BSc CSIT (TU) Science Mathematics I (BSc CSIT, MTH112) Question Paper 2081 Nepal

Definitions

Limit: A function $f(x)$ is said to have a 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

Continuity: A function $f$ is continuous at $x = a$ if all three conditions hold:

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

Evaluation of $\lim_{x\to 0}\frac{\sin x}{x}$

Using the standard squeeze argument $\cos x < \dfrac{\sin x}{x} < 1$ for $0 < |x| < \tfrac{\pi}{2}$, and since $\lim_{x\to 0}\cos x = 1$, by the Sandwich theorem

(Equivalently, by L'Hôpital's rule on the $\tfrac{0}{0}$ form: $\lim_{x\to 0}\dfrac{\cos x}{1} = 1$.)

Continuity of $f(x) = \frac{\sin x}{x}$ at $x = 0$

At $x = 0$ the expression $\frac{\sin x}{x}$ is undefined (it gives $\tfrac{0}{0}$). Hence condition (1) fails and $f$ is not continuous at $x = 0$ as originally written; the discontinuity is removable because $\lim_{x\to 0} f(x) = 1$ exists.

If we redefine

then $f(0) = 1 = \lim_{x\to 0} f(x)$, so the redefined function is continuous at $x = 0$.

Leibnitz's Theorem

If $u$ and $v$ are functions of $x$ that are $n$-times differentiable, then the $n$th derivative of their product is

where $u_k = \dfrac{d^k u}{dx^k}$ and $v_k = \dfrac{d^k v}{dx^k}$.

Finding $y_n$ for $y = x^2 e^x$

Take $u = e^x$ and $v = x^2$ (choose the polynomial as $v$ since its higher derivatives vanish).

Derivatives of $u = e^x$: $\;u_{n} = e^x,\quad u_{n-1} = e^x,\quad u_{n-2} = e^x.$

Derivatives of $v = x^2$: $\;v = x^2,\quad v_1 = 2x,\quad v_2 = 2,\quad v_3 = v_4 = \cdots = 0.$

By Leibnitz's theorem, only the first three terms survive:

Substituting:

Setting up the area

The curve $y = x^2$ lies above the $x$-axis on $[1,3]$, so the required area between the curve, the $x$-axis and the lines $x = 1$, $x = 3$ is

Evaluation

Result

Using the identity $1 - \cos x = 2\sin^2\dfrac{x}{2}$:

(Alternatively, by L'Hôpital twice: $\lim \dfrac{\sin x}{2x} = \lim \dfrac{\cos x}{2} = \tfrac12$.)

Let $y = \tan^{-1} x$, so that $\tan y = x$.

Differentiating both sides with respect to $x$:

Statement of Rolle's Theorem

If a function $f$ 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 $f(x) = x^2 - 5x + 6$ on $[2,3]$

Condition (i): $f$ is a polynomial, hence continuous on $[2,3]$.

Condition (ii): $f$ is a polynomial, hence differentiable on $(2,3)$.

Condition (iii):

so $f(2) = f(3) = 0$.

All three hypotheses hold, so Rolle's theorem applies.

Finding $c$: $f'(x) = 2x - 5$. Setting $f'(c) = 0$:

Since $c = 2.5 \in (2,3)$, Rolle's theorem is verified.

The slope of the tangent is $\dfrac{dy}{dx}$.

At the point $(1,3)$: slope $m = 2(1) + 2 = 4$. (Check: $y(1) = 1 + 2 = 3$, so the point lies on the curve.)

Using the point-slope form $y - y_1 = m(x - x_1)$:

Complete the square in the denominator:

Using the standard result $\displaystyle\int \frac{dt}{t^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{t}{a} + C$ with $t = x+3$ and $a = 2$:

The equation is variable-separable. Write $e^{x-y} = \dfrac{e^x}{e^y}$:

Integrating both sides:

where $C$ is an arbitrary constant of integration.

With $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$, the cross product is

Expanding along the first row:

Treat $y$ and $z$ as constants and differentiate $u = x^3 + y^3 + z^3 - 3xyz$ with respect to $x$:

Evaluate the inner integral with respect to $y$ first (treating $x$ constant):

Now integrate with respect to $x$: