Sum local area inequalities, then shift the axis · source reference

Printed question

(i) Prove $\frac d{dx}(\tan^{-1}x)=1/(1+x^2)$. [2]

The diagram sketches $y=1/(1+x^2)$, with a shaded rectangle from $n-1$ to $n$.

(ii) Considering the rectangle and the area between graph and $x$-axis on $n-1\le x\le n$, $n\ge1$, show $1/(1+n^2)<\tan^{-1}n-\tan^{-1}(n-1)$. [2]

(iii) Deduce $1/2+1/5+1/10+\cdots+1/(1+n^2)<\tan^{-1}n$. [2]

R is bounded by $y=1/(1+x^2)$, $x=3$ and $y=1/2$. Write the equation after translation 3 units in negative $x$-direction. Hence or otherwise find its volume rotating completely about $x=3$. [4]

[2015/NYJC/Prelim/II/4 (modified)]

Printed working

(i)

Let $y=\tan^{-1}x$. Then $\tan y=x$. Differentiate:

$$\sec^2y\frac{dy}{dx}=1,\qquad(1+\tan^2y)\frac{dy}{dx}=1,\qquad\frac{dy}{dx}=\frac1{1+x^2}\quad\text{(shown)}.$$

(ii)

From graph: area of rectangle $<$ area under curve on $[n-1,n]$.

$$\frac1{1+n^2}[n-(n-1)]<\int_{n-1}^n\frac1{1+x^2}dx,$$ $$\frac1{1+n^2}<[\tan^{-1}x]_{n-1}^n=\tan^{-1}n-\tan^{-1}(n-1)\quad\text{(shown)}.$$

(iii)

Comparing rectangles with area on $0\le x\le n$,

$$\frac1{1+1^2}+\frac1{1+2^2}+\frac1{1+3^2}+\cdots+\frac1{1+n^2}<\int_0^n\frac1{1+x^2}dx,$$ $$\frac12+\frac15+\frac1{10}+\cdots+\frac1{1+n^2}<[\tan^{-1}x]_0^n=\tan^{-1}n\quad\text{(shown)}.$$

Alternative solution, using (ii):

$$\begin{aligned}\frac1{1+1^2}&<\tan^{-1}1-\tan^{-1}0&&n=1,\\\frac1{1+2^2}&<\tan^{-1}2-\tan^{-1}1&&n=2,\\&\vdots\\\frac1{1+(n-1)^2}&<\tan^{-1}(n-1)-\tan^{-1}(n-2),\\\frac1{1+n^2}&<\tan^{-1}n-\tan^{-1}(n-1).\end{aligned}$$

Adding cancels all adjacent arctangents, giving the required inequality since $\tan^{-1}0=0$.

Final rotation task

After translation: $y=1/[1+(x+3)^2]$.

$$(x+3)^2=1/y-1\Rightarrow x=-3\pm\sqrt{1/y-1}.$$

The minus branch is not the curve enclosing R; use $x=-3+\sqrt{1/y-1}$. The printed description calls the translated region “enclosed by $y=1/[1+(x+3)^2]$, line $y=1/2$ and x-axis”, then rotates it about the $y$-axis. The stated x-axis is a printed boundary error: the diagram uses the $y$-axis; its region lies between levels $1/10$ and $1/2$.

$$V=\pi\int_{1/10}^{1/2}x^2dy=\pi\int_{1/10}^{1/2}(-3+\sqrt{1/y-1})^2dy\approx2.5998=2.60\text{ or }0.828\pi\quad(3\text{ s.f.}).$$

Teacher notes

The unshifted diagram labels $y=1/(1+x^2)$, $1$, $x=1$, $x=3$, $y=1/2$, R and rotation arrow at $x=3$. Shifted diagram labels $y=1/[1+(x+3)^2]$, $x=-3,-2$, peak 1, $y=1/2$, R and rotation about $y$-axis. The repeated telescoping cancellation strokes are represented by the cancellation explanation above.

Printed answer key

The short key labels the final rotation task “15. (iii)”: $y=1/[1+(x+3)^2]$; $2.60$ units$^3$ (3 s.f.). The original question prints no part label for this task; its actual (iii) is the finite-sum inequality.

Source diagrams in native form