Eric Nicholas K's answer to Nelson Loo's Junior College 2 H2 Maths Singapore question.

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
Not 100% sure about this.
J
J
3 years ago
Your steps are alright, just that the sketch of the graph is of tan x instead of tan-¹ x
J
J
3 years ago
If the graph is meant to be tan x indeed, then for the limits it should be tan x instead of tan-¹x
Eric Nicholas K
Eric Nicholas K
3 years ago
It was intended to be tan x, to show him that tan (pi/2) and tan (-pi/2) is infinite, so by converse, inverse tangent of this “infinite” is pi/2 and so on.

However, the integral has to be continuous (non-broken) so the limits of the integral has to fall within the principal values of -pi/2 and pi/2.
J
J
3 years ago
When x = π/2, tan-¹ x = tan-¹ (π/2) ≈ 1.004 rad
When x = -π/2, tan-¹ x = tan-¹ (-π/2) ≈ -1.004 rad


The correct expressions would be :

lim tan x = -∞
x→π/2 ⁺

lim tan x = ∞
x→π/2 ⁻

lim tan x = -∞
x→-π/2 ⁺

lim tan x = ∞
x→-π/2 ⁻
Eric Nicholas K
Eric Nicholas K
3 years ago
Whoops

I just realised I wrote tan inverse (pi/2)- = infinity when I meant tan (pi/2)- = infinity