The solution set of the equation `e^(4x) - 5e^(2x) + 4 = 0` over `R` is
A. `{1, 4}`
B. `{– 4, – 1})`
C. `{– 2, – 1, 1, 2})`
D. `{– log_e(2), 0, log_e(2)}`
E. `{0, log_e(2)}`
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The solution set of the equation `e^(4x) - 5e^(2x) + 4 = 0` over `R` is
A. `{1, 4}`
B. `{– 4, – 1})`
C. `{– 2, – 1, 1, 2})`
D. `{– log_e(2), 0, log_e(2)}`
E. `{0, log_e(2)}`
`E`
`text(Solve for)\ \ x\ \ text(on CAS:)`
`x = 0 \ or \ x = log_e(2)`
`=> E`
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 graph of `f(x) = (x - 1)^2 - 2, x in [– 2, 2]`, is shown below. The graph intersects the `x`-axis where `x = a.`
Find the value of `a.` (1 mark)
Note: other parts of this question are out of the syllabus and not included.
`1 – sqrt 2`
`text(Find)\ \ x text(-intercept:)`
`(x – 1)^2 – 2` | `= 0` |
`(x – 1)^2` | `= 2` |
`x – 1` | `= +- sqrt 2` |
`x` | `= 1 +- sqrt 2,\ \ a < 0` |
`:. a` | `= 1 – sqrt 2` |
The range of the function `f:\ text{(−1, 2]} -> R,\ \ f(x) = -x^2 + 2x-3` is
`C`