BSc CSIT (TU) Science Mathematics I (BSc CSIT, MTH112) Question Paper 2078 Nepal

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Question

1long10 marks

Define limit and continuity. Evaluate (\lim_{x \to 0}\frac{e^x - 1}{x}) and (\lim_{x \to 0}\frac{a^x - 1}{x}).

limits

Answer 1

Definitions

Limit. A function $f(x)$ has limit $L$ as $x \to a$ , written $\lim_{x\to a} f(x) = L$ , if for every $\varepsilon > 0$ there exists $\delta > 0$ such that $0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon$ . Intuitively, $f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close (but not equal) to $a$ .

Continuity. A function $f$ is continuous at $x = a$ if:

$f(a)$ is defined,
$\lim_{x\to a} f(x)$ exists, and
$\lim_{x\to a} f(x) = f(a)$ .

A function is continuous on an interval if it is continuous at every point of the interval.

Evaluation 1: $\displaystyle\lim_{x\to 0}\frac{e^x - 1}{x}$

Using the expansion $e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$ :

\frac{e^x - 1}{x} = \frac{x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{x} = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots

Taking $x \to 0$ , all terms after the first vanish:

\boxed{\lim_{x\to 0}\frac{e^x - 1}{x} = 1.}

Evaluation 2: $\displaystyle\lim_{x\to 0}\frac{a^x - 1}{x}$ ( $a>0$ )

Write $a^x = e^{x\ln a}$ . Put $t = x\ln a$ (so $t\to 0$ as $x\to 0$ ):

\frac{a^x - 1}{x} = \frac{e^{x\ln a} - 1}{x} = \ln a \cdot \frac{e^{t} - 1}{t}.

Using the previous result $\dfrac{e^t-1}{t}\to 1$ :

\boxed{\lim_{x\to 0}\frac{a^x - 1}{x} = \ln a.}

(As a check, putting $a = e$ gives $\ln e = 1$ , consistent with the first limit.)

Answer 2

To prove: $(1-x^2)y_{n+2} - (2n+1)x\,y_{n+1} - n^2 y_n = 0$ , where $y = (\sin^{-1}x)^2$ .

Step 1 — First derivative.

y_1 = 2\sin^{-1}x \cdot \frac{1}{\sqrt{1-x^2}}.

Therefore $\sqrt{1-x^2}\,y_1 = 2\sin^{-1}x$ , and squaring,

(1-x^2)y_1^{\,2} = 4(\sin^{-1}x)^2 = 4y. \qquad (\ast)

Step 2 — Differentiate $(\ast)$ . Differentiating both sides with respect to $x$ :

(1-x^2)\cdot 2y_1 y_2 + (-2x)y_1^{\,2} = 4y_1.

Dividing throughout by $2y_1$ (assuming $y_1 \neq 0$ ):

(1-x^2)y_2 - x\,y_1 = 2. \qquad (\ast\ast)

Step 3 — Apply Leibnitz's theorem. Differentiate $(\ast\ast)$ $n$ times using Leibnitz's rule. For the product $(1-x^2)y_2$ :

D^n\big[(1-x^2)y_2\big] = (1-x^2)y_{n+2} + n(-2x)y_{n+1} + \frac{n(n-1)}{2}(-2)y_n.

For the product $x\,y_1$ :

D^n\big[x\,y_1\big] = x\,y_{n+1} + n\,y_n.

The right side $2$ is constant, so its $n$ -th derivative (for $n\ge 1$ ) is $0$ .

Step 4 — Combine.

(1-x^2)y_{n+2} - 2nx\,y_{n+1} - n(n-1)y_n - \big(x\,y_{n+1} + n\,y_n\big) = 0.

Grouping like terms:

(1-x^2)y_{n+2} - (2n+1)x\,y_{n+1} - \big(n(n-1)+n\big)y_n = 0.

Since $n(n-1)+n = n^2$ ,

\boxed{(1-x^2)y_{n+2} - (2n+1)x\,y_{n+1} - n^2 y_n = 0.}

Hence proved.

Answer 3

Area enclosed between $y = x^2$ and $y = x$ .

Step 1 — Points of intersection. Set $x^2 = x \Rightarrow x^2 - x = 0 \Rightarrow x(x-1)=0$ , so $x = 0$ and $x = 1$ . The curves meet at $(0,0)$ and $(1,1)$ .

Step 2 — Identify the upper curve. On $0 < x < 1$ , take $x = \tfrac12$ : line gives $y = 0.5$ , parabola gives $y = 0.25$ . So the line $y = x$ lies above the parabola $y = x^2$ on $[0,1]$ .

Step 3 — Set up the integral.

A = \int_0^1 \big(x - x^2\big)\,dx.

Step 4 — Integrate.

A = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{3-2}{6} = \frac{1}{6}.

\boxed{A = \frac{1}{6}\ \text{square units}.}

Answer 4

Use $\dfrac{\sin\theta}{\theta}\to 1$ as $\theta\to 0$ . Multiply and divide to introduce the angles:

\frac{\sin 3x}{\sin 5x} = \frac{\dfrac{\sin 3x}{3x}\cdot 3x}{\dfrac{\sin 5x}{5x}\cdot 5x} = \frac{\dfrac{\sin 3x}{3x}}{\dfrac{\sin 5x}{5x}}\cdot\frac{3x}{5x}.

As $x\to 0$ , both fractions tend to $1$ , leaving

\lim_{x\to 0}\frac{\sin 3x}{\sin 5x} = 1\cdot 1\cdot\frac{3}{5} = \boxed{\dfrac{3}{5}}.

Answer 5

Let $y = \tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)$ . Substitute $x = \tan\theta$ , so $\theta = \tan^{-1}x$ . Then

\frac{2x}{1-x^2} = \frac{2\tan\theta}{1-\tan^2\theta} = \tan 2\theta.

Hence

y = \tan^{-1}(\tan 2\theta) = 2\theta = 2\tan^{-1}x

(valid for $-1 < x < 1$ ). Differentiating,

\frac{dy}{dx} = 2\cdot\frac{1}{1+x^2} = \boxed{\dfrac{2}{1+x^2}}.

Answer 6

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 $f(x) = \sin x$ on $[0,\pi]$

Continuity: $\sin x$ is continuous everywhere, hence on $[0,\pi]$ . ✓
Differentiability: $\sin x$ is differentiable everywhere, hence on $(0,\pi)$ . ✓
Equal endpoints: $f(0) = \sin 0 = 0$ and $f(\pi) = \sin\pi = 0$ , so $f(0)=f(\pi)$ . ✓

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

f'(x) = \cos x = 0 \implies x = \frac{\pi}{2}.

Since $c = \dfrac{\pi}{2}\in(0,\pi)$ , the theorem is verified. $\blacksquare$

Answer 7

For $y = x^3 - 6x^2 + 9x$ :

y' = 3x^2 - 12x + 9,\qquad y'' = 6x - 12.

Set $y'' = 0$ : $6x - 12 = 0 \Rightarrow x = 2$ .

Check sign change of $y''$ : for $x<2$ , $y''<0$ (concave down); for $x>2$ , $y''>0$ (concave up). Concavity changes, so $x = 2$ is a genuine point of inflection.

Find $y$ at $x = 2$ : $y = 8 - 24 + 18 = 2$ .

\boxed{\text{Point of inflection: } (2,\,2).}

Answer 8

Use the Wallis reduction formula for $\displaystyle\int_0^{\pi/2}\cos^n x\,dx$ with even $n$ :

\int_0^{\pi/2}\cos^n x\,dx = \frac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2}\cdot\frac{\pi}{2}.

For $n = 6$ :

\int_0^{\pi/2}\cos^6 x\,dx = \frac{5\cdot 3\cdot 1}{6\cdot 4\cdot 2}\cdot\frac{\pi}{2} = \frac{15}{48}\cdot\frac{\pi}{2} = \frac{5}{16}\cdot\frac{\pi}{2}.

\boxed{\int_0^{\pi/2}\cos^6 x\,dx = \dfrac{5\pi}{32}.}

Answer 9

This is a linear first-order ODE $\dfrac{dy}{dx} + P(x)y = Q(x)$ with $P = 2x$ , $Q = x$ .

Integrating factor:

\text{IF} = e^{\int 2x\,dx} = e^{x^2}.

Multiply through and integrate:

\frac{d}{dx}\big(y\,e^{x^2}\big) = x\,e^{x^2}.

Integrate the right side with $u = x^2$ , $du = 2x\,dx$ :

y\,e^{x^2} = \int x e^{x^2}\,dx = \frac{1}{2}e^{x^2} + C.

Solve for $y$ :

\boxed{y = \frac{1}{2} + C\,e^{-x^2}.}

Answer 10

The (scalar) projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|}$ .

Dot product:

\vec a\cdot\vec b = (1)(2) + (2)(-1) + (-1)(3) = 2 - 2 - 3 = -3.

Magnitude of $\vec b$ :

|\vec b| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4+1+9} = \sqrt{14}.

Projection:

\boxed{\text{proj} = \frac{\vec a\cdot\vec b}{|\vec b|} = \frac{-3}{\sqrt{14}} = -\frac{3}{\sqrt{14}}\approx -0.802.}

The negative sign indicates the projection is directed opposite to $\vec b$ .

Answer 11

Let $z = f(u)$ where $u = \dfrac{x}{y}$ . By the chain rule:

\frac{\partial z}{\partial x} = f'(u)\,\frac{\partial u}{\partial x} = f'(u)\cdot\frac{1}{y},

\frac{\partial z}{\partial y} = f'(u)\,\frac{\partial u}{\partial y} = f'(u)\cdot\left(-\frac{x}{y^2}\right).

Now form the required combination:

x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = x\cdot\frac{f'(u)}{y} + y\cdot\left(-\frac{x f'(u)}{y^2}\right) = \frac{x f'(u)}{y} - \frac{x f'(u)}{y} = 0.

\boxed{x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = 0.}

Hence proved. (This is consistent with Euler's theorem, since $f(x/y)$ is a homogeneous function of degree $0$ .)

Answer 12

Evaluate the double integral (the limits are constants, so the order is immaterial):

I = \int_0^a\int_0^b xy\,dx\,dy.

Inner integral (over $x$ ):

\int_0^b xy\,dx = y\left[\frac{x^2}{2}\right]_0^b = \frac{b^2 y}{2}.

Outer integral (over $y$ ):

I = \int_0^a \frac{b^2 y}{2}\,dy = \frac{b^2}{2}\left[\frac{y^2}{2}\right]_0^a = \frac{b^2}{2}\cdot\frac{a^2}{2}.

\boxed{I = \frac{a^2 b^2}{4}.}

BSc CSIT (TU) Science Mathematics I (BSc CSIT, MTH112) Question Paper 2078 Nepal

Section A: Long Answer Questions

Definitions

Evaluation 1: $\displaystyle\lim_{x\to 0}\frac{e^x - 1}{x}$

Evaluation 2: $\displaystyle\lim_{x\to 0}\frac{a^x - 1}{x}$ ( $a>0$ )

To prove: $(1-x^2)y_{n+2} - (2n+1)x\,y_{n+1} - n^2 y_n = 0$ , where $y = (\sin^{-1}x)^2$ .

Area enclosed between $y = x^2$ and $y = x$ .

Section B: Short Answer Questions

Rolle's Theorem (statement)

Verification for $f(x) = \sin x$ on $[0,\pi]$

Frequently asked questions

Section A: Long Answer Questions

Definitions

Evaluation 1: lim⁡x→0ex−1x\displaystyle\lim_{x\to 0}\frac{e^x - 1}{x}x→0lim​xex−1​

Evaluation 2: lim⁡x→0ax−1x\displaystyle\lim_{x\to 0}\frac{a^x - 1}{x}x→0lim​xax−1​ (a>0a>0a>0)

To prove: (1−x2)yn+2−(2n+1)x yn+1−n2yn=0(1-x^2)y_{n+2} - (2n+1)x\,y_{n+1} - n^2 y_n = 0(1−x2)yn+2​−(2n+1)xyn+1​−n2yn​=0, where y=(sin⁡−1x)2y = (\sin^{-1}x)^2y=(sin−1x)2.

Area enclosed between y=x2y = x^2y=x2 and y=xy = xy=x.

Section B: Short Answer Questions

Rolle's Theorem (statement)

Verification for f(x)=sin⁡xf(x) = \sin xf(x)=sinx on [0,π][0,\pi][0,π]

Frequently asked questions

Evaluation 1: $\displaystyle\lim_{x\to 0}\frac{e^x - 1}{x}$

Evaluation 2: $\displaystyle\lim_{x\to 0}\frac{a^x - 1}{x}$ ( $a>0$ )

To prove: $(1-x^2)y_{n+2} - (2n+1)x\,y_{n+1} - n^2 y_n = 0$ , where $y = (\sin^{-1}x)^2$ .

Area enclosed between $y = x^2$ and $y = x$ .

Verification for $f(x) = \sin x$ on $[0,\pi]$