Let \(f(x)=e^{x-1}\).
Given that the product function \(f(x)\times g(x)=e^{(x-1)^2}\), the rule for the function \(g\) is
- \(g(x)=e^{x-1}\)
- \(g(x)=e^{(x-2)(x-1)}\)
- \(g(x)=e^{(x+2)(x-1)}\)
- \(g(x)=e^{x(x-2)}\)
- \(g(x)=e^{x(x-3)}\)
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Let \(f(x)=e^{x-1}\).
Given that the product function \(f(x)\times g(x)=e^{(x-1)^2}\), the rule for the function \(g\) is
\(B\)
\(f(x)\times g(x)\) | \(=e^{(x-1)^2}\) |
\(e^{x-1}\times g(x)\) | \(=e^{(x-1)^2}\) |
\( g(x)\) | \(=\dfrac{e^{(x-1)^2}}{e^{(x-1)}}\) |
\(=e^{(x-1)^2}\times e^{(x-1)^-1}\) | |
\(=e^{x^2-3x+2}\) | |
\(=e^{(x-2)(x-1)}\) |
\(\Rightarrow B\)
Consider the functions `f(x) = sqrt{x+2}` and `g(x) = sqrt{1-2x}`, defined over their maximal domains.
The maximal domain of the function `h = f + g` is.
`D`
`f(x) \ = \ sqrt{x + 2} \ => \ text{domain} \ x ≥ -2`
`g(x) \ = \ sqrt{1-2x} \ => \ text{domain} \ x ≤ 1/2`
`text{Intersection of domains = domain} \ h(x)`
`:. \ h(x) ∈ [-2, 1/2]`
`=> D`
The function `f: R -> R, \ f(x)` is a polynomial function of degree 4. Part of the graph of `f` is shown below.
The graph of `f` touches the `x`-axis at the origin.
Let `g` be a function with the same rule as `f`.
Let `h: D -> R, \ h(x) = log_e (g(x)) - log_e (x^3 + x^2)`, where `D` is the maximal domain of `h`.
a. | `y` | `= ax^2 (x – 1)(x + 1)` |
`= ax^2 (x^2 – 1)\ …\ (1)` |
`text(Substitute)\ (1/ sqrt 2, 1)\ text{into (1):}`
`1 = a ⋅ (1/sqrt 2)^2 ((1/sqrt 2)^2 – 1)`
`1 = a(1/2)(-1/2)`
`a = -4`
`:. f(x) = -4x^2(x^2 – 1)`
b. | `g(x) > 0` | `=> -4x^2 (x^2 – 1) > 0` |
`=> x in (-1, 1)\ text(\{0})` |
`text(and)`
`x^3 + x^2 > 0 => text(true for)\ \ x in (-1 , 1)\ text(\{0})`
`:. D:\ x in (-1, 1)\ text(\{0})`
c. | `h(x)` | `= log_e ((-4x^2(x^2 – 1))/(x^3 + x^2))` |
`= log_e ((-4x^2(x + 1)(x – 1))/(x^2(x + 1)))` | ||
`= log_e (4(1 – x))\ \ text(where)\ \ x in (-1, 1)\ text(\{0})` |
`text(As)\ \ x -> -1,\ \ h(x) -> log_e 8 = 3 log_e 2`
`text(As)\ \ x -> 1,\ \ h(x) -> -oo`
`text(As)\ \ x -> 0,\ \ h(x) -> log_e 4 = 2 log_e 2`
`text{(}h(x)\ text(undefined when)\ \ x = 0 text{)}`
`:.\ text(Range)\ \ h(x) in (-oo, 3 log_e 2)\ text(\{) 2 log_e 2 text(})`
The function `f` has the property `f (x + f (x)) = f (2x)` for all non-zero real numbers `x`.
Which one of the following is a possible rule for the function?
`C`
`text(By trial and error,)`
`text(Consider option C:)`
`x + f(x)` | `= x + x = 2x` |
`f(2x)` | `=2x` |
`:. f(x + f(x))` | `= f(2x)` |
`=> C`
Let `h:(−1,1) -> R`, `h(x) = 1/(x - 1)`.
Which one of the following statements about `h` is not true?
`E`
`text(By trial and error, consider option)\ E:`
`h(x) = 1/(x – 1)`
`(h(x))^2 = 1/(x – 1)^2=1/(x^2-2x+1)`
`h(x^2)=1/(x^2-1) != (h(x))^2`
`=> E`
Let `f: R -> R,\ f(x) = e^x + e^(–x).`
For all `u in R,\ f(2u)` is equal to
`C`
`text(Solution 1)`
`text(Define)\ \ f(x) = e^x + e^-x`
`text(Enter each functional equation)`
`[text(i.e.)\ \ f(2u) = (f(u))^2 – 2]`
`text(until CAS output is “true”)`
`=> C`
`text(Solution 2)`
`f(2u)` | `=e^(2u) + e^(-2u)` |
`(f(u))^2` | `=(e^u + e^(-u))^2` |
`=e^(2u) + 2 + e^(-2u)` | |
`:. f(2u) = (f(u))^2-2`
`=>C`
Let `f: R -> R,\ f (x) = x^2`
Which one of the following is not true?
`D`
`text(Solution 1)`
`text(Consider option)\ D:`
`f(x-y)` | `=(x-y)^2` |
`=x^2 -2xy+y^2` | |
`f(x)-f(y)` | `= x^2-y^2` |
`:.f(x-y)` | `!=f(x)-f(y)` |
`=>D`
`text(Solution 2)`
`text(Define)\ \ f(x) = x^2`
`text(Enter each functional equation on CAS)`
`text(until output does NOT read “true”.)`
`=> D`
The function `f` has the property `f(x) - f(y) = (y - x)\ f(xy)` for all non-zero real numbers `x` and `y`.
Which one of the following is a possible rule for the function?
`D`
`text(Solution 1)`
`text(Consider option)\ D:`
`text(LHS)\ = 1/x – 1/y`
`text(RHS)\ =(y-x) xx 1/(xy) = 1/x – 1/y =\ text(LHS)`
`=>D`
`text{Solution 2 (using technology)}`
`text(Define each specific function on CAS)`
`[text(i.e.)\ \ f(x) = 1/x]`
`text(Enter functional equation)\ \ f(x) – f(y) = (y – x)\ f(xy)`
`text(unitl CAS output is “TRUE”)`
`=> D`
If the equation `f(2x) - 2f(x) = 0` is true for all real values of `x`, then the rule for `f` could be
`C`
`text(We need)\ \ f(2x)=2\ f(x),`
`text(Consider)\ C,`
`f(x)` | `=2x,` |
`f(2x)` | `= 2(2x)` |
`= 2\ f(x)` |
`text(CAS can be used to quickly prove other answers)`
`text(do not satisfy equation.)`
`=> C`
If `f: text{(−∞, 1)} -> R,\ \ f(x) = 2 log_e (1 - x)\ \ text(and)\ \ g: text{[−1, ∞)} -> R, g(x) = 3 sqrt (x + 1),` then the maximal domain of the function `f + g` is
`A`
`text(Consider)\ \ f(x) = 2 log_e (1 – x):`
`(1-x)` | `>0` |
`:. x` | `<1` |
`text(Consider)\ \ g(x) = 3 sqrt (x + 1):`
`(x+1)` | `>=0` |
`:. x` | `>= -1` |
`:.\ text(The maximal domain of)\ \ f + g\ \ text{is [−1, 1)}.`
`=> A`