TU BCA Mathematics I Master Guide: Calculus, Limits, Derivatives & Maxima-Minima (Solved Board Problems)
Author: Bhuban Subedi | Subject: Mathematics I (CAMT104) | Semester: First Semester
For first-semester Tribhuvan University BCA students, Mathematics I (CAMT104) is often considered one of the toughest theoretical subjects. The transition from +2 mathematics to university calculus requires a solid conceptual understanding of limits, continuous functions, differentiation rules, and optimization techniques.
In TU board exams, Mathematics I carries 60 marks in the external final paper, where Group B and Group C consistently feature numerical problems on L’Hôpital’s rule, successive differentiation (Leibnitz’s Theorem), mean value theorems, and curve optimization.
In this guide, I will break down the highest-weightage calculus topics with step-by-step solved board examination questions.
1. Limits & Continuity: Evaluating Indeterminate Forms
In TU exams, limit evaluations predominantly involve the indeterminate forms $\frac{0}{0}$, $\frac{\infty}{\infty}$, $0 \times \infty$, and $1^\infty$.
L’Hôpital’s Rule:
If $\lim_{x \to a} \frac{f(x)}{g(x)}$ results in $\frac{0}{0}$ or $\frac{\infty}{\infty}$, and both functions are differentiable:
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$
Solved TU Board Problem 1: Indeterminate Form $\frac{0}{0}$
Problem: Evaluate $\lim_{x \to 0} \frac{e^x – e^{-x} – 2x}{x – \sin x}$.
Solution:
1. Substitute $x = 0$:
$$\frac{e^0 – e^0 – 2(0)}{0 – \sin 0} = \frac{1 – 1 – 0}{0 – 0} = \frac{0}{0} \quad \text{(Indeterminate Form)}$$
-
Apply L’Hôpital’s Rule (differentiating numerator and denominator with respect to $x$):
$$\lim_{x \to 0} \frac{\frac{d}{dx}(e^x – e^{-x} – 2x)}{\frac{d}{dx}(x – \sin x)} = \lim_{x \to 0} \frac{e^x + e^{-x} – 2}{1 – \cos x}$$ -
Substitute $x = 0$:
$$\frac{1 + 1 – 2}{1 – 1} = \frac{0}{0} \quad \text{(Still Indeterminate Form)}$$ -
Apply L’Hôpital’s Rule a second time:
$$\lim_{x \to 0} \frac{e^x – e^{-x}}{\sin x}$$ -
Substitute $x = 0$:
$$\frac{1 – 1}{0} = \frac{0}{0} \quad \text{(Indeterminate Form)}$$ -
Apply L’Hôpital’s Rule a third time:
$$\lim_{x \to 0} \frac{e^x + e^{-x}}{\cos x} = \frac{e^0 + e^0}{\cos 0} = \frac{1 + 1}{1} = \mathbf{2}$$
Final Answer: 2
2. Successive Differentiation & Leibnitz’s Theorem
When finding the $n^{\text{th}}$ derivative of the product of two functions $u(x)$ and $v(x)$, Leibnitz’s Theorem is applied:
$$(uv)n = \sum v_2 + \dots + u v_n$$}^n \binom{n}{k} u_{n-k} v_k = u_n v + \binom{n}{1} u_{n-1} v_1 + \binom{n}{2} u_{n-2
+-------------------------------------------------------------------------------+
| BHUBAN'S TU EXAM TIP: LEIBNITZ'S RULE |
| Always choose 'v' as the polynomial term (e.g., x^2, x^3) because its higher |
| derivatives eventually become zero (e.g., v = x^2, v1 = 2x, v2 = 2, v3 = 0). |
| This eliminates all higher terms after k = 2! |
+-------------------------------------------------------------------------------+
Solved TU Board Problem 2: Leibnitz’s Theorem
Problem: If $y = x^2 e^{3x}$, find $y_n$ (the $n^{\text{th}}$ derivative of $y$).
Solution:
1. Let $u = e^{3x}$ and $v = x^2$.
2. Compute individual derivatives:
– $u = e^{3x} \implies u_n = 3^n e^{3x}$
– $u_{n-1} = 3^{n-1} e^{3x}$
– $u_{n-2} = 3^{n-2} e^{3x}$
– $v = x^2 \implies v_1 = 2x, \quad v_2 = 2, \quad v_3 = 0$
-
Apply Leibnitz’s Theorem formula:
$$y_n = u_n v + n u_{n-1} v_1 + \frac{n(n-1)}{2} u_{n-2} v_2$$ -
Substitute terms:
$$y_n = (3^n e^{3x})(x^2) + n (3^{n-1} e^{3x})(2x) + \frac{n(n-1)}{2} (3^{n-2} e^{3x})(2)$$ -
Factor out $3^{n-2} e^{3x}$:
$$y_n = 3^{n-2} e^{3x} \left[ 9x^2 + 6nx + n(n-1) \right]$$
3. Rolle’s Theorem & Lagrange’s Mean Value Theorem (LMVT)
+-------------------+---------------------------------------------------------------+
| Theorem | Conditions & Conclusion |
+-------------------+---------------------------------------------------------------+
| **Rolle's** | 1. Continuous in $[a, b]$ |
| **Theorem** | 2. Differentiable in $(a, b)$ |
| | 3. $f(a) = f(b)$ |
| | **Conclusion:** There exists $c \in (a, b)$ such that $f'(c) = 0$.|
+-------------------+---------------------------------------------------------------+
| **Lagrange's** | 1. Continuous in $[a, b]$ |
| **MVT** | 2. Differentiable in $(a, b)$ |
| | **Conclusion:** There exists $c \in (a, b)$ such that: |
| | $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ |
+-------------------+---------------------------------------------------------------+
Solved TU Board Problem 3: Rolle’s Theorem Verification
Problem: Verify Rolle’s Theorem for $f(x) = x^2 – 4x + 3$ on the closed interval $[1, 3]$.
Solution:
1. Continuity: $f(x)$ is a polynomial function, hence it is continuous on $[1, 3]$.
2. Differentiability: $f'(x) = 2x – 4$ exists for all $x \in (1, 3)$, hence $f(x)$ is differentiable on $(1, 3)$.
3. Boundary Values:
– $f(1) = (1)^2 – 4(1) + 3 = 1 – 4 + 3 = 0$
– $f(3) = (3)^2 – 4(3) + 3 = 9 – 12 + 3 = 0$
– Since $f(1) = f(3) = 0$, all three conditions of Rolle’s theorem are satisfied.
- Finding Point $c$:
$$f'(c) = 0 \implies 2c – 4 = 0 \implies 2c = 4 \implies \mathbf{c = 2}$$ - Since $c = 2 \in (1, 3)$, Rolle’s Theorem is fully verified!
4. Maxima, Minima & Optimization
To find the local maximum and minimum values of a differentiable function $f(x)$:
1. Find the first derivative $f'(x)$ and set $f'(x) = 0$ to determine critical points.
2. Compute the second derivative $f”(x)$:
– If $f”(c) < 0 \implies$ Local Maximum at $x = c$.
– If $f”(c) > 0 \implies$ Local Minimum at $x = c$.
– If $f”(c) = 0 \implies$ Test fails (inflection point or higher derivative test required).
Frequently Asked Questions (FAQ)
Q1: What is the passing mark for TU BCA Mathematics I?
In the 60-mark external board examination, you must score at least 24 marks (40%) to pass. The internal assessment contributes an additional 40 marks.
Q2: Are calculators allowed in the Mathematics I exam?
Yes, standard non-programmable scientific calculators (such as Casio fx-991ES Plus / fx-991CW) are permitted.
