MM 1131.9 : MATHEMATICS – I

First Semester B.C.A. Degree Examination

Kerala University | 2013 Admission Batch

πŸ“š Complete Exam-Style Answer Reference | All Sections Covered

Based on Question Papers: 2016 (Jan), 2016 (Dec), 2018, 2021 | Max. Marks: 80

πŸ“‹ Table of Contents

⭐ SECTION IV β€” 15 Marks Each (Answer Any 2)
Question 32 Appears Every Year
Find the Laplace transform of cos at and sin at.
15 Marks | Section IV

Definition of Laplace Transform

The Laplace transform of a function \( f(t) \) is defined as:

\[ \mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st} f(t)\, dt, \quad s > 0 \]

Part 1: Laplace Transform of \( e^{at} \)

We first derive this fundamental result which is needed below.

Step 1: Apply definition \[ \mathcal{L}\{e^{at}\} = \int_0^{\infty} e^{-st} \cdot e^{at}\, dt = \int_0^{\infty} e^{-(s-a)t}\, dt \]
Step 2: Evaluate the integral \[ = \left[ \frac{e^{-(s-a)t}}{-(s-a)} \right]_0^{\infty} = 0 - \frac{1}{-(s-a)} = \frac{1}{s-a}, \quad s > a \]
Result: \( \mathcal{L}\{e^{at}\} = \dfrac{1}{s-a} \)

Part 2: Laplace Transform of \( \cos at \) using Euler's formula

We know from Euler's formula:

\[ e^{iat} = \cos at + i \sin at \]

Taking Laplace transform of both sides:

Step 1: \[ \mathcal{L}\{e^{iat}\} = \mathcal{L}\{\cos at\} + i\,\mathcal{L}\{\sin at\} \]
Step 2: Apply the formula \( \mathcal{L}\{e^{at}\} = \dfrac{1}{s-a} \) with \( a \to ia \) \[ \mathcal{L}\{e^{iat}\} = \frac{1}{s - ia} \]
Step 3: Rationalise by multiplying numerator and denominator by \( (s + ia) \) \[ \frac{1}{s - ia} = \frac{s + ia}{(s - ia)(s + ia)} = \frac{s + ia}{s^2 + a^2} \]
Step 4: Separate real and imaginary parts \[ \frac{s + ia}{s^2 + a^2} = \frac{s}{s^2 + a^2} + i\cdot\frac{a}{s^2 + a^2} \]
Step 5: Equate real and imaginary parts \[ \mathcal{L}\{\cos at\} = \frac{s}{s^2 + a^2}, \qquad \mathcal{L}\{\sin at\} = \frac{a}{s^2 + a^2} \]
Final Answers:
\[ \mathcal{L}\{\cos at\} = \frac{s}{s^2 + a^2}, \quad s > 0 \] \[ \mathcal{L}\{\sin at\} = \frac{a}{s^2 + a^2}, \quad s > 0 \]
πŸ“ Exam Tip: If only 2 marks or 4 marks are asked, just state the formula directly. For 15 marks, always show the full derivation using Euler's formula and rationalisation.
Question 32(a) High Frequency
State and prove the Second Shifting Theorem for Laplace Transform. Also evaluate \( \mathcal{L}^{-1}[\tan^{-1}(1/s)] \).
15 Marks | Section IV

Statement of Second Shifting Theorem

If \( \mathcal{L}\{f(t)\} = F(s) \), then

\[ \mathcal{L}\{f(t-a)\cdot u(t-a)\} = e^{-as} F(s), \quad a > 0 \] where \( u(t-a) \) is the unit step function defined as: \[ u(t-a) = \begin{cases} 0 & t < a \\ 1 & t \geq a \end{cases} \]

Proof

Step 1: Apply the definition of Laplace transform \[ \mathcal{L}\{f(t-a)\cdot u(t-a)\} = \int_0^{\infty} e^{-st} f(t-a)\cdot u(t-a)\, dt \]
Step 2: Since \( u(t-a) = 0 \) for \( t < a \) and \( 1 \) for \( t \geq a \), limits change: \[ = \int_a^{\infty} e^{-st} f(t-a)\, dt \]
Step 3: Substitute \( \tau = t - a \), so \( t = \tau + a \), \( dt = d\tau \). When \( t = a \), \( \tau = 0 \); as \( t \to \infty \), \( \tau \to \infty \). \[ = \int_0^{\infty} e^{-s(\tau + a)} f(\tau)\, d\tau = \int_0^{\infty} e^{-s\tau} e^{-sa} f(\tau)\, d\tau \]
Step 4: Factor out the constant \( e^{-sa} \) \[ = e^{-as} \int_0^{\infty} e^{-s\tau} f(\tau)\, d\tau = e^{-as} \cdot F(s) \]
Hence proved: \( \mathcal{L}\{f(t-a)\cdot u(t-a)\} = e^{-as} F(s) \)

Part (b): Evaluate \( \mathcal{L}^{-1}\!\left[\tan^{-1}\!\left(\dfrac{1}{s}\right)\right] \)

We use the formula: if \( \mathcal{L}\{f(t)\} = F(s) \), then \( \mathcal{L}\left\{\dfrac{f(t)}{t}\right\} = \int_s^{\infty} F(u)\, du \).

Step 1: Differentiate \( F(s) = \tan^{-1}(1/s) \) with respect to \( s \) \[ \frac{d}{ds}\left[\tan^{-1}\!\left(\frac{1}{s}\right)\right] = \frac{1}{1 + (1/s)^2} \cdot \left(-\frac{1}{s^2}\right) = \frac{1}{1 + 1/s^2} \cdot \left(-\frac{1}{s^2}\right) \] \[ = \frac{s^2}{s^2 + 1} \cdot \left(-\frac{1}{s^2}\right) = -\frac{1}{s^2 + 1} \]
Step 2: Use the inverse formula: \( \mathcal{L}\{t \cdot f(t)\} = -\dfrac{d}{ds}F(s) \)
So if \( F(s) = \tan^{-1}(1/s) \), then \( -F'(s) = \dfrac{1}{s^2+1} \).

We know \( \mathcal{L}^{-1}\!\left[\dfrac{1}{s^2+1}\right] = \sin t \).
Step 3: Since \( \mathcal{L}\{t\cdot f(t)\} = -F'(s) \), we have \( t \cdot f(t) = \sin t \), so: \[ f(t) = \frac{\sin t}{t} \]
Result: \( \mathcal{L}^{-1}\!\left[\tan^{-1}\!\left(\dfrac{1}{s}\right)\right] = \dfrac{\sin t}{t} \)
Question 32 alt High Frequency
State Leibnitz's theorem. If \( y = (\sin^{-1}x)^2 \), prove that \((1-x^2)y_{n+2} - (2n+1)xy_{n+1} - n^2 y_n = 0\).
15 Marks | Section IV

Statement of Leibnitz's Theorem

If \( u \) and \( v \) are functions of \( x \), each possessing derivatives up to \( n \)-th order, then the \( n \)-th derivative of their product is:

\[ (uv)_n = u_n v + \binom{n}{1} u_{n-1} v_1 + \binom{n}{2} u_{n-2} v_2 + \cdots + u\, v_n \] \[ = \sum_{r=0}^{n} \binom{n}{r} u_{n-r}\, v_r \]
where \( u_r = \dfrac{d^r u}{dx^r} \) and \( v_r = \dfrac{d^r v}{dx^r} \).

Application: Prove the recurrence for \( y = (\sin^{-1}x)^2 \)

Step 1: Find \( y_1 \) \[ y_1 = 2(\sin^{-1}x) \cdot \frac{1}{\sqrt{1-x^2}} \]
Step 2: Remove the square root β€” multiply both sides by \( \sqrt{1-x^2} \) \[ y_1 \sqrt{1-x^2} = 2\sin^{-1}x \]
Step 3: Square both sides \[ y_1^2 (1-x^2) = 4(\sin^{-1}x)^2 = 4y \]
Step 4: Differentiate once with respect to \( x \) \[ 2y_1 y_2 (1-x^2) + y_1^2 (-2x) = 4y_1 \]

Divide throughout by \( 2y_1 \) (assuming \( y_1 \neq 0 \)):

\[ y_2(1-x^2) - xy_1 = 2 \quad \cdots (1) \]
Step 5: Differentiate (1) n times using Leibnitz's theorem

For \( (1-x^2)y_2 \): differentiating \( n \) times,

\[ (1-x^2)y_{n+2} + n(-2x)y_{n+1} + \frac{n(n-1)}{2}(-2)y_n \]

For \( -xy_1 \): differentiating \( n \) times,

\[ -xy_{n+1} - ny_n \]

RHS: \( 2 \) differentiates to \( 0 \) for \( n \geq 1 \).

Step 6: Combine all terms \[ (1-x^2)y_{n+2} - 2nxy_{n+1} - n(n-1)y_n - xy_{n+1} - ny_n = 0 \] \[ (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - n^2 y_n = 0 \]
Hence proved: \( (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - n^2 y_n = 0 \)
Question 32 alt 2020 Paper
When \(n\) is a positive integer, find a reduction formula for \(\mathcal{L}[t^n]\) and hence evaluate \(\mathcal{L}[t^n]\).
15 Marks | Section IV
Step 1: Apply definition of Laplace transform \[ \mathcal{L}\{t^n\} = \int_0^{\infty} e^{-st} t^n\, dt \]
Step 2: Integrate by parts β€” let \( u = t^n \), \( dv = e^{-st}dt \) \[ = \left[ t^n \cdot \frac{e^{-st}}{-s} \right]_0^{\infty} + \frac{n}{s} \int_0^{\infty} e^{-st} t^{n-1}\, dt \]
Step 3: The boundary term vanishes (since \( e^{-st} \to 0 \) faster than \( t^n \to \infty \)) \[ = 0 + \frac{n}{s} \int_0^{\infty} e^{-st} t^{n-1}\, dt = \frac{n}{s} \mathcal{L}\{t^{n-1}\} \]
Reduction Formula: \( \mathcal{L}\{t^n\} = \dfrac{n}{s} \mathcal{L}\{t^{n-1}\} \)
Step 4: Apply the reduction formula repeatedly \[ \mathcal{L}\{t^n\} = \frac{n}{s} \cdot \frac{n-1}{s} \cdot \frac{n-2}{s} \cdots \frac{1}{s} \cdot \mathcal{L}\{t^0\} \] Since \( \mathcal{L}\{1\} = \dfrac{1}{s} \): \[ = \frac{n(n-1)(n-2)\cdots 1}{s^n} \cdot \frac{1}{s} \]
Final Result: \[ \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, \quad s > 0 \]
πŸ“ Example: \( \mathcal{L}\{t^3\} = \dfrac{6}{s^4} \), \quad \( \mathcal{L}\{t^5\} = \dfrac{120}{s^6} \)
Question 33(a) Appears Every Year
If \(p\) is a prime number, prove that \(a^p \equiv a \pmod{p}\) for any integer \(a\). (Fermat's Little Theorem)
15 Marks | Section IV

Statement

If \( p \) is a prime and \( a \) is any integer, then \( a^p \equiv a \pmod{p} \).

Equivalently, if \( p \nmid a \) (p does not divide a), then \( a^{p-1} \equiv 1 \pmod{p} \).

Proof by Mathematical Induction on \(a\)

Base Case: \( a = 1 \) \[ 1^p = 1 \equiv 1 \pmod{p} \quad \checkmark \]
Inductive Hypothesis: Assume \( a^p \equiv a \pmod{p} \) is true for some positive integer \( a \).
Inductive Step: Prove for \( (a+1) \)

We need to show: \( (a+1)^p \equiv (a+1) \pmod{p} \)

By Binomial Theorem:

\[ (a+1)^p = \sum_{k=0}^{p} \binom{p}{k} a^k = a^p + \binom{p}{1}a^{p-1} + \binom{p}{2}a^{p-2} + \cdots + \binom{p}{p-1}a + 1 \]
Key Lemma: For \( 1 \leq k \leq p-1 \), \( \binom{p}{k} \equiv 0 \pmod{p} \)

Proof of Lemma: \( \binom{p}{k} = \dfrac{p!}{k!(p-k)!} \). Since \( p \) is prime, \( p \) divides \( p! \) in the numerator. Since \( 1 \leq k \leq p-1 \), neither \( k! \) nor \( (p-k)! \) contains \( p \) as a factor. Hence \( p \mid \binom{p}{k} \) for \( 1 \leq k \leq p-1 \). ∎

Continuing the inductive step: \[ (a+1)^p \equiv a^p + 0 + 0 + \cdots + 0 + 1 \equiv a^p + 1 \pmod{p} \] By inductive hypothesis, \( a^p \equiv a \pmod{p} \), therefore: \[ (a+1)^p \equiv a + 1 \pmod{p} \quad \checkmark \]
Hence by Mathematical Induction, \( a^p \equiv a \pmod{p} \) for all positive integers \( a \), and hence for all integers \( a \). ∎
πŸ“ Application Example (often asked as part (b)): Compute remainder of \( 8^{103} \div 13 \).
By Fermat: \( 8^{12} \equiv 1 \pmod{13} \). Now \( 103 = 12 \times 8 + 7 \), so \( 8^{103} = (8^{12})^8 \cdot 8^7 \equiv 1^8 \cdot 8^7 \pmod{13} \).
\( 8^2 = 64 = 4 \times 13 + 12 \equiv 12 \equiv -1 \pmod{13} \).
\( 8^7 = 8^6 \cdot 8 = (8^2)^3 \cdot 8 \equiv (-1)^3 \cdot 8 = -8 \equiv 5 \pmod{13} \).
Remainder = 5.
Question 33(b) High Frequency
Prove that \( 5^{2n+2} - 24n - 25 \) is divisible by 576.
15 Marks (part b) | Section IV

We prove this by Mathematical Induction.

Step 1: Base Case \( n = 1 \) \[ 5^{2(1)+2} - 24(1) - 25 = 5^4 - 24 - 25 = 625 - 49 = 576 = 1 \times 576 \] So the result is true for \( n = 1 \). βœ“
Step 2: Inductive Hypothesis Assume \( 576 \mid (5^{2k+2} - 24k - 25) \) for some positive integer \( k \).
That is, \( 5^{2k+2} - 24k - 25 = 576m \) for some integer \( m \).
So \( 5^{2k+2} = 576m + 24k + 25 \quad \cdots (1) \)
Step 3: Inductive Step β€” Prove for \( n = k+1 \) We need to show: \( 576 \mid (5^{2(k+1)+2} - 24(k+1) - 25) \)
i.e., \( 576 \mid (5^{2k+4} - 24k - 48 - 25) \)
i.e., \( 576 \mid (5^{2k+4} - 24k - 49) \)
Step 4: Express in terms of the previous step \[ 5^{2k+4} - 24k - 49 = 5^2 \cdot 5^{2k+2} - 24k - 49 = 25 \cdot 5^{2k+2} - 24k - 49 \] Substitute from (1): \( 5^{2k+2} = 576m + 24k + 25 \) \[ = 25(576m + 24k + 25) - 24k - 49 \] \[ = 25 \times 576m + 600k + 625 - 24k - 49 \] \[ = 576 \times 25m + 576k + 576 \] \[ = 576(25m + k + 1) \]
Since \( (25m + k + 1) \) is an integer, we have \( 576 \mid (5^{2k+4} - 24(k+1) - 25) \).

Hence by Mathematical Induction, \( 5^{2n+2} - 24n - 25 \) is divisible by 576 for all positive integers \( n \). ∎
Question 33(b) alt 2018 Paper
Prove that \( n^5 - 1 \equiv 0 \pmod{30} \) ... actually: prove \( n^5 \equiv n \pmod{30} \) (i.e. \( 30 \mid n^5 - n \) for all integers \( n \)).
8 Marks | Section III

Since \( 30 = 2 \times 3 \times 5 \), it is enough to show \( 2 \mid n^5-n \), \( 3 \mid n^5-n \), and \( 5 \mid n^5-n \).

Using Fermat's Little Theorem:
For any prime \( p \) and integer \( n \): \( n^p \equiv n \pmod p \).

Therefore:
β€’ \( n^2 \equiv n \pmod 2 \Rightarrow n^5 = (n^2)^2 \cdot n \equiv n^2 \cdot n = n^3 \equiv n^2 \cdot n \equiv \cdots \equiv n \pmod 2 \)
β€’ \( n^3 \equiv n \pmod 3 \Rightarrow n^5 = n^3 \cdot n^2 \equiv n \cdot n^2 = n^3 \equiv n \pmod 3 \)
β€’ \( n^5 \equiv n \pmod 5 \) β€” directly by Fermat's theorem with \( p = 5 \).
Conclusion: Since \( 2 \mid n^5-n \), \( 3 \mid n^5-n \), \( 5 \mid n^5-n \), and \( \gcd(2,3,5)=1 \) pairwise: \[ 2 \times 3 \times 5 = 30 \mid n^5 - n \]
Hence \( n^5 \equiv n \pmod{30} \) for all integers \( n \). ∎
Question 34 High Frequency
Solve the non-homogeneous equation \( y'' - y' - 2y = 10\cos x \)
15 Marks | Section IV

Step 1: Write as operator form and find Auxiliary Equation (A.E.)

Operator form: \( (D^2 - D - 2)y = 10\cos x \)

Auxiliary equation: \( m^2 - m - 2 = 0 \)

Solve A.E.: \[ m^2 - m - 2 = (m-2)(m+1) = 0 \Rightarrow m = 2, \; m = -1 \] Two distinct real roots.

Step 2: Complementary Function (C.F.)

\[ y_c = C_1 e^{2x} + C_2 e^{-x} \]

Step 3: Particular Integral (P.I.) for \( 10\cos x \)

Use the formula: For \( f(D)y = a\cos bx \), replace \( D^2 \to -b^2 \)
Here \( b = 1 \), so replace \( D^2 \to -1 \): \[ P.I. = \frac{10\cos x}{D^2 - D - 2} = \frac{10\cos x}{-1 - D - 2} = \frac{10\cos x}{-3 - D} \]
Rationalise: multiply numerator and denominator by \( (-3 + D) \): \[ P.I. = \frac{10(-3+D)\cos x}{(-3-D)(-3+D)} = \frac{10(-3\cos x + D\cos x)}{9 - D^2} \] Now \( D\cos x = -\sin x \) and replace \( D^2 \to -1 \) again: \[ = \frac{10(-3\cos x - \sin x)}{9 - (-1)} = \frac{10(-3\cos x - \sin x)}{10} \]
Simplify: \[ P.I. = -3\cos x - \sin x \]

Step 4: General Solution

\[ y = y_c + P.I. = C_1 e^{2x} + C_2 e^{-x} - 3\cos x - \sin x \]
Question 34 alt 2018 Paper
Solve \( (D^2 - 4D - 5)y = e^{3x} + 4\cos 3x \)
15 Marks | Section IV
Step 1: Auxiliary Equation \[ m^2 - 4m - 5 = 0 \Rightarrow (m-5)(m+1) = 0 \Rightarrow m = 5, \; m = -1 \]
Step 2: Complementary Function \[ y_c = C_1 e^{5x} + C_2 e^{-x} \]
Step 3: P.I. for \( e^{3x} \) β€” use \( \dfrac{e^{ax}}{f(D)} = \dfrac{e^{ax}}{f(a)} \) (provided \( f(a) \neq 0 \)) \[ P.I._1 = \frac{e^{3x}}{D^2 - 4D - 5} = \frac{e^{3x}}{9 - 12 - 5} = \frac{e^{3x}}{-8} = -\frac{e^{3x}}{8} \]
Step 4: P.I. for \( 4\cos 3x \) β€” replace \( D^2 \to -9 \) (since \( b = 3 \)) \[ P.I._2 = \frac{4\cos 3x}{D^2 - 4D - 5} = \frac{4\cos 3x}{-9 - 4D - 5} = \frac{4\cos 3x}{-14 - 4D} \] Rationalise with \( (-14 + 4D) \): \[ = \frac{4(-14 + 4D)\cos 3x}{(-14)^2 - 16D^2} = \frac{4(-14\cos 3x + 4D\cos 3x)}{196 - 16(-9)} \] \( D\cos 3x = -3\sin 3x \): \[ = \frac{4(-14\cos 3x - 12\sin 3x)}{196 + 144} = \frac{4(-14\cos 3x - 12\sin 3x)}{340} \] \[ = \frac{-56\cos 3x - 48\sin 3x}{340} = \frac{-14\cos 3x - 12\sin 3x}{85} \]
General Solution: \[ y = C_1 e^{5x} + C_2 e^{-x} - \frac{e^{3x}}{8} - \frac{14\cos 3x + 12\sin 3x}{85} \]
Question 33 (2016) 2016 Paper
Solve the non-homogeneous equation \( y'' - 4y = e^{-2x} - 2x \), given \( y(0) = 0, y'(0) = 0 \).
15 Marks | Section IV
Step 1: A.E. β†’ \( m^2 - 4 = 0 \Rightarrow m = \pm 2 \)
Step 2: C.F. = \( C_1 e^{2x} + C_2 e^{-2x} \)
Step 3: P.I. for \( e^{-2x} \)
\( f(-2) = (-2)^2 - 4 = 0 \) β€” so normal formula fails! Use the special case:
\[ P.I._1 = \frac{e^{-2x}}{D^2-4} = x \cdot \frac{e^{-2x}}{2D\big|_{D=-2}} = x \cdot \frac{e^{-2x}}{2(-2)} = -\frac{x e^{-2x}}{4} \]
Step 4: P.I. for \( -2x \) β€” treat as polynomial \[ P.I._2 = \frac{-2x}{D^2-4} = \frac{-2x}{-4(1 - D^2/4)} = \frac{x}{2}\left(1 + \frac{D^2}{4} + \cdots\right) = \frac{x}{2} \] (Higher terms vanish since \( D^2 x = 0 \))
Step 5: General solution \[ y = C_1 e^{2x} + C_2 e^{-2x} - \frac{xe^{-2x}}{4} + \frac{x}{2} \]
Step 6: Apply initial conditions \( y(0) = 0, y'(0) = 0 \)
At \( x = 0 \): \( C_1 + C_2 = 0 \Rightarrow C_1 = -C_2 \)
\( y' = 2C_1 e^{2x} - 2C_2 e^{-2x} - \frac{e^{-2x}}{4} + \frac{xe^{-2x}}{2} + \frac{1}{2} \)
At \( x = 0 \): \( 2C_1 - 2C_2 - \frac{1}{4} + \frac{1}{2} = 0 \)
\( 2C_1 - 2(-C_1) = -\frac{1}{4} \Rightarrow 4C_1 = -\frac{1}{4} \Rightarrow C_1 = -\frac{1}{16} \), \( C_2 = \frac{1}{16} \)
\[ y = -\frac{e^{2x}}{16} + \frac{e^{-2x}}{16} - \frac{xe^{-2x}}{4} + \frac{x}{2} \]
Question 35 Appears Every Year
Solve the following LPP graphically:
Maximize \( z = 5x_1 + 7x_2 \) subject to: \( x_1 + x_2 \leq 4 \), \( 3x_1 + 8x_2 \leq 24 \), \( 10x_1 + 7x_2 \leq 35 \), \( x_1, x_2 \geq 0 \).
15 Marks | Section IV

Step 1: Find boundary lines (convert ≀ to =)

ConstraintLine equationx₁-interceptxβ‚‚-intercept
\( x_1 + x_2 \leq 4 \)\( x_1 + x_2 = 4 \)(4, 0)(0, 4)
\( 3x_1 + 8x_2 \leq 24 \)\( 3x_1 + 8x_2 = 24 \)(8, 0)(0, 3)
\( 10x_1 + 7x_2 \leq 35 \)\( 10x_1 + 7x_2 = 35 \)(3.5, 0)(0, 5)

Step 2: Find corner points of feasible region

A = (0, 0) β€” origin

B = (3.5, 0) β€” x₁ intercept of constraint 3

C: Intersection of \( 10x_1 + 7x_2 = 35 \) and \( x_1 + x_2 = 4 \):

From \( x_1 + x_2 = 4 \Rightarrow x_1 = 4 - x_2 \). Substitute:
\( 10(4-x_2) + 7x_2 = 35 \Rightarrow 40 - 10x_2 + 7x_2 = 35 \Rightarrow -3x_2 = -5 \Rightarrow x_2 = 5/3 \)
\( x_1 = 4 - 5/3 = 7/3 \)
C = (7/3, 5/3)

D: Intersection of \( x_1 + x_2 = 4 \) and \( 3x_1 + 8x_2 = 24 \):

\( x_1 = 4 - x_2 \). Substitute: \( 3(4-x_2) + 8x_2 = 24 \Rightarrow 12 + 5x_2 = 24 \Rightarrow x_2 = 12/5 = 2.4 \)
\( x_1 = 4 - 2.4 = 1.6 \)
D = (1.6, 2.4)

E = (0, 3) β€” xβ‚‚ intercept of constraint 2

Step 3: Evaluate objective function at each corner point

Corner Point\(x_1\)\(x_2\)\(z = 5x_1 + 7x_2\)
A000
B3.5017.5
C7/35/335/3 + 35/3 = 70/3 β‰ˆ 23.33
D1.62.48 + 16.8 = 24.8
E0321
Maximum value of \( z = 24.8 \) occurs at \( (x_1, x_2) = (1.6, 2.4) \)
Question 35 alt 2016 Paper
Use graphical method to solve the LPP: Minimize \( z = 20x_1 + 10x_2 \) subject to: \( x_1 + 2x_2 \leq 40 \), \( 3x_1 + x_2 \geq 30 \), \( 4x_1 + 3x_2 \geq 60 \), \( x_1, x_2 \geq 0 \).
15 Marks | Section IV

Step 1: Boundary lines

Constraintx₁-interceptxβ‚‚-interceptType
\( x_1 + 2x_2 = 40 \)(40, 0)(0, 20)≀ (upper bound)
\( 3x_1 + x_2 = 30 \)(10, 0)(0, 30)β‰₯ (lower bound)
\( 4x_1 + 3x_2 = 60 \)(15, 0)(0, 20)β‰₯ (lower bound)

Step 2: Find corner points of feasible region

A: Intersection of \( 3x_1 + x_2 = 30 \) and \( x_2 = 0 \): β†’ A = (10, 0)

B: Intersection of \( 3x_1 + x_2 = 30 \) and \( 4x_1 + 3x_2 = 60 \):

From first: \( x_2 = 30 - 3x_1 \). Substitute: \( 4x_1 + 3(30-3x_1) = 60 \Rightarrow 4x_1 + 90 - 9x_1 = 60 \Rightarrow -5x_1 = -30 \Rightarrow x_1 = 6 \)
\( x_2 = 30 - 18 = 12 \). B = (6, 12)

C: Intersection of \( 4x_1 + 3x_2 = 60 \) and \( x_1 + 2x_2 = 40 \):

Multiply second by 4: \( 4x_1 + 8x_2 = 160 \). Subtract: \( 5x_2 = 100 \Rightarrow x_2 = 20, \; x_1 = 0 \). C = (0, 20)

Step 3: Evaluate z at corner points

Corner Point\(x_1\)\(x_2\)\(z = 20x_1 + 10x_2\)
A100200
B612120 + 120 = 240
C020200
Minimum value of \( z = 200 \) occurs at both A = (10, 0) and C = (0, 20)
(All points on the line AC also give z = 200 β€” this is a case of alternate optima.)
Question 34 alt High Frequency
State and prove Lagrange's Mean Value Theorem.
15 Marks | Section IV

Statement

If a function \( f(x) \) is:

Then there exists at least one point \( c \in (a, b) \) such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

Proof

Step 1: Define auxiliary function Let \( g(x) = f(x) - f(a) - \dfrac{f(b)-f(a)}{b-a}(x-a) \)
Step 2: Check conditions for Rolle's Theorem

\( g(x) \) is continuous on \([a,b]\) and differentiable on \((a,b)\) since \( f \) is.

At \( x = a \): \( g(a) = f(a) - f(a) - 0 = 0 \)

At \( x = b \): \( g(b) = f(b) - f(a) - \dfrac{f(b)-f(a)}{b-a}(b-a) = f(b) - f(a) - [f(b)-f(a)] = 0 \)

So \( g(a) = g(b) = 0 \). βœ“

Step 3: Apply Rolle's Theorem to \( g(x) \) By Rolle's Theorem, there exists \( c \in (a,b) \) such that \( g'(c) = 0 \).
Step 4: Compute \( g'(x) \) \[ g'(x) = f'(x) - \frac{f(b)-f(a)}{b-a} \] Setting \( g'(c) = 0 \): \[ f'(c) - \frac{f(b)-f(a)}{b-a} = 0 \] \[ \boxed{f'(c) = \frac{f(b)-f(a)}{b-a}} \]
Hence Lagrange's Mean Value Theorem is proved. ∎
SECTION III β€” 4 Marks Each (Answer Any 6 from Q23–Q31)
Question 26
Find the general solution of \( y'' + 10y' + 29y = 0 \)
4 Marks | Section III
Auxiliary Equation: \( m^2 + 10m + 29 = 0 \) \[ m = \frac{-10 \pm \sqrt{100 - 116}}{2} = \frac{-10 \pm \sqrt{-16}}{2} = \frac{-10 \pm 4i}{2} = -5 \pm 2i \] Complex roots: \( \alpha = -5 \), \( \beta = 2 \)
\[ y = e^{-5x}(C_1 \cos 2x + C_2 \sin 2x) \]
Question 28 / 18
Find the gcd(34, 126) and express it as a linear combination of 34 and 126.
4 Marks | Section III
Euclidean Algorithm: \[ 126 = 3 \times 34 + 24 \] \[ 34 = 1 \times 24 + 10 \] \[ 24 = 2 \times 10 + 4 \] \[ 10 = 2 \times 4 + 2 \] \[ 4 = 2 \times 2 + 0 \] The last non-zero remainder is 2.
Back-substitution to find linear combination: \[ 2 = 10 - 2 \times 4 \] \[ = 10 - 2(24 - 2 \times 10) = 5 \times 10 - 2 \times 24 \] \[ = 5(34 - 24) - 2 \times 24 = 5 \times 34 - 7 \times 24 \] \[ = 5 \times 34 - 7(126 - 3 \times 34) = 5 \times 34 - 7 \times 126 + 21 \times 34 \] \[ = 26 \times 34 - 7 \times 126 \]
\( \gcd(34, 126) = 2 = 26 \times 34 - 7 \times 126 \)
Question 18 (2018 paper)
Find the gcd(58, 86) and express it as a linear combination.
4 Marks
\[ 86 = 1 \times 58 + 28 \] \[ 58 = 2 \times 28 + 2 \] \[ 28 = 14 \times 2 + 0 \] \( \gcd(58, 86) = 2 \)
Back-substitute: \( 2 = 58 - 2 \times 28 = 58 - 2(86 - 58) = 3 \times 58 - 2 \times 86 \)
\( \gcd(58, 86) = 2 = 3 \times 58 - 2 \times 86 \)
Question 18 / 19
Calculate \( \phi(5040) \), where \( \phi \) is the Euler phi-function.
4 Marks | Section III
Step 1: Prime factorisation of 5040 \[ 5040 = 7! = 2^4 \times 3^2 \times 5 \times 7 \]
Step 2: Apply Euler's formula \[ \phi(n) = n \prod_{p \mid n} \left(1 - \frac{1}{p}\right) \] \[ \phi(5040) = 5040 \times \left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right) \] \[ = 5040 \times \frac{1}{2} \times \frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \] \[ = 5040 \times \frac{48}{210} = 5040 \times \frac{8}{35} = \frac{40320}{35} = 1152 \]
\( \phi(5040) = 1152 \)
Question 24
Find the Laplace transform of \( \cosh(at) \). Also find \( \mathcal{L}\{e^{at}\} \).
4 Marks | Section III
Using definition: \( \cosh(at) = \dfrac{e^{at} + e^{-at}}{2} \) \[ \mathcal{L}\{\cosh(at)\} = \frac{1}{2}\left[\mathcal{L}\{e^{at}\} + \mathcal{L}\{e^{-at}\}\right] = \frac{1}{2}\left[\frac{1}{s-a} + \frac{1}{s+a}\right] \] \[ = \frac{1}{2} \cdot \frac{(s+a)+(s-a)}{(s-a)(s+a)} = \frac{1}{2} \cdot \frac{2s}{s^2-a^2} \]
\( \mathcal{L}\{e^{at}\} = \dfrac{1}{s-a}, \quad \mathcal{L}\{\cosh(at)\} = \dfrac{s}{s^2-a^2} \)
Question 12 / 12
Verify Lagrange's Mean Value Theorem for \( f(x) = x^2 + 2x + 9, \; x \in [1, 5] \).
4 Marks | Section II/III
Check conditions: \( f(x) \) is a polynomial β€” continuous and differentiable everywhere. βœ“
Compute values: \[ f(1) = 1 + 2 + 9 = 12, \quad f(5) = 25 + 10 + 9 = 44 \] \[ \frac{f(5) - f(1)}{5 - 1} = \frac{44 - 12}{4} = \frac{32}{4} = 8 \]
Find c: \( f'(x) = 2x + 2 \). Set \( f'(c) = 8 \): \[ 2c + 2 = 8 \Rightarrow c = 3 \] Since \( c = 3 \in (1, 5) \) βœ“
MVT is verified. \( c = 3 \in (1, 5) \) such that \( f'(3) = 8 = \dfrac{f(5)-f(1)}{5-1} \)
Question 26
Find the general solution of \( y'' + 2ky' + k^2 y = 0 \)
4 Marks
A.E.: \( m^2 + 2km + k^2 = 0 \Rightarrow (m+k)^2 = 0 \Rightarrow m = -k, -k \) Repeated real root.
\[ y = (C_1 + C_2 x)e^{-kx} \]
Question 13
Solve the differential equation \( y'' - y' - 6y = 0 \)
2 Marks
A.E.: \( m^2 - m - 6 = 0 \Rightarrow (m-3)(m+2) = 0 \Rightarrow m = 3, -2 \)
\[ y = C_1 e^{3x} + C_2 e^{-2x} \]
SECTION II β€” 2 Marks Each (Answer Any 8 from Q11–Q22)
All 2-Mark Questions
Complete answers for all 2-mark questions
2 Marks each

Q15 / Q16: Find \( \mathcal{L}\{e^{at}\} \) and \( \mathcal{L}\{e^{-at}\} \)

\( \mathcal{L}\{e^{at}\} = \dfrac{1}{s-a} \), \quad \( \mathcal{L}\{e^{-at}\} = \dfrac{1}{s+a} \)

Q16: Find \( \mathcal{L}^{-1}\!\left[\dfrac{s}{s^2+36}\right] \)

Using \( \mathcal{L}^{-1}\!\left[\dfrac{s}{s^2+a^2}\right] = \cos at \), with \( a = 6 \):
\( \mathcal{L}^{-1}\!\left[\dfrac{s}{s^2+36}\right] = \cos 6t \)

Q16 alt: Find \( \mathcal{L}^{-1}\!\left[\dfrac{2}{s^2+4^2}\right] \)

\( \mathcal{L}^{-1}\!\left[\dfrac{2}{s^2+16}\right] = \sin 4t \) (using \( \mathcal{L}^{-1}\left[\dfrac{a}{s^2+a^2}\right] = \sin at \) with \( a=4 \))

Q16: Evaluate \( \mathcal{L}^{-1}\!\left[\dfrac{2s+1}{s(s+1)}\right] \)

Partial fractions: \[ \frac{2s+1}{s(s+1)} = \frac{A}{s} + \frac{B}{s+1} \] Multiply through: \( 2s+1 = A(s+1) + Bs \)
At \( s=0 \): \( 1 = A \). At \( s=-1 \): \( -1 = -B \Rightarrow B = 1 \).
So \( \dfrac{1}{s} + \dfrac{1}{s+1} \).
\( \mathcal{L}^{-1}\!\left[\dfrac{2s+1}{s(s+1)}\right] = 1 + e^{-t} \)

Q17 / Q17: Find gcd(βˆ’8, βˆ’36)

\( \gcd(-8, -36) = \gcd(8, 36) \). Using Euclidean: \( 36 = 4 \times 8 + 4 \), \( 8 = 2 \times 4 \).
\( \gcd(-8, -36) = 4 \)

Q18: Calculate \( \phi(1575) \)

\( 1575 = 3^2 \times 5^2 \times 7 \)
\( \phi(1575) = 1575 \times (1-1/3)(1-1/5)(1-1/7) = 1575 \times \frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \)
\( = 1575 \times \frac{48}{105} = 1575 \times \frac{16}{35} = 720 \)
\( \phi(1575) = 720 \)

Q19: Find remainder of \( 8^{103} \div 13 \)

By Fermat's theorem: \( 8^{12} \equiv 1 \pmod{13} \).
\( 103 = 12 \times 8 + 7 \). So \( 8^{103} \equiv 8^7 \pmod{13} \).
\( 8^2 = 64 \equiv 12 \equiv -1 \pmod{13} \).
\( 8^6 = (8^2)^3 \equiv (-1)^3 = -1 \pmod{13} \).
\( 8^7 = 8^6 \times 8 \equiv -8 \equiv 5 \pmod{13} \).
Remainder = 5

Q20: Show that 41 divides \( 2^{20} - 1 \)

By Fermat's theorem: \( 2^{40} \equiv 1 \pmod{41} \).
\( 2^{20} - 1 \). Note \( 2^{20} \equiv x \pmod{41} \). Since \( (2^{20})^2 = 2^{40} \equiv 1 \pmod{41} \), we have \( 2^{20} \equiv \pm 1 \pmod{41} \).
Check: \( 2^{10} = 1024 = 24 \times 41 + 40 \equiv -1 \pmod{41} \). So \( 2^{20} \equiv (-1)^2 = 1 \pmod{41} \).
Therefore \( 2^{20} - 1 \equiv 0 \pmod{41} \). βœ“
Hence 41 divides \( 2^{20} - 1 \). ∎

Q15: Find the Laplace transform of \( f(t) = (t^2+1)^2 \)

\( (t^2+1)^2 = t^4 + 2t^2 + 1 \)
\( \mathcal{L}\{t^4\} = \dfrac{4!}{s^5} = \dfrac{24}{s^5} \), \quad \( \mathcal{L}\{2t^2\} = \dfrac{2 \cdot 2!}{s^3} = \dfrac{4}{s^3} \), \quad \( \mathcal{L}\{1\} = \dfrac{1}{s} \)
\( \mathcal{L}\{(t^2+1)^2\} = \dfrac{24}{s^5} + \dfrac{4}{s^3} + \dfrac{1}{s} \)

Q17: Find \( \text{lcm}(272, 1479) \)

First find \( \gcd(272, 1479) \):
\( 1479 = 5 \times 272 + 119 \)
\( 272 = 2 \times 119 + 34 \)
\( 119 = 3 \times 34 + 17 \)
\( 34 = 2 \times 17 + 0 \)
\( \gcd = 17 \)
\( \text{lcm} = \dfrac{272 \times 1479}{17} = 16 \times 1479 = 23664 \)
\( \text{lcm}(272, 1479) = 23664 \)

Q11: Show that \( \cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y \)

RHS \( = \dfrac{e^x+e^{-x}}{2} \cdot \dfrac{e^y+e^{-y}}{2} + \dfrac{e^x-e^{-x}}{2} \cdot \dfrac{e^y-e^{-y}}{2} \)
\( = \dfrac{e^{x+y}+e^{x-y}+e^{-x+y}+e^{-(x+y)}}{4} + \dfrac{e^{x+y}-e^{x-y}-e^{-x+y}+e^{-(x+y)}}{4} \)
\( = \dfrac{2e^{x+y}+2e^{-(x+y)}}{4} = \dfrac{e^{x+y}+e^{-(x+y)}}{2} = \cosh(x+y) \) βœ“

Q12: Verify Lagrange's MVT for \( f(x) = 2x^3 - 3x^2 - x, \; x \in [1,2] \)

\( f(1) = 2 - 3 - 1 = -2 \), \( f(2) = 16 - 12 - 2 = 2 \)
\( \dfrac{f(2)-f(1)}{2-1} = \dfrac{4}{1} = 4 \)
\( f'(x) = 6x^2 - 6x - 1 \). Set \( f'(c) = 4 \):
\( 6c^2 - 6c - 1 = 4 \Rightarrow 6c^2 - 6c - 5 = 0 \)
\( c = \dfrac{6 \pm \sqrt{36 + 120}}{12} = \dfrac{6 \pm \sqrt{156}}{12} \approx \dfrac{6 + 12.49}{12} \approx 1.54 \in (1,2) \) βœ“
MVT verified. \( c \approx 1.54 \in (1,2) \)
SECTION I β€” 1 Mark Each (All 10 Compulsory)
Questions 1–10
All one-mark compulsory answers
1 Mark each Γ— 10 = 10 Marks
QQuestionAnswer
1Derivative of \( \log_a x \)\( \dfrac{1}{x \ln a} \)
2\( \sinh x \) in exponential form\( \dfrac{e^x - e^{-x}}{2} \)
3If \( \text{sech}\, x = \dfrac{4}{5} \), find \( \cosh x \)\( \cosh x = \dfrac{5}{4} \)
4State Rolle's TheoremIf f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then βˆƒ c ∈ (a,b) such that fβ€²(c) = 0.
5Degree and order of \( \dfrac{dy}{dx} + (x^{1/2} - y) = 0 \)Order = 1, Degree = 1
6General solution of \( \dfrac{dy}{dx} = 2x \)\( y = x^2 + C \)
7Solve \( y' = -xy \)\( y = Ce^{-x^2/2} \)
8When are two integers relatively prime?When their gcd = 1.
9Modulus of \( 3 - 2i \)\( |3-2i| = \sqrt{9+4} = \sqrt{13} \)
10Identify objective function and constraints in the LPP: Maximize z = 12x₁ + 7xβ‚‚, s.t. 2x₁+x₂≀9, x₁+2x₂≀28Objective function: z = 12x₁ + 7xβ‚‚. Constraints: 2x₁+xβ‚‚ ≀ 9, x₁+2xβ‚‚ ≀ 28, x₁,xβ‚‚ β‰₯ 0.
β€”Define Laplace transform of f(t)\( \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t)\, dt \), \( s > 0 \)
β€”State Wilson's TheoremIf p is prime, then (pβˆ’1)! ≑ βˆ’1 (mod p).
β€”Principal amplitude of βˆ’i\( \arg(-i) = -\pi/2 \)
β€”If cosech x = βˆ’4/3, find sinh x\( \sinh x = -3/4 \)
β€”Prove L(1) = 1/s\( \mathcal{L}\{1\} = \int_0^\infty e^{-st} dt = \left[\frac{e^{-st}}{-s}\right]_0^\infty = \frac{1}{s} \)
πŸ“ Quick Formula Sheet β€” Memorise These!
Master Reference
All key formulas at a glance

Laplace Transforms

\(f(t)\)\(\mathcal{L}\{f(t)\} = F(s)\)
\(1\)\(\dfrac{1}{s}\)
\(t^n\)\(\dfrac{n!}{s^{n+1}}\)
\(e^{at}\)\(\dfrac{1}{s-a}\)
\(\sin at\)\(\dfrac{a}{s^2+a^2}\)
\(\cos at\)\(\dfrac{s}{s^2+a^2}\)
\(\sinh at\)\(\dfrac{a}{s^2-a^2}\)
\(\cosh at\)\(\dfrac{s}{s^2-a^2}\)
\(e^{at} f(t)\)\(F(s-a)\)
\(f(t-a)u(t-a)\)\(e^{-as}F(s)\) (2nd Shift)
\(t^n f(t)\)\((-1)^n F^{(n)}(s)\)
\(f(t)/t\)\(\int_s^\infty F(u)\,du\)

ODE General Solutions (Standard Cases)

Roots of A.E.C.F. Form
Real distinct: \(m_1, m_2\)\(C_1 e^{m_1 x} + C_2 e^{m_2 x}\)
Real equal: \(m, m\)\((C_1 + C_2 x)e^{mx}\)
Complex: \(\alpha \pm i\beta\)\(e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)\)

Particular Integral Formulas

RHSP.I. Formula
\(e^{ax}\)\(\dfrac{e^{ax}}{f(a)}\), if \(f(a) \neq 0\). If \(f(a)=0\): \(x\cdot\dfrac{e^{ax}}{f'(a)}\)
\(\sin bx\) or \(\cos bx\)Replace \(D^2 \to -b^2\) in \(f(D)\); rationalise if needed
Polynomial \(p(x)\)Expand \([f(D)]^{-1}\) as power series in \(D\), apply to \(p(x)\)

Number Theory Quick Reference

Fermat's Little Theorem: If p is prime, \( a^p \equiv a \pmod p \) for any integer \(a\).
Equivalently: if \( p \nmid a \), then \( a^{p-1} \equiv 1 \pmod p \).

Euler's Phi Function: \( \phi(n) = n \displaystyle\prod_{p \mid n}\left(1 - \frac{1}{p}\right) \)

Euclidean Algorithm: \( \gcd(a,b) = \gcd(b, a \bmod b) \)

Wilson's Theorem: \( (p-1)! \equiv -1 \pmod p \) for prime \(p\)

Hyperbolic Functions

FunctionDefinitionIdentity
\(\sinh x\)\(\dfrac{e^x - e^{-x}}{2}\)\(\cosh^2 x - \sinh^2 x = 1\)
\(\cosh x\)\(\dfrac{e^x + e^{-x}}{2}\)\(\tanh^2 x + \text{sech}^2 x = 1\)
\(\tanh x\)\(\dfrac{e^x - e^{-x}}{e^x + e^{-x}}\)\(\coth^2 x - \text{csch}^2 x = 1\)

Lagrange's MVT

\( f'(c) = \dfrac{f(b) - f(a)}{b - a} \) for some \( c \in (a, b) \)

LPP Method Summary

Graphical Method Steps:
1. Convert each constraint to equality and find two points (x-intercept, y-intercept)
2. Plot lines and shade feasible region (satisfying ALL constraints including x₁,xβ‚‚ β‰₯ 0)
3. Identify corner points of feasible region (find intersections algebraically)
4. Evaluate objective function z at each corner point
5. For Maximize: pick the largest z. For Minimize: pick the smallest z.

✊ All the best for your exam!

MM 1131.9 Mathematics – I | Kerala University BCA First Semester