BSc CSIT (TU) Science B.Sc. Mathematics (Model) Question Paper 2079 Nepal

Bijective function. A function $f : X \to Y$ is bijective if it is both injective (one-to-one) and surjective (onto): for all $x_1, x_2 \in X$, $f(x_1)=f(x_2) \Rightarrow x_1=x_2$, and for every $y \in Y$ there exists $x \in X$ with $f(x)=y$.

Example: $f : \mathbb{R} \to \mathbb{R}$, $f(x)=2x+3$ is bijective (one-to-one and onto).

Is composition commutative? No. In general $g \circ f \neq f \circ g$. For example, with $f(x)=x+1$ and $g(x)=x^2$ on $\mathbb{R}$, $(g\circ f)(x)=(x+1)^2$ while $(f\circ g)(x)=x^2+1$, which are not equal. (Moreover $f\circ g$ may not even be defined unless the domains/codomains match.)

Proof that $g\circ f$ is bijective.

Injective: Suppose $(g\circ f)(x_1)=(g\circ f)(x_2)$, i.e. $g(f(x_1))=g(f(x_2))$. Since $g$ is injective, $f(x_1)=f(x_2)$; since $f$ is injective, $x_1=x_2$. Hence $g\circ f$ is injective.

Surjective: Let $z \in Z$. Since $g$ is surjective, there is $y \in Y$ with $g(y)=z$. Since $f$ is surjective, there is $x \in X$ with $f(x)=y$. Then $(g\circ f)(x)=g(f(x))=g(y)=z$. Hence $g\circ f$ is surjective.

Therefore $g\circ f$ is bijective and its inverse exists.

Proof that $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$. For any $z \in Z$,

and similarly $(f^{-1}\circ g^{-1})\circ(g\circ f)(x)=x$ for all $x\in X$. Since the composition with $f^{-1}\circ g^{-1}$ gives the identity on both sides, by uniqueness of inverses $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$. $\blacksquare$

(a) Interior point. A point $x$ of a set $A \subseteq \mathbb{R}$ is an interior point of $A$ if there exists $\varepsilon > 0$ such that the open interval $(x-\varepsilon, x+\varepsilon) \subseteq A$.

Interior points of $\{1,2,3,4,5,6\}$: This is a finite set; no interval $(x-\varepsilon, x+\varepsilon)$ around any of its points lies inside the set. Hence the set has no interior points (its interior is empty).

Finite intersection of open sets is open. Let $G_1, G_2, \dots, G_n$ be open and $G=\bigcap_{i=1}^{n} G_i$. Let $x\in G$. Then $x\in G_i$ for each $i$, so there exist $\varepsilon_i>0$ with $(x-\varepsilon_i, x+\varepsilon_i)\subseteq G_i$. Put $\varepsilon=\min\{\varepsilon_1,\dots,\varepsilon_n\}>0$. Then $(x-\varepsilon, x+\varepsilon)\subseteq G_i$ for all $i$, so $(x-\varepsilon, x+\varepsilon)\subseteq G$. Thus every point of $G$ is interior, and $G$ is open. $\blacksquare$

Infinite intersection need not be open. Consider $G_n=\left(-\tfrac{1}{n}, \tfrac{1}{n}\right)$, each open. Then

which is not open. Hence the intersection of infinitely many open sets need not be open.

OR (b) Adherent point and closure. A point $x$ is an adherent point (point of closure) of $X$ if every open interval $(x-\varepsilon, x+\varepsilon)$ contains at least one point of $X$. The closure $\bar{X}$ is the set of all adherent points of $X$.

Adherent points of $\{0\}\cup(1,7)$: the closure is $\{0\}\cup[1,7]$, i.e. $0$ together with the closed interval $[1,7]$ (the endpoints $1$ and $7$ are adherent).

Proof: $X$ is closed $\iff$ it contains all its adherent points. ($\Rightarrow$) Suppose $X$ is closed, so $\mathbb{R}\setminus X$ is open. If $x$ is adherent to $X$ but $x\notin X$, then $x\in\mathbb{R}\setminus X$, which is open, so some $(x-\varepsilon, x+\varepsilon)\subseteq \mathbb{R}\setminus X$ contains no point of $X$, contradicting adherence. Hence $X$ contains all its adherent points. ($\Leftarrow$) Suppose $X$ contains all its adherent points. Let $x\in\mathbb{R}\setminus X$. Then $x$ is not adherent, so some $(x-\varepsilon,x+\varepsilon)$ contains no point of $X$, i.e. it lies in $\mathbb{R}\setminus X$. Thus $\mathbb{R}\setminus X$ is open and $X$ is closed. $\blacksquare$

Convergence of a series. A series $\sum x_n$ converges to $s$ if the sequence of partial sums $s_n = x_1 + x_2 + \cdots + x_n$ converges to $s$, i.e. $\lim_{n\to\infty} s_n = s$.

Cauchy's general principle of convergence. $\sum x_n$ converges $\iff$ the sequence $\{s_n\}$ is a Cauchy sequence (since $\mathbb{R}$ is complete). For $m>n$, $s_m - s_n = x_{n+1}+\cdots+x_m$. So $\{s_n\}$ is Cauchy $\iff$ for each $\varepsilon>0$ there exists $N$ such that $|s_m-s_n| = |x_{n+1}+\cdots+x_m| < \varepsilon$ whenever $m>n>N$. Hence the series converges iff this criterion holds. $\blacksquare$

(i) Harmonic series $\sum \frac{1}{n}$. Take $m=2n$. Then

Choosing $\varepsilon=\tfrac12$, the criterion fails, so the harmonic series diverges.

(ii) $\sum \frac{1}{n^2}$. For $m>n$,

Given $\varepsilon>0$, choose $N>\tfrac1\varepsilon$; then for $m>n>N$ the tail is $<\varepsilon$. So $\sum\frac{1}{n^2}$ converges.

OR — Cauchy's root test. Let $\sum u_n$, $u_n>0$, with $\lim (u_n)^{1/n}=l$.

(i) If $l<1$, choose $r$ with $l<r<1$. For large $n$, $(u_n)^{1/n}<r$, so $u_n<r^n$. Since $\sum r^n$ converges ($0<r<1$), by comparison $\sum u_n$ converges.

(ii) If $l>1$, then for infinitely many $n$, $(u_n)^{1/n}>1$, so $u_n>1$; thus $u_n \not\to 0$ and the series diverges.

(iii) If $l=1$, the test is inconclusive: e.g. $\sum\frac1n$ diverges while $\sum\frac1{n^2}$ converges, both with $l=1$.

Application. Here $u_n=\left(\frac{n}{n+1}\right)^{n-1}x^{n-1}$ (general term $\left(\frac{n}{n+1}\right)^{n}x^{n}$ form). Then

so $l=|x|$. Therefore the series converges for $|x|<1$, diverges for $|x|>1$, and for $|x|=1$ the term $\left(\frac{n}{n+1}\right)^n \to e^{-1}\neq 0$, so it diverges. $\blacksquare$

Intermediate Value Theorem. Let $f:[a,b]\to\mathbb{R}$ be continuous and let $k$ lie strictly between $f(a)$ and $f(b)$. Then there exists $c\in(a,b)$ with $f(c)=k$.

Proof. Assume WLOG $f(a)<k<f(b)$. Define $g(x)=f(x)-k$, continuous on $[a,b]$ with $g(a)<0<g(b)$. Let

$S$ is nonempty ($a\in S$) and bounded above by $b$; let $c=\sup S$. By continuity of $g$:

• If $g(c)<0$, then $g<0$ on a neighbourhood of $c$, so points slightly larger than $c$ lie in $S$, contradicting $c=\sup S$.
• If $g(c)>0$, then $g>0$ on a neighbourhood of $c$, so $c$ is not a limit of points of $S$, again a contradiction.

Hence $g(c)=0$, i.e. $f(c)=k$. $\blacksquare$

Geometrical meaning. As $x$ moves from $a$ to $b$, the graph of a continuous $f$ is an unbroken curve from $(a,f(a))$ to $(b,f(b))$; it must cross every horizontal line $y=k$ lying between the two heights at least once. So the curve attains every intermediate value.

Application. Let $f(x)=x^3-x-1$, continuous on $[1,2]$. Then

Since $f$ is continuous and $0$ lies between $f(1)$ and $f(2)$, by the IVT there exists $c\in(1,2)$ with $f(c)=0$. Hence the equation $x^3-x-1=0$ has a root in $[1,2]$. $\blacksquare$

Upper and lower Riemann sums. Let $f$ be bounded on $[a,b]$ and let $P=\{a=x_0<x_1<\cdots<x_n=b\}$ be a partition. On each subinterval $[x_{i-1},x_i]$ put

The upper sum and lower sum are

The upper and lower integrals are $\overline{\int}_a^b f=\inf_P U(P,f)$ and $\underline{\int}_a^b f=\sup_P L(P,f)$; $f$ is Riemann integrable iff these are equal.

Riemann's criterion. $f$ is integrable on $[a,b]$ $\iff$ for every $\varepsilon>0$ there is a partition $P$ with $U(P,f)-L(P,f)<\varepsilon$.

Proof. ($\Leftarrow$, sufficiency) Suppose for each $\varepsilon>0$ such a $P$ exists. Since always

we get $0\le \overline{\int}_a^b f - \underline{\int}_a^b f \le U(P,f)-L(P,f)<\varepsilon$. As $\varepsilon>0$ is arbitrary, $\overline{\int}_a^b f=\underline{\int}_a^b f$, so $f$ is integrable.

($\Rightarrow$, necessity) Suppose $f$ is integrable, so $\overline{\int}=\underline{\int}=I$. Given $\varepsilon>0$, by definition of inf and sup there exist partitions $P_1,P_2$ with

Let $P=P_1\cup P_2$ be their common refinement. Refinement decreases upper sums and increases lower sums, so $U(P,f)\le U(P_1,f)$ and $L(P,f)\ge L(P_2,f)$. Hence

Thus the required partition exists. $\blacksquare$

Supremum and infimum. For $S\subseteq\mathbb{R}$ bounded above, the supremum $\sup S$ is the least upper bound: $u=\sup S$ if (i) $x\le u$ for all $x\in S$, and (ii) for every $\varepsilon>0$ there is $x\in S$ with $x>u-\varepsilon$. Similarly the infimum $\inf S$ is the greatest lower bound. Examples: for $S=(0,1)$, $\sup S=1$ and $\inf S=0$ (neither attained).

Completeness axiom. Every nonempty subset of $\mathbb{R}$ that is bounded above has a least upper bound (supremum) in $\mathbb{R}$.

Density of rationals. Claim: if $a<b$ then there exists $r\in\mathbb{Q}$ with $a<r<b$.

Proof. Since $b-a>0$, by the Archimedean property there is $n\in\mathbb{N}$ with $n(b-a)>1$, i.e. $\tfrac1n<b-a$. Again by the Archimedean property the set of integers $m$ with $m>na$ is nonempty and bounded below, so it has a least element $m$; thus

Then $na<m\le na+1$, so dividing by $n$,

Hence $r=\frac{m}{n}$ is rational and $a<r<b$. $\blacksquare$

OR — Countable and uncountable sets. A set is countable if it is finite or in bijection with $\mathbb{N}$ (e.g. $\mathbb{Z}$, $\mathbb{Q}$ are countable). A set is uncountable if it is not countable (e.g. $\mathbb{R}$, or $(0,1)$).

Cantor's diagonal proof that $(0,1)$ is uncountable. Suppose, for contradiction, $(0,1)$ is countable, so its elements can be listed as $x_1,x_2,x_3,\dots$ Write each in decimal:

(using the non-terminating representation to avoid ambiguity). Construct $y=0.b_1 b_2 b_3\dots$ where $b_k$ is chosen with $b_k\neq a_{kk}$ and $b_k\in\{1,\dots,8\}$ (avoiding $0$ and $9$). Then $y\in(0,1)$, but $y$ differs from each $x_k$ in the $k$-th digit, so $y\neq x_k$ for all $k$ — contradicting that the list contained every element of $(0,1)$. Hence $(0,1)$ is uncountable. $\blacksquare$

Monotonic sequence. A sequence $\{a_n\}$ is monotonically increasing if $a_n \le a_{n+1}$ for all $n$, and monotonically decreasing if $a_n \ge a_{n+1}$ for all $n$. A sequence is monotonic if it is either increasing or decreasing. Example: $a_n=n$ is increasing; $a_n=\frac1n$ is decreasing.

Monotone Convergence Theorem (increasing, bounded above). Let $\{a_n\}$ be increasing and bounded above. By the completeness axiom the set $S=\{a_n : n\in\mathbb{N}\}$ has a least upper bound $L=\sup S$.

Proof of convergence to $L$. Let $\varepsilon>0$. Since $L$ is the least upper bound, $L-\varepsilon$ is not an upper bound, so there exists $N$ with $a_N > L-\varepsilon$. Because $\{a_n\}$ is increasing, for all $n\ge N$,

so $|a_n - L|<\varepsilon$. Hence $a_n \to L = \sup S$. Thus the sequence converges to its least upper bound. $\blacksquare$

Test $\left\{\dfrac{n}{n+1}\right\}$. Compute

Since $a_{n+1}>a_n$ for all $n$, the sequence is monotonically increasing (and bounded above by $1$, with $\lim a_n = 1$).

Squeeze (Sandwich) Theorem. Suppose $f(x)\le g(x)\le h(x)$ for all $x$ in a deleted neighbourhood of $c$, and $\lim_{x\to c} f(x)=\lim_{x\to c} h(x)=L$. Then $\lim_{x\to c} g(x)=L$.

Proof. Let $\varepsilon>0$. Since $\lim_{x\to c} f(x)=L$, there is $\delta_1>0$ such that $0<|x-c|<\delta_1 \Rightarrow |f(x)-L|<\varepsilon$, i.e. $L-\varepsilon<f(x)$. Since $\lim_{x\to c} h(x)=L$, there is $\delta_2>0$ such that $0<|x-c|<\delta_2 \Rightarrow h(x)<L+\varepsilon$. Let $\delta=\min\{\delta_1,\delta_2\}$ (also small enough that the inequality $f\le g\le h$ holds). Then for $0<|x-c|<\delta$,

so $|g(x)-L|<\varepsilon$. Hence $\lim_{x\to c} g(x)=L$. $\blacksquare$

Application. For $x\neq 0$, $\left|\sin\frac1x\right|\le 1$, so

Since $\lim_{x\to 0}(-x^2)=0$ and $\lim_{x\to 0} x^2 = 0$, by the Squeeze Theorem

Rolle's Theorem. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists $c\in(a,b)$ with $f'(c)=0$.

Proof. Since $f$ is continuous on the closed bounded interval $[a,b]$, by the Extreme Value Theorem it attains a maximum $M$ and minimum $m$ on $[a,b]$.

• Case 1: $M=m$. Then $f$ is constant, so $f'(x)=0$ for every $x\in(a,b)$; any interior point works.
• Case 2: $M>m$. Since $f(a)=f(b)$, at least one of the extreme values is attained at an interior point $c\in(a,b)$. As $f$ is differentiable at $c$ and $c$ is a local extremum, by Fermat's theorem $f'(c)=0$.

In either case there exists $c\in(a,b)$ with $f'(c)=0$. $\blacksquare$

Geometrical interpretation. Since $f(a)=f(b)$, the chord joining the endpoints of the graph is horizontal. Rolle's theorem says there is at least one interior point $c$ where the tangent to the curve is horizontal (parallel to the $x$-axis), i.e. $f'(c)=0$.

Is continuity at the end-points necessary? Yes. All three hypotheses are essential. If $f$ fails to be continuous at an end-point the conclusion can fail. Example: define $f$ on $[0,1]$ by $f(x)=x$ for $0\le x<1$ and $f(1)=0$. Then $f(0)=f(1)=0$ and $f$ is differentiable on $(0,1)$, but $f$ is not continuous at the end-point $x=1$, and $f'(x)=1\neq 0$ everywhere on $(0,1)$ — so no such $c$ exists. This shows continuity on the closed interval (including the end-points) is required.

Primitive vs. integral. A primitive (antiderivative) of $f$ is a function $G$ with $G'(x)=f(x)$ on the interval; this is the indefinite integral $\int f(x)\,dx = G(x)+C$. The (definite) integral $\int_a^b f(x)\,dx$ is a number — the limit of Riemann sums — representing (signed) area. Example: for $f(x)=2x$, a primitive is $G(x)=x^2$ (a function), whereas the definite integral $\int_0^1 2x\,dx = 1$ (a number).

Continuity of $F(x)=\int_a^x f$. Since $f$ is integrable on $[a,b]$, it is bounded: $|f(t)|\le M$ for some $M>0$. For $x, y\in[a,b]$ with (say) $y>x$,

Given $\varepsilon>0$, take $\delta=\dfrac{\varepsilon}{M}$; then $|y-x|<\delta \Rightarrow |F(y)-F(x)|<\varepsilon$. Hence $F$ is (uniformly) continuous on $[a,b]$. $\blacksquare$

OR — Fundamental theorems of calculus.

First FTC. If $f$ is continuous on $[a,b]$ and $F(x)=\int_a^x f(t)\,dt$, then $F$ is differentiable on $(a,b)$ and $F'(x)=f(x)$.

Second FTC. If $f$ is continuous on $[a,b]$ and $G$ is any primitive of $f$ (i.e. $G'=f$), then

Proof of the second FTC. Let $G$ be a primitive of $f$, and set $F(x)=\int_a^x f(t)\,dt$. By the first FTC, $F'(x)=f(x)=G'(x)$, so $(F-G)'=0$ on $(a,b)$; hence $F(x)-G(x)=k$ (constant). At $x=a$, $F(a)=0$, so $k=-G(a)$, giving $F(x)=G(x)-G(a)$. Putting $x=b$,

Evaluation. With $f(x)=\sqrt{x+1}$, a primitive is $G(x)=\frac{2}{3}(x+1)^{3/2}$. Then