BSc CSIT (TU) Science Mathematics I (BSc CSIT, MTH112) Question Paper 2077 Nepal

Continuity at a Point

A function $f(x)$ is said to be continuous at a point $x = a$ if the following three conditions hold:

1. $f(a)$ is defined (i.e. $a$ lies in the domain of $f$).
2. $\lim_{x \to a} f(x)$ exists (left-hand limit $=$ right-hand limit).
3. $\lim_{x \to a} f(x) = f(a)$.

If any one of these fails, $f$ is discontinuous at $x = a$.

Continuity of $f(x) = \dfrac{x^2 - 1}{x - 1}$ at $x = 1$

At $x = 1$ the denominator is $1 - 1 = 0$, so $f(1) = \dfrac{0}{0}$ is undefined. Hence condition (1) already fails and $f$ is discontinuous at $x = 1$.

Now examine the limit. For $x \neq 1$:

Therefore

The limit exists and equals 2, but $f(1)$ does not exist. This is a removable discontinuity.

Redefining $f$ to Make it Continuous

Since the only obstruction is the missing value at $x = 1$, we plug the hole by defining $f(1) = 2$:

Now $F(1) = 2 = \lim_{x \to 1} F(x)$, so all three conditions are satisfied and $F$ is continuous at $x = 1$ (indeed $F(x) = x+1$ everywhere).

Fundamental Theorem of Integral Calculus

Statement. If $f$ is continuous on $[a,b]$ and $F$ is an antiderivative of $f$ (i.e. $F'(x) = f(x)$ for all $x \in [a,b]$), then

Proof. Define the area function

Using the definition of the derivative and the additivity of the integral,

By the Mean Value Theorem for integrals, there exists $c \in [x, x+h]$ with $\int_x^{x+h} f(t)\,dt = f(c)\,h$. Hence

since $c \to x$ as $h \to 0$ and $f$ is continuous. Thus $\Phi$ is an antiderivative of $f$.

If $F$ is any other antiderivative, then $F(x) = \Phi(x) + C$ for some constant $C$. Therefore

This proves the theorem. $\blacksquare$

Evaluation of $\displaystyle\int_1^2 (3x^2 + 2x)\,dx$

An antiderivative is $F(x) = x^3 + x^2$. Hence

Solving $\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = e^{2x}$

This is a linear non-homogeneous ODE with constant coefficients. The general solution is $y = y_c + y_p$.

Complementary Function (CF)

The auxiliary equation is

Since the roots are real and distinct,

Particular Integral (PI)

Write $D = \dfrac{d}{dx}$. Then

Substituting $D = 2$ gives $2^2 - 5(2) + 6 = 4 - 10 + 6 = 0$, so the operator vanishes — this is the case of resonance (since $e^{2x}$ is part of the CF). Apply the rule for repeated/overlapping roots:

Substituting $D = 2$ in $2D - 5$ gives $2(2) - 5 = -1$. Hence

General Solution

where $C_1, C_2$ are arbitrary constants.

Check: With $y_p = -xe^{2x}$, one finds $y_p'' - 5y_p' + 6y_p = e^{2x}$, confirming the result.

Evaluate $\displaystyle\lim_{x \to 0} \frac{\tan x - x}{x^3}$

Use the Maclaurin expansion $\tan x = x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + \cdots$.

Then

so

Taking the limit as $x \to 0$ (higher-order terms vanish):

(Equivalently, applying L'Hôpital's rule three times to the $\tfrac{0}{0}$ form gives the same answer $\tfrac{1}{3}$.)

Differentiate $y = \log(\sin x)$

Using the chain rule with $\dfrac{d}{dx}\log u = \dfrac{1}{u}\dfrac{du}{dx}$ and $u = \sin x$:

Taylor's Theorem

If $f(x)$ is a function having continuous derivatives up to order $n$ in a neighbourhood of $x = a$, then $f$ can be expanded about $x = a$ as

where $R_n$ is the remainder term (Lagrange's form: $R_n = \dfrac{(x-a)^n}{n!}f^{(n)}(a+\theta(x-a))$, $0<\theta<1$).

Maclaurin Series

The Maclaurin series is the special case $a = 0$:

Expansion of $e^x$

For $f(x) = e^x$, every derivative is $e^x$, so $f^{(n)}(0) = e^0 = 1$ for all $n$. Hence

This series is valid for all real $x$.

Radius of Curvature of $y = x^2$ at $(1,1)$

The radius of curvature for a Cartesian curve $y = f(x)$ is

For $y = x^2$:

At the point $(1,1)$, i.e. $x = 1$:

Therefore

Evaluate $\displaystyle\int e^x \sin x\, dx$

Let $I = \int e^x \sin x\, dx$. Apply integration by parts twice.

First integration by parts ($u = \sin x,\ dv = e^x dx$):

Second integration by parts on $\int e^x \cos x\, dx$ ($u = \cos x,\ dv = e^x dx$):

Substitute back:

Solving for $I$:

Solve $(x^2 + y^2)\,dx - 2xy\,dy = 0$

This is a homogeneous differential equation (both coefficients are homogeneous of degree 2). Rewrite:

Substitution. Let $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Then

So

Separate variables:

Integrate. Note $\int \dfrac{2v}{1-v^2}\,dv = -\ln|1 - v^2|$ (since $\frac{d}{dv}(1-v^2) = -2v$):

Thus $\ln|1 - v^2| = -\ln|x| - \ln C_1$, giving $(1 - v^2)\,x = C$ (a constant).

Back-substitute $v = \dfrac{y}{x}$:

where $C$ is an arbitrary constant.

Scalar Triple Product of $\hat{i}+\hat{j},\ \hat{j}+\hat{k},\ \hat{k}+\hat{i}$

The scalar triple product $[\vec{a}\ \vec{b}\ \vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c})$ equals the determinant of the component matrix.

With $\vec{a} = \hat{i}+\hat{j} = (1,1,0)$, $\vec{b} = \hat{j}+\hat{k} = (0,1,1)$, $\vec{c} = \hat{k}+\hat{i} = (1,0,1)$:

Expanding along the first row:

Since the value is non-zero, the three vectors are non-coplanar (linearly independent).

Show $\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0$ for $u = \log(x^2 + y^2)$

Let $r^2 = x^2 + y^2$, so $u = \log(x^2 + y^2)$.

First derivative w.r.t. $x$:

Second derivative w.r.t. $x$ (quotient rule):

By symmetry (swap $x \leftrightarrow y$):

Adding:

Hence $u$ satisfies Laplace's equation. $\blacksquare$

Evaluate $\displaystyle\int_0^1 \int_0^1 (x^2 + y^2)\,dx\,dy$

Inner integral (with respect to $x$, treating $y$ as constant):

Outer integral (with respect to $y$):