ISC Class 12 Mathematics Board Exam Question Paper 2019 with Solutions

ISC Class 12 Mathematics Board Exam Question Paper with Solutions

SECTION A

Question 1

(i) If \( f : R \rightarrow R^{+} \), \( f(x) = x^{3} \) and \( g : R \rightarrow R \), \( g(x) = 2x^{2} + 1 \), and \( R \) is the set of real numbers, then find \( fog(x) \) and \( gof(x) \). [1 Mark]

Answer:
Given \( f(x) = x^{3} \) and \( g(x) = 2x^{2} + 1 \).
\( fog(x) = f(g(x)) = f(2x^{2} + 1) = (2x^{2} + 1)^{3} \)
\( gof(x) = g(f(x)) = g(x^{3}) = 2(x^{3})^{2} + 1 = 2x^{6} + 1 \)

Teacher's Note:
a) To find composite functions, substitute the inner function into the outer function step-by-step.
b) Ensure the domain and codomain conditions are understood, though computation requires direct substitution.

(ii) Solve: \( \sin(2 \tan^{-1} x) = 1 \) [1 Mark]

Answer:
Given \( \sin(2 \tan^{-1} x) = 1 \)
\( 2 \tan^{-1} x = \sin^{-1}(1) = \frac{\pi}{2} \)
\( \tan^{-1} x = \frac{\pi}{4} \)
\( x = \tan\left(\frac{\pi}{4}\right) = 1 \)

Teacher's Note:
a) Use inverse trigonometric properties to simplify the equation stepwise.
b) Remember that \( \sin\left(\frac{\pi}{2}\right) = 1 \), which leads to the principal value of the angle.

(iii) Using determinants, find the values of \( k \); if the area of triangle with vertices \( (-2, 0) \), \( (0, 4) \) and \( (0, k) \) is \( 4 \) square units. [1 Mark]

Answer:
Area of triangle is given by: \[ \Delta = \frac{1}{2} \begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix} = \pm 4 \] \[ \frac{1}{2} \left[ -2(4 - k) - 0 + 1(0) \right] = \pm 4 \] \[ \frac{1}{2} (-8 + 2k) = \pm 4 \] Case 1: \( -8 + 2k = 8 \implies 2k = 16 \implies k = 8 \)
Case 2: \( -8 + 2k = -8 \implies 2k = 0 \implies k = 0 \)
Therefore, \( k = 0 \) or \( k = 8 \).

Teacher's Note:
a) Always consider both positive and negative values when equating absolute area to a given magnitude.
b) The official key lists only \( k = 8 \), but \( k = 0 \) is also a valid solution since area can be positive or negative before taking the absolute value.

(iv) Show that \( (A + A') \) is a symmetric matrix, if \( A = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix} \). [1 Mark]

Answer:
Given \( A = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix} \)
Transpose of \( A \), \( A' = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} \)
Let \( P = A + A' = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix} + \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ 7 & 10 \end{bmatrix} \)
Now, \( P' = \begin{bmatrix} 4 & 7 \\ 7 & 10 \end{bmatrix}' = \begin{bmatrix} 4 & 7 \\ 7 & 10 \end{bmatrix} = P \)
Since \( P' = P \), \( (A + A') \) is a symmetric matrix.

Teacher's Note:
a) A matrix \( P \) is symmetric if \( P' = P \).
b) Matrix addition is commutative, ensuring \( (A + A')' = A' + A'' = A' + A = A + A' \).

(v) \( f(x) = \frac{x^{2} - 9}{x - 3} \) is not defined at \( x = 3 \). What value should be assigned to \( f(3) \) for continuity of \( f(x) \) at \( x = 3 \)? [1 Mark]

Answer:
For continuity at \( x = 3 \), \( f(3) = \lim_{x \to 3} f(x) \)
\( \lim_{x \to 3} \frac{x^{2} - 9}{x - 3} = \lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 3 + 3 = 6 \)
Therefore, \( f(3) \) should be assigned the value \( 6 \).

Teacher's Note:
a) Removable discontinuity can be resolved by defining the function at the point equal to its limit.
b) Factorize the numerator to cancel out the term causing division by zero.

(vi) Prove that the function \( f(x) = x^{3} - 6x^{2} + 12x + 5 \) is increasing on \( R \). [1 Mark]

Answer:
Given \( f(x) = x^{3} - 6x^{2} + 12x + 5 \)
Differentiating with respect to \( x \):
\( f'(x) = 3x^{2} - 12x + 12 = 3(x^{2} - 4x + 4) = 3(x - 2)^{2} \)
Since \( (x - 2)^{2} \ge 0 \) for all \( x \in R \), we have \( f'(x) \ge 0 \) for all \( x \in R \).
Therefore, the function is increasing on \( R \).

Teacher's Note:
a) A function is increasing if its first derivative is greater than or equal to zero for all points in the domain.
b) Expressing the derivative as a complete square makes the sign test trivial.

(vii) Evaluate \( \int \frac{\sec^{2} x}{\csc^{2} x} dx \) [1 Mark]

Answer:
\( \int \frac{\sec^{2} x}{\csc^{2} x} dx = \int \frac{\frac{1}{\cos^{2} x}}{\frac{1}{\sin^{2} x}} dx = \int \frac{\sin^{2} x}{\cos^{2} x} dx = \int \tan^{2} x dx \)
\( = \int (\sec^{2} x - 1) dx = \tan x - x + c \)

Teacher's Note:
a) Convert secant and cosecant into sine and cosine to simplify the integrand.
b) Use the trigonometric identity \( \tan^{2} x = \sec^{2} x - 1 \) to integrate directly.

(viii) Using L'Hopital's Rule evaluate \( \lim_{x \to 0} \frac{8^{x} - 4^{x}}{4x} \) [1 Mark]

Answer:
Given limit is of the form \( \frac{0}{0} \) as \( x \to 0 \).
Applying L'Hopital's Rule by differentiating numerator and denominator with respect to \( x \):
\( \lim_{x \to 0} \frac{8^{x} \ln 8 - 4^{x} \ln 4}{4} \)
Substitute \( x = 0 \):
\( = \frac{8^{0} \ln 8 - 4^{0} \ln 4}{4} = \frac{\ln 8 - \ln 4}{4} = \frac{\ln(8 / 4)}{4} = \frac{\ln 2}{4} \)

Teacher's Note:
a) Verify that the limit yields an indeterminate form before applying L'Hopital's Rule.
b) Recall that \( \frac{d}{dx}(a^{x}) = a^{x} \ln a \).

(ix) Two balls are drawn from containing 3 white, 5 red and 2 black ball, one by one without replacement. What is the probability that least one ball is red? [1 Mark]

Answer:
Total balls = \( 3 + 5 + 2 = 10 \).
Probability of getting no red ball in two draws = Probability that both balls are non-red (white or black, total \( 3 + 2 = 5 \) balls).
\( P(\text{no red}) = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9} \)
Probability that at least one ball is red = \( 1 - P(\text{no red}) = 1 - \frac{2}{9} = \frac{7}{9} \)

Teacher's Note:
a) Use the complementary probability method (\( 1 - P(\text{none}) \)) to solve "at least one" problems efficiently.
b) The official key shows an incorrect calculation of \( \frac{1}{2} \); the correct probability is \( \frac{7}{9} \) because non-red balls are 5 out of 10 in the first draw and 4 out of 9 in the second draw. .

(x) If event A and B are independent, such that \( P(A) = \frac{3}{5}, P(B) = \frac{2}{3} \)... [1 Mark]

Answer:
For independent events A and B:
\( P(A \cap B) = P(A) \cdot P(B) = \frac{3}{5} \times \frac{2}{3} = \frac{2}{5} \)
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{3}{5} + \frac{2}{3} - \frac{2}{5} = \frac{9 + 10 - 6}{15} = \frac{13}{15} \)

[Note: The question text in the paper was cut off mid-sentence, but standard textbook continuation asks for \( P(A \cap B) \) or \( P(A \cup B) \).]

Teacher's Note:
a) For independent events, the intersection probability is simply the product of their individual probabilities.
b) Apply the addition theorem of probability to find the union.

Question 2 [4]
If \( f : A \rightarrow A \) and \( A = R - \left\{\frac{8}{5}\right\} \), show that the function \( f(x) = \frac{8x + 3}{5x - 8} \) is one-one onto. Hence, find \( f^{-1} \)

Answer:
1. One-one check:
Let \( f(x_1) = f(x_2) \implies \frac{8x_1 + 3}{5x_1 - 8} = \frac{8x_2 + 3}{5x_2 - 8} \)
\( (8x_1 + 3)(5x_2 - 8) = (8x_2 + 3)(5x_1 - 8) \)
\( 40x_1x_2 - 64x_1 + 15x_2 - 24 = 40x_1x_2 - 64x_2 + 15x_1 - 24 \)
\( -64x_1 + 15x_2 = -64x_2 + 15x_1 \)
\( 79x_1 = 79x_2 \implies x_1 = x_2 \). Hence, \( f \) is one-one.
2. Onto check and inverse:
Let \( y = \frac{8x + 3}{5x - 8} \)
\( y(5x - 8) = 8x + 3 \implies 5xy - 8y = 8x + 3 \)
\( 5xy - 8x = 8y + 3 \implies x(5y - 8) = 8y + 3 \)
\( x = \frac{8y + 3}{5y - 8} \)
Since for every \( y \in A \), there exists \( x = \frac{8y + 3}{5y - 8} \in A \), the function is onto.
Thus, \( f^{-1}(x) = \frac{8x + 3}{5x - 8} \).

Teacher's Note:
a) A function is one-one if \( f(x_1) = f(x_2) \implies x_1 = x_2 \), and onto if every element in the codomain has a pre-image.
b) This particular function is its own inverse (self-inverse).

Question 3 [4]
(a) Solve for \( x \): \( \tan^{-1}\left(\frac{x - 1}{x - 2}\right) + \tan^{-1}\left(\frac{x + 1}{x + 2}\right) = \frac{\pi}{4} \)

Answer:
Using the formula \( \tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A + B}{1 - AB}\right) \):
\( \tan^{-1}\left( \frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \left(\frac{x - 1}{x - 2}\right)\left(\frac{x + 1}{x + 2}\right)} \right) = \frac{\pi}{4} \)
\( \frac{(x - 1)(x + 2) + (x + 1)(x - 2)}{(x - 2)(x + 2) - (x - 1)(x + 1)} = \tan\left(\frac{\pi}{4}\right) = 1 \)
\( \frac{(x^{2} + x - 2) + (x^{2} - x - 2)}{(x^{2} - 4) - (x^{2} - 1)} = 1 \)
\( \frac{2x^{2} - 4}{-3} = 1 \)
\( 2x^{2} - 4 = -3 \implies 2x^{2} = 1 \implies x^{2} = \frac{1}{2} \)
\( x = \pm \frac{1}{\sqrt{2}} \)

Teacher's Note:
a) Apply the sum formula for inverse tangent carefully with algebraic fractions.
b) Verify that the obtained solutions lie within the domain of the given inverse trigonometric expression.

OR

(b) If \( \sec^{-1} x = \csc^{-1} y \), show that \( \frac{1}{x^{2}} + \frac{1}{y^{2}} = 1 \) [4 Marks]

Answer:
Let \( \sec^{-1} x = \csc^{-1} y = t \)
Then \( x = \sec t \implies \cos t = \frac{1}{x} \)
And \( y = \csc t \implies \sin t = \frac{1}{y} \)
We know that \( \cos^{2} t + \sin^{2} t = 1 \)
Substituting the values: \( \left(\frac{1}{x}\right)^{2} + \left(\frac{1}{y}\right)^{2} = 1 \)
\( \frac{1}{x^{2}} + \frac{1}{y^{2}} = 1 \). Hence proved.

Teacher's Note:
a) Convert inverse secant and cosecant into standard trigonometric ratios using parameter substitution.
b) Use the fundamental Pythagorean identity \( \sin^{2} t + \cos^{2} t = 1 \) to complete the proof.

Question 4 [4]
Using properties of determinants prove that:
\[ \begin{vmatrix} x & x(x^{2} + 1) & x + 1 \\ y & y(y^{2} + 1) & y + 1 \\ z & z(z^{2} + 1) & z + 1 \end{vmatrix} = (x - y)(y - z)(z - x)(x + y + z) \]

Answer:
Let \( \Delta = \begin{vmatrix} x & x^{3} + x & x + 1 \\ y & y^{3} + y & y + 1 \\ z & z^{3} + z & z + 1 \end{vmatrix} \)
Apply row operations \( R_{2} \rightarrow R_{2} - R_{1} \) and \( R_{3} \rightarrow R_{3} - R_{1} \):
\( \Delta = \begin{vmatrix} x & x^{3} + x & x + 1 \\ y - x & y^{3} - x^{3} + y - x & y - x \\ z - x & z^{3} - x^{3} + z - x & z - x \end{vmatrix} \)
Take out \( (y - x) \) from \( R_2 \) and \( (z - x) \) from \( R_3 \):
\( \Delta = (x - y)(z - x) \begin{vmatrix} x & x^{3} + x & x + 1 \\ 1 & y^{2} + xy + x^{2} + 1 & 1 \\ 1 & z^{2} + xz + x^{2} + 1 & 1 \end{vmatrix} \)
Apply \( R_{3} \rightarrow R_{3} - R_{2} \):
\( \Delta = (x - y)(z - x)(y - z) \begin{vmatrix} x & x^{3} + x & x + 1 \\ 1 & y^{2} + xy + x^{2} + 1 & 1 \\ 0 & z + y + x & 0 \end{vmatrix} \)Expanding along the third row: gives \( (x - y)(y - z)(z - x)(x + y + z) \). Hence proved.

Teacher's Note:
a) Use elementary row operations to create zeros and extract common factors.
b) Factorize cubic differences like \( y^{3} - x^{3} \) into \( (y - x)(y^{2} + xy + x^{2}) \) to simplify the determinant.

Question 5 [4]
(a) Show that the function \( f(x) = |x - 4|, x \in R \) is continuous, but not differentiable at \( x = 4 \).

Answer:
1. Continuity at \( x = 4 \):
\( \lim_{x \to 4^{-}} f(x) = \lim_{h \to 0} |4 - h - 4| = \lim_{h \to 0} |-h| = 0 \)
\( \lim_{x \to 4^{+}} f(x) = \lim_{h \to 0} |4 + h - 4| = \lim_{h \to 0} |h| = 0 \)
\( f(4) = |4 - 4| = 0 \)
Since LHL = RHL = \( f(4) \), \( f(x) \) is continuous at \( x = 4 \).
2. Differentiability at \( x = 4 \):
Left Hand Derivative (LHD) = \( \lim_{h \to 0} \frac{f(4 - h) - f(4)}{-h} = \lim_{h \to 0} \frac{|-h| - 0}{-h} = \lim_{h \to 0} \frac{h}{-h} = -1 \)
Right Hand Derivative (RHD) = \( \lim_{h \to 0} \frac{f(4 + h) - f(4)}{h} = \lim_{h \to 0} \frac{|h| - 0}{h} = \lim_{h \to 0} \frac{h}{h} = 1 \)
Since LHD \( \neq \) RHD (\( -1 \neq 1 \)), \( f(x) \) is not differentiable at \( x = 4 \).

Teacher's Note:
a) Modulus functions are continuous everywhere but have sharp corners (cusps) where derivatives fail to exist.
b) Always test differentiability using first principles (left-hand and right-hand limits of the derivative).

OR

(b) Verify the Lagrange's Mean Value Theorem for the function \( f(x) = x + \frac{1}{x} \) in the interval \( [1, 3] \). [4 Marks]

Answer:
Given \( f(x) = x + \frac{1}{x} \) on \( [1, 3] \).
1. \( f(x) \) is continuous on \( [1, 3] \) and differentiable on \( (1, 3) \).
2. By Lagrange's Mean Value Theorem, there exists at least one \( c \in (1, 3) \) such that: \[ f'(c) = \frac{f(3) - f(1)}{3 - 1} \] \( f'(x) = 1 - \frac{1}{x^{2}} \implies f'(c) = 1 - \frac{1}{c^{2}} \)
\( f(3) = 3 + \frac{1}{3} = \frac{10}{3} \)
\( f(1) = 1 + 1 = 2 \)
\( \frac{f(3) - f(1)}{3 - 1} = \frac{\frac{10}{3} - 2}{2} = \frac{\frac{4}{3}}{2} = \frac{2}{3} \)
Now, \( 1 - \frac{1}{c^{2}} = \frac{2}{3} \implies \frac{1}{c^{2}} = 1 - \frac{2}{3} = \frac{1}{3} \)
\( c^{2} = 3 \implies c = \pm \sqrt{3} \)
Since \( c \in (1, 3) \), we take \( c = \sqrt{3} \). Hence verified.

Teacher's Note:
a) State the conditions of LMVT (continuity on closed interval and differentiability on open interval) before applying it.
b) Reject negative values of \( c \) if they fall outside the given interval.

Question 6 [4]
If \( y = e^{\sin^{-1} x} \) and \( z = e^{-\cos^{-1} x} \) prove that \( \frac{dy}{dz} = e^{\pi/2} \)

Answer:
Given \( y = e^{\sin^{-1} x} \) and \( z = e^{-\cos^{-1} x} \).
We know that \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \implies \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \).
Thus, \( z = e^{-(\frac{\pi}{2} - \sin^{-1} x)} = e^{-\frac{\pi}{2} + \sin^{-1} x} = e^{-\frac{\pi}{2}} \cdot e^{\sin^{-1} x} = e^{-\frac{\pi}{2}} y \)
Differentiating \( y \) with respect to \( x \): \( \frac{dy}{dx} = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1 - x^{2}}} \)
Differentiating \( z \) with respect to \( x \): \( \frac{dz}{dx} = e^{-\cos^{-1} x} \cdot \frac{1}{\sqrt{1 - x^{2}}} \)
Now, \( \frac{dy}{dz} = \frac{\frac{dy}{dx}}{\frac{dz}{dx}} = \frac{e^{\sin^{-1} x}}{e^{-\cos^{-1} x}} = e^{\sin^{-1} x + \cos^{-1} x} = e^{\pi/2} \). Hence proved.

Teacher's Note:
a) Use parametric differentiation by finding \( \frac{dy}{dx} \) and \( \frac{dz}{dx} \).
b) The identity \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) simplifies the exponent directly.

Question 7 [4]
A \( 13 \) m long ladder is leaning against a vertical wall at a certain height from the ground level. The bottom of the ladder is pulled away from the wall, along the ground, at the rate of \( 2 \) m/s. How fast is the height on the wall decreasing when the foot of the ladder is \( 5 \) m from the wall?

[Figure: Right-angled triangle showing wall of height \( y \), ground distance \( x = 5 \) m, and ladder length \( 13 \) m]

Answer:
Let \( x \) be the distance of the foot of the ladder from the wall and \( y \) be the height of the ladder on the wall.
Given \( \frac{dx}{dt} = 2 \) m/s and ladder length \( z = 13 \) m.
By Pythagoras theorem: \( x^{2} + y^{2} = 13^{2} = 169 \)
When \( x = 5 \) m, \( 5^{2} + y^{2} = 169 \implies 25 + y^{2} = 169 \implies y^{2} = 144 \implies y = 12 \) m.
Differentiating \( x^{2} + y^{2} = 169 \) with respect to time \( t \):
\( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \implies x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \)
Substitute \( x = 5 \), \( y = 12 \), and \( \frac{dx}{dt} = 2 \):
\( 5(2) + 12 \frac{dy}{dt} = 0 \implies 10 + 12 \frac{dy}{dt} = 0 \)
\( \frac{dy}{dt} = -\frac{10}{12} = -\frac{5}{6} \) m/s.
Therefore, the height is decreasing at the rate of \( \frac{5}{6} \) m/s.

Teacher's Note:
a) Set up the geometric relation using the Pythagorean theorem before differentiating implicitly with respect to time.
b) A negative sign in rate of change indicates decrease; state the final rate as positive with the word "decreasing".

Question 8 [4]
(a) Evaluate: \( \int \frac{x(1 + x^{2})}{1 + x^{4}} dx \)

Answer:
\( \int \frac{x(1 + x^{2})}{1 + x^{4}} dx = \int \frac{x + x^{3}}{1 + x^{4}} dx = \int \frac{x}{1 + x^{4}} dx + \int \frac{x^{3}}{1 + x^{4}} dx \)
For the first integral, put \( x^{2} = t \implies 2x dx = dt \implies x dx = \frac{1}{2} dt \):
\( \int \frac{x}{1 + x^{4}} dx = \frac{1}{2} \int \frac{dt}{1 + t^{2}} = \frac{1}{2} \tan^{-1}(t) = \frac{1}{2} \tan^{-1}(x^{2}) \)
For the second integral, put \( 1 + x^{4} = u \implies 4x^{3} dx = du \implies x^{3} dx = \frac{1}{4} du \):
\( \int \frac{x^{3}}{1 + x^{4}} dx = \frac{1}{4} \int \frac{du}{u} = \frac{1}{4} \ln|u| = \frac{1}{4} \ln(1 + x^{4}) \)
Combining both: \( \frac{1}{2} \tan^{-1}(x^{2}) + \frac{1}{4} \ln(1 + x^{4}) + c \)

Teacher's Note:
a) Split the fraction into two parts to apply separate substitutions.
b) Recognize standard integral forms like \( \int \frac{1}{1 + t^{2}} dt = \tan^{-1} t \).

OR

(b) Evaluate: \( \int_{-6}^{3} |x + 3| dx \) [4 Marks]

Answer:
The integrand \( |x + 3| = -(x + 3) \) for \( x \in [-6, -3] \) and \( (x + 3) \) for \( x \in [-3, 3] \).
Splitting the integral at \( x = -3 \):
\( \int_{-6}^{3} |x + 3| dx = \int_{-6}^{-3} -(x + 3) dx + \int_{-3}^{3} (x + 3) dx \)
\( = \left[ -\frac{(x + 3)^{2}}{2} \right]_{-6}^{-3} + \left[ \frac{(x + 3)^{2}}{2} \right]_{-3}^{3} \)
\( = \left( 0 - \left(-\frac{(-3 + 3)^{2}}{2}\right) \right) + \left( \frac{(3 + 3)^{2}}{2} - 0 \right) \)
Wait, let's evaluate properly:
First part: \( \int_{-6}^{-3} -(x + 3) dx = \left[ -\frac{x^{2}}{2} - 3x \right]_{-6}^{-3} = \left( -\frac{9}{2} + 9 \right) - \left( -18 + 18 \right) = \frac{9}{2} \)
Second part: \( \int_{-3}^{3} (x + 3) dx = \left[ \frac{x^{2}}{2} + 3x \right]_{-3}^{3} = \left( \frac{9}{2} + 9 \right) - \left( \frac{9}{2} - 9 \right) = 18 \)
Total integral = \( \frac{9}{2} + 18 = \frac{45}{2} \)

Teacher's Note:
a) Always split definite integrals of modulus functions at points where the expression inside the modulus is zero.
b) Evaluate each region independently and sum the results.

Question 9 [4]
Solve the differential equation: \( \frac{dy}{dx} = \frac{x + y + 2}{2(x + y) - 1} \)

Answer:
Let \( x + y = v \implies 1 + \frac{dy}{dx} = \frac{dv}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1 \)
Substitute into the differential equation:
\( \frac{dv}{dx} - 1 = \frac{v + 2}{2v - 1} \)
\( \frac{dv}{dx} = \frac{v + 2}{2v - 1} + 1 = \frac{v + 2 + 2v - 1}{2v - 1} = \frac{3v + 1}{2v - 1} \)
Separating variables: \( \frac{2v - 1}{3v + 1} dv = dx \)
\( \left( \frac{2}{3} - \frac{5}{3(3v + 1)} \right) dv = dx \)
Integrating both sides:
\( \frac{2}{3}v - \frac{5}{9} \ln|3v + 1| = x + c \)
Substitute back \( v = x + y \):
\( \frac{2}{3}(x + y) - \frac{5}{9} \ln|3(x + y) + 1| = x + c \)

Teacher's Note:
a) Use substitution \( v = x + y \) when the right-hand side is a function of the linear combination \( (x + y) \).
b) Perform polynomial division on improper rational functions of \( v \) before integration.

Question 10 [4]
Bag A contains 4 white balls and 3 black balls, while Bag B contains 3 white balls and 5 black balls. Two balls are drawn from Bag A and placed in bag B. Then, what is the probability of drawing a white ball from Bag B?

Answer:
Bag A: 4W, 3B (Total 7 balls).
Bag B: 3W, 5B (Total 8 balls).
Two balls are drawn from Bag A and transferred to Bag B. There are three cases for the transferred balls:
Case 1: Both are white (2W).
\( P(2W) = \frac{{}^{4}C_{2}}{{}^{7}C_{2}} = \frac{6}{21} = \frac{2}{7} \)
Bag B now has: \( 3 + 2 = 5 \) white balls and 5 black balls (Total 10 balls).
Probability of drawing a white ball from B = \( \frac{5}{10} \).
Contribution = \( \frac{2}{7} \times \frac{5}{10} = \frac{10}{70} \)
Case 2: Both are black (2B).
\( P(2B) = \frac{{}^{3}C_{2}}{{}^{7}C_{2}} = \frac{3}{21} = \frac{1}{7} \)
Bag B now has: 3 white balls and \( 5 + 2 = 7 \) black balls (Total 10 balls).
Probability of drawing a white ball from B = \( \frac{3}{10} \).
Contribution = \( \frac{1}{7} \times \frac{3}{10} = \frac{3}{70} \)
Case 3: One white and one black (1W, 1B).
\( P(1W, 1B) = \frac{{}^{4}C_{1} \times {}^{3}C_{1}}{{}^{7}C_{2}} = \frac{4 \times 3}{21} = \frac{12}{21} = \frac{4}{7} \)
Bag B now has: \( 3 + 1 = 4 \) white balls and \( 5 + 1 = 6 \) black balls (Total 10 balls).
Probability of drawing a white ball from B = \( \frac{4}{10} \).
Contribution = \( \frac{4}{7} \times \frac{4}{10} = \frac{16}{70} \)
Total Probability = \( \frac{10}{70} + \frac{3}{70} + \frac{16}{70} = \frac{29}{70} \)

Teacher's Note:
a) Break down the problem into mutually exclusive cases based on the composition of transferred balls.
b) The official key contains calculation fragmentation; total probability using law of total probability is correctly \( \frac{29}{70} \). .

Question 11 [6]
Solve the following system of linear equations using matrix method:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 9 \]
\[ \frac{2}{x} + \frac{5}{y} + \frac{7}{z} = 52 \]
\[ \frac{2}{x} + \frac{1}{y} - \frac{1}{z} = 0 \]

Answer:
Let \( \frac{1}{x} = a, \frac{1}{y} = b, \frac{1}{z} = c \). The system becomes:
\( a + b + c = 9 \)
\( 2a + 5b + 7c = 52 \)
\( 2a + b - c = 0 \)
In matrix form \( AX = B \):
\[ \begin{bmatrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 9 \\ 52 \\ 0 \end{bmatrix} \] Determinant of \( A \):
\( |A| = 1(-5 - 7) - 1(-2 - 14) + 1(2 - 10) = -12 + 16 - 8 = -4 \neq 0 \)
Adjoint of \( A \):
\( \text{Adj}(A) = \begin{bmatrix} -12 & 2 & 2 \\ 16 & -3 & -5 \\ 8 & 1 & 3 \end{bmatrix} \)
\( X = A^{-1}B = \frac{1}{-4} \begin{bmatrix} -12 & 2 & 2 \\ 16 & -3 & -5 \\ 8 & 1 & 3 \end{bmatrix} \begin{bmatrix} 9 \\ 52 \\ 0 \end{bmatrix} \)
\[ = -\frac{1}{4} \begin{bmatrix} -108 + 104 \\ 144 - 156 \\ 72 + 52 \end{bmatrix} = -\frac{1}{4} \begin{bmatrix} -4 \\ -12 \\ 12 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ -3 \end{bmatrix} \] Wait, let's re-verify row 3 product: \( 8(9) + 1(52) + 3(0) = 72 + 52 = 124 \), divided by \( -4 \) is \( -31 \). Let's check original equation 3: \( 2(1) + 1(3) - 1(z) = 0 \implies 2 + 3 - c = 0 \implies c = 5 \).
Let's check \( \text{Adj}(A) \) element \( A_{33} \): cofactor of \(-1\) is \( 5 - 2 = 3 \).
Thus \( a = 1, b = 3, c = 5 \).
Therefore, \( x = 1, y = \frac{1}{3}, z = \frac{1}{5} \).

Teacher's Note:
a) Substitute reciprocal variables to convert non-linear equations into a standard linear system.
b) Always substitute back to find original variables \( x, y, z \) after solving for \( a, b, c \).

Question 12 [6]
(a) The volume of a closed rectangular metal box with a square base is \( 4096 \text{ cm}^{3} \). The cost of polishing the outer surface of the box is Rs \( 4 \) per \( \text{cm}^{2} \). Find the dimensions of the box for the minimum cost of polishing it.

[Figure: Cuboid with square base of side \( a \) and height \( b \)]

Answer:
Let the side of the square base be \( a \) and height be \( b \).
Volume \( V = a^{2}b = 4096 \implies b = \frac{4096}{a^{2}} \).
Total surface area \( S = 2a^{2} + 4ab \).
Cost \( C = 4 \times S = 4(2a^{2} + 4ab) = 8a^{2} + 16ab \).
Substitute \( b = \frac{4096}{a^{2}} \):
\( C(a) = 8a^{2} + 16a\left(\frac{4096}{a^{2}}\right) = 8a^{2} + \frac{65536}{a} \)
Differentiating with respect to \( a \):
\( C'(a) = 16a - \frac{65536}{a^{2}} \)
For critical points, set \( C'(a) = 0 \):
\( 16a = \frac{65536}{a^{2}} \implies 16a^{3} = 65536 \implies a^{3} = 4096 \implies a = 16 \) cm.
Second derivative: \( C''(a) = 16 + \frac{131072}{a^{3}} \)
At \( a = 16 \), \( C''(16) > 0 \) (local minimum).
Height \( b = \frac{4096}{16^{2}} = \frac{4096}{256} = 16 \) cm.
Dimensions of the box are \( 16 \) cm \( \times \) \( 16 \) cm \( \times \) \( 16 \) cm.

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

OR

(b) Find the point on the straight line \( 2x + 3y = 6 \), which is closest to the origin. [6 Marks]

[Figure: Line \( 2x + 3y = 6 \) intercepting axes at \( (3,0) \) and \( (0,2) \), with perpendicular from origin to point \( (\alpha, \beta) \)]

Answer:
Let point \( P(\alpha, \beta) \) lie on the line \( 2\alpha + 3\beta = 6 \implies \beta = \frac{6 - 2\alpha}{3} \).
Distance squared from origin \( D = \alpha^{2} + \beta^{2} = \alpha^{2} + \left(\frac{6 - 2\alpha}{3}\right)^{2} = \alpha^{2} + \frac{36 - 24\alpha + 4\alpha^{2}}{9} = \frac{13\alpha^{2} - 24\alpha + 36}{9} \)
Differentiating with respect to \( \alpha \):
\( \frac{dD}{d\alpha} = \frac{26\alpha - 24}{9} \)
Set \( \frac{dD}{d\alpha} = 0 \implies 26\alpha = 24 \implies \alpha = \frac{12}{13} \).
Corresponding \( \beta = \frac{6 - 2(\frac{12}{13})}{3} = \frac{6 - \frac{24}{13}}{3} = \frac{\frac{54}{13}}{3} = \frac{18}{13} \).
Second derivative is positive, hence minimum.
The closest point is \( \left(\frac{12}{13}, \frac{18}{13}\right) \).

Teacher's Note:
a) Minimize the square of the distance to avoid square root calculations in calculus.
b) The official key has minor typography in final coordinates; the correct coordinates are \( \left(\frac{12}{13}, \frac{18}{13}\right) \). .

Question 13 [6]
Evaluate: \( \int_{0}^{\pi} \frac{x \tan x}{\sec x + \tan x} dx \)

Answer:
Let \( I = \int_{0}^{\pi} \frac{x \tan x}{\sec x + \tan x} dx \) ---(1)
Using property \( \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx \):
\( I = \int_{0}^{\pi} \frac{(\pi - x) \tan(\pi - x)}{\sec(\pi - x) + \tan(\pi - x)} dx = \int_{0}^{\pi} \frac{(\pi - x)(-\tan x)}{-\sec x - \tan x} dx = \int_{0}^{\pi} \frac{(\pi - x) \tan x}{\sec x + \tan x} dx \) ---(2)
Adding (1) and (2):
\( 2I = \int_{0}^{\pi} \frac{x \tan x + (\pi - x) \tan x}{\sec x + \tan x} dx = \int_{0}^{\pi} \frac{\pi \tan x}{\sec x + \tan x} dx \)
\( 2I = \pi \int_{0}^{\pi} \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} dx = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \sin x} dx \)
Multiply numerator and denominator by \( (1 - \sin x) \):
\( 2I = \pi \int_{0}^{\pi} \frac{\sin x(1 - \sin x)}{1 - \sin^{2} x} dx = \pi \int_{0}^{\pi} \frac{\sin x - \sin^{2} x}{\cos^{2} x} dx = \pi \int_{0}^{\pi} (\tan x \sec x - \tan^{2} x) dx \)
\( 2I = \pi \left[ \sec x - (\tan x - x) \right]_{0}^{\pi} = \pi \left[ \sec x - \tan x + x \right]_{0}^{\pi} \)
Substitute limits: \( \pi \left( (-1 - 0 + \pi) - (1 - 0 + 0) \right) = \pi (\pi - 2) \)
Therefore, \( I = \frac{\pi}{2}(\pi - 2) \).

Teacher's Note:
a) King's property \( \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx \) is the standard approach for definite integrals with \( x \) in the numerator.
b) Rationalize the denominator by multiplying with the conjugate expression to simplify trigonometric integration.

Question 14 [6]
(a) Given three identical Boxes A, B and C, Box A contains 2 gold and 1 silver coin, Box B contains 1 gold and 2 silver coins and Box C contains 3 silver coins. A person chooses a Box at random and takes out a coin. If the coin drawn is of silver, find the probability that it has been drawn from the Box which has the remaining two coins also of silver.

Answer:
Let \( E_1, E_2, E_3 \) be the events of choosing Box A, B, and C respectively.
\( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \).
Let S be the event of drawing a silver coin.
\( P(S | E_1) = \frac{1}{3} \) (Box A has 1 silver out of 3 coins)
\( P(S | E_2) = \frac{2}{3} \) (Box B has 2 silver out of 3 coins)
\( P(S | E_3) = \frac{3}{3} = 1 \) (Box C has 3 silver out of 3 coins - this is the box with remaining two coins also of silver).
We need to find \( P(E_3 | S) \) using Bayes' Theorem:
\[ P(E_3 | S) = \frac{P(E_3) P(S | E_3)}{P(E_1) P(S | E_1) + P(E_2) P(S | E_2) + P(E_3) P(S | E_3)} \] \[ = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{3} + \frac{1}{3} \times 1} \] \[ = \frac{1}{\frac{1}{3} + \frac{2}{3} + 1} = \frac{1}{\frac{3}{3} + 1} = \frac{1}{1 + 1} = \frac{1}{2} \] Wait, let's re-add denominators: \( \frac{1}{9} + \frac{2}{9} + \frac{3}{9} = \frac{6}{9} = \frac{2}{3} \).
\( P(E_3 | S) = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \).

Teacher's Note:
a) Apply Bayes' Theorem when conditional probabilities in reverse are required.
b) The official key shows \( \frac{1}{4} \) due to arithmetic slips; the correct application yields \( \frac{1}{2} \). .

OR

(b) Determine the binomial distribution where mean is 9 and standard deviation is \( \frac{3}{2} \). Also, find the probability of obtaining at most one success. [6 Marks]

Answer:
Given Mean \( np = 9 \) and Standard Deviation \( \sqrt{npq} = \frac{3}{2} \).
Squaring the standard deviation: \( npq = \left(\frac{3}{2}\right)^{2} = \frac{9}{4} \)
Substitute \( np = 9 \): \( 9q = \frac{9}{4} \implies q = \frac{1}{4} \)
Since \( p = 1 - q = 1 - \frac{1}{4} = \frac{3}{4} \).
Find \( n \): \( np = 9 \implies n\left(\frac{3}{4}\right) = 9 \implies n = 9 \times \frac{4}{3} = 12 \).
Binomial distribution is given by \( P(X = r) = {}^{12}C_{r} \left(\frac{3}{4}\right)^{r} \left(\frac{1}{4}\right)^{12-r} \).
Probability of at most one success \( P(X \le 1) = P(X = 0) + P(X = 1) \):
\( P(X = 0) = {}^{12}C_{0} \left(\frac{3}{4}\right)^{0} \left(\frac{1}{4}\right)^{12} = \left(\frac{1}{4}\right)^{12} \)
\( P(X = 1) = {}^{12}C_{1} \left(\frac{3}{4}\right)^{1} \left(\frac{1}{4}\right)^{11} = 12 \times \frac{3}{4} \times \left(\frac{1}{4}\right)^{11} = 9 \left(\frac{1}{4}\right)^{11} = \frac{36}{4^{12}} \)
Total probability = \( \left(\frac{1}{4}\right)^{12} + \frac{36}{4^{12}} = \frac{37}{4^{12}} \).

Teacher's Note:
a) Use the standard relations for mean (\( np \)) and variance (\( npq \)) in binomial distributions.
b) "At most one success" means summing probabilities for \( r = 0 \) and \( r = 1 \).

Question 15 [3×2]
(a) If \( \vec{a} \) and \( \vec{b} \) are perpendicular vectors, \( |\vec{a} + \vec{b}| = 13 \) and \( |\vec{a}| = 5 \), find the value of \( |\vec{b}| \).

Answer:
Since \( \vec{a} \) and \( \vec{b} \) are perpendicular, \( \vec{a} \cdot \vec{b} = 0 \).
Given \( |\vec{a} + \vec{b}| = 13 \implies |\vec{a} + \vec{b}|^{2} = 13^{2} = 169 \)
\( |\vec{a}|^{2} + |\vec{b}|^{2} + 2(\vec{a} \cdot \vec{b}) = 169 \)
\( 5^{2} + |\vec{b}|^{2} + 2(0) = 169 \)
\( 25 + |\vec{b}|^{2} = 169 \implies |\vec{b}|^{2} = 169 - 25 = 144 \)
\( |\vec{b}| = 12 \)

Teacher's Note:
a) The dot product of two perpendicular vectors is always zero.
b) Square the magnitude of the vector sum to expand it using dot product properties.

(b) Find the length ofarket the perpendicular from origin to the plane \( \vec{r} \cdot (3\hat{i} - 4\hat{j} - 12\hat{k}) + 39 = 0 \) [3 Marks]

Answer:
Equation of plane in Cartesian form: \( 3x - 4y - 12z + 39 = 0 \) or \( 3x - 4y - 12z = -39 \).
Length of perpendicular from origin \( (0, 0, 0) \) to the plane: \[ d = \frac{|3(0) - 4(0) - 12(0) + 39|}{\sqrt{3^{2} + (-4)^{2} + (-12)^{2}}} \] \[ d = \frac{|39|}{\sqrt{9 + 16 + 144}} = \frac{39}{\sqrt{169}} = \frac{39}{13} = 3 \] units.

Teacher's Note:
a) Use the perpendicular distance formula from a point to a plane: \( d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \).
b) Ensure the constant term is on the correct side or handled with absolute value appropriately.

(c) Find the angle between the two lines \( 2x = 3y = -z \) and \( \frac{x}{2} = \frac{y}{3} = \frac{z}{1} \) [3 Marks]

Answer:
Rewrite the first line equation in standard symmetric form: \( \frac{x}{1/2} = \frac{y}{1/3} = \frac{z}{-1} \) or multiplying by 6: \( \frac{x}{3} = \frac{y}{2} = \frac{z}{-6} \). Direction ratios are \( \vec{u_1} = (3, 2, -6) \).
Second line equation: \( \frac{x}{2} = \frac{y}{3} = \frac{z}{1} \). Direction ratios are \( \vec{u_2} = (2, 3, 1) \).
Let \( \theta \) be the angle between the lines:
\[ \cos \theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \] \[ \cos \theta = \frac{3(2) + 2(3) + (-6)(1)}{\sqrt{3^{2} + 2^{2} + (-6)^{2}} \sqrt{2^{2} + 3^{2} + 1^{2}}} \] \[ \cos \theta = \frac{6 + 6 - 6}{\sqrt{49} \sqrt{14}} = \frac{6}{7\sqrt{14}} \] \( \theta = \cos^{-1}\left(\frac{6}{7\sqrt{14}}\right) \).

Teacher's Note:
a) Convert non-standard line equations into symmetric form \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \) to identify direction ratios.
b) Use the dot product formula of direction ratios to find the angle between lines.

Question 16 [4]
(a) If \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k} \), prove that \( \vec{a} \) and \( \vec{a} \times \vec{b} \) are perpendicular.

Answer:
First, find \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix} \] \( = \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 - (-4)) = 1\hat{i} + 11\hat{j} + 7\hat{k} \)
Now, check the dot product of \( \vec{a} \) and \( (\vec{a} \times \vec{b}) \):
\( \vec{a} \cdot (\vec{a} \times \vec{b}) = (1)(1) + (-2)(11) + (3)(7) = 1 - 22 + 21 = 0 \)
Since the dot product is zero, \( \vec{a} \) and \( \vec{a} \times \vec{b} \) are perpendicular.

Teacher's Note:
a) The cross product of two vectors is always orthogonal to both individual vectors.
b) Verify orthogonality by demonstrating that their dot product equals zero.

OR

(b) If \( \vec{a} \) and \( \vec{b} \) are not collinear vectors, find the value of \( x \) such that the vectors \( \vec{\alpha} = (x - 2)\vec{a} + \vec{b} \) and \( \vec{\beta} = (3 + 2x)\vec{a} - 2\vec{b} \) are collinear. [4 Marks]

Answer:
For non-collinear vectors \( \vec{a} \) and \( \vec{b} \), two vectors are collinear if one is a scalar multiple of the other.
\( \vec{\alpha} = k \vec{\beta} \implies (x - 2)\vec{a} + \vec{b} = k(3 + 2x)\vec{a} - 2k\vec{b} \)
Equating coefficients of \( \vec{a} \) and \( \vec{b} \):
\( x - 2 = k(3 + 2x) \) ---(1)
\( 1 = -2k \implies k = -\frac{1}{2} \) ---(2)
Substitute \( k = -\frac{1}{2} \) into equation (1):
\( x - 2 = -\frac{1}{2}(3 + 2x) \)
\( 2(x - 2) = -(3 + 2x) \)
\( 2x - 4 = -3 - 2x \)
\( 4x = 1 \implies x = \frac{1}{4} \)

Teacher's Note:
a) Two vectors are collinear if their respective component coefficients are proportional.
b) Equating coefficients of independent non-collinear vectors provides simultaneous equations to solve for unknowns.

Question 17 [4]
(a) Find the equation of the plane through the intersection of the planes \( 2x + 2y - 3z - 7 = 0 \) and \( 2x + 5y + 3z - 7 = 9 \) (Note: second plane equation is \( 2x + 5y + 3z - 9 = 0 \) in standard form) such that the intercepts made by the resulting plane on the \( x \)-axis and the \( z \)-axis are equal.

Answer:
Equation of the plane through the intersection of the two planes:
\( (2x + 2y - 3z - 7) + \lambda(2x + 5y + 3z - 9) = 0 \)
\( (2 + 2\lambda)x + (2 + 5\lambda)y + (-3 + 3\lambda)z - (7 + 9\lambda) = 0 \)
For \( x \)-intercept (\( y = 0, z = 0 \)): \( a = \frac{7 + 9\lambda}{2 + 2\lambda} \)
For \( z \)-intercept (\( x = 0, y = 0 \)): \( c = \frac{7 + 9\lambda}{-3 + 3\lambda} \)
Given that \( x \)-intercept equals \( z \)-intercept (\( a = c \)):
\( \frac{7 + 9\lambda}{2 + 2\lambda} = \frac{7 + 9\lambda}{-3 + 3\lambda} \)
Assuming \( 7 + 9\lambda \neq 0 \):
\( \frac{1}{2(1 + \lambda)} = \frac{1}{3(\lambda - 1)} \)
\( 3(\lambda - 1) = 2(\lambda + 1) \)
\( 3\lambda - 3 = 2\lambda + 2 \implies \lambda = 5 \)
Substitute \( \lambda = 5 \) into the plane equation:
\( (2 + 10)x + (2 + 25)y + (-3 + 15)z - (7 + 45) = 0 \)
\( 12x + 27y + 12z - 52 = 0 \).

Teacher's Note:
a) Use the family of planes formula \( P_1 + \lambda P_2 = 0 \) for planes passing through the line of intersection.
b) Set intercepts by putting other variables to zero and equate them according to the given condition.

OR

(b) Find the equation of the lines passing through the point \( (2, 1, 3) \) and perpendicular to the lines \( \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \) and \( \frac{x}{-3} = \frac{y}{2} = \frac{z}{5} \). [4 Marks]

Answer:
Let direction ratios of the required line be \( a, b, c \).
Equation of line passing through \( (2, 1, 3) \) is \( \frac{x - 2}{a} = \frac{y - 1}{b} = \frac{z - 3}{c} \).
Since it is perpendicular to the given lines with direction ratios \( (1, 2, 3) \) and \( (-3, 2, 5) \):
\( 1a + 2b + 3c = 0 \) ---(1)
\( -3a + 2b + 5c = 0 \) ---(2)
Solving by cross-multiplication:
\( \frac{a}{10 - 6} = \frac{b}{-9 - 5} = \frac{c}{2 - (-6)} \)
\( \frac{a}{4} = \frac{b}{-14} = \frac{c}{8} \implies \frac{a}{2} = \frac{b}{-7} = \frac{c}{4} \)
Therefore, direction ratios are \( 2, -7, 4 \).
Equation of the line is \( \frac{x - 2}{2} = \frac{y - 1}{-7} = \frac{z - 3}{4} \).

Teacher's Note:
a) Use the perpendicularity condition \( aa_1 + bb_1 + cc_1 = 0 \) to set up equations for direction ratios.
b) Simplify direction ratios by dividing through by their common factor.

Question 18 [6]
Draw a rough sketch and find the area bounded by the curve \( x^{2} = y \) and \( x + y = 2 \).

[Figure: Parabola \( x^2 = y \) intersecting line \( x + y = 2 \) at points \( (-2, 4) \) and \( (1, 1) \)]

Answer:
Find intersection points of parabola \( y = x^{2} \) and line \( y = 2 - x \):
\( x^{2} = 2 - x \implies x^{2} + x - 2 = 0 \implies (x + 2)(x - 1) = 0 \implies x = -2, x = 1 \).
The area bounded between the line and parabola from \( x = -2 \) to \( x = 1 \) is:
\( A = \int_{-2}^{1} ((2 - x) - x^{2}) dx \)
\( = \left[ 2x - \frac{x^{2}}{2} - \frac{x^{3}}{3} \right]_{-2}^{1} \)
Upper limit \( x = 1 \): \( 2(1) - \frac{1}{2} - \frac{1}{3} = 2 - \frac{5}{6} = \frac{7}{6} \)
Lower limit \( x = -2 \): \( 2(-2) - \frac{4}{2} - \frac{-8}{3} = -4 - 2 + \frac{8}{3} = -6 + \frac{8}{3} = -\frac{10}{3} = -\frac{20}{6} \)
Area = \( \frac{7}{6} - \left(-\frac{20}{6}\right) = \frac{27}{6} = \frac{9}{2} \) square units.

Teacher's Note:
a) Always find the points of intersection first to determine the exact limits of integration.
b) Integrate (upper curve minus lower curve) across the bounded interval.

SECTION C (20 Marks)

Question 19 [3×2]
(a) A company produces a commodity with Rs 24000 as fixed cost. The variable cost estimated to be 25% of the total revenue received on selling the product, is at the rate of Rs 8 per unit. Find the break-even point.

Answer:
Fixed Cost (FC) = Rs 24,000.
Selling Price per unit = Rs 8.
Let \( x \) be the number of units produced and sold.
Total Revenue (TR) = \( 8x \).
Variable Cost (VC) = 25% of TR = \( 0.25 \times 8x = 2x \).
Total Cost (TC) = FC + VC = \( 24000 + 2x \).
At break-even point, Total Revenue = Total Cost:
\( 8x = 24000 + 2x \)
\( 6x = 24000 \implies x = 4,000 \) units.

Teacher's Note:
a) Break-even point occurs where Total Revenue equals Total Cost (\( TR = TC \)).
b) Express variable cost accurately as a percentage of total revenue as given in the problem statement.

(b) The total cost function for a production is given by \( C(x) = \frac{3x^{2}}{4} - 7x + 27 \). Find the number of units produced for which \( MC = AC \) (MC = Marginal Cost and AC = Average Cost). [3 Marks]

Answer:
Given \( C(x) = \frac{3}{4}x^{2} - 7x + 27 \).
Average Cost \( AC = \frac{C(x)}{x} = \frac{3}{4}x - 7 + \frac{27}{x} \).
Marginal Cost \( MC = C'(x) = \frac{3}{2}x - 7 \).
Given condition \( MC = AC \):
\( \frac{3}{2}x - 7 = \frac{3}{4}x - 7 + \frac{27}{x} \)
\( \frac{3}{2}x - \frac{3}{4}x = \frac{27}{x} \)
\( \frac{3}{4}x = \frac{27}{x} \)
\( 3x^{2} = 108 \implies x^{2} = 36 \implies x = 6 \) units (ignoring negative quantity).

Teacher's Note:
a) Average cost is \( \frac{C(x)}{x} \) and marginal cost is the first derivative \( C'(x) \).
b) Equate both expressions and solve algebraically for production units \( x \).

(c) If \( \bar{x} = 18, \bar{y} = 100, \sigma_x = 14, \sigma_y = 20 \) and correlation coefficient \( r_{xy} = 0.8 \) find the regression equation of \( y \) on \( x \). [3 Marks]

Answer:
Regression coefficient of \( y \) on \( x \), \( b_{yx} = r \cdot \frac{\sigma_y}{\sigma_x} = 0.8 \times \frac{20}{14} = \frac{4}{5} \times \frac{20}{14} = \frac{16}{14} = \frac{8}{7} \).
Regression equation of \( y \) on \( x \) is given by:
\( y - \bar{y} = b_{yx}(x - \bar{x}) \)
\( y - 100 = \frac{8}{7}(x - 18) \)
\( 7(y - 100) = 8(x - 18) \)
\( 7y - 700 = 8x - 144 \)
\( 8x - 7y + 556 = 0 \).

Teacher's Note:
a) Use the formula \( b_{yx} = r \frac{\sigma_y}{\sigma_x} \) for the regression coefficient of \( y \) on \( x \).
b) Substitute mean values and regression coefficient into the standard linear regression formula.

Question 20 [4]
(a) The following results were obtained with respect to two variables \( x \) and \( y \): \( \sum x = 15, \sum y = 25, \sum xy = 83, \sum y^{2} = 135 \) and \( n = 5 \)
(i) Find the regression coefficient \( b_{xy} \).
(ii) Find the regression equation of \( x \) on \( y \).

Answer:
(i) Regression coefficient \( b_{xy} \):
\[ b_{xy} = \frac{n \sum xy - (\sum x)(\sum y)}{n \sum y^{2} - (\sum y)^{2}} \] Substitute given values: \( \sum x = 15, \sum y = 25, \sum xy = 83, \sum y^{2} = 135, n = 5 \):
\[ b_{xy} = \frac{5(83) - (15)(25)}{5(135) - (25)^{2}} = \frac{415 - 375}{675 - 625} = \frac{40}{50} = \frac{4}{5} = 0.8 \] (ii) Regression equation of \( x \) on \( y \):
Mean \( \bar{x} = \frac{\sum x}{n} = \frac{15}{5} = 3 \), Mean \( \bar{y} = \frac{\sum y}{n} = \frac{25}{5} = 5 \).
\( x - \bar{x} = b_{xy}(y - \bar{y}) \)
\( x - 3 = 0.8(y - 5) \)
\( x - 3 = \frac{4}{5}(y - 5) \)
\( 5(x - 3) = 4(y - 5) \implies 5x - 15 = 4y - 20 \)
\( 5x - 4y + 5 = 0 \).

Teacher's Note:
a) Ensure the correct formula is used for \( b_{xy} \) (denominator contains \( \sum y^2 \) terms).
b) Calculate means \( \bar{x} \) and \( \bar{y} \) before formulating the regression line.

OR

(b) Find the equation of the regression line of \( y \) on \( x \), if the observations \( (x, y) \) are as follows: \( (1, 4), (2, 8), (3, 2), (4, 12), (5, 10), (6, 14), (7, 16), (8, 6), (9, 18) \). Also, find the estimated value of \( y \) when \( x = 4 \). [4 Marks]

Answer:
Number of observations \( n = 9 \).
\( \sum x = 45, \sum y = 90, \sum xy = 530, \sum x^{2} = 285 \).
Means: \( \bar{x} = \frac{45}{9} = 5 \), \( \bar{y} = \frac{90}{9} = 10 \).
Regression coefficient \( b_{yx} \):
\[ b_{yx} = \frac{n \sum xy - (\sum x)(\sum y)}{n \sum x^{2} - (\sum x)^{2}} \] \[ b_{yx} = \frac{9(530) - (45)(90)}{9(285) - (45)^{2}} = \frac{4770 - 4050}{2565 - 2025} = \frac{720}{540} = \frac{4}{3} \] Regression equation of \( y \) on \( x \):
\( y - \bar{y} = b_{yx}(x - \bar{x}) \)
\( y - 10 = \frac{4}{3}(x - 5) \)
\( 3(y - 10) = 4(x - 5) \implies 3y - 30 = 4x - 20 \)
\( 4x - 3y + 10 = 0 \)
When \( x = 4 \):
\( 4(4) - 3y + 10 = 0 \implies 16 - 3y + 10 = 0 \implies 3y = 26 \implies y = \frac{26}{3} = 8.67 \).

Teacher's Note:
a) Tabulate summation columns (\( x, y, xy, x^2 \) accurately to prevent arithmetic mistakes).
b) Substitute \( x = 4 \) directly into the derived regression equation to estimate \( y \).

Question 21 [4]
(a) The cost function of a product is given by \( C(x) = \frac{x^{2}}{3} - 45x^{2} - 900x + 36 \) (Note: standard textbook form usually has \( \frac{x^3}{3} - 45x^2 \)... let's use given derivative) where \( x \) is the number of units produced. How many units should be produced to minimize the marginal cost?

Answer:
Assuming standard cost function \( C(x) = \frac{x^{3}}{3} - 45x^{2} + 900x + 36 \):
Marginal Cost \( MC = C'(x) = x^{2} - 900x + 900 \) or based on standard quadratic: let \( MC = x^{2} - 45x + 900 \).
To minimize marginal cost, differentiate \( MC \) with respect to \( x \) and set to zero:
Let \( M(x) = x^{2} - 90x + 900 \) (or derivative of given cost).
For minimum marginal cost, \( M'(x) = 0 \).
Using given terms from standard problem: \( C(x) = \frac{x^{3}}{3} - 45x^{2} + 900x + 36 \implies MC = x^{2} - 90x + 900 \).
\( \frac{d(MC)}{dx} = 2x - 90 = 0 \implies x = 45 \) units.
Second derivative \( \frac{d^2(MC)}{dx^2} = 2 > 0 \), confirming minimum at \( x = 45 \).

Teacher's Note:
a) Marginal cost is the first derivative of the total cost function.
b) To minimize marginal cost, find the derivative of the marginal cost function and set it equal to zero.

OR

(b) The marginal cost function of \( x \) units of a product is given by \( MC = 3x^{2} - 10x + 3 \). The cost of producing one unit is Rs 7. Find the total cost function and average cost function. [4 Marks]

Answer:
Total Cost \( C(x) = \int MC \, dx = \int (3x^{2} - 10x + 3) dx \)
\( C(x) = x^{3} - 5x^{2} + 3x + c \)
Given that the cost of producing one unit is Rs 7 (i.e., \( C(1) = 7 \)):
\( C(1) = (1)^{3} - 5(1)^{2} + 3(1) + c = 7 \)
\( 1 - 5 + 3 + c = 7 \implies -1 + c = 7 \implies c = 8 \)
Total Cost Function \( C(x) = x^{3} - 5x^{2} + 3x + 8 \).
Average Cost Function \( AC = \frac{C(x)}{x} = \frac{x^{3} - 5x^{2} + 3x + 8}{x} = x^{2} - 5x + 3 + \frac{8}{x} \).

Teacher's Note:
a) Integrate the marginal cost function to find the total cost function along with constant of integration \( c \).
b) Use the given boundary condition (\( C(1) = 7 \)) to evaluate \( c \).

Question 22 [6]
A carpenter has 90, 80 and 50 running feet respectively of teak wood, plywood and rosewood which is used to produce product A and product B. Each unit of product A requires 2, 1 and 1 running feet and each unit of product B requires 1, 2 and 1 running feet of teak wood, plywood and rosewood respectively. If product A is sold for Rs 48 per unit and product B is sold for Rs 40 per unit, how many units of product A and product B should be produced and sold by the carpenter, in order to obtain the maximum gross income?
Formulate the above as a Linear Programming Problem and solve it, indicating clearly the feasible region in the graph.

[Figure: LPP feasible region with corner points (0,0), (0,40), (20,30), (40,10), (45,0)]

Answer:
Let \( x \) be units of product A and \( y \) be units of product B.
Objective function: Maximize \( Z = 48x + 40y \)
Subject to constraints:
1) Teak wood: \( 2x + y \le 90 \)
2) Plywood: \( x + 2y \le 80 \)
3) Rosewood: \( x + y \le 50 \)
4) Non-negativity: \( x \ge 0, y \ge 0 \)
Corner points of the feasible region:
- \( O(0, 0) \implies Z = 0 \)
- \( A(0, 40) \implies Z = 48(0) + 40(40) = 1600 \)
- \( B(20, 30) \) (Intersection of \( 2x + y = 90 \) and \( x + 2y = 80 \)): \( Z = 48(20) + 40(30) = 960 + 1200 = 2160 \)
Wait, let's solve \( 2x + y = 90 \) and \( x + y = 50 \): subtract gives \( x = 40, y = 10 \).
Let's check point \( C(40, 10) \): \( Z = 48(40) + 40(10) = 1920 + 400 = 2320 \)
- Corner point \( D(45, 0) \implies Z = 48(45) + 40(0) = 2160 \).
Maximum gross income is Rs 2,320 obtained at \( x = 40 \) units of product A and \( y = 10 \) units of product B.

Teacher's Note:
a) Formulate LPP constraints carefully from resource availability and unit requirements.
b) Evaluate the objective function at all corner points of the feasible polygon to determine the absolute maximum.