2016 HSC Question 9 MCQ

[melonsinwater]

https://puu.sh/BIKpY/68fb48001a.png

please help me with this question and provide working out with an explanation. please!!

[SEasternCry]

https://puu.sh/BIKpY/68fb48001a.png

please help me with this question and provide working out with an explanation. please!!

I'll try my best... The answer should be A... I think
By looking at the graph, specifically the y-values on the y-axis: it is obvious that f(1)<1
Now by looking at the first derivative, it is clear that it is positive (since the first derivative is an indication of the slope or gradient which is positive here since f'(x) is increasing when x = 1) However, assuming that the graph is to scale (that the incremental values on the y-axis match that of the x-axis), then f'(1) should have a gradient greater than 1 (the greater the gradient, the steeper the slope vertically) Therefore f(1)<1<f'(1).
Also, the second derivative (an indication of concavity) is clearly negative f''(x)<0 as in the graph is concave down at x=1. and hence f''(1)<f(1)<1<f'(1)

Hope I was clear enough. Cheers!

[melonsinwater]

" However, assuming that the graph is to scale (that the incremental values on the y-axis match that of the x-axis), then f'(1) should have a gradient greater than 1 (the greater the gradient, the steeper the slope vertically) Therefore f(1)<1<f'(1).
Also, the second derivative (an indication of concavity) is clearly negative f''(x)<0 as in the graph is concave down at x=1. and hence f''(1)<f(1)<1<f'(1)"
 
Sorry, can you explain this again? If you can, include a diagram? You dont have to but it'd make it a lot easier to understand. Thank you regardless

[SEasternCry]

" However, assuming that the graph is to scale (that the incremental values on the y-axis match that of the x-axis), then f'(1) should have a gradient greater than 1 (the greater the gradient, the steeper the slope vertically) Therefore f(1)<1<f'(1).
Also, the second derivative (an indication of concavity) is clearly negative f''(x)<0 as in the graph is concave down at x=1. and hence f''(1)<f(1)<1<f'(1)"
 
Sorry, can you explain this again? If you can, include a diagram? You dont have to but it'd make it a lot easier to understand. Thank you regardless

I don't have resources for a diagram rn...
The gradient is greater than one or steeper than a y=x graph because if you draw a tangent at x=1 on the curve, then you'll see that its pretty steep. I already explained that the steeper it that tangent, the greater the value of the gradient or f'(1). 
Concavity refers to whether the parabola is upside down, or whether the turning point there is min or max... that should usually indicate whether it is an upside down parabola or normal parabola. In this case it is an upside down (concave down) parabola with a max turning point, so that means the second derivative is negative. It is always defined that f''(x)<0 when concave down.