1. If y+x = pi/2, what is the smallest value of tan(x) +cot(y)?
a) 0
b) 1
c) pi/2
d) 2
2. f(x) = ax^3 + bx^2 + cx +d. Where a, b, c, and d belongs to R, a does not equal 0 is increasing on R if
a) a > 0
b) ab > 0c) b^2-3ac < 0, a <0
d) b^2-3ac < 0, a > 03. The second derivative of f(x) is everywhere positive. f has a minimum value at x = a. The tangent line to f at x = b is given by the equation y = mx + c. Which of the following statements is true?a) f(x) >= mx + c b) f(x) > mx + c c) f(x) <= mx + c d) f(x) < mx + c
4. At some point (a, f(a)) on the graph of f(x) = -1 + 2x - x^2 the tangent to this graph goes through the origin. Which point is it?a) (0,-1)
b) (1,0), (2,-1)
c) (1,0), (-1,-4)
d) (-1,-4), (2,-1)
Explain, please?
2020-06-01 5:59 am
回答 (2)
2020-06-01 6:49 am
Hints: 1. This can be directly answered via highschool trig algebra.
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2020-06-01 6:46 am
1. If y+x = pi/2, what is the smallest value of tan(x) +cot(y)?
y = (pi/2) - x
cot(y) = cot((pi/2) - x) = tan(x)
tan(x) + cot(y) = tan(x) + tan(x) = 2*tan(x)
To minimize that, we need to minimize tan(x), but that's impossible as tan(x) has no minimum; it can approach negative infinity. So there is no answer.
2.
f(x) = ax^3 + bx^2 + cx + d
f'(x) = 3ax^2 + 2bx + c
If f is increasing for all x, then f'(x) >= 0 for all x. Right away, this means that a > 0, so that f'(x) can be positive when x approaches positive or negative infinity. But we also need 3ax^2 + 2bx + c, a quadratic with all real coefficients, to have at most one unique root, so its discriminant is either zero or negative:
(2b)^2 - 4*3a*c <= 0
4b^2 - 12ac <= 0
b^2 - 3ac <= 0
Answer (d) is close, but it says b^2 - 3ac < 0. It's also allowed that b^2 - 3ac = 0. Remember, "increasing on R" means if you pick any TWO values for x, the higher value for x will have the higher value for y. If the discriminant above is zero, that means that's only ONE value of x where the tangent line of f is horizontal -- that does NOT allow you to pick TWO points where a higher x value does not translate into a higher y value. So answer (d) is close, but not quite perfect.
y = (pi/2) - x
cot(y) = cot((pi/2) - x) = tan(x)
tan(x) + cot(y) = tan(x) + tan(x) = 2*tan(x)
To minimize that, we need to minimize tan(x), but that's impossible as tan(x) has no minimum; it can approach negative infinity. So there is no answer.
2.
f(x) = ax^3 + bx^2 + cx + d
f'(x) = 3ax^2 + 2bx + c
If f is increasing for all x, then f'(x) >= 0 for all x. Right away, this means that a > 0, so that f'(x) can be positive when x approaches positive or negative infinity. But we also need 3ax^2 + 2bx + c, a quadratic with all real coefficients, to have at most one unique root, so its discriminant is either zero or negative:
(2b)^2 - 4*3a*c <= 0
4b^2 - 12ac <= 0
b^2 - 3ac <= 0
Answer (d) is close, but it says b^2 - 3ac < 0. It's also allowed that b^2 - 3ac = 0. Remember, "increasing on R" means if you pick any TWO values for x, the higher value for x will have the higher value for y. If the discriminant above is zero, that means that's only ONE value of x where the tangent line of f is horizontal -- that does NOT allow you to pick TWO points where a higher x value does not translate into a higher y value. So answer (d) is close, but not quite perfect.
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