UPSC 2023 Maths Optional Paper 2 Q8b — Step-by-Step Solution

15 marks · Section B

Regula Falsi (False Position) · Numerical Analysis · Read the full method →

Question

Compute a root of the equation $\log_{10}(2x+1)-x^2+3=0$ in the interval $[0,3]$ by Regula-Falsi method, correct to 6 decimal places.

Regula-Falsi (false-position); linear interpolation across the bracket; replace endpoint with same sign as $f(x^{*})$ .

Let $f(x)=\log_{10}(2x+1)-x^{2}+3$ .

f(0)=\log_{10}(1)-0+3=3>0,\qquad f(3)=\log_{10}7-9+3=0.845098-6=-5.154902<0.

So a root lies in $[0,3]$ .

The Regula-Falsi update with bracket $[a,b]$ where $f(a)f(b)<0$ is

x^{*}=\frac{a\,f(b)-b\,f(a)}{f(b)-f(a)}.

After computing $f(x^{*})$ , replace whichever endpoint shares its sign.

$n$	$a$	$b$	$f(a)$	$f(b)$	$x_n=x^{*}$	$f(x_n)$	new bracket
1	0.000000	3	3.000000	$-5.154902$	1.103635	$+2.288126$	$[1.103635,3]$
2	1.103635	3	2.288126	$-5.154902$	1.686606	$+0.796157$	$[1.686606,3]$
3	1.686606	3	0.796157	$-5.154902$	1.862353	$+0.206016$	$[1.862353,3]$
4	1.862353	3	0.206016	$-5.154902$	1.906049	$+0.049312$	$[1.906049,3]$
5	1.906049	3	0.049312	$-5.154902$	1.916477	$+0.011332$	$[1.916477,3]$
6	1.916477	3	0.011332	$-5.154902$	1.918861	$+0.002617$	$[1.918861,3]$
7	1.918861	3	0.002617	$-5.154902$	1.919472	$+0.000387$	$[1.919472,3]$
8	1.919472	3	0.000387	$-5.154902$	1.919548	$+0.000057$	$[1.919548,3]$
9	1.919548	3	0.000057	$-5.154902$	1.919570	$+0.000009$	$[1.919570,3]$
10	1.919570	3	0.000009	$-5.154902$	1.919573	$+0.000001$	—

Sample arithmetic (iteration 1):

x_1=\frac{0\cdot(-5.154902)-3\cdot 3}{-5.154902-3}=\frac{-9}{-8.154902}=1.103635.

$f(1.103635)=\log_{10}(3.207270)-1.218010+3=0.506136-1.218010+3=2.288126$ .

Sample arithmetic (iteration 4):

x_4=\frac{1.862353\cdot(-5.154902)-3\cdot 0.206016}{-5.154902-0.206016}=\frac{-9.600061-0.618048}{-5.360918}=\frac{-10.218109}{-5.360918}=1.906049.

By iteration 10, $|x_{n}-x_{n-1}|<5\times 10^{-6}$ and $|f(x_n)|<10^{-6}$ . So the root, correct to 6 decimal places, is

\boxed{\;x\approx 1.919573.\;}