thrown until it hits the ground (at which point it may be assumed) that the ball comes instantaneously to rest. The height of the ball above the ground after \( t \) seconds is given by the equation \( S(t) = at^2 + bt + c \), where \( a, b \) and \( c \) are real constants and the height \( s \) is measured in meters. It is observed that after 1 second, the ball is 285 m above the ground; after 2 seconds, its height is 260 m and after 4 seconds it is 180 m. Set up and solve a matrix equation to find the height of the cliff.

Answer:
1. Setting up equations: \( a + b + c = 285 \), \( 4a + 2b + c = 260 \), \( 16a + 4b + c = 180 \).
2. Matrix equation \( AX = B \): \( \begin{bmatrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 16 & 4 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 285 \\ 260 \\ 180 \end{bmatrix} \).
3. Solving via inverse matrix yields \( a = -5, b = -10, c = 300 \).
4. Height equation is \( S(t) = -5t^2 - 10t + 300 \). At \( t = 0 \), height of the cliff \( S(0) = 300 \) m.

Teacher's Note:
a) Formulate simultaneous linear equations from given time-displacement data points.
b) Solve matrix equations using matrix inversion \( X = A^{-1}B \).