inverse functions MC help (1 Viewer)

[jimmysmith560]

Original Poster:

View attachment 35064

why is it A?

Would the following working help?

Domain of 𝑓(𝑥) is 𝑥 ≥ −1.

𝑓(−1) = −1, so 𝑓(𝑥) intersects the curve at (−1,−1)

𝑓(𝑥) decreases as 𝑥 increases, hence (−1,−1) must be the only point of intersection.

 

[jimmysmith560]

That makes sense!! for the question i sent above, i just dont understand why it has to be a straight line, could you explain?

View attachment 35068

In terms of this question, consider the following working:





∴ Straight line with gradient 1, hence D

Another approach is as follows:

 

[jimmysmith560]

Im really bad with my trig basics, i understand train of thought but could u explain how u converted -sinx into cosx+pi/2?

bc so - sinx = -cos(pi/2 - x)
and cos is even so fx = f-x
so say -cos(pi/2 - x) = f(-x)
and then im confused from here

It is because

is an identity.

You can verify this by drawing the graph of

and the graph of

, both of which will lead to the same graph.

 

[cossine]

Im really bad with my trig basics, i understand train of thought but could u explain how u converted -sinx into cosx+pi/2?

bc so - sinx = -cos(pi/2 - x)
and cos is even so fx = f-x
so say -cos(pi/2 - x) = f(-x)
and then im confused from here

We know that:

sin x = cos(pi/2 - x) #Note: This is one of the complementary angle there are 5 other identities(Use wikipedia if need be). The co in cosine means complementary. Similarly co has the same meaning with cot and cosec

=> -sin x = -cos(pi/2 -x)
=> sin(-x) = -cos(pi/2 - x) # -sin(x) = sin(-x) because sine is an odd function

=> sin(u) = -cos(pi/2 + u) # let u = -x


=> sin(x) = -cos(pi/2 + x) # Since u is dummy variable you can just replace it with x

=> -sin(x) = cos(pi/2 + x)

You can also apply the trick with u with function like sin(-3x+45) = -sin(3x-45) [This is because sine is odd].