BE Computer Engineering (Pokhara University) Calculus I (PU, MTH 110) Question Paper 2078 Nepal

(a) ε–δ definition and proof that $\lim_{x\to 3}(4x-5)=7$

Definition. We say $\lim_{x\to a} f(x)=L$ if for every $\varepsilon>0$ there exists a $\delta>0$ such that

$$0<|x-a|<\delta \;\Longrightarrow\; |f(x)-L|<\varepsilon.$$

Proof. Here $f(x)=4x-5$, $a=3$, $L=7$. Consider

$$|f(x)-7|=|4x-5-7|=|4x-12|=4|x-3|.$$

Given $\varepsilon>0$, choose $\delta=\dfrac{\varepsilon}{4}$. Then whenever $0<|x-3|<\delta$,

$$|f(x)-7|=4|x-3|<4\delta=4\cdot\frac{\varepsilon}{4}=\varepsilon.$$

Hence by definition $\lim_{x\to 3}(4x-5)=7$. $\blacksquare$

(b) Continuity at $x=2$

For $x\neq 2$,

$$f(x)=\frac{x^2-x-2}{x-2}=\frac{(x-2)(x+1)}{x-2}=x+1.$$

So

$$\lim_{x\to 2} f(x)=\lim_{x\to 2}(x+1)=3.$$

But $f(2)=5$. Since $\lim_{x\to 2} f(x)=3\neq 5=f(2)$, $f$ is discontinuous at $x=2$.

Because the limit exists (and is finite) but differs from the function value, this is a removable discontinuity. It is removed by redefining $f(2)=3$, which makes $f(x)=x+1$ continuous everywhere.

(c) Derivative of $y=\sqrt{2x+1}$ from first principles

$$\frac{dy}{dx}=\lim_{h\to 0}\frac{\sqrt{2(x+h)+1}-\sqrt{2x+1}}{h}.$$

Rationalise the numerator:

$$=\lim_{h\to 0}\frac{\big(2(x+h)+1\big)-\big(2x+1\big)}{h\big(\sqrt{2x+2h+1}+\sqrt{2x+1}\big)}=\lim_{h\to 0}\frac{2h}{h\big(\sqrt{2x+2h+1}+\sqrt{2x+1}\big)}.$$

Cancelling $h$ and letting $h\to 0$:

$$\frac{dy}{dx}=\frac{2}{2\sqrt{2x+1}}=\frac{1}{\sqrt{2x+1}}.$$

Slope of the tangent at $x=4$:

$$\left.\frac{dy}{dx}\right|_{x=4}=\frac{1}{\sqrt{2(4)+1}}=\frac{1}{\sqrt{9}}=\frac{1}{3}.$$

(a) Rolle's and Lagrange Mean Value Theorems; verification

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 $f'(c)=0$.

Lagrange Mean Value Theorem (MVT). If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in(a,b)$ such that

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

Verification for $f(x)=x^3-3x$ on $[-2,2]$. $f$ is a polynomial, hence continuous on $[-2,2]$ and differentiable on $(-2,2)$.

$$f(-2)=(-8)-(-6)=-2,\qquad f(2)=8-6=2.$$ $$\frac{f(2)-f(-2)}{2-(-2)}=\frac{2-(-2)}{4}=\frac{4}{4}=1.$$

$f'(x)=3x^2-3$. Set $f'(c)=1$:

$$3c^2-3=1\;\Rightarrow\;3c^2=4\;\Rightarrow\;c^2=\frac{4}{3}\;\Rightarrow\;c=\pm\frac{2}{\sqrt 3}\approx\pm1.155.$$

Both values lie in $(-2,2)$, so the MVT is verified with $c=\pm\dfrac{2}{\sqrt 3}$.

(b) Minimum surface area of a closed cylinder with $V=500\,\text{cm}^3$

Volume constraint: $\pi r^2 h=500\Rightarrow h=\dfrac{500}{\pi r^2}$. Total surface area (two ends + side):

$$S=2\pi r^2+2\pi r h=2\pi r^2+2\pi r\cdot\frac{500}{\pi r^2}=2\pi r^2+\frac{1000}{r}.$$

Differentiate and set to zero:

$$\frac{dS}{dr}=4\pi r-\frac{1000}{r^2}=0\;\Rightarrow\;4\pi r^3=1000\;\Rightarrow\;r^3=\frac{250}{\pi}.$$ $$r=\left(\frac{250}{\pi}\right)^{1/3}\approx 4.301\,\text{cm}.$$

Since $\dfrac{d^2S}{dr^2}=4\pi+\dfrac{2000}{r^3}>0$, this is a minimum. Then

$$h=\frac{500}{\pi r^2}\approx\frac{500}{\pi(4.301)^2}\approx 8.603\,\text{cm}.$$

Note $h=2r$ at the optimum. Dimensions: $r\approx 4.30\,\text{cm}$, $h\approx 8.60\,\text{cm}$.

(c) $\displaystyle\lim_{x\to 0}\frac{e^x-1-x}{x^2}$ by L'Hôpital

At $x=0$ the form is $\dfrac{0}{0}$, so L'Hôpital applies:

$$\lim_{x\to 0}\frac{e^x-1-x}{x^2}=\lim_{x\to 0}\frac{e^x-1}{2x}.$$

This is again $\dfrac{0}{0}$; apply L'Hôpital once more:

$$=\lim_{x\to 0}\frac{e^x}{2}=\frac{e^0}{2}=\frac{1}{2}.$$

Hence the limit is $\dfrac{1}{2}$.

(a) Reduction formula for $I_n=\int\sin^n x\,dx$ and $\int_0^{\pi/2}\sin^4 x\,dx$

Write $\sin^n x=\sin^{n-1}x\cdot\sin x$ and integrate by parts with $u=\sin^{n-1}x$, $dv=\sin x\,dx$:

$$du=(n-1)\sin^{n-2}x\cos x\,dx,\qquad v=-\cos x.$$ $$I_n=-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}x\cos^2 x\,dx.$$

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

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

That is $I_n=-\sin^{n-1}x\cos x+(n-1)I_{n-2}-(n-1)I_n$, so

$$nI_n=-\sin^{n-1}x\cos x+(n-1)I_{n-2}\;\Longrightarrow\;\boxed{I_n=-\frac{\sin^{n-1}x\cos x}{n}+\frac{n-1}{n}I_{n-2}.}$$

Definite case. For $\displaystyle J_n=\int_0^{\pi/2}\sin^n x\,dx$, the boundary term $-\frac{\sin^{n-1}x\cos x}{n}$ vanishes at both $0$ and $\pi/2$, giving $J_n=\frac{n-1}{n}J_{n-2}$. With $J_0=\frac{\pi}{2}$:

$$J_4=\frac{3}{4}J_2=\frac{3}{4}\cdot\frac{1}{2}J_0=\frac{3}{8}\cdot\frac{\pi}{2}=\frac{3\pi}{16}.$$

Hence $\displaystyle\int_0^{\pi/2}\sin^4 x\,dx=\frac{3\pi}{16}$.

(b) $\displaystyle\int\frac{2x+3}{(x-1)(x^2+1)}\,dx$ by partial fractions

Let

$$\frac{2x+3}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}.$$

Then $2x+3=A(x^2+1)+(Bx+C)(x-1)$.

• $x=1$: $5=2A\Rightarrow A=\dfrac{5}{2}$.
• Coefficient of $x^2$: $0=A+B\Rightarrow B=-\dfrac{5}{2}$.
• Constant: $3=A-C\Rightarrow C=A-3=\dfrac{5}{2}-3=-\dfrac{1}{2}$.

So

$$\int\!\left[\frac{5/2}{x-1}+\frac{-\tfrac{5}{2}x-\tfrac{1}{2}}{x^2+1}\right]dx=\frac{5}{2}\ln|x-1|-\frac{5}{2}\int\frac{x}{x^2+1}\,dx-\frac{1}{2}\int\frac{dx}{x^2+1}.$$ $$=\frac{5}{2}\ln|x-1|-\frac{5}{4}\ln(x^2+1)-\frac{1}{2}\tan^{-1}x+C.$$

(a) Integral test and convergence of $\sum \frac{1}{n(\ln n)^2}$

Integral test. If $f$ is positive, continuous, and decreasing on $[N,\infty)$ and $a_n=f(n)$, then $\sum_{n=N}^{\infty} a_n$ and $\int_N^{\infty} f(x)\,dx$ converge or diverge together.

Take $f(x)=\dfrac{1}{x(\ln x)^2}$ on $[2,\infty)$, which is positive, continuous, and decreasing. With $u=\ln x$, $du=\dfrac{dx}{x}$:

$$\int_2^{\infty}\frac{dx}{x(\ln x)^2}=\int_{\ln 2}^{\infty}\frac{du}{u^2}=\left[-\frac{1}{u}\right]_{\ln 2}^{\infty}=\frac{1}{\ln 2}.$$

The integral is finite, so by the integral test the series converges.

(b) Radius and interval of convergence of $\sum_{n=1}^{\infty}\frac{(x-2)^n}{n\,3^n}$

Apply the ratio test with $a_n=\dfrac{(x-2)^n}{n\,3^n}$:

$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(x-2)^{n+1}}{(n+1)3^{n+1}}\cdot\frac{n\,3^n}{(x-2)^n}\right|=\frac{n}{n+1}\cdot\frac{|x-2|}{3}\;\xrightarrow{n\to\infty}\;\frac{|x-2|}{3}.$$

Convergence requires $\dfrac{|x-2|}{3}<1\Rightarrow|x-2|<3$. Thus the radius of convergence is $R=3$, with $-1<x<5$.

Endpoints:

• $x=5$: series is $\sum\dfrac{3^n}{n\,3^n}=\sum\dfrac{1}{n}$ — the harmonic series, diverges.
• $x=-1$: series is $\sum\dfrac{(-3)^n}{n\,3^n}=\sum\dfrac{(-1)^n}{n}$ — alternating harmonic, converges (by the alternating series test).

Interval of convergence: $[-1,5)$, i.e. $-1\le x<5$, with $R=3$.

(a) $\displaystyle\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}$

Factor: $\tan x-\sin x=\sin x\left(\dfrac{1}{\cos x}-1\right)=\sin x\cdot\dfrac{1-\cos x}{\cos x}$.

$$\frac{\tan x-\sin x}{x^3}=\frac{\sin x}{x}\cdot\frac{1-\cos x}{x^2}\cdot\frac{1}{\cos x}.$$

Using $\dfrac{\sin x}{x}\to 1$, $\dfrac{1-\cos x}{x^2}\to\dfrac12$, $\cos x\to 1$:

$$\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}=1\cdot\frac12\cdot 1=\frac{1}{2}.$$

(b) $\displaystyle\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{2x}$

Recall $\left(1+\dfrac{3}{x}\right)^{x}\to e^3$ as $x\to\infty$. Therefore

$$\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{2x}=\left[\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{x}\right]^{2}=(e^3)^2=e^{6}.$$

(a) $y=x^x$ by logarithmic differentiation

Take logs: $\ln y=x\ln x$. Differentiate both sides w.r.t. $x$:

$$\frac{1}{y}\frac{dy}{dx}=\ln x+x\cdot\frac{1}{x}=\ln x+1.$$ $$\frac{dy}{dx}=y(\ln x+1)=x^x(\ln x+1).$$

(b) Implicit differentiation of $x^3+y^3=6xy$ at $(3,3)$

Differentiate w.r.t. $x$ (product rule on $6xy$):

$$3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.$$

Collect $\dfrac{dy}{dx}$:

$$\big(3y^2-6x\big)\frac{dy}{dx}=6y-3x^2\;\Rightarrow\;\frac{dy}{dx}=\frac{6y-3x^2}{3y^2-6x}=\frac{2y-x^2}{y^2-2x}.$$

At $(3,3)$:

$$\frac{dy}{dx}=\frac{2(3)-3^2}{3^2-2(3)}=\frac{6-9}{9-6}=\frac{-3}{3}=-1.$$

Related rates for a spherical balloon

Given $\dfrac{dr}{dt}=2\,\text{cm/s}$ and $r=10\,\text{cm}$.

(a) Rate of change of volume. $V=\dfrac{4}{3}\pi r^3$, so

$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi(10)^2(2)=800\pi\approx 2513.3\,\text{cm}^3/\text{s}.$$

(b) Rate of change of surface area. $S=4\pi r^2$, so

$$\frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi(10)(2)=160\pi\approx 502.7\,\text{cm}^2/\text{s}.$$

(a) $\displaystyle\int x^2 e^x\,dx$ by parts

Let $u=x^2,\;dv=e^x dx\Rightarrow du=2x\,dx,\;v=e^x$:

$$\int x^2 e^x\,dx=x^2 e^x-2\int x e^x\,dx.$$

For $\int x e^x\,dx$, take $u=x,\;dv=e^x dx$: $\int x e^x\,dx=x e^x-\int e^x dx=x e^x-e^x$. Hence

$$\int x^2 e^x\,dx=x^2 e^x-2(x e^x-e^x)+C=e^x\big(x^2-2x+2\big)+C.$$

(b) $\displaystyle\int\frac{dx}{\sqrt{9-4x^2}}$

Write $9-4x^2=4\!\left(\dfrac{9}{4}-x^2\right)$, so $\sqrt{9-4x^2}=2\sqrt{(3/2)^2-x^2}$. Using $\int\dfrac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\dfrac{x}{a}$ with $a=\dfrac32$:

$$\int\frac{dx}{\sqrt{9-4x^2}}=\frac{1}{2}\int\frac{dx}{\sqrt{(3/2)^2-x^2}}=\frac{1}{2}\sin^{-1}\!\frac{x}{3/2}+C=\frac{1}{2}\sin^{-1}\!\frac{2x}{3}+C.$$

(a) $\displaystyle I=\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$

Using $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ with $a=\dfrac{\pi}{2}$ (so $\sin x\leftrightarrow\cos x$):

$$I=\int_0^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx.$$

Adding the two expressions for $I$:

$$2I=\int_0^{\pi/2}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx=\int_0^{\pi/2}1\,dx=\frac{\pi}{2}.$$

Therefore $I=\dfrac{\pi}{4}$.

(b) Area between $y=x^2$ and $y=x+2$

Intersections: $x^2=x+2\Rightarrow x^2-x-2=0\Rightarrow(x-2)(x+1)=0\Rightarrow x=-1,2$. On $[-1,2]$ the line lies above the parabola, so

$$A=\int_{-1}^{2}\big[(x+2)-x^2\big]\,dx=\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}.$$

At $x=2$: $2+4-\tfrac{8}{3}=\tfrac{10}{3}$. At $x=-1$: $\tfrac12-2+\tfrac13=-\tfrac{7}{6}$.

$$A=\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}\;\text{square units}.$$

(a) $\displaystyle\int_1^{\infty}\frac{dx}{x^2}$

$$\int_1^{\infty}\frac{dx}{x^2}=\lim_{b\to\infty}\int_1^{b}x^{-2}\,dx=\lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^{b}=\lim_{b\to\infty}\left(1-\frac{1}{b}\right)=1.$$

The limit is finite, so the integral converges to $1$.

(b) $\displaystyle\int_0^{1}\frac{dx}{\sqrt{1-x}}$

The integrand is unbounded near $x=1$, so it is improper there:

$$\int_0^{1}\frac{dx}{\sqrt{1-x}}=\lim_{t\to 1^-}\int_0^{t}(1-x)^{-1/2}\,dx=\lim_{t\to 1^-}\Big[-2\sqrt{1-x}\Big]_0^{t}.$$ $$=\lim_{t\to 1^-}\big(-2\sqrt{1-t}+2\sqrt{1}\big)=0+2=2.$$

The limit exists, so the integral converges to $2$.

(a) Convergence of $a_n=\dfrac{3n^2+2n}{n^2+1}$

Divide numerator and denominator by $n^2$:

$$a_n=\frac{3+\tfrac{2}{n}}{1+\tfrac{1}{n^2}}\xrightarrow{n\to\infty}\frac{3+0}{1+0}=3.$$

The sequence converges to $3$.

(b) Ratio test for $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^n}$

Let $a_n=\dfrac{n!}{n^n}$. Then

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac{(n+1)\,n^n}{(n+1)^{n+1}}=\frac{n^n}{(n+1)^{n}}=\left(\frac{n}{n+1}\right)^{n}=\frac{1}{\left(1+\tfrac{1}{n}\right)^{n}}.$$

As $n\to\infty$, $\left(1+\tfrac{1}{n}\right)^n\to e$, so

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{1}{e}\approx 0.368<1.$$

Since the ratio limit is less than $1$, the series converges.

(a) Cardioid $r=1+\cos\theta$

This is a heart-shaped curve. Key points: $\theta=0\Rightarrow r=2$ (rightmost), $\theta=\pi/2\Rightarrow r=1$, $\theta=\pi\Rightarrow r=0$ (the cusp at the pole), $\theta=3\pi/2\Rightarrow r=1$. The curve passes through $(2,0)$, has a cusp at the origin pointing left, and bulges to the right. Symmetry: since replacing $\theta$ by $-\theta$ leaves $r$ unchanged ($\cos(-\theta)=\cos\theta$), the cardioid is symmetric about the polar axis (the $x$-axis).

(b) Area of one loop of $r=2\sin 2\theta$

The rose $r=2\sin 2\theta$ has four petals. One loop is traced as $\theta$ goes from $0$ to $\pi/2$ (where $\sin 2\theta\ge 0$). The polar area is

$$A=\frac{1}{2}\int_{0}^{\pi/2} r^2\,d\theta=\frac{1}{2}\int_0^{\pi/2}4\sin^2 2\theta\,d\theta=2\int_0^{\pi/2}\sin^2 2\theta\,d\theta.$$

Using $\sin^2 2\theta=\dfrac{1-\cos 4\theta}{2}$:

$$A=2\cdot\frac{1}{2}\int_0^{\pi/2}(1-\cos 4\theta)\,d\theta=\left[\theta-\frac{\sin 4\theta}{4}\right]_0^{\pi/2}=\frac{\pi}{2}-0=\frac{\pi}{2}.$$

Hence the area enclosed by one loop is $\dfrac{\pi}{2}$ square units.