ISC Class 12 Mathematics Board Exam Question Paper 2018 with Solutions

ISC Class 12 Mathematics Board Exam Question Paper with Solutions

 

SECTION - A (80 Marks)

 

Question 1

 

(i) The binary operation \(*: R \times R \to R\) is defined as \(a*b = 2a + b\). Find \((2*3)*4\). [1 Mark]

Answer:
Given \(a*b = 2a + b\)
\((2*3)*4 = (2(2) + 3)*4\)
\(= (4 + 3)*4\)
\(= 7*4\)
\(= 2(7) + 4\)
\(= 14 + 4 = 18\)

Teacher's Note:
a) Apply the binary operation formula step by step starting from the innermost parenthesis.
b) Ensure proper substitution of values into \(2a + b\) without calculation errors.

 

(ii) If \(A = \begin{pmatrix} 5 & a \\ b & 0 \end{pmatrix}\) and \(A\) is symmetric matrix, show that \(a = b\). [1 Mark]

Answer:
Given \(A = \begin{pmatrix} 5 & a \\ b & 0 \end{pmatrix}\) is a symmetric matrix.
Since \(A\) is symmetric, \(A = A^{T}\).
\(\begin{pmatrix} 5 & a \\ b & 0 \end{pmatrix} = \begin{pmatrix} 5 & b \\ a & 0 \end{pmatrix}\)
Comparing the corresponding elements, we get \(a = b\).

Teacher's Note:
a) Recall that a matrix \(A\) is symmetric if and only if \(A = A^{T}\).
b) Equate corresponding elements of the matrix and its transpose to find the relation.

 

(iii) Solve : \(3\tan^{-1} x + \cot^{-1} x = \pi\). [1 Mark]

Answer:
\(3\tan^{-1} x + \cot^{-1} x = \pi\)
\(2\tan^{-1} x + \tan^{-1} x + \cot^{-1} x = \pi\)
Since \(\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}\), we get:
\(2\tan^{-1} x + \frac{\pi}{2} = \pi\)
\(2\tan^{-1} x = \pi - \frac{\pi}{2}\)
\(2\tan^{-1} x = \frac{\pi}{2}\)
\(\tan^{-1} x = \frac{\pi}{4}\)
\(x = \tan\left(\frac{\pi}{4}\right) = 1\)

Teacher's Note:
a) Use the standard inverse trigonometric identity \(\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}\).
b) Isolate \(\tan^{-1} x\) and evaluate to find \(x\).

 

(iv) Without expanding at any stage, find the value of: \(\begin{vmatrix} a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z \end{vmatrix}\). [1 Mark]

Answer:
Let \(\Delta = \begin{vmatrix} a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z \end{vmatrix}\)
Apply operation \(R_{2} \to R_{2} - 2R_{3}\):
\(\Delta = \begin{vmatrix} a & b & c \\ a & b & c \\ x & y & z \end{vmatrix}\)
Since two rows (\(R_{1}\) and \(R_{2}\)) are identical, the value of the determinant is \(0\).

Teacher's Note:
a) Use elementary row operations to create identical rows.
b) State the property that a determinant with two identical rows is zero.

 

(v) Find the value of constant \(k\) so that the function \(f(x)\) defined as:
\(f(x) = \begin{cases} \frac{x^{2}-2x-3}{x+1}, & x \neq -1 \\ k, & x = -1 \end{cases}\)
is continuous at \(x = -1\). [1 Mark]

Answer:
Given \(f(x)\) is continuous at \(x = -1\), therefore:
\(f(-1) = \lim_{x \to -1} f(x)\)
\(k = \lim_{x \to -1} \frac{x^{2}-2x-3}{x+1}\)
\(k = \lim_{x \to -1} \frac{(x-3)(x+1)}{x+1}\)
Since \(x \to -1 \implies x + 1 \neq 0\):
\(k = \lim_{x \to -1} (x - 3) = -1 - 3 = -4\)
\(k = -4\)

Teacher's Note:
a) For continuity at a point, the value of the function must equal the limit of the function at that point.
b) Factorize the numerator and cancel the common term to avoid indeterminate form \(\frac{0}{0}\).

 

(vi) Find the approximate change in the volume \(V\) of a cube of side \(x\) metres caused by decreasing the side by \(1\%\). [1 Mark]

Answer:
Volume of a cube \(V = x^{3}\)
\(\frac{dV}{dx} = 3x^{2}\)
Given decrease in side = \(1\%\), so \(\delta x = x \times \left(-\frac{1}{100}\right) = -\frac{x}{100}\)
Change in volume \(\delta V \approx \frac{dV}{dx} \cdot \delta x\)
\(\delta V = (3x^{2})\left(-\frac{x}{100}\right) = -\frac{3}{100}x^{3} = -\frac{3}{100}V\)
Thus, the volume decreases by \(3\%\).

Teacher's Note:
a) Use differentials to approximate the change: \(\delta V \approx \frac{dV}{dx} \delta x\).
b) Express the final change as a percentage of the original volume.

 

(vii) Evaluate : \(\int \frac{x^{3} + 5x^{2} + 4x + 1}{x^{2}} dx\). [1 Mark]

Answer:
\(I = \int \left(\frac{x^{3}}{x^{2}} + \frac{5x^{2}}{x^{2}} + \frac{4x}{x^{2}} + \frac{1}{x^{2}}\right) dx\)
\(I = \int \left(x + 5 + \frac{4}{x} + x^{-2}\right) dx\)
\(I = \frac{x^{2}}{2} + 5x + 4\ln|x| - \frac{1}{x} + c\)

Teacher's Note:
a) Divide each term in the numerator individually by the denominator before integrating.
b) Do not forget to include the constant of integration \(c\).

 

(viii) Find the differential equation of the family of concentric circles \(x^{2} + y^{2} = a^{2}\). [1 Mark]

Answer:
Given equation of the family of concentric circles: \(x^{2} + y^{2} = a^{2}\)
Differentiating both sides with respect to \(x\):
\(2x + 2y \frac{dy}{dx} = 0\)
\(x + y \frac{dy}{dx} = 0\)

Teacher's Note:
a) Differentiate the given equation with respect to \(x\) to eliminate the arbitrary constant \(a\).
b) Simplify the resulting equation to obtain the first-order differential equation.

 

(ix) If \(A\) and \(B\) are events such that \(P(A) = \frac{1}{2}\), \(P(B) = \frac{1}{3}\) and \(P(A \cap B) = \frac{1}{4}\), then find:
(a) \(P(A/B)\)
(b) \(P(B/A)\) [1 Mark]

Answer:
Given \(P(A) = \frac{1}{2}\), \(P(B) = \frac{1}{3}\), \(P(A \cap B) = \frac{1}{4}\)
(a) \(P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{1/3} = \frac{3}{4}\)
(b) \(P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{1/4}{1/2} = \frac{1}{2}\)

Teacher's Note:
a) Use the conditional probability formula \(P(A/B) = \frac{P(A \cap B)}{P(B)}\).
b) Ensure correct substitution of fractions and simplify carefully.

 

(x) In a race, the probabilities of A and B winning the race are \(\frac{1}{3}\) and \(\frac{1}{6}\) respectively. Find the probability of neither of them winning the race. [1 Mark]

Answer:
Let \(E_{1}\) be the event that A wins the race and \(E_{2}\) be the event that B wins the race.
\(P(E_{1}) = \frac{1}{3}\), \(P(E_{2}) = \frac{1}{6}\)
The events are independent. The probability of neither winning is \(P(E_{1}^{\prime} \cap E_{2}^{\prime}) = P(E_{1}^{\prime}) \cdot P(E_{2}^{\prime})\)
\(P(E_{1}^{\prime}) = 1 - P(E_{1}) = 1 - \frac{1}{3} = \frac{2}{3}\)
\(P(E_{2}^{\prime}) = 1 - P(E_{2}) = 1 - \frac{1}{6} = \frac{5}{6}\)
Required probability \(= \frac{2}{3} \times \frac{5}{6} = \frac{10}{18} = \frac{5}{9}\)

Teacher's Note:
a) Use the complement rule to find the probability of each person not winning.
b) Since winning the race by independent participants are independent events, multiply their probabilities of losing.

 

Question 2

If the function \(f(x) = \sqrt{2x-3}\) is invertible then find its inverse. Hence prove that \((fof^{-1})(x) = x\). [4 Marks]

Answer:
Let \(y = f(x) = \sqrt{2x-3}\)
\(y^{2} = 2x - 3\)
\(2x = y^{2} + 3\)
\(x = \frac{y^{2} + 3}{2}\)
Therefore, \(f^{-1}(x) = \frac{x^{2} + 3}{2}\).
Now, L.H.S. \(= (fof^{-1})(x) = f(f^{-1}(x))\)
\(= f\left(\frac{x^{2}+3}{2}\right) = \sqrt{2\left(\frac{x^{2}+3}{2}\right) - 3}\)
\(= \sqrt{x^{2} + 3 - 3} = \sqrt{x^{2}} = x\)
Hence proved that \((fof^{-1})(x) = x\).

Teacher's Note:
a) To find the inverse, express \(x\) in terms of \(y\) and replace \(y\) with \(x\).
b) Verify the property by substituting the inverse function back into the original function.

 

Question 3

If \(\tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi\), prove that \(a + b + c = abc\). [4 Marks]

Answer:
Given \(\tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi\)
\(\tan^{-1} a + \tan^{-1} b = \pi - \tan^{-1} c\)
Taking tangent on both sides:
\(\tan(\tan^{-1} a + \tan^{-1} b) = \tan(\pi - \tan^{-1} c)\)
\(\frac{a + b}{1 - ab} = -\tan(\tan^{-1} c)\)
\(\frac{a + b}{1 - ab} = -c\)
\(a + b = -c(1 - ab)\)
\(a + b = -c + abc\)
\(a + b + c = abc\)
Hence proved.

Teacher's Note:
a) Shift one term to the right side and apply the tangent formula for the sum of two inverse tangents.
b) Use the property \(\tan(\pi - \theta) = -\tan\theta\) to simplify the equation.

 

Question 4

Use properties of determinants to solve for \(x\):
\(\begin{vmatrix} x+a & b & c \\ c & x+b & a \\ a & b & x+c \end{vmatrix} = 0\) and \(x \neq 0\). [4 Marks]

Answer:
Given \(\begin{vmatrix} x+a & b & c \\ c & x+b & a \\ a & b & x+c \end{vmatrix} = 0\)
Applying \(C_{1} \to C_{1} + C_{2} + C_{3}\):
\(\begin{vmatrix} x+a+b+c & b & c \\ x+a+b+c & x+b & a \\ x+a+b+c & b & x+c \end{vmatrix} = 0\)
Taking \((x+a+b+c)\) common from \(C_{1}\):
\((x+a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & x+b & a \\ 1 & b & x+c \end{vmatrix} = 0\)
Applying \(R_{2} \to R_{2} - R_{1}\) and \(R_{3} \to R_{3} - R_{1}\):
\((x+a+b+c) \begin{vmatrix} 1 & b & c \\ 0 & x & a-c \\ 0 & 0 & x \end{vmatrix} = 0\)
Expanding along \(C_{1}\):
\((x+a+b+c)(1)(x \cdot x - 0) = 0\)
\((x+a+b+c)x^{2} = 0\)
Since \(x \neq 0\), we have \(x^{2} \neq 0\), which gives:
\(x + a + b + c = 0 \implies x = -(a+b+c)\).

Teacher's Note:
a) Use column operations to make elements uniform in a column, then factor out the common term.
b) Apply row operations to reduce the determinant to an upper triangular form before expanding.

 

Question 5

(a) Show that the function \(f(x) = \begin{cases} x^{2}, & x \leq 1 \\ 1, & x > 1 \end{cases}\) is continuous at \(x = 1\) but not differentiable. [4 Marks]

Answer:
Continuity at \(x = 1\):
\(f(1) = 1^{2} = 1\)
\(\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} x^{2} = 1\)
\(\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} 1 = 1\)
Since \(\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{+}} f(x) = f(1) = 1\), \(f(x)\) is continuous at \(x = 1\).
Differentiability at \(x = 1\):
Right-Hand Derivative (R.H.D.):
\(RHD = \lim_{x \to 1^{+}} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^{+}} \frac{1 - 1}{x - 1} = 0\)
Left-Hand Derivative (L.H.D.):
\(LHD = \lim_{x \to 1^{-}} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^{-}} \frac{x^{2} - 1}{x - 1} = \lim_{x \to 1^{-}} (x + 1) = 2\)
Since \(LHD \neq RHD\) (\(2 \neq 0\)), \(f(x)\) is not differentiable at \(x = 1\).

Teacher's Note:
a) Check continuity by comparing left-hand limit, right-hand limit, and the value of the function.
b) Check differentiability by evaluating Left-Hand Derivative and Right-Hand Derivative separately using limits.

OR

(b) Verify Rolle's theorem for the following function: \(f(x) = e^{-x} \sin x\) on \([0, \pi]\). [4 Marks]

Answer:
Given \(f(x) = e^{-x} \sin x\) on \([0, \pi]\).
(i) \(f(x)\) is continuous on \([0, \pi]\) as it is the product of continuous functions.\br />(ii) \(f(x)\) is differentiable on \((0, \pi)\).\br />(iii) \(f(0) = e^{0} \sin 0 = 0\) and \(f(\pi) = e^{-\pi} \sin \pi = 0\), so \(f(0) = f(\pi)\).\br />Thus, all conditions of Rolle's theorem are satisfied. Hence, there exists at least one \(c \in (0, \pi)\) such that \(f^{\prime}(c) = 0\).\br />\(f^{\prime}(x) = e^{-x} \cos x - e^{-x} \sin x = e^{-x}(\cos x - \sin x)\)
\(f^{\prime}(c) = e^{-c}(\cos c - \sin c) = 0\)
Since \(e^{-c} \neq 0\), we have \(\cos c - \sin c = 0 \implies \tan c = 1\)
Since \(c \in (0, \pi)\), \(c = \frac{\pi}{4}\).
Thus, Rolle's theorem is verified.

Teacher's Note:
a) Verify the three main conditions of Rolle's theorem: continuity on closed interval, differentiability on open interval, and \(f(a) = f(b)\).
b) Find \(f^{\prime}(c) = 0\) to determine the value of \(c\) and ensure it lies within the given open interval.

 

Question 6

If \(x = \tan\left(\frac{1}{a} \log y\right)\), prove that \((1+x^{2})\frac{d^{2}y}{dx^{2}} + (2x-a)\frac{dy}{dx} = 0\). [4 Marks]

Answer:
Given \(x = \tan\left(\frac{1}{a} \log y\right)\)
\(\frac{1}{a} \log y = \tan^{-1} x\)
\(\log y = a \tan^{-1} x\)
\(y = e^{a \tan^{-1} x}\)
Differentiating both sides with respect to \(x\):
\(\frac{dy}{dx} = e^{a \tan^{-1} x} \cdot \frac{a}{1 + x^{2}}\)
\(\frac{dy}{dx} = \frac{ay}{1 + x^{2}}\)
\((1 + x^{2})\frac{dy}{dx} = ay\)
Differentiating again with respect to \(x\):
\((1 + x^{2})\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx}(2x) = a\frac{dy}{dx}\)
Since \(ay = (1+x^{2})\frac{dy}{dx}\), substituting or rearranging gives:
\((1 + x^{2})\frac{d^{2}y}{dx^{2}} + 2x\frac{dy}{dx} - a\frac{dy}{dx} = 0\)
\((1 + x^{2})\frac{d^{2}y}{dx^{2}} + (2x - a)\frac{dy}{dx} = 0\)
Hence proved.

Teacher's Note:
a) Simplify the expression by taking inverse tangent and converting the logarithmic function into an exponential form before differentiation.
b) Differentiate twice and rearrange terms carefully to match the required second-order differential equation.

 

Question 7

Evaluate : \(\int \tan^{-1}\sqrt{x} dx\). [4 Marks]

Answer:
Let \(I = \int \tan^{-1}\sqrt{x} dx\)
Put \(\sqrt{x} = t \implies x = t^{2} \implies dx = 2t dt\)
\(I = \int \tan^{-1}(t) \cdot 2t dt = 2 \int t \tan^{-1}t dt\)
Using integration by parts:
\(I = 2 \left[ \tan^{-1}t \cdot \frac{t^{2}}{2} - \int \frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} dt \right]\)
\(I = t^{2}\tan^{-1}t - \int \frac{t^{2}}{1+t^{2}} dt\)
\(I = t^{2}\tan^{-1}t - \int \left(1 - \frac{1}{1+t^{2}}\right) dt\)
\(I = t^{2}\tan^{-1}t - t + \tan^{-1}t + c\)
\(I = (t^{2} + 1)\tan^{-1}t - t + c\)
Substituting back \(t = \sqrt{x}\):
\(I = (x + 1)\tan^{-1}\sqrt{x} - \sqrt{x} + c\)

Teacher's Note:
a) Use substitution first (\(\sqrt{x} = t\)) to simplify the inverse trigonometric function argument.
b) Apply integration by parts, followed by algebraic adjustment (\(+1\) and \(-1\)) in the rational function.

 

Question 8

(a) Find the points on the curve \(y = 4x^{3} - 3x + 5\) at which the equation of the tangent is parallel to the x-axis. [4 Marks]

Answer:
Given curve: \(y = 4x^{3} - 3x + 5\) ---(i)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = 12x^{2} - 3\)
Since the tangent is parallel to the x-axis, its slope is zero:
\(\frac{dy}{dx} = 0 \implies 12x^{2} - 3 = 0\)
\(12x^{2} = 3 \implies x^{2} = \frac{1}{4} \implies x = \pm\frac{1}{2}\)
When \(x = \frac{1}{2}\):
\(y = 4\left(\frac{1}{2}\right)^{3} - 3\left(\frac{1}{2}\right) + 5 = 4\left(\frac{1}{8}\right) - \frac{3}{2} + 5 = \frac{1}{2} - \frac{3}{2} + 5 = 4\)
Point is \(\left(\frac{1}{2}, 4\right)\).
When \(x = -\frac{1}{2}\):
\(y = 4\left(-\frac{1}{2}\right)^{3} - 3\left(-\frac{1}{2}\right) + 5 = -\frac{1}{2} + \frac{3}{2} + 5 = 6\)
Point is \(\left(-\frac{1}{2}, 6\right)\).

Teacher's Note:
a) Recall that if a tangent is parallel to the x-axis, its slope \(\frac{dy}{dx} = 0\).
b) Substitute the values of \(x\) back into the original equation of the curve to find the corresponding y-coordinates.

OR

(b) Water is dripping out from a conical funnel of semi-vertical angle \(\frac{\pi}{4}\) at the uniform rate of \(2\text{ cm}^{2}/\text{sec}\) in the surface, through a tiny hole at the vertex of the bottom. When the slant height of the water level is \(4\text{ cm}\), find the rate of decrease of the slant height of the water. [4 Marks]

Answer:
Let \(r\) be the radius, \(h\) be the height, and \(V\) be the volume of the water cone.
\(V = \frac{1}{3}\pi r^{2}h\)
Given semi-vertical angle \(\theta = 45^{\circ}\). Let \(l\) be the slant height.
\(r = l \sin 45^{\circ} = \frac{l}{\sqrt{2}}\) and \(h = l \cos 45^{\circ} = \frac{l}{\sqrt{2}}\)
\(V = \frac{1}{3}\pi \left(\frac{l}{\sqrt{2}}\right)^{2}\left(\frac{l}{\sqrt{2}}\right) = \frac{\pi l^{3}}{6\sqrt{2}}\)
Given \(\frac{dV}{dt} = -2\text{ cm}^{3}/\text{sec}\).
\(\frac{dV}{dt} = \frac{\pi}{6\sqrt{2}} \cdot 3l^{2} \frac{dl}{dt} = \frac{\pi l^{2}}{2\sqrt{2}} \frac{dl}{dt}\)
\(-2 = \frac{\pi l^{2}}{2\sqrt{2}} \frac{dl}{dt} \implies \frac{dl}{dt} = \frac{-4\sqrt{2}}{\pi l^{2}}\)
At \(l = 4\text{ cm}\):
\(\frac{dl}{dt} = \frac{-4\sqrt{2}}{\pi (4)^{2}} = -\frac{\sqrt{2}}{4\pi}\text{ cm/sec}\).
Thus, the rate of decrease of the slant height is \(\frac{\sqrt{2}}{4\pi}\text{ cm/sec}\).

Teacher's Note:
a) Express the volume of the cone purely in terms of the slant height \(l\) using trigonometric relations.
b) Differentiate with respect to time \(t\) and substitute the given values to find the required rate of change.

 

Question 9

(a) Solve : \(\sin x \frac{dy}{dx} - y = \sin x \cdot \tan\frac{x}{2}\). [5 Marks]

Answer:
Given differential equation:
\(\sin x \frac{dy}{dx} - y = \sin x \cdot \tan\frac{x}{2}\)
Dividing by \(\sin x\):
\(\frac{dy}{dx} - y \csc x = \tan\frac{x}{2}\) ---(i)
This is a linear differential equation of the form \(\frac{dy}{dx} + Py = Q\), where \(P = -\csc x\) and \(Q = \tan\frac{x}{2}\).
\(\text{I.F.} = e^{\int P dx} = e^{\int -\csc x dx} = e^{-\ln(\csc x - \cot x)} = (\csc x - \cot x)^{-1} = \csc x + \cot x\)
Solution is given by:
\(y(\text{I.F.}) = \int Q(\text{I.F.}) dx\)
\(y(\csc x + \cot x) = \int \tan\frac{x}{2}(\csc x + \cot x) dx\)
Using half-angle formulas, \(\csc x + \cot x = \cot\frac{x}{2}\):
\(y(\csc x + \cot x) = \int \tan\frac{x}{2} \cdot \cot\frac{x}{2} dx = \int 1 dx = x + c\)
\(y(\csc x + \cot x) = x + c\)

Teacher's Note:
a) Reduce the differential equation to standard linear form by dividing through by \(\sin x\).
b) Evaluate the integrating factor and simplify trigonometric identities carefully during integration.

OR

(b) The population of a town grows at the rate of \(10\%\) per year. Using differential equation, find how long will it take for the population to grow \(4\) times. [5 Marks]

Answer:
Let \(x\) be the population at time \(t\).
\(\frac{dx}{dt} \propto x \implies \frac{dx}{dt} = rx\), where \(r = 10\% = 0.10\).
\(\frac{dx}{x} = 0.10 dt\)
Integrating both sides:
\(\ln x = 0.10t + c \implies x = e^{0.10t + c} = Ke^{0.10t}\), where \(K = e^{c}\) is initial population \(x_{0}\).
Given \(x = 4x_{0}\):
\(4x_{0} = x_{0}e^{0.10t} \implies 4 = e^{0.10t}\)
Taking natural logarithm on both sides:
\(\ln 4 = 0.10t \implies 2\ln 2 = 0.10t\)
\(t = \frac{2 \times 0.69314}{0.10} = 13.86\text{ years}\).

Teacher's Note:
a) Set up the differential equation for exponential growth: \(\frac{dx}{dt} = rx\).
b) Use the initial condition and the given growth rate to solve for time \(t\).

 

Question 10

(a) Using matrices, solve the following system of equations:
\(2x - 3y + 5z = 11\)
\(3x + 2y - 4z = -5\)
\(x + y - 2z = -3\) [6 Marks]

Answer:
The given system can be written as \(AX = B\), where:
\(A = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix}\), \(X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\), \(B = \begin{pmatrix} 11 \\ -5 \\ -3 \end{pmatrix}\)
\(|A| = 2(-4 + 4) - (-3)(-6 + 4) + 5(3 - 2) = 2(0) + 3(-2) + 5(1) = -6 + 5 = -1 \neq 0\).
Since \(|A| \neq 0\), \(A^{-1}\) exists and \(X = A^{-1}B\).
Finding cofactors of matrix \(A\):
\(A_{11} = 0\), \(A_{12} = 2\), \(A_{13} = 1\)
\(A_{21} = -1\), \(A_{22} = -9\), \(A_{23} = -5\)
\(A_{31} = 2\), \(A_{32} = 23\), \(A_{33} = 13\)
\(\text{adj } A = \begin{pmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{pmatrix}\)
\(A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{-1}\begin{pmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{pmatrix} = \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix}\)
\(X = A^{-1}B = \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix} \begin{pmatrix} 11 \\ -5 \\ -3 \end{pmatrix} = \begin{pmatrix} 0 - 5 + 6 \\ -22 - 45 + 69 \\ -11 - 25 + 39 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\)
Therefore, \(x = 1\), \(y = 2\), \(z = 3\).

Teacher's Note:
a) Express the linear equations in matrix form \(AX = B\).
b) Compute the determinant, find the matrix of cofactors, transpose to get the adjoint, and multiply by \(B\) to find the solution vector.

OR

(b) Using elementary transformation, find the inverse of the matrix : \(\begin{pmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix}\). [6 Marks]

Answer:
Let \(A = \begin{pmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix}\)
We know \(A = IA\):
\(\begin{pmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A\)
Applying row operations to reduce \(A\) to identity matrix:
Interchange \(R_{1} \leftrightarrow R_{2}\):
\(\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} A\)
\(R_{3} \to R_{3} - 3R_{1}\):
\(\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{pmatrix} A\)
\(R_{3} \to R_{3} + 5R_{2}\):
\(\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{pmatrix} A\)
\(R_{2} \to R_{2} - R_{3}\) (adjusting rows):
\(R_{3} \to \frac{1}{2}R_{3}\), \(R_{1} \to R_{1} - 2R_{2}\), etc., finally resulting in:
\(A^{-1} = \frac{1}{2}\begin{pmatrix} 1 & -1 & 1 \\ -8 & 6 & -2 \\ 5 & -3 & 1 \end{pmatrix}\)

Teacher's Note:
a) Start with the matrix equation \(A = IA\) and apply elementary row transformations systematically.
b) Create zeros below the principal diagonal first, then clear upper elements to obtain the identity matrix on the left.

 

Question 11

A speaks truth in \(60\%\) of the cases, while B is in \(40\%\) of the cases. In what percent of cases are they likely to contradict each other in stating the same fact ? [4 Marks]

Answer:
Probability that A speaks truth \(P(A) = \frac{60}{100}\), Probability that A lies \(P(A^{\prime}) = \frac{40}{100}\)
Probability that B speaks truth \(P(B) = \frac{40}{100}\), Probability that B lies \(P(B^{\prime}) = \frac{60}{100}\)
They contradict each other if one speaks the truth and the other lies:
\(\text{Contradiction} = P(A) \cdot P(B^{\prime}) + P(A^{\prime}) \cdot P(B)\)
\(= \left(\frac{60}{100} \times \frac{60}{100}\right) + \left(\frac{40}{100} \times \frac{40}{100}\right)\)
\(= \frac{3600}{10000} + \frac{1600}{10000} = \frac{5200}{10000} = \frac{52}{100}\)
Percentage of cases they are likely to contradict each other \(= \frac{52}{100} \times 100 = 52\%\).

Teacher's Note:
a) Two people contradict each other when one speaks the truth and the other speaks falsely.
b) Add the probabilities of these two mutually exclusive cases and convert the final fraction into a percentage.

 

Question 12

A cone is inscribed in a sphere of radius \(12\text{ cm}\). If the volume of the cone is maximum, find its height. [4 Marks]

Answer:
Let \(R = 12\text{ cm}\) be the radius of the sphere. Let \(x\) be the distance from the center of the sphere to the base of the cone.
Height of the cone \(h = R + x = 12 + x\).
Radius of the base of the cone \(r = \sqrt{R^{2} - x^{2}} = \sqrt{12^{2} - x^{2}} = \sqrt{144 - x^{2}}\).
Volume of the cone \(V = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi (144 - x^{2})(12 + x) = \frac{1}{3}\pi (1728 + 144x - 12x^{2} - x^{3})\).
Differentiating with respect to \(x\):
\(\frac{dV}{dx} = \frac{1}{3}\pi (144 - 24x - 3x^{2})\)
For maximum volume, \(\frac{dV}{dx} = 0 \implies 144 - 24x - 3x^{2} = 0 \implies x^{2} + 8x - 48 = 0\)
\((x + 12)(x - 4) = 0 \implies x = 4\) (since \(x = -12\) is inadmissible).
Second derivative \(\frac{d^{2}V}{dx^{2}} = \frac{1}{3}\pi (-24 - 6x)\). At \(x = 4\), \(\frac{d^{2}V}{dx^{2}} = \frac{1}{3}\pi (-24 - 24) = -16\pi < 0\) (maximum).
Height of the cone \(= R + x = 12 + 4 = 16\text{ cm}\).

Teacher's Note:
a) Express both the radius and height of the inscribed cone in terms of the sphere's radius and a variable distance \(x\).
b) Use the first and second derivative tests to find the value of \(x\) that maximizes the volume.

 

Question 13

(a) Evaluate : \(\int \frac{x-1}{\sqrt{x^{2}-x}} dx\). [4 Marks]

Answer:
Let \(I = \int \frac{x-1}{\sqrt{x^{2}-x}} dx\)
Let numerator be expressed as \(A \cdot \frac{d}{dx}(x^{2}-x) + B\):
\(x - 1 = A(2x - 1) + B \implies 2A = 1 \implies A = \frac{1}{2}\)
and \(-A + B = -1 \implies -\frac{1}{2} + B = -1 \implies B = -\frac{1}{2}\)
\(I = \int \frac{\frac{1}{2}(2x-1) - \frac{1}{2}}{\sqrt{x^{2}-x}} dx = \frac{1}{2}\int \frac{2x-1}{\sqrt{x^{2}-x}} dx - \frac{1}{2}\int \frac{dx}{\sqrt{(x - \frac{1}{2})^{2} - (\frac{1}{2})^{2}}}\)
\(I = \frac{1}{2}(2\sqrt{x^{2}-x}) - \frac{1}{2}\ln\left| (x - \frac{1}{2}) + \sqrt{x^{2}-x} \right| + c\)
\(I = \sqrt{x^{2}-x} - \frac{1}{2}\ln\left| x - \frac{1}{2} + \sqrt{x^{2}-x} \right| + c\)

Teacher's Note:
a) Express the numerator as \(A(\text{derivative of quadratic}) + B\) to split the integral into standard forms.
b) Use standard integration formulas for radicals and logarithmic forms.

OR

(b) Evaluate : \(\int_{0}^{\pi/2} \frac{\cos^{2}x}{1 + \sin x \cos x} dx\). [4 Marks]

Answer:
Let \(I = \int_{0}^{\pi/2} \frac{\cos^{2}x}{1 + \sin x \cos x} dx\) ---(1)
Using property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx\):
\(I = \int_{0}^{\pi/2} \frac{\cos^{2}(\frac{\pi}{2}-x)}{1 + \sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2}-x)} dx\)
\(I = \int_{0}^{\pi/2} \frac{\sin^{2}x}{1 + \cos x \sin x} dx\) ---(2)
Adding (1) and (2):
\(2I = \int_{0}^{\pi/2} \frac{\cos^{2}x + \sin^{2}x}{1 + \sin x \cos x} dx = \int_{0}^{\pi/2} \frac{1}{1 + \sin x \cos x} dx\)
\(2I = \int_{0}^{\pi/2} \frac{\sec^{2}x}{\sec^{2}x + \tan x} dx = \int_{0}^{\pi/2} \frac{\sec^{2}x}{1 + \tan^{2}x + \tan x} dx\)
Put \(\tan x = t \implies \sec^{2}x dx = dt\). Limits: when \(x = 0, t = 0\); when \(x = \frac{\pi}{2}, t = \infty\).
\(2I = \int_{0}^{\infty} \frac{dt}{1 + t^{2} + t} = \int_{0}^{\infty} \frac{dt}{(t + \frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2}}\)
\(2I = \frac{2}{\sqrt{3}} \left[ \tan^{-1}\left(\frac{t + 1/2}{\sqrt{3}/2}\right) \right]_{0}^{\infty} = \frac{2}{\sqrt{3}}\left(\frac{\pi}{2} - \frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}\left(\frac{\pi}{3}\right) = \frac{2\pi}{3\sqrt{3}}\)
\(I = \frac{\pi}{3\sqrt{3}}\).

Teacher's Note:
a) Apply the definite integral property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx\) and add the integrals.
b) Convert the resulting trigonometric integral into an algebraic one using substitution and complete the square.

 

Question 14

From a lot of \(6\) items containing \(2\) defective items, a sample of \(4\) items are drawn at random. Let the random variable \(X\) denote the number of defective items in the sample.
If the sample is drawn without replacement, find:
(a) The probability distribution of \(X\)
(b) Mean of \(X\)
(c) Variance of \(X\) [4 Marks]

Answer:
Total items = \(6\), Defective = \(2\), Non-defective = \(4\). Sample size = \(4\).
Possible values of \(X\) (number of defective items): \(0, 1, 2\).
(a) Probability distribution:
\(P(X = 0) = \frac{{}^{2}C_{0} \times {}^{4}C_{4}}{{}^{6}C_{4}} = \frac{1 \times 1}{15} = \frac{1}{15}\)
\(P(X = 1) = \frac{{}^{2}C_{1} \times {}^{4}C_{3}}{{}^{6}C_{4}} = \frac{2 \times 4}{15} = \frac{8}{15}\)
\(P(X = 2) = \frac{{}^{2}C_{2} \times {}^{4}C_{2}}{{}^{6}C_{4}} = \frac{1 \times 6}{15} = \frac{6}{15}\)

Probability Distribution Table:

\(X\)	\(P(X)\)	\(X P(X)\)	\(X^{2} P(X)\)
0	\(1/15\)	0	0
1	\(8/15\)	\(8/15\)	\(8/15\)
2	\(6/15\)	\(12/15\)	\(24/15\)

(b) Mean \(\mu = \sum X P(X) = 0 + \frac{8}{15} + \frac{12}{15} = \frac{20}{15} = \frac{4}{3}\).
(c) Variance \(\sigma^{2} = \sum X^{2} P(X) - (\sum X P(X))^{2} = \frac{32}{15} - \left(\frac{4}{3}\right)^{2} = \frac{32}{15} - \frac{16}{9} = \frac{96 - 80}{45} = \frac{16}{45} \approx 0.35\).

Teacher's Note:
a) Use combinations to calculate the probabilities for each possible value of the random variable \(X\).
b) Use the standard formulas for mean \(\sum X P(X)\) and variance \(\sum X^{2}P(X) - (\mu)^{2}\).

 

SECTION - B (20 Marks)

 

Question 15

(a) Find \(\lambda\) if the scalar projection of \(\vec{a} = \lambda\hat{i} + \hat{j} + 4\hat{k}\) on \(\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}\) is \(4\) units. [2 Marks]

Answer:
Scalar projection of \(\vec{a}\) on \(\vec{b}\) is given by \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 4\).
\(\vec{a} \cdot \vec{b} = (\lambda)(2) + (1)(6) + (4)(3) = 2\lambda + 6 + 12 = 2\lambda + 18\)
\(|\vec{b}| = \sqrt{2^{2} + 6^{2} + 3^{2}} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7\)
\(\frac{2\lambda + 18}{7} = 4\)
\(2\lambda + 18 = 28 \implies 2\lambda = 10 \implies \lambda = 5\).

Teacher's Note:
a) Use the formula for scalar projection of vector \(\vec{a}\) on \(\vec{b}\), which is \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\).
b) Substitute the given magnitude and solve the linear equation for \(\lambda\).

(b) The Cartesian equation of line is : \(2x - 3 = 3y + 1 = 5 - 6z\). Find the vector equation of a line passing through \((7, -5, 0)\) and parallel to the given line. [2 Marks]

Answer:
Given Cartesian equation: \(2x - 3 = 3y + 1 = 5 - 6z\)
Rewrite in standard symmetrical form: \(\frac{x - 3/2}{1/2} = \frac{y + 1/3}{1/3} = \frac{z - 5/6}{-1/6}\)
Multiplying by \(6\): \(\frac{x - 3/2}{-3} = \frac{y + 1/3}{-2} = \frac{z - 5/6}{1}\) (or directly dividing coefficients of \(x, y, z\) to get direction ratios \(-3, -2, 1\)).
Direction ratios of the given line are \(-3, -2, 1\).
The vector equation of a line passing through position vector \(\vec{a} = 7\hat{i} - 5\hat{j} + 0\hat{k}\) and parallel to \(\vec{b} = -3\hat{i} - 2\hat{j} + \hat{k}\) is:
\(\vec{r} = (7\hat{i} - 5\hat{j}) + \lambda(-3\hat{i} - 2\hat{j} + \hat{k})\).

Teacher's Note:
a) Convert the Cartesian equation into standard symmetrical form by making the coefficients of \(x, y, z\) unity.
b) Use the point-direction form for the vector equation of a line: \(\vec{r} = \vec{a} + \lambda \vec{b}\).

(c) Find the equation of the plane through the intersection of the planes \(\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 9\) and \(\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3\) and passing through the origin. [3 Marks]

Answer:
Cartesian equations of the given planes are:
\(x + 3y - z - 9 = 0\)
\(2x - y + z - 3 = 0\)
Equation of the plane passing through the intersection of these planes is:
\((x + 3y - z - 9) + \lambda(2x - y + z - 3) = 0\)
Since the plane passes through the origin \((0, 0, 0)\):
\((0 + 0 - 0 - 9) + \lambda(0 - 0 + 0 - 3) = 0\)
\(-9 - 3\lambda = 0 \implies 3\lambda = -9 \implies \lambda = -3\)
Substituting \(\lambda = -3\) back into the equation:
\((x + 3y - z - 9) - 3(2x - y + z - 3) = 0\)
\(x + 3y - z - 9 - 6x + 3y - 3z + 9 = 0\)
\(-5x + 6y - 4z = 0 \implies 5x - 6y + 4z = 0\).

Teacher's Note:
a) Use the family of planes formula \(P_{1} + \lambda P_{2} = 0\).
b) Substitute the coordinates of the origin \((0, 0, 0)\) to find the value of \(\lambda\).

 

Question 16

(a) If A, B, C are three non-collinear points with position vectors \(\vec{a}, \vec{b}, \vec{c}\) respectively, then show that the length of the perpendicular from C on AB is \(\frac{|\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a}|}{|\vec{b}-\vec{a}|}\). [4 Marks]

Answer:
Let \(ABC\) be a triangle with position vectors \(\vec{a}, \vec{b}, \vec{c}\) for vertices \(A, B, C\) respectively.
Area of \(\Delta ABC = \frac{1}{2} |\vec{AB}| \times CM\), where \(CM\) is the perpendicular from \(C\) on \(AB\).
Also, Area of \(\Delta ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}|\)
\(= \frac{1}{2} |(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a})|\)
\(= \frac{1}{2} |\vec{b}\times\vec{c} - \vec{b}\times\vec{a} - \vec{a}\times\vec{c} + \vec{a}\times\vec{a}|\)
\(= \frac{1}{2} |\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a}|\)
Equating the two expressions for area:
\(\frac{1}{2} |\vec{b}-\vec{a}| \cdot CM = \frac{1}{2} |\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a}|\)
\(CM = \frac{|\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a}|}{|\vec{b}-\vec{a}|}\)
Hence proved.

Teacher's Note:
a) Express the area of triangle \(ABC\) in two different ways: using base and height, and using cross product of position vectors.
b) Equate both expressions to derive the required perpendicular distance formula.

OR

(b) Show that the four points A, B, C and D with position vectors \(4\hat{i} + 5\hat{j} + \hat{k}\), \(-\hat{j} - \hat{k}\), \(3\hat{i} + 9\hat{j} + 4\hat{k}\) and \(4(-\hat{i} + \hat{j} + \hat{k})\) respectively, are coplanar. [4 Marks]

Answer:
Let position vectors be:\br />\(\vec{a} = 4\hat{i} + 5\hat{j} + \hat{k}\)
\(\vec{b} = -\hat{j} - \hat{k}\)
\(\vec{c} = 3\hat{i} + 9\hat{j} + 4\hat{k}\)
\(\vec{d} = -4\hat{i} + 4\hat{j} + 4\hat{k}\)
Finding vectors \(\vec{AB}, \vec{AC}, \vec{AD}\):
\(\vec{AB} = \vec{b} - \vec{a} = -4\hat{i} - 6\hat{j} - 2\hat{k}\)
\(\vec{AC} = \vec{c} - \vec{a} = -\hat{i} + 4\hat{j} + 3\hat{k}\)
\(\vec{AD} = \vec{d} - \vec{a} = -8\hat{i} - \hat{j} + 3\hat{k}\)
Points are coplanar if scalar triple product \([\vec{AB} \ \vec{AC} \ \vec{AD}] = 0\):
\(\begin{vmatrix} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix}\)
\(= -4(12 - (-3)) - (-6)(-3 - (-24)) + (-2)(1 - (-32))\)
\(= -4(15) + 6(21) - 2(33) = -60 + 126 - 66 = 0\).
Since the scalar triple product is zero, the vectors are coplanar, and hence the four points are coplanar.

Teacher's Note:
a) Form three vectors from the four points using a common initial point (e.g., \(\vec{AB}, \vec{AC}, \vec{AD}\)).
b) Compute their scalar triple product using a determinant and show that it equals zero.

 

Question 17

(a) Draw a rough sketch of the curve and find the area of the region bounded by curve \(y^{2} = 8x\) and the line \(x = 2\text{ [Figure: Rough sketch showing a parabola symmetric about the x-axis, intersecting the line x=2 at (2,4) and (2,-4)]}\). [5 Marks]

Answer:
Curve: \(y^{2} = 8x\) is a parabola opening to the right.
Line: \(x = 2\).
Points of intersection with \(x = 2\): \(y^{2} = 8(2) = 16 \implies y = \pm 4\).\br />Required Area \(= 2 \int_{0}^{2} y dx = 2 \int_{0}^{2} \sqrt{8x} dx\)
\(= 2\sqrt{8} \int_{0}^{2} x^{1/2} dx = 4\sqrt{2} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{2}\)
\(= \frac{8\sqrt{2}}{3} \left[ 2^{3/2} - 0 \right] = \frac{8\sqrt{2}}{3} \times 2\sqrt{2} = \frac{32}{3}\text{ sq. units}\).

Teacher's Note:
a) Identify the curves and determine the limits of integration from the intersection points.
b) Use symmetry across the x-axis to simplify the definite integral calculation.

OR

(b) Sketch the graph of \(y = |x + 4|\). Using integration, find the area of the region bounded by the curve \(y = |x + 4|\) and \(x = -6\) and \(x = 0\text{ [Figure: V-shaped graph of y = |x+4| with vertex at (-4,0), bounded between x=-6 and x=0]}\). [5 Marks]

Answer:
\(y = |x + 4| = \begin{cases} -(x+4), & x < -4 \\ x+4, & x \geq -4 \end{cases}\)
Required Area \(= \int_{-6}^{-4} -(x+4) dx + \int_{-4}^{0} (x+4) dx\)
\(= \left[ -\frac{x^{2}}{2} - 4x \right]_{-6}^{-4} + \left[ \frac{x^{2}}{2} + 4x \right]_{-4}^{0}\)
\(= \left[ \left(-\frac{(-4)^{2}}{2} - 4(-4)\right) - \left(-\frac{(-6)^{2}}{2} - 4(-6)\right) \right] + \left[ 0 - \left(\frac{(-4)^{2}}{2} + 4(-4)\right) \right]\)
\(= [(-8 + 16) - (-18 + 24)] + [0 - (8 - 16)]\)
\(= [8 - 6] + [8] = 2 + 8 = 10\text{ sq. units}\).

Teacher's Note:
a) Split the absolute value function into two integrals at the critical point where the expression inside the modulus is zero (\(x = -4\)).
b) Integrate each part separately and sum the areas.

 

Question 18

Find the image of a point having position vector : \(3\hat{i} - 2\hat{j} + \hat{k}\) in the plane \(\vec{r} \cdot (3\hat{i} - \hat{j} + 4\hat{k}) = 2\text{ [Figure: Diagram showing point A, its image M, and the foot of perpendicular B on the plane]}\). [6 Marks]

Answer:
Let \(A(3, -2, 1)\) be the given point and let the equation of the plane be \(3x - y + 4z = 2\).
Equation of the line passing through \(A\) and normal to the plane is:
\(\frac{x - 3}{3} = \frac{y + 2}{-1} = \frac{z - 1}{4} = \lambda\)
Any general point on this line is \(B(3\lambda + 3, -\lambda - 2, 4\lambda + 1)\).
Since \(B\) lies on the plane \(3x - y + 4z = 2\):
\(3(3\lambda + 3) - (-\lambda - 2) + 4(4\lambda + 1) = 2\)
\(9\lambda + 9 + \lambda + 2 + 16\lambda + 4 = 2\)
\(26\lambda + 15 = 2 \implies 26\lambda = -13 \implies \lambda = -\frac{1}{2}\).
Coordinates of the foot of the perpendicular \(B\):
\(x = 3\left(-\frac{1}{2}\right) + 3 = \frac{3}{2}\)
\(y = -\left(-\frac{1}{2}\right) - 2 = -\frac{3}{2}\)
\(z = 4\left(-\frac{1}{2}\right) + 1 = -1\)
Let \((x_{1}, y_{1}, z_{1})\) be the image point of \(A\). Since \(B\) is the midpoint of segment joining \(A\) and its image:
\(\frac{3 + x_{1}}{2} = \frac{3}{2} \implies x_{1} = 0\)
\(\frac{-2 + y_{1}}{2} = -\frac{3}{2} \implies y_{1} = -1\)
\(\frac{1 + z_{1}}{2} = -1 \implies z_{1} = -3\)
Thus, the image point is \((0, -1, -3)\), or in vector form \(-\hat{j} - 3\hat{k}\).

Teacher's Note:
a) Find the equation of the line passing through the given point and normal to the plane.
b) Determine the intersection point (foot of the perpendicular) and use the midpoint formula to find the image coordinates.