DP Mathematics HL Questionbank

Concept of function f:x↦f(x) : domain, range; image (value)
Path: |
Description
[N/A]Directly related questions
- 18M.1.hl.TZ2.10c: The function h is defined by h(x)=√x, for x ≥ 0. State the...
- 18M.1.hl.TZ2.10a: Find the inverse function f−1, stating its domain.
- 18M.2.hl.TZ2.10a.iv: Explain why f is not a function for...
- 18M.2.hl.TZ2.10a.iii: Explain why f has no inverse on the given domain.
- 18M.2.hl.TZ2.10a.ii: With reference to your graph, explain why f is a function on the given domain.
- 18M.2.hl.TZ2.10a.i: Sketch the graph of y=f(x)...
- 16M.2.hl.TZ1.5b.ii: Write down the range of f.
- 16M.2.hl.TZ1.5b.i: Sketch the graph y=f(x).
- 16M.2.hl.TZ1.5a: Prove that f is an even function.
- 16M.2.hl.TZ1.2b: If f(x)=x+2 and (g∘f)(x)=x2+4x−2 write down g(x).
- 16M.2.hl.TZ1.2a: Express x2+4x−2 in the form (x+a)2+b where a, b∈Z.
- 16N.2.hl.TZ0.2: Find the acute angle between the planes with equations x+y+z=3 and 2x−z=2.
- 17N.1.hl.TZ0.11a: Determine whether fn is an odd or even function, justifying your answer.
- 12N.1.hl.TZ0.12d: (i) State Fn(0) and Fn(1) . (ii) Show that Fn(x)<x ,...
- 09N.1.hl.TZ0.4: Consider the function f , where f(x)=arcsin(lnx). (a) Find the domain of f . (b)...
- 13M.1.hl.TZ1.12d: Find the range of f.
- 11N.1.hl.TZ0.9b: Hence determine the range of the function f:x→x+1x2+x+1.
- 11M.1.hl.TZ1.10a: Find the largest possible domain of the function g .
- 09N.2.hl.TZ0.9: (a) Given that the domain of g is x⩾a , find the least value of a such...
- 11M.1.hl.TZ1.8b: Find the coordinates of the point where the graph of y=f(x) and the graph of...
- 11M.1.hl.TZ1.8a: (i) Find (g∘f)(x) and write down the domain of the...
- 15M.1.hl.TZ1.9b: Find the range of g∘f.
- 15M.1.hl.TZ1.6b: Given that f(x) can be written in the form f(x)=A+B2x−1, find the values...
- 15M.1.hl.TZ2.13b: Hence show that √2−1<1√2.
- 15M.1.hl.TZ2.13a: Show that 1√n+√n+1=√n+1−√n where...
- 15N.1.hl.TZ0.12d: Find the range of f.
- 15N.2.hl.TZ0.12a: The functions u and v are defined as u(x)=x−3, v(x)=2x where...