Mathematics and Statistics - July 2017
Time: 3 Hours | Total Marks: 80
i. The inverse of the matrix \( \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \) is _______.
- (A) \( \frac{1}{5} \begin{bmatrix} 3 & -1 \\ -2 & 1 \end{bmatrix} \)
- (B) \( \frac{1}{5} \begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix} \)
- (C) \( \frac{1}{5} \begin{bmatrix} -3 & 1 \\ 5 & -2 \end{bmatrix} \)
- (D) \( \frac{1}{5} \begin{bmatrix} 3 & -1 \\ 5 & 2 \end{bmatrix} \)
Answer: (B)
Let \( A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \).
\( |A| = (1)(3) - (-1)(2) = 3 + 2 = 5 \).
\( A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{5} \begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix} \).
ii. If \( \bar{a} = 3\hat{i} - \hat{j} + 4\hat{k} \), \( \bar{b} = 2\hat{i} + 3\hat{j} - \hat{k} \), \( \bar{c} = -5\hat{i} + 2\hat{j} + 3\hat{k} \), then \( \bar{a} \cdot (\bar{b} \times \bar{c}) = \) _______.
- (A) 100
- (B) 101
- (C) 110
- (D) 109
Answer: (C)
\( \bar{a} \cdot (\bar{b} \times \bar{c}) = [\bar{a} \bar{b} \bar{c}] = \begin{vmatrix} 3 & -1 & 4 \\ 2 & 3 & -1 \\ -5 & 2 & 3 \end{vmatrix} \)
\( = 3(9 - (-2)) - (-1)(6 - 5) + 4(4 - (-15)) \)
\( = 3(11) + 1(1) + 4(19) = 33 + 1 + 76 = 110 \).
iii. If a line makes angles \( 90^\circ, 135^\circ, 45^\circ \) with the X, Y, and Z axes respectively, then its direction cosines are _______.
- (A) \( 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \)
- (B) \( 0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \)
- (C) \( 1, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \)
- (D) \( 0, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \)
Answer: (B)
\( l = \cos 90^\circ = 0 \)
\( m = \cos 135^\circ = \cos(180^\circ - 45^\circ) = -\cos 45^\circ = -\frac{1}{\sqrt{2}} \)
\( n = \cos 45^\circ = \frac{1}{\sqrt{2}} \)
Direction cosines are \( 0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \).
i. If the line \( \bar{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + \hat{j} + 2\hat{k}) \) is parallel to the plane \( \bar{r} \cdot (3\hat{i} - 2\hat{j} + p\hat{k}) = 10 \), find the value of p.
The line is parallel to the vector \( \bar{b} = 2\hat{i} + \hat{j} + 2\hat{k} \).
The normal vector to the plane is \( \bar{n} = 3\hat{i} - 2\hat{j} + p\hat{k} \).
Since the line is parallel to the plane, the normal to the plane is perpendicular to the line. Thus, \( \bar{b} \cdot \bar{n} = 0 \).
\( (2)(3) + (1)(-2) + (2)(p) = 0 \)
\( 6 - 2 + 2p = 0 \)
\( 4 + 2p = 0 \Rightarrow 2p = -4 \Rightarrow p = -2 \).
ii. If a line makes angles \( \alpha, \beta, \gamma \) with co-ordinate axes, prove that \( \cos 2\alpha + \cos 2\beta + \cos 2\gamma + 1 = 0 \).
We know that \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \).
L.H.S = \( \cos 2\alpha + \cos 2\beta + \cos 2\gamma + 1 \)
Using the identity \( \cos 2\theta = 2\cos^2\theta - 1 \):
\( = (2\cos^2\alpha - 1) + (2\cos^2\beta - 1) + (2\cos^2\gamma - 1) + 1 \)
\( = 2(\cos^2\alpha + \cos^2\beta + \cos^2\gamma) - 3 + 1 \)
\( = 2(1) - 2 \)
\( = 0 \) = R.H.S. (Proved)
iii. Write the negations of the following statements:
a. \( \forall n \in N, n + 7 > 6 \)
b. The kitchen is neat and tidy.
a. The negation of a universal quantifier (\( \forall \)) is an existential quantifier (\( \exists \)).
Negation: \( \exists n \in N, n + 7 \le 6 \).
b. Let p: The kitchen is neat. q: The kitchen is tidy.
Statement is \( p \land q \). Negation is \( \sim(p \land q) \equiv \sim p \lor \sim q \).
Negation: "The kitchen is not neat or the kitchen is not tidy."
iv. Find the angle between the lines whose direction ratios are 4, –3, 5 and 3, 4, 5.
Let the direction ratios be \( a_1=4, b_1=-3, c_1=5 \) and \( a_2=3, b_2=4, c_2=5 \).
Let \( \theta \) be the angle between the lines.
\( \cos \theta = \left| \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} \right| \)
\( a_1a_2 + b_1b_2 + c_1c_2 = 4(3) + (-3)(4) + 5(5) = 12 - 12 + 25 = 25 \).
\( \sqrt{a_1^2+b_1^2+c_1^2} = \sqrt{16+9+25} = \sqrt{50} = 5\sqrt{2} \).
\( \sqrt{a_2^2+b_2^2+c_2^2} = \sqrt{9+16+25} = \sqrt{50} = 5\sqrt{2} \).
\( \cos \theta = \frac{25}{5\sqrt{2} \cdot 5\sqrt{2}} = \frac{25}{50} = \frac{1}{2} \).
\( \theta = \cos^{-1}(\frac{1}{2}) = 60^\circ \) or \( \frac{\pi}{3} \).
v. If \( \bar{a}, \bar{b}, \bar{c} \) are position vectors of the points A, B, C respectively such that \( 3\bar{a} + 5\bar{b} - 8\bar{c} = \bar{0} \), find the ratio in which A divides BC.
Given \( 3\bar{a} + 5\bar{b} - 8\bar{c} = \bar{0} \).
To find the ratio in which A divides BC, we express \( \bar{a} \) in terms of \( \bar{b} \) and \( \bar{c} \).
\( 3\bar{a} = 8\bar{c} - 5\bar{b} \)
\( \bar{a} = \frac{8\bar{c} - 5\bar{b}}{3} = \frac{8\bar{c} - 5\bar{b}}{8 - 5} \).
This is the section formula for external division \( \bar{r} = \frac{m\bar{c} - n\bar{b}}{m - n} \).
Comparing, \( m = 8 \) and \( n = 5 \).
Therefore, point A divides segment BC externally in the ratio 8 : 5.
i. If \( \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4} \), then find the value of ‘x’.
Using \( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \):
\( \tan^{-1}\left(\frac{2x + 3x}{1 - (2x)(3x)}\right) = \frac{\pi}{4} \)
\( \frac{5x}{1 - 6x^2} = \tan(\frac{\pi}{4}) = 1 \)
\( 5x = 1 - 6x^2 \Rightarrow 6x^2 + 5x - 1 = 0 \)
\( 6x^2 + 6x - x - 1 = 0 \)
\( 6x(x+1) - 1(x+1) = 0 \)
\( (6x - 1)(x + 1) = 0 \)
\( x = \frac{1}{6} \) or \( x = -1 \).
Check: If \( x = -1 \), L.H.S is \( \tan^{-1}(-2) + \tan^{-1}(-3) \), which is negative. R.H.S is positive. So \( x = -1 \) is rejected.
Therefore, \( x = \frac{1}{6} \).
ii. Write the converse, inverse and contrapositive of the following statement: "If it rains then the match will be cancelled."
Let p: It rains, q: The match will be cancelled.
Given statement: \( p \rightarrow q \).
Converse (\( q \rightarrow p \)): If the match is cancelled then it rains.
Inverse (\( \sim p \rightarrow \sim q \)): If it does not rain then the match will not be cancelled.
Contrapositive (\( \sim q \rightarrow \sim p \)): If the match is not cancelled then it does not rain.
iii. Find p and q, if the equation \( px^2 - 8xy + 3y^2 + 14x + 2y + q = 0 \) represents a pair of perpendicular lines.
Comparing with \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \):
\( a = p, b = 3, c = q \)
\( 2h = -8 \Rightarrow h = -4 \)
\( 2g = 14 \Rightarrow g = 7 \)
\( 2f = 2 \Rightarrow f = 1 \)
1. Condition for perpendicular lines: \( a + b = 0 \)
\( p + 3 = 0 \Rightarrow p = -3 \).
2. Condition for pair of lines: \( \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \)
\( \begin{vmatrix} -3 & -4 & 7 \\ -4 & 3 & 1 \\ 7 & 1 & q \end{vmatrix} = 0 \)
\( -3(3q - 1) - (-4)(-4q - 7) + 7(-4 - 21) = 0 \)
\( -9q + 3 + 4(-4q - 7) + 7(-25) = 0 \)
\( -9q + 3 - 16q - 28 - 175 = 0 \)
\( -25q - 200 = 0 \Rightarrow -25q = 200 \Rightarrow q = -8 \).
Answer: \( p = -3, q = -8 \).
i. Find the equation of the plane passing through the intersection of the planes \( 3x + 2y – z + 1 = 0 \) and \( x + y + z – 2 = 0 \) and the point (2, 2, 1).
The equation of the required plane is \( (3x + 2y - z + 1) + \lambda(x + y + z - 2) = 0 \).
Since it passes through (2, 2, 1), substitute x=2, y=2, z=1:
\( (3(2) + 2(2) - 1 + 1) + \lambda(2 + 2 + 1 - 2) = 0 \)
\( (6 + 4) + \lambda(3) = 0 \)
\( 10 + 3\lambda = 0 \Rightarrow \lambda = -\frac{10}{3} \).
Substitute \( \lambda \) back into the equation:
\( (3x + 2y - z + 1) - \frac{10}{3}(x + y + z - 2) = 0 \)
Multiply by 3:
\( 3(3x + 2y - z + 1) - 10(x + y + z - 2) = 0 \)
\( 9x + 6y - 3z + 3 - 10x - 10y - 10z + 20 = 0 \)
\( -x - 4y - 13z + 23 = 0 \)
Or \( x + 4y + 13z - 23 = 0 \).
ii. Prove section formula \( \bar{r} = \frac{m\bar{b} + n\bar{a}}{m+n} \). Hence find R dividing A(1, –2, 1) and B(1, 4, –2) internally in ratio 2 : 1.
Part 1: Proof
Let A and B be two points with position vectors \( \bar{a} \) and \( \bar{b} \). Let R(\( \bar{r} \)) divide AB internally in ratio m:n.
Since R is on AB, \( \vec{AR} \) and \( \vec{RB} \) are collinear and in the same direction.
\( \frac{AR}{RB} = \frac{m}{n} \Rightarrow n(\vec{AR}) = m(\vec{RB}) \)
\( n(\bar{r} - \bar{a}) = m(\bar{b} - \bar{r}) \)
\( n\bar{r} - n\bar{a} = m\bar{b} - m\bar{r} \)
\( m\bar{r} + n\bar{r} = m\bar{b} + n\bar{a} \)
\( \bar{r}(m + n) = m\bar{b} + n\bar{a} \)
\( \bar{r} = \frac{m\bar{b} + n\bar{a}}{m + n} \). (Proved)
Part 2: Calculation
\( \bar{a} = \hat{i} - 2\hat{j} + \hat{k} \)
\( \bar{b} = \hat{i} + 4\hat{j} - 2\hat{k} \)
\( m = 2, n = 1 \)
\( \bar{r} = \frac{2(\hat{i} + 4\hat{j} - 2\hat{k}) + 1(\hat{i} - 2\hat{j} + \hat{k})}{2 + 1} \)
\( = \frac{(2\hat{i} + 8\hat{j} - 4\hat{k}) + (\hat{i} - 2\hat{j} + \hat{k})}{3} \)
\( = \frac{3\hat{i} + 6\hat{j} - 3\hat{k}}{3} \)
\( = \hat{i} + 2\hat{j} - \hat{k} \).
Coordinates of R are (1, 2, -1).
iii. The angles of the \( \Delta ABC \) are in A.P. and \( b : c = \sqrt{3} : \sqrt{2} \) then find \( \angle A, \angle B, \angle C \).
Angles A, B, C are in A.P. means \( 2B = A + C \).
Also \( A + B + C = 180^\circ \).
\( 3B = 180^\circ \Rightarrow B = 60^\circ \).
Using Sine Rule: \( \frac{b}{\sin B} = \frac{c}{\sin C} \)
\( \frac{\sqrt{3}}{\sin 60^\circ} = \frac{\sqrt{2}}{\sin C} \)
\( \frac{\sqrt{3}}{\sqrt{3}/2} = \frac{\sqrt{2}}{\sin C} \)
\( 2 = \frac{\sqrt{2}}{\sin C} \)
\( \sin C = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \)
\( C = 45^\circ \) or \( 135^\circ \).
Since \( B = 60^\circ \), if \( C = 135^\circ \), sum \( > 180 \). So \( C = 45^\circ \).
\( A = 180 - (B + C) = 180 - (60 + 45) = 75^\circ \).
Angles are \( 75^\circ, 60^\circ, 45^\circ \).
i. Find the vector equation and cartesian equation of a line passing through the points A(3, 4, –7) and B(6, –1, 1).
Vector Equation:
Line passes through \( \bar{a} = 3\hat{i} + 4\hat{j} - 7\hat{k} \).
Direction vector \( \bar{d} = \bar{b} - \bar{a} = (6-3)\hat{i} + (-1-4)\hat{j} + (1-(-7))\hat{k} = 3\hat{i} - 5\hat{j} + 8\hat{k} \).
Equation: \( \bar{r} = (3\hat{i} + 4\hat{j} - 7\hat{k}) + \lambda(3\hat{i} - 5\hat{j} + 8\hat{k}) \).
Cartesian Equation:
Direction ratios are 3, -5, 8.
\( \frac{x - 3}{3} = \frac{y - 4}{-5} = \frac{z + 7}{8} \).
ii. Find the general solution of \( \cot x + \tan x = 2 \csc x \).
\( \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \frac{2}{\sin x} \)
\( \frac{\cos^2 x + \sin^2 x}{\sin x \cos x} = \frac{2}{\sin x} \)
\( \frac{1}{\sin x \cos x} = \frac{2}{\sin x} \)
Since \( \csc x \) exists, \( \sin x \neq 0 \). We can cancel \( \sin x \) from denominator (assuming valid domain).
\( \frac{1}{\cos x} = 2 \Rightarrow \cos x = \frac{1}{2} \).
\( \cos x = \cos \frac{\pi}{3} \).
General solution: \( x = 2n\pi \pm \frac{\pi}{3}, n \in Z \).
iii. Express the following switching circuit in symbolic form of logic. Construct its switching table and write your conclusion from it:
Top branch: S1 and S2'
Bottom branch: S1' and S2]
Let p: switch \( S_1 \) is closed. q: switch \( S_2 \) is closed.
\( \sim p \): switch \( S_1' \) is closed. \( \sim q \): switch \( S_2' \) is closed.
Top branch: \( p \land \sim q \). Bottom branch: \( \sim p \land q \).
Branches are in parallel.
Symbolic Form: \( L \equiv (p \land \sim q) \lor (\sim p \land q) \).
Switching Table:
| p | q | ~p | ~q | p ∧ ~q | ~p ∧ q | (p ∧ ~q) ∨ (~p ∧ q) |
|---|---|---|---|---|---|---|
| 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 |
Conclusion: The lamp glows if and only if exactly one of the switches is closed (Exclusive OR).
i. If \( A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \), verify that \( A(\text{adj } A) = |A|I \).
Step 1: Find |A|
\( |A| = 1(0 - 0) - (-1)(9 - (-2)) + 2(0 - 0) \)
\( |A| = 0 + 1(11) + 0 = 11 \).
So \( |A|I = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} \).
Step 2: Find adj A
Cofactors:
\( C_{11} = 0, C_{12} = -11, C_{13} = 0 \)
\( C_{21} = -(-3) = 3, C_{22} = 1, C_{23} = -1 \)
\( C_{31} = 2, C_{32} = -(-8) = 8, C_{33} = 3 \)
\( \text{adj } A = \text{Cofactor Matrix}^T = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \)
Step 3: Multiply A and adj A
\( A(\text{adj } A) = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \)
Row 1: \( 0 + 11 + 0 = 11 \); \( 3 - 1 - 2 = 0 \); \( 2 - 8 + 6 = 0 \)
Row 2: \( 0 + 0 + 0 = 0 \); \( 9 + 0 + 2 = 11 \); \( 6 + 0 - 6 = 0 \)
Row 3: \( 0 + 0 + 0 = 0 \); \( 3 + 0 - 3 = 0 \); \( 2 + 0 + 9 = 11 \)
Result = \( \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = |A|I \). Verified.
ii. L.P.P. Problem: Maximize Profit. Bicycles and Tricycles. Machines A and B.
Let \( x \) be number of bicycles and \( y \) be number of tricycles.
Maximize \( Z = 180x + 220y \).
Constraints:
Machine A: \( 6x + 4y \le 120 \Rightarrow 3x + 2y \le 60 \)
Machine B: \( 3x + 10y \le 180 \)
\( x \ge 0, y \ge 0 \).
Points of intersection:
\( 3x + 2y = 60 \) (1)
\( 3x + 10y = 180 \) (2)
Subtracting (1) from (2): \( 8y = 120 \Rightarrow y = 15 \).
\( 3x + 30 = 60 \Rightarrow 3x = 30 \Rightarrow x = 10 \).
Intersection point C(10, 15).
Feasible Region Corners:
A(0, 0): \( Z = 0 \)
B(20, 0): \( Z = 180(20) = 3600 \)
D(0, 18): \( Z = 220(18) = 3960 \)
C(10, 15): \( Z = 180(10) + 220(15) = 1800 + 3300 = 5100 \)
Maximum Profit is ₹ 5100 when 10 bicycles and 15 tricycles are manufactured.
iii. If \( \theta \) is acute angle between lines \( ax^2 + 2hxy + by^2 = 0 \), prove \( \tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right| \). Hence find acute angle for \( x^2 - 4xy + y^2 = 0 \).
Proof: The equation represents two lines passing through origin \( y = m_1x \) and \( y = m_2x \).
Slopes sum \( m_1 + m_2 = \frac{-2h}{b} \), product \( m_1m_2 = \frac{a}{b} \).
\( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right| = \left| \frac{\sqrt{(m_1+m_2)^2 - 4m_1m_2}}{1 + m_1m_2} \right| \)
\( = \left| \frac{\sqrt{\frac{4h^2}{b^2} - \frac{4a}{b}}}{1 + \frac{a}{b}} \right| = \left| \frac{\frac{2}{b}\sqrt{h^2 - ab}}{\frac{a+b}{b}} \right| = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right| \).
Numerical: \( x^2 - 4xy + y^2 = 0 \)
\( a = 1, b = 1, 2h = -4 \Rightarrow h = -2 \).
\( \tan \theta = \left| \frac{2\sqrt{(-2)^2 - (1)(1)}}{1+1} \right| = \frac{2\sqrt{3}}{2} = \sqrt{3} \).
\( \theta = 60^\circ \) or \( \frac{\pi}{3} \).
i. Given \( f(x) = 2x, x < 0 \) and \( 0, x \ge 0 \). Then f(x) is _______.
- (A) discontinuous and not differentiable at x = 0
- (B) continuous and differentiable at x = 0
- (C) discontinuous and differentiable at x = 0
- (D) continuous and not differentiable at x = 0
Answer: (D)
L.H.L = 0, R.H.L = 0. Continuous.
L.H.D = \( \frac{d}{dx}(2x) = 2 \). R.H.D = 0. Not differentiable.
ii. If \( \int_0^\alpha (3x^2 + 2x + 1) dx = 14 \), then \( \alpha = \) _______.
- (A) 1
- (B) 2
- (C) -1
- (D) -2
Answer: (B)
\( [x^3 + x^2 + x]_0^\alpha = 14 \)
\( \alpha^3 + \alpha^2 + \alpha - 14 = 0 \).
Put \( \alpha = 2 \): \( 8 + 4 + 2 - 14 = 0 \). Correct.
iii. The function \( f(x) = x^3 - 3x^2 + 3x - 100 \) is _______.
- (A) increasing
- (B) decreasing
- (C) increasing and decreasing
- (D) neither increasing nor decreasing
Answer: (A)
\( f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x-1)^2 \).
\( f'(x) \ge 0 \) for all real x. Increasing.
i. Differentiate \( 3^x \) w.r.t. \( \log_3 x \).
Let \( u = 3^x \) and \( v = \log_3 x = \frac{\ln x}{\ln 3} \).
\( \frac{du}{dx} = 3^x \ln 3 \).
\( \frac{dv}{dx} = \frac{1}{x \ln 3} \).
\( \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{3^x \ln 3}{1/(x \ln 3)} = x \cdot 3^x (\ln 3)^2 \).
ii. Check whether the conditions of Rolle’s theorem are satisfied by \( f(x) = (x-1)(x-2)(x-3), x \in [1, 3] \).
1. f(x) is a polynomial, so it is continuous on [1, 3].
2. f(x) is differentiable on (1, 3).
3. \( f(1) = (0)(-1)(-2) = 0 \).
4. \( f(3) = (2)(1)(0) = 0 \).
Since \( f(1) = f(3) \), all conditions of Rolle's theorem are satisfied.
iii. Evaluate: \( \int \frac{\sqrt{\tan x}}{\sin x \cos x} dx \).
Divide numerator and denominator by \( \cos^2 x \):
\( I = \int \frac{\sqrt{\tan x} / \cos^2 x}{(\sin x \cos x) / \cos^2 x} dx = \int \frac{\sec^2 x \sqrt{\tan x}}{\tan x} dx = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx \).
Put \( \tan x = t \Rightarrow \sec^2 x dx = dt \).
\( I = \int t^{-1/2} dt = 2t^{1/2} + c = 2\sqrt{\tan x} + c \).
iv. Find area bounded by \( x^2 = 16y \), \( y=2, y=6 \) and Y-axis in first quadrant.
\( x^2 = 16y \Rightarrow x = 4\sqrt{y} \) (First quadrant \( x > 0 \)).
Area \( A = \int_2^6 x dy = \int_2^6 4\sqrt{y} dy \).
\( A = 4 \left[ \frac{2}{3}y^{3/2} \right]_2^6 = \frac{8}{3} (6^{3/2} - 2^{3/2}) \).
\( A = \frac{8}{3} (6\sqrt{6} - 2\sqrt{2}) \).
v. Given \( X \sim B(n, p) \), \( n=10, p=0.4 \). Find E(X) and Var(X).
\( E(X) = np = 10(0.4) = 4 \).
\( q = 1 - p = 0.6 \).
\( Var(X) = npq = 10(0.4)(0.6) = 2.4 \).
i. If \( f(x) = \frac{(5^{\sin x} - 1)^2}{x \log(1+2x)}, x \neq 0 \) is continuous at x=0, find f(0).
\( f(0) = \lim_{x \to 0} \frac{(5^{\sin x} - 1)^2}{x \log(1+2x)} \)
Divide numerator and denominator by \( (\sin x)^2 \) and adjust terms:
\( = \lim_{x \to 0} \frac{\left(\frac{5^{\sin x}-1}{\sin x}\right)^2 \cdot (\sin x)^2}{x \cdot \frac{\log(1+2x)}{2x} \cdot 2x} \)
\( = \lim_{x \to 0} \frac{(\log 5)^2 \cdot (\frac{\sin x}{x})^2 \cdot x^2}{x \cdot (1) \cdot 2x} \)
\( = \frac{(\log 5)^2 \cdot (1)^2 \cdot x^2}{2x^2} = \frac{(\log 5)^2}{2} \).
ii. Find expected value and variance from PMF table.
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.08 | 0.15 | 0.45 | 0.27 | 0.05 |
\( E(X) = \sum xP(x) = 0(0.08) + 1(0.15) + 2(0.45) + 3(0.27) + 4(0.05) \)
\( = 0 + 0.15 + 0.90 + 0.81 + 0.20 = 2.06 \).
\( E(X^2) = \sum x^2P(x) = 0 + 1(0.15) + 4(0.45) + 9(0.27) + 16(0.05) \)
\( = 0 + 0.15 + 1.80 + 2.43 + 0.80 = 5.18 \).
\( Var(X) = E(X^2) - [E(X)]^2 = 5.18 - (2.06)^2 = 5.18 - 4.2436 = 0.9364 \).
iii. 80% families own TV. Sample of 5. Find Prob (a) 3 own TV (b) at least 2 own TV.
\( p = 0.8, q = 0.2, n = 5 \).
a. P(X = 3):
\( = \binom{5}{3} (0.8)^3 (0.2)^2 = 10 (0.512)(0.04) = 0.2048 \).
b. P(X >= 2):
\( = 1 - [P(0) + P(1)] \)
\( P(0) = (0.2)^5 = 0.00032 \)
\( P(1) = 5(0.8)(0.2)^4 = 5(0.8)(0.0016) = 0.0064 \)
\( P(X \ge 2) = 1 - (0.00032 + 0.0064) = 1 - 0.00672 = 0.99328 \).
i. Find approximate value of \( \cos(60^\circ 30') \).
Let \( f(x) = \cos x \). \( a = 60^\circ = \frac{\pi}{3} \). \( h = 30' = 0.5^\circ \).
\( h \) in radians = \( 0.5 \times 0.0175 = 0.00875 \).
\( f(a) = \cos 60^\circ = 0.5 \). \( f'(x) = -\sin x \). \( f'(a) = -\sin 60^\circ = -0.8660 \).
\( f(a+h) \approx f(a) + hf'(a) \)
\( = 0.5 + (0.00875)(-0.8660) \)
\( = 0.5 - 0.0075775 = 0.4924 \). (Approx)
ii. Bacteria growth problem. \( N_0 = 1000 \), doubles in 1 hour. Find number after 2.5 hours.
\( \frac{dN}{dt} = kN \Rightarrow N = c e^{kt} \).
At \( t=0, N=1000 \Rightarrow c=1000 \). So \( N = 1000 e^{kt} \).
At \( t=1, N=2000 \Rightarrow 2000 = 1000 e^k \Rightarrow e^k = 2 \Rightarrow k = \ln 2 \).
At \( t=2.5 \):
\( N = 1000 e^{2.5 \ln 2} = 1000 (e^{\ln 2})^{2.5} = 1000 (2)^{2.5} \)
\( = 1000 \cdot 2^2 \cdot \sqrt{2} = 1000 \cdot 4 \cdot 1.414 = 5656 \).
i. If f(x) is continuous on [-4, 2], show \( a + b = -\frac{7}{6} \). f(x) = 6b - 3ax (-4 to -2) and 4x+1 (-2 to 2).
Since f is continuous at \( x = -2 \):
\( \lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x) \)
\( 6b - 3a(-2) = 4(-2) + 1 \)
\( 6b + 6a = -8 + 1 \)
\( 6(a + b) = -7 \)
\( a + b = -\frac{7}{6} \). (Shown)
iii. P(X=x): 0.1, 0.3, 0.4, 0.2 for x=1,2,3,4. Find \( P(X \ge 2) \) and CDF.
\( P(X \ge 2) \):
\( P(2) + P(3) + P(4) = 0.3 + 0.4 + 0.2 = 0.9 \).
CDF \( F(x) \):
F(1) = 0.1
F(2) = 0.1 + 0.3 = 0.4
F(3) = 0.4 + 0.4 = 0.8
F(4) = 0.8 + 0.2 = 1.0
i. Solve \( \frac{dy}{dx} - y = e^x \). Find particular solution for x=0, y=1.
Integrating Factor (I.F.) = \( e^{\int -1 dx} = e^{-x} \).
Solution: \( y(I.F.) = \int Q(I.F.) dx + c \)
\( y e^{-x} = \int e^x e^{-x} dx = \int 1 dx = x + c \).
\( y = e^x(x+c) \).
Put x=0, y=1: \( 1 = e^0(0+c) \Rightarrow c=1 \).
Particular Solution: \( y = e^x(x+1) \).
ii. Derivative of inverse function theorem. Show for \( y = \sin^{-1} x \), \( \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \).
Let \( y = \sin^{-1} x \), so \( x = \sin y \).
\( \frac{dx}{dy} = \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2} \). (Since y is in principal range)
\( \frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{\sqrt{1-x^2}} \), for \( |x| < 1 \).
iii. Evaluate \( \int \frac{8}{(x+2)(x^2+4)} dx \).
Using Partial Fractions: \( \frac{8}{(x+2)(x^2+4)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+4} \).
\( 8 = A(x^2+4) + (Bx+C)(x+2) \).
Put \( x = -2 \): \( 8 = A(8) \Rightarrow A=1 \).
Compare \( x^2 \) coeffs: \( A+B=0 \Rightarrow B=-1 \).
Compare constants: \( 4A + 2C = 8 \Rightarrow 4 + 2C = 8 \Rightarrow C=2 \).
Integral = \( \int \frac{1}{x+2} dx + \int \frac{-x+2}{x^2+4} dx \)
\( = \ln|x+2| - \int \frac{x}{x^2+4} dx + 2\int \frac{1}{x^2+4} dx \)
\( = \ln|x+2| - \frac{1}{2}\ln(x^2+4) + 2 \cdot \frac{1}{2}\tan^{-1}(\frac{x}{2}) + c \).
\( = \ln|x+2| - \frac{1}{2}\ln(x^2+4) + \tan^{-1}(\frac{x}{2}) + c \).