You should not round off values prematurely. Here, rounding your answers to 2 dp before rounding the final answers to 2 dp is a very bad habit to cultivate. In the O Levels, premature approximation may result in an inaccurate answer. If I recall, one mark may be deducted for premature approximation, unless the question specifically asks you to do so..

MM, do take note that should you wish to not write the full line in the calculator, you should use at least 4 or 5 decimal places long in your sub-workings before obtaining your final answer which is to 2 decimal places.

Actually your own answer itself has too many decimal places also.

The max one should use for working is 2 dp/sf above the required degree of accuracy for the final answer

Technically we should put as many as we can.

The reason why schools teach the students to write two or three digits beyond is because the remaining digits are unlikely to cause rounding errors.

In real-life precision things like engineering, exchange rates, stocks etc, even five or six decimal places matter a lot to those who invest large sums of money, or the design of buildings for example where even slight inaccuracies may affect the design of the bulding, so every decimal counts.

Penalisation will result for having too many places as per Cambridge's mark scheme. MM would have been deducted 1 mark if he uses that.

Students are allowed to use the calculator value in their calculations. It's just not to be shown in the presentation for clarity.

And also if you noticed in real life situations that you mentioned, it the same. no one actually uses that many figures in presentations due to lack of clarity. Figures are always rounded up or truncated.

The accuracy is left to the calculations where the computing software takes care of everything.

Thank you.

Thank you

Thank you sir. One more doubt
How to integrate 1/ax^2 +bx

I assume you mean the entire denominator is (ax² + bx), so that the fraction of interest is 1 / (ax² + bx) instead of 1 / (ax²), then plus bx.

Now this is going to be slightly tricky, as we can't simply put the fraction as (ax² + bx)^-1 and integrate.

What we can do, however, is to factorise the denominator into x (ax + b), so that we get

1 / [x (ax + b)]

before we attempt to split them into their respective partial fractions.

Let 1 / [x (ax + b)] = P / x plus Q / (ax + b)

Multiplying throughout by x (ax + b),

1 = P (ax + b) + Qx

Comparing constants,
1 = Pb
P = 1/b

Comparing terms in x,
0x = Pax + Qx
0 = Pa + Q
Q = - Pa
Q = -a/b

So, our fraction becomes

[1 / bx] - a / [b (ax + b)]

which can be written as

(1/b) / x - (a/b) / (ax + b)

Integrating this gets us

(1/b) ln x - (a/b) [ln (ax + b)] divided by a + const
= (1/b) ln x - (1/b) ln (ax + b) + const
= (1/b) [ln x - ln (ax + b] + const

Thank you sir for your explanation