BSc CSIT (TU) Science Mathematics I (BSc CSIT, MTH112) Question Paper 2080 Nepal

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Question

1long10 marks

Define the derivative of a function from first principles. Find the derivative of (\sin x) using the first-principle method.

differentiation

Answer 1

Derivative from First Principles

The derivative of a function $f(x)$ at a point $x$ is defined as the limit of the difference quotient:

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

provided this limit exists. Geometrically it gives the slope of the tangent to the curve $y=f(x)$ at the point.

Derivative of $\sin x$ by First Principles

Let $f(x) = \sin x$ . Then

f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}

Using the identity $\sin C - \sin D = 2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$ with $C = x+h$ , $D = x$ :

\sin(x+h) - \sin x = 2\cos\!\left(x + \tfrac{h}{2}\right)\sin\!\left(\tfrac{h}{2}\right)

Therefore

f'(x) = \lim_{h \to 0} \frac{2\cos\!\left(x + \tfrac{h}{2}\right)\sin\!\left(\tfrac{h}{2}\right)}{h} = \lim_{h \to 0} \cos\!\left(x + \tfrac{h}{2}\right)\cdot \frac{\sin(h/2)}{h/2}

As $h \to 0$ , $\cos\!\left(x + \tfrac{h}{2}\right) \to \cos x$ and using the standard limit $\displaystyle\lim_{t\to 0}\frac{\sin t}{t} = 1$ (with $t = h/2$ ):

f'(x) = \cos x \cdot 1 = \cos x

\boxed{\dfrac{d}{dx}(\sin x) = \cos x}

Answer 2

Reduction Formula for $\int \sin^n x\, dx$

Let $I_n = \int \sin^n x \, dx = \int \sin^{n-1} x \cdot \sin x \, dx$ .

Integrate by parts with $u = \sin^{n-1}x$ and $dv = \sin x\, dx$ , so $du = (n-1)\sin^{n-2}x\cos x\,dx$ and $v = -\cos x$ :

I_n = -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\cos^2 x \, dx

Write $\cos^2 x = 1 - \sin^2 x$ :

I_n = -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\,dx - (n-1)\int \sin^n x\,dx

I_n = -\sin^{n-1}x\cos x + (n-1)I_{n-2} - (n-1)I_n

Bringing $I_n$ terms together, $I_n[1 + (n-1)] = -\sin^{n-1}x\cos x + (n-1)I_{n-2}$ , i.e. $n I_n = -\sin^{n-1}x\cos x + (n-1)I_{n-2}$ :

\boxed{\,I_n = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}I_{n-2}\,}

Definite Integral Form

For the definite integral over $[0,\pi/2]$ the boundary term $\left[-\dfrac{\sin^{n-1}x\cos x}{n}\right]_0^{\pi/2} = 0$ (since $\cos(\pi/2)=0$ and $\sin 0 = 0$ ), so

J_n = \int_0^{\pi/2}\sin^n x\,dx = \frac{n-1}{n}\,J_{n-2}

Evaluation of $\int_0^{\pi/2}\sin^5 x\,dx$

With $n=5$ (odd), recurse down to $J_1$ :

J_5 = \frac{4}{5}J_3 = \frac{4}{5}\cdot\frac{2}{3}J_1

and $J_1 = \int_0^{\pi/2}\sin x\,dx = [-\cos x]_0^{\pi/2} = 1$ . Hence

J_5 = \frac{4}{5}\cdot\frac{2}{3}\cdot 1 = \frac{8}{15}

\boxed{\int_0^{\pi/2}\sin^5 x\,dx = \frac{8}{15}}

Answer 3

Solving the Homogeneous Equation

\frac{dy}{dx} = \frac{y}{x} + \tan\frac{y}{x}

The right side is a function of $y/x$ , so use the substitution

y = vx \quad\Rightarrow\quad \frac{dy}{dx} = v + x\frac{dv}{dx}

Substituting (with $y/x = v$ ):

v + x\frac{dv}{dx} = v + \tan v

x\frac{dv}{dx} = \tan v

Separate variables:

\frac{dv}{\tan v} = \frac{dx}{x} \quad\Rightarrow\quad \cot v \, dv = \frac{dx}{x}

Integrate both sides:

\int \cot v\, dv = \int \frac{dx}{x}

\log|\sin v| = \log|x| + \log C

\log|\sin v| = \log|Cx|

Therefore $\sin v = Cx$ . Replacing $v = y/x$ :

\boxed{\sin\!\left(\frac{y}{x}\right) = Cx}

where $C$ is the arbitrary constant of integration.

Answer 4

Evaluation

\lim_{x \to 0}\frac{\log(1+x)}{x}

This is of the indeterminate form $\frac{0}{0}$ . Using the series expansion $\log(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$ :

\frac{\log(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - \cdots \to 1 \quad\text{as } x\to 0

Alternatively, by L'Hospital's rule, $\displaystyle\lim_{x\to 0}\frac{1/(1+x)}{1} = 1$ .

\boxed{\lim_{x \to 0}\frac{\log(1+x)}{x} = 1}

Answer 5

$n$ -th Derivative of $y = x^n \log x$

Differentiate once:

y_1 = n x^{n-1}\log x + x^n\cdot\frac{1}{x} = n x^{n-1}\log x + x^{n-1}

Multiply by $x$ : $x y_1 = n x^n \log x + x^n = n y + x^n$ , i.e.

x y_1 = n y + x^n.

A standard result (provable by induction) gives the $n$ -th derivative:

\boxed{y_n = \frac{d^n}{dx^n}\big(x^n\log x\big) = n!\left(\log x + 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right)}

That is, $y_n = n!\left(\log x + \sum_{k=1}^{n}\frac{1}{k}\right)$ .

Check (n = 1): $y_1 = 1!(\log x + 1) = \log x + 1$ , which matches $y_1$ above for the case $n=1$ ( $y = x\log x$ ). The result follows by repeatedly differentiating and collecting the harmonic terms.

Answer 6

Cauchy's Mean Value Theorem (Statement)

If $f(x)$ and $g(x)$ are continuous on $[a,b]$ , differentiable on $(a,b)$ , and $g'(x)\neq 0$ on $(a,b)$ , then there exists at least one point $c\in(a,b)$ such that

\frac{f'(c)}{g'(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}.

Verification for $f(x)=x^2,\ g(x)=x$ on $[1,2]$

Both are polynomials, hence continuous on $[1,2]$ and differentiable on $(1,2)$ ; also $g'(x)=1\neq 0$ . So the theorem applies.

Derivatives: $f'(x)=2x,\ g'(x)=1$ , so $\dfrac{f'(c)}{g'(c)} = 2c$ .

Right-hand side:

\frac{f(2)-f(1)}{g(2)-g(1)} = \frac{4-1}{2-1} = 3.

Set equal: $2c = 3 \Rightarrow c = \dfrac{3}{2}$ .

Since $c = 1.5 \in (1,2)$ , the theorem is verified.

\boxed{c = \tfrac{3}{2} \in (1,2)}

Answer 7

Maxima and Minima of $f(x)=\sin x + \cos x$

$f'(x) = \cos x - \sin x$ . Set $f'(x)=0$ :

\cos x = \sin x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4},\ \frac{5\pi}{4},\ \dots

Second derivative: $f''(x) = -\sin x - \cos x = -(\sin x + \cos x) = -f(x)$ .

At $x=\pi/4$ : $f(\pi/4) = \tfrac{1}{\sqrt2}+\tfrac{1}{\sqrt2} = \sqrt2$ and $f''(\pi/4) = -\sqrt2 < 0$ → maximum.

At $x=5\pi/4$ : $f(5\pi/4) = -\tfrac{1}{\sqrt2}-\tfrac{1}{\sqrt2} = -\sqrt2$ and $f''(5\pi/4) = \sqrt2 > 0$ → minimum.

\boxed{\text{Maximum value} = \sqrt2,\qquad \text{Minimum value} = -\sqrt2}

(Equivalently, $f(x)=\sqrt2\sin\!\big(x+\tfrac{\pi}{4}\big)$ , whose range is $[-\sqrt2,\ \sqrt2]$ .)

Answer 8

Evaluation by Partial Fractions

Let $t = x^2$ . Resolve $\dfrac{x^2}{(x^2+1)(x^2+4)} = \dfrac{t}{(t+1)(t+4)}$ :

\frac{t}{(t+1)(t+4)} = \frac{A}{t+1} + \frac{B}{t+4}

Then $t = A(t+4) + B(t+1)$ . Put $t=-1$ : $-1 = 3A \Rightarrow A = -\tfrac{1}{3}$ . Put $t=-4$ : $-4 = -3B \Rightarrow B = \tfrac{4}{3}$ .

So in terms of $x$ :

\frac{x^2}{(x^2+1)(x^2+4)} = -\frac{1}{3}\cdot\frac{1}{x^2+1} + \frac{4}{3}\cdot\frac{1}{x^2+4}

Integrate using $\int\dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a}$ :

\int \frac{x^2}{(x^2+1)(x^2+4)}dx = -\frac{1}{3}\tan^{-1}x + \frac{4}{3}\cdot\frac{1}{2}\tan^{-1}\frac{x}{2} + C

\boxed{\;= -\frac{1}{3}\tan^{-1}x + \frac{2}{3}\tan^{-1}\frac{x}{2} + C\;}

Answer 9

Solving the Linear ODE

\frac{dy}{dx} + y\cot x = \cos x

This is linear of the form $\dfrac{dy}{dx} + Py = Q$ with $P=\cot x$ , $Q=\cos x$ .

Integrating factor:

\text{I.F.} = e^{\int \cot x\,dx} = e^{\log|\sin x|} = \sin x

Multiply through by $\sin x$ :

\frac{d}{dx}(y\sin x) = \cos x\sin x = \tfrac{1}{2}\sin 2x

Integrate:

y\sin x = \int \tfrac{1}{2}\sin 2x\, dx = -\frac{1}{4}\cos 2x + C

Therefore

\boxed{\,y\sin x = -\frac{1}{4}\cos 2x + C\,}

or equivalently $y = \dfrac{C - \tfrac14\cos 2x}{\sin x}$ .

Answer 10

Work Done by a Force

The work done is the dot product of the force and the displacement:

W = \vec{F}\cdot\vec{d}

With $\vec{F} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{d} = \hat{i} + \hat{j} + \hat{k}$ :

W = (2)(1) + (3)(1) + (1)(1) = 2 + 3 + 1 = 6

\boxed{W = 6\ \text{units}}

Answer 11

Partial Derivatives of $u = \tan^{-1}(y/x)$

Let $u = \tan^{-1}\dfrac{y}{x}$ . Using $\dfrac{d}{dt}\tan^{-1}t = \dfrac{1}{1+t^2}$ with $t = y/x$ and $1 + (y/x)^2 = \dfrac{x^2+y^2}{x^2}$ :

With respect to $x$ ( $\partial(y/x)/\partial x = -y/x^2$ ):

\frac{\partial u}{\partial x} = \frac{1}{1+\frac{y^2}{x^2}}\cdot\left(-\frac{y}{x^2}\right) = \frac{x^2}{x^2+y^2}\cdot\left(-\frac{y}{x^2}\right) = \boxed{-\frac{y}{x^2+y^2}}

With respect to $y$ ( $\partial(y/x)/\partial y = 1/x$ ):

\frac{\partial u}{\partial y} = \frac{1}{1+\frac{y^2}{x^2}}\cdot\frac{1}{x} = \frac{x^2}{x^2+y^2}\cdot\frac{1}{x} = \boxed{\frac{x}{x^2+y^2}}

Answer 12

Evaluating the Double Integral

I = \int_0^1 \int_0^{\sqrt{1-x^2}} dy\, dx

Inner integral (over $y$ ):

\int_0^{\sqrt{1-x^2}} dy = \sqrt{1-x^2}

Outer integral:

I = \int_0^1 \sqrt{1-x^2}\, dx

Using $\displaystyle\int \sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$ with $a=1$ :

I = \left[\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}x\right]_0^1 = \left(0 + \frac{1}{2}\cdot\frac{\pi}{2}\right) - 0 = \frac{\pi}{4}

\boxed{I = \frac{\pi}{4}}

This is expected, since the region is the quarter of the unit disk in the first quadrant, whose area is $\tfrac{1}{4}\pi(1)^2 = \tfrac{\pi}{4}$ .

BSc CSIT (TU) Science Mathematics I (BSc CSIT, MTH112) Question Paper 2080 Nepal

Section A: Long Answer Questions

Derivative from First Principles

Derivative of $\sin x$ by First Principles

Reduction Formula for $\int \sin^n x\, dx$

Definite Integral Form

Evaluation of $\int_0^{\pi/2}\sin^5 x\,dx$

Solving the Homogeneous Equation

Section B: Short Answer Questions

Evaluation

$n$ -th Derivative of $y = x^n \log x$

Cauchy's Mean Value Theorem (Statement)

Verification for $f(x)=x^2,\ g(x)=x$ on $[1,2]$

Maxima and Minima of $f(x)=\sin x + \cos x$

Evaluation by Partial Fractions

Solving the Linear ODE

Work Done by a Force

Partial Derivatives of $u = \tan^{-1}(y/x)$

Evaluating the Double Integral

Frequently asked questions

Section A: Long Answer Questions

Derivative from First Principles

Derivative of sin⁡x\sin xsinx by First Principles

Reduction Formula for ∫sin⁡nx dx\int \sin^n x\, dx∫sinnxdx

Definite Integral Form

Evaluation of ∫0π/2sin⁡5x dx\int_0^{\pi/2}\sin^5 x\,dx∫0π/2​sin5xdx

Solving the Homogeneous Equation

Section B: Short Answer Questions

Evaluation

nnn-th Derivative of y=xnlog⁡xy = x^n \log xy=xnlogx

Cauchy's Mean Value Theorem (Statement)

Verification for f(x)=x2, g(x)=xf(x)=x^2,\ g(x)=xf(x)=x2, g(x)=x on [1,2][1,2][1,2]

Maxima and Minima of f(x)=sin⁡x+cos⁡xf(x)=\sin x + \cos xf(x)=sinx+cosx

Evaluation by Partial Fractions

Solving the Linear ODE

Work Done by a Force

Partial Derivatives of u=tan⁡−1(y/x)u = \tan^{-1}(y/x)u=tan−1(y/x)

Evaluating the Double Integral

Frequently asked questions

Derivative of $\sin x$ by First Principles

Reduction Formula for $\int \sin^n x\, dx$

Evaluation of $\int_0^{\pi/2}\sin^5 x\,dx$

$n$ -th Derivative of $y = x^n \log x$

Verification for $f(x)=x^2,\ g(x)=x$ on $[1,2]$

Maxima and Minima of $f(x)=\sin x + \cos x$

Partial Derivatives of $u = \tan^{-1}(y/x)$