BSc CSIT (TU) Science Mathematics I (BSc CSIT, MTH112) Question Paper 2079 Nepal

Q: Where can I find the BSc CSIT (TU) Mathematics I (BSc CSIT, MTH112) question paper 2079?

The full BSc CSIT (TU) Mathematics I (BSc CSIT, MTH112) 2079 (regular) question paper is available free on Kekkei. You can read every question online and attempt the paper under timed exam conditions.

Q: Does the Mathematics I (BSc CSIT, MTH112) 2079 paper come with solutions?

Yes. Every question on this Mathematics I (BSc CSIT, MTH112) past paper includes a step-by-step solution, plus instant AI feedback when you attempt it on Kekkei.

Q: How many marks is the BSc CSIT (TU) Mathematics I (BSc CSIT, MTH112) 2079 paper?

The BSc CSIT (TU) Mathematics I (BSc CSIT, MTH112) 2079 paper carries 60 full marks and is meant to be completed in 180 minutes, across 12 questions.

Q: Is practising this Mathematics I (BSc CSIT, MTH112) past paper free?

Yes — reading and attempting this Mathematics I (BSc CSIT, MTH112) past paper on Kekkei is completely free.

Question

1long10 marks

State and prove the mean value theorem (Lagrange's). Verify it for (f(x) = \sqrt{x}) on [1,4].

mean-value-theorem

Answer 1

Lagrange's Mean Value Theorem (MVT)

Statement. If a function $f$ is

continuous on the closed interval $[a,b]$ , and
differentiable on the open interval $(a,b)$ ,

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

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

Proof (using Rolle's Theorem)

Define an auxiliary function

\phi(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}\,(x - a).

$\phi$ is continuous on $[a,b]$ (sum/difference of continuous functions).
$\phi$ is differentiable on $(a,b)$ .
At the endpoints:

\phi(a) = f(a) - f(a) - 0 = 0,

\phi(b) = f(b) - f(a) - \frac{f(b)-f(a)}{b-a}(b-a) = 0.

Since $\phi(a) = \phi(b) = 0$ , by Rolle's Theorem there exists $c \in (a,b)$ with $\phi'(c) = 0$ . Now

\phi'(x) = f'(x) - \frac{f(b) - f(a)}{b - a},

so $\phi'(c) = 0$ gives

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

Verification for $f(x) = \sqrt{x}$ on $[1,4]$

$f(x) = \sqrt{x}$ is continuous on $[1,4]$ and differentiable on $(1,4)$ , so MVT applies.

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

Also $f'(x) = \dfrac{1}{2\sqrt{x}}$ . Set $f'(c) = \tfrac{1}{3}$ :

\frac{1}{2\sqrt{c}} = \frac{1}{3} \;\Rightarrow\; 2\sqrt{c} = 3 \;\Rightarrow\; \sqrt{c} = \frac{3}{2} \;\Rightarrow\; c = \frac{9}{4} = 2.25.

Since $c = 2.25 \in (1,4)$ , the theorem is verified.

Answer 2

Evaluate $\displaystyle \int \frac{dx}{(x^2 + a^2)^2}$

Use the substitution $x = a\tan\theta$ , so that $dx = a\sec^2\theta\,d\theta$ and

x^2 + a^2 = a^2\tan^2\theta + a^2 = a^2\sec^2\theta.

Then

\int \frac{dx}{(x^2+a^2)^2} = \int \frac{a\sec^2\theta\,d\theta}{(a^2\sec^2\theta)^2} = \int \frac{a\sec^2\theta}{a^4\sec^4\theta}\,d\theta = \frac{1}{a^3}\int \cos^2\theta\,d\theta.

Using $\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}$ :

= \frac{1}{a^3}\cdot \frac{1}{2}\left(\theta + \frac{\sin 2\theta}{2}\right) = \frac{1}{2a^3}\left(\theta + \frac{\sin 2\theta}{2}\right).

Now convert back. Since $\tan\theta = \dfrac{x}{a}$ , we have $\theta = \tan^{-1}\dfrac{x}{a}$ , and

\sin\theta = \frac{x}{\sqrt{x^2+a^2}}, \quad \cos\theta = \frac{a}{\sqrt{x^2+a^2}},

\sin 2\theta = 2\sin\theta\cos\theta = \frac{2ax}{x^2 + a^2}.

Therefore

\boxed{\int \frac{dx}{(x^2+a^2)^2} = \frac{1}{2a^3}\tan^{-1}\frac{x}{a} + \frac{x}{2a^2(x^2+a^2)} + C.}

Answer 3

Solve $\dfrac{d^2y}{dx^2} + 4y = \cos 2x$

Complementary function (CF)

Auxiliary equation: $m^2 + 4 = 0 \Rightarrow m = \pm 2i$ . Hence

y_c = C_1\cos 2x + C_2\sin 2x.

Particular integral (PI)

y_p = \frac{1}{D^2 + 4}\cos 2x.

The operator $D^2 + 4$ applied to $\cos 2x$ gives $-2^2 + 4 = 0$ (failure case, since $\cos 2x$ is part of the CF). Use the standard result for resonance:

\frac{1}{D^2 + a^2}\cos ax = \frac{x}{2a}\sin ax.

With $a = 2$ :

y_p = \frac{x}{2\cdot 2}\sin 2x = \frac{x}{4}\sin 2x.

General solution

\boxed{y = C_1\cos 2x + C_2\sin 2x + \frac{x}{4}\sin 2x.}

Answer 4

Evaluate $\displaystyle\lim_{x \to 1}\frac{x^n - 1}{x - 1}$

This is a $\tfrac{0}{0}$ form. Using the standard limit $\displaystyle\lim_{x\to a}\frac{x^n - a^n}{x - a} = n a^{n-1}$ with $a = 1$ :

\lim_{x \to 1}\frac{x^n - 1}{x - 1} = n\cdot 1^{\,n-1} = n.

(Equivalently, factor $x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + 1)$ , cancel $(x-1)$ , and put $x=1$ in the $n$ -term sum to get $n$ .)

\boxed{\lim_{x \to 1}\frac{x^n - 1}{x - 1} = n.}

Answer 5

If $y = e^{m\sin^{-1}x}$ , find $\dfrac{dy}{dx}$

Differentiate using the chain rule. Let $u = m\sin^{-1}x$ , so $y = e^{u}$ and $\dfrac{dy}{dx} = e^{u}\dfrac{du}{dx}$ .

Since $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$ ,

\frac{du}{dx} = \frac{m}{\sqrt{1 - x^2}}.

Therefore

\boxed{\frac{dy}{dx} = \frac{m\,e^{m\sin^{-1}x}}{\sqrt{1 - x^2}} = \frac{m\,y}{\sqrt{1 - x^2}}.}

Answer 6

Maclaurin expansion of $\log(1 + x)$

The Maclaurin series is $f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \cdots$

For $f(x) = \log(1+x)$ :

f(0)=0,\; f'(x)=\tfrac{1}{1+x}\Rightarrow f'(0)=1,\; f''(x)=-\tfrac{1}{(1+x)^2}\Rightarrow f''(0)=-1,

f'''(x)=\tfrac{2}{(1+x)^3}\Rightarrow f'''(0)=2,\quad f^{(4)}(0)=-6,\ \dots

In general $f^{(n)}(0) = (-1)^{n-1}(n-1)!$ , so the $n$ -th term is $\dfrac{(-1)^{n-1}(n-1)!}{n!}x^n = \dfrac{(-1)^{n-1}}{n}x^n$ . Hence

\boxed{\log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^n,}

valid for $-1 < x \le 1$ .

Answer 7

Asymptotes of $y = \dfrac{2x^2 - 1}{x^2 - 4}$

Vertical asymptotes

Set the denominator to zero: $x^2 - 4 = 0 \Rightarrow x = \pm 2$ . The numerator is nonzero there, so

x = 2 \quad\text{and}\quad x = -2

are the vertical asymptotes.

Horizontal asymptote

The degrees of numerator and denominator are equal, so

\lim_{x\to\pm\infty} y = \frac{2}{1} = 2 \;\Rightarrow\; y = 2.

(There is no oblique asymptote since the degree of the numerator does not exceed that of the denominator.)

\boxed{\text{Asymptotes: } x = 2,\; x = -2,\; y = 2.}

Answer 8

Evaluate $\displaystyle\int \frac{1}{1 + \cos x}\,dx$

Use the half-angle identity $1 + \cos x = 2\cos^2\dfrac{x}{2}$ :

\int \frac{dx}{1 + \cos x} = \int \frac{dx}{2\cos^2\frac{x}{2}} = \frac{1}{2}\int \sec^2\frac{x}{2}\,dx.

Since $\displaystyle\int \sec^2\frac{x}{2}\,dx = 2\tan\frac{x}{2}$ ,

= \frac{1}{2}\cdot 2\tan\frac{x}{2} + C.

\boxed{\int \frac{dx}{1 + \cos x} = \tan\frac{x}{2} + C.}

Answer 9

Solve $(1 + x^2)\dfrac{dy}{dx} + 2xy = 4x^2$

Notice the left side is an exact derivative. Dividing through is optional because

\frac{d}{dx}\big[(1+x^2)y\big] = (1+x^2)\frac{dy}{dx} + 2xy.

So the equation becomes

\frac{d}{dx}\big[(1+x^2)y\big] = 4x^2.

Integrating both sides:

(1+x^2)y = \int 4x^2\,dx = \frac{4x^3}{3} + C.

\boxed{y = \frac{1}{1+x^2}\left(\frac{4x^3}{3} + C\right).}

(Equivalently, in standard linear form $\dfrac{dy}{dx} + \dfrac{2x}{1+x^2}y = \dfrac{4x^2}{1+x^2}$ the integrating factor is $e^{\int \frac{2x}{1+x^2}dx} = 1+x^2$ , giving the same result.)

Answer 10

Given $\vec{a}\times\vec{b} = \vec{0}$ and $\vec{a}\cdot\vec{b} = 0$

$\vec{a}\times\vec{b} = \vec{0}$ means $|\vec a||\vec b|\sin\theta = 0$ , so $\vec a$ and $\vec b$ are parallel (or one is the zero vector).
$\vec{a}\cdot\vec{b} = 0$ means $|\vec a||\vec b|\cos\theta = 0$ , so $\vec a$ and $\vec b$ are perpendicular (or one is the zero vector).

Two nonzero vectors cannot be simultaneously parallel and perpendicular. Hence both conditions hold together only if at least one of $\vec a$ , $\vec b$ is the zero vector:

\boxed{\vec a = \vec 0 \quad\text{or}\quad \vec b = \vec 0.}

Answer 11

Find $\dfrac{\partial u}{\partial x}$ if $u = x^y$

When taking the partial derivative with respect to $x$ , treat $y$ as a constant. Then $u = x^y$ is a power function of $x$ , so by the power rule

\boxed{\frac{\partial u}{\partial x} = y\,x^{\,y-1}.}

(For contrast, $\dfrac{\partial u}{\partial y} = x^y \log x$ , where $x$ is held constant.)

Answer 12

Change the order of integration in $\displaystyle\int_0^1\!\int_x^1 f(x,y)\,dy\,dx$

Identify the region

For the given limits: $x$ runs from $0$ to $1$ , and for each $x$ , $y$ runs from $y = x$ up to $y = 1$ . The region $R$ is therefore the triangle bounded by

y = x,\quad y = 1,\quad x = 0,

i.e. $\{(x,y): 0 \le x \le y,\; 0 \le y \le 1\}$ — the triangle with vertices $(0,0)$ , $(0,1)$ and $(1,1)$ .

Reverse the order

Describing the same region with $y$ as the outer variable: $y$ ranges from $0$ to $1$ , and for each fixed $y$ , $x$ ranges from $0$ to $y$ (since $y \ge x$ ). Hence

\boxed{\int_0^1\!\int_x^1 f(x,y)\,dy\,dx = \int_0^1\!\int_0^y f(x,y)\,dx\,dy.}

BSc CSIT (TU) Science Mathematics I (BSc CSIT, MTH112) Question Paper 2079 Nepal

Section A: Long Answer Questions

Lagrange's Mean Value Theorem (MVT)

Proof (using Rolle's Theorem)

Verification for $f(x) = \sqrt{x}$ on $[1,4]$

Evaluate $\displaystyle \int \frac{dx}{(x^2 + a^2)^2}$

Solve $\dfrac{d^2y}{dx^2} + 4y = \cos 2x$

Complementary function (CF)

Particular integral (PI)

General solution

Section B: Short Answer Questions

Evaluate $\displaystyle\lim_{x \to 1}\frac{x^n - 1}{x - 1}$

If $y = e^{m\sin^{-1}x}$ , find $\dfrac{dy}{dx}$

Maclaurin expansion of $\log(1 + x)$

Asymptotes of $y = \dfrac{2x^2 - 1}{x^2 - 4}$

Vertical asymptotes

Horizontal asymptote

Evaluate $\displaystyle\int \frac{1}{1 + \cos x}\,dx$

Solve $(1 + x^2)\dfrac{dy}{dx} + 2xy = 4x^2$

Given $\vec{a}\times\vec{b} = \vec{0}$ and $\vec{a}\cdot\vec{b} = 0$

Find $\dfrac{\partial u}{\partial x}$ if $u = x^y$

Change the order of integration in $\displaystyle\int_0^1\!\int_x^1 f(x,y)\,dy\,dx$

Identify the region

Reverse the order

Frequently asked questions

Section A: Long Answer Questions

Lagrange's Mean Value Theorem (MVT)

Proof (using Rolle's Theorem)

Verification for f(x)=xf(x) = \sqrt{x}f(x)=x​ on [1,4][1,4][1,4]

Evaluate ∫dx(x2+a2)2\displaystyle \int \frac{dx}{(x^2 + a^2)^2}∫(x2+a2)2dx​

Solve d2ydx2+4y=cos⁡2x\dfrac{d^2y}{dx^2} + 4y = \cos 2xdx2d2y​+4y=cos2x

Complementary function (CF)

Particular integral (PI)

General solution

Section B: Short Answer Questions

Evaluate lim⁡x→1xn−1x−1\displaystyle\lim_{x \to 1}\frac{x^n - 1}{x - 1}x→1lim​x−1xn−1​

If y=emsin⁡−1xy = e^{m\sin^{-1}x}y=emsin−1x, find dydx\dfrac{dy}{dx}dxdy​

Maclaurin expansion of log⁡(1+x)\log(1 + x)log(1+x)

Asymptotes of y=2x2−1x2−4y = \dfrac{2x^2 - 1}{x^2 - 4}y=x2−42x2−1​

Vertical asymptotes

Horizontal asymptote

Evaluate ∫11+cos⁡x dx\displaystyle\int \frac{1}{1 + \cos x}\,dx∫1+cosx1​dx

Solve (1+x2)dydx+2xy=4x2(1 + x^2)\dfrac{dy}{dx} + 2xy = 4x^2(1+x2)dxdy​+2xy=4x2

Given a⃗×b⃗=0⃗\vec{a}\times\vec{b} = \vec{0}a×b=0 and a⃗⋅b⃗=0\vec{a}\cdot\vec{b} = 0a⋅b=0

Find ∂u∂x\dfrac{\partial u}{\partial x}∂x∂u​ if u=xyu = x^yu=xy

Change the order of integration in ∫01 ⁣∫x1f(x,y) dy dx\displaystyle\int_0^1\!\int_x^1 f(x,y)\,dy\,dx∫01​∫x1​f(x,y)dydx

Identify the region

Reverse the order

Frequently asked questions

Verification for $f(x) = \sqrt{x}$ on $[1,4]$

Evaluate $\displaystyle \int \frac{dx}{(x^2 + a^2)^2}$

Solve $\dfrac{d^2y}{dx^2} + 4y = \cos 2x$

Evaluate $\displaystyle\lim_{x \to 1}\frac{x^n - 1}{x - 1}$

If $y = e^{m\sin^{-1}x}$ , find $\dfrac{dy}{dx}$

Maclaurin expansion of $\log(1 + x)$

Asymptotes of $y = \dfrac{2x^2 - 1}{x^2 - 4}$

Evaluate $\displaystyle\int \frac{1}{1 + \cos x}\,dx$

Solve $(1 + x^2)\dfrac{dy}{dx} + 2xy = 4x^2$

Given $\vec{a}\times\vec{b} = \vec{0}$ and $\vec{a}\cdot\vec{b} = 0$

Find $\dfrac{\partial u}{\partial x}$ if $u = x^y$

Change the order of integration in $\displaystyle\int_0^1\!\int_x^1 f(x,y)\,dy\,dx$