BSc CSIT (TU) Science Mathematics I (BSc CSIT, MTH112) Question Paper 2074 Nepal

Theorem: Every differentiable function is continuous

Statement. If a function $f$ is differentiable at a point $x = c$, then $f$ is continuous at $x = c$.

Proof. Since $f$ is differentiable at $c$, the derivative

exists (is finite). For $x \neq c$ write

Taking limits as $x \to c$,

Hence $\lim_{x\to c} f(x) = f(c)$, which is exactly the condition that $f$ is continuous at $c$. $\blacksquare$

The converse is not true: continuity does not imply differentiability, as the next example shows.

Continuity and differentiability of $f(x) = |x|$ at $x = 0$

Here $f(x) = |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}$ and $f(0) = 0$.

Continuity at $x = 0$:

Both one-sided limits equal $f(0) = 0$, so $f$ is continuous at $x = 0$.

Differentiability at $x = 0$: compute the left- and right-hand derivatives.

Since $Lf'(0) = -1 \neq +1 = Rf'(0)$, the derivative does not exist; $f$ is not differentiable at $x = 0$.

Conclusion. $f(x)=|x|$ is continuous but not differentiable at $x=0$, confirming that continuity does not guarantee differentiability (geometrically, the graph has a sharp corner at the origin).

Standard integral 1: $\displaystyle\int \frac{dx}{a^2 + x^2}$

Put $x = a\tan\theta$, so $dx = a\sec^2\theta\, d\theta$ and $a^2 + x^2 = a^2\sec^2\theta$.

Standard integral 2: $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}}$

Put $x = a\sin\theta$, so $dx = a\cos\theta\, d\theta$ and $\sqrt{a^2 - x^2} = a\cos\theta$.

Application: $\displaystyle\int \frac{dx}{x^2 + 4x + 8}$

Complete the square in the denominator:

Let $t = x + 2$, $dt = dx$, with $a = 2$. Using result 1,

Area bounded by $y^2 = 4ax$ and its latus rectum

Setup. For the parabola $y^2 = 4ax$ (vertex at origin, opening to the right), the focus is at $(a, 0)$ and the latus rectum is the vertical chord through the focus, i.e. the line $x = a$. We want the area enclosed between the parabola and this line.

By symmetry about the $x$-axis, the required area is twice the area above the $x$-axis. On the parabola $y = \sqrt{4ax} = 2\sqrt{a}\,\sqrt{x}$, with $x$ running from $0$ (vertex) to $a$ (latus rectum).

Evaluate:

Result.

(Equivalently, since the length of the latus rectum is $4a$ and it spans height $2\cdot 2a = 4a$, the enclosed region is two-thirds of the bounding rectangle $a \times 4a = 4a^2$, giving $\frac{2}{3}(4a^2) = \frac{8}{3}a^2$.)

Evaluate $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$

Geometric (sandwich) argument. For $0 < x < \dfrac{\pi}{2}$, comparing the areas of triangle $OAB$, the circular sector $OAB$, and triangle $OAC$ in a unit circle gives

Dividing throughout by $\sin x > 0$:

The function $\frac{\sin x}{x}$ is even, so the same inequality holds for $x<0$. As $x\to 0$, $\cos x \to 1$, and by the Squeeze (Sandwich) Theorem

Differentiate $y = e^{ax}\sin bx$

Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$ with $u = e^{ax}$, $v = \sin bx$:

(This can also be written as $\frac{dy}{dx} = \sqrt{a^2+b^2}\;e^{ax}\sin(bx + \phi)$ where $\tan\phi = b/a$.)

Rolle's Theorem

Statement. If $f$ is (i) continuous on $[a,b]$, (ii) differentiable on $(a,b)$, and (iii) $f(a) = f(b)$, then there exists at least one point $c \in (a,b)$ such that $f'(c) = 0$.

Verification for $f(x) = x^2 - 4x + 3$ on $[1,3]$

Condition (i): $f$ is a polynomial, hence continuous on $[1,3]$. ✓

Condition (ii): $f$ is a polynomial, hence differentiable on $(1,3)$. ✓

Condition (iii):

So $f(1) = f(3) = 0$. ✓

All three hypotheses hold, so Rolle's theorem applies. To find $c$:

Since $c = 2 \in (1,3)$, Rolle's theorem is verified, with $f'(2) = 0$.

Asymptotes of $\displaystyle y = \frac{x^2 + 1}{x - 1}$

Vertical asymptote. The denominator vanishes at $x = 1$, where the numerator $x^2+1 = 2 \neq 0$. Hence $y \to \pm\infty$ as $x \to 1$, giving the vertical asymptote

Oblique (slant) asymptote. Since the degree of the numerator exceeds that of the denominator by one, divide:

As $x \to \pm\infty$, the remainder $\dfrac{2}{x-1} \to 0$, so

is the oblique asymptote.

Horizontal asymptote: none (the curve grows without bound). The asymptotes of the curve are therefore $x = 1$ and $y = x + 1$.

Evaluate $\displaystyle\int_0^{\pi/2}\sin^4 x\, dx$ by reduction formula

The reduction formula for $I_n = \displaystyle\int_0^{\pi/2}\sin^n x\, dx$ is

With $n = 4$:

The base case is

Therefore

(Equivalently, by Wallis' formula for even $n$: $\dfrac{3\cdot 1}{4\cdot 2}\cdot\dfrac{\pi}{2} = \dfrac{3\pi}{16}$.)

Solve $\displaystyle\frac{dy}{dx} + y\tan x = \sec x$

This is a linear first-order ODE $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ with $P = \tan x$, $Q = \sec x$.

Integrating factor:

Multiply through by the IF (or use $\frac{d}{dx}(y\cdot\text{IF}) = Q\cdot\text{IF}$):

Integrate both sides:

General solution:

Angle between $\vec a = \hat i + 2\hat j + 3\hat k$ and $\vec b = 3\hat i - 2\hat j + \hat k$

Dot product:

Magnitudes:

Angle $\theta$ from $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$:

Partial derivatives of $u = x^2 + y^2$

With respect to $x$ (treat $y$ as constant):

With respect to $y$ (treat $x$ as constant):

Hence $\dfrac{\partial u}{\partial x} = 2x$ and $\dfrac{\partial u}{\partial y} = 2y$.

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

Inner integral (integrate w.r.t. $y$ from $0$ to $2$, $x$ constant):

Outer integral (integrate w.r.t. $x$ from $0$ to $1$):