answer:If there exists an x in mathbb{R} such that x^{2}-mx+1 leqslant 0 holds true, then we need to consider the discriminant of the quadratic expression. The discriminant Delta = m^{2} - 4. For the quadratic expression to be less than or equal to zero, the discriminant must be greater than or equal to zero. Computing the discriminant, we have m^{2} - 4 geqslant 0. Solving for m, we get m geqslant 2 or m leqslant -2. Therefore, the range of values for m is boxed{m geqslant 2 text{ or } m leqslant -2}.

answer:Since the two roots of the quadratic equation 3x^2+kx+12=0 are equal, the discriminant must equal zero. For a quadratic given by ax^2+bx+c=0, the discriminant is Delta = b^2-4ac. Let's apply this formula to our equation: [ Delta = k^2 - 4 cdot 3 cdot 12 = k^2 - 144 ] To have equal roots, the discriminant must be zero: [ k^2 - 144 = 0 ] Solving for k gives us: [ k^2 = 144 ] Taking the square root of both sides, we obtain: [ k = pm sqrt{144} implies k = pm 12 ] Thus, the values of k that satisfy the condition are: [ boxed{k = 12} quad text{or} quad boxed{k = -12} ]

answer:(1) Substituting the point (0, -sqrt{3}) into the function's expression gives us [ -sqrt{3} = 2asin^2 0 + 2sin 0 cos 0 - a, ] which simplifies to a = sqrt{3}. (2) With a = sqrt{3}, the function f(x) becomes [ f(x) = 2sqrt{3}sin^2 x + 2sin 2x - sqrt{3}. ] Applying the double-angle identity sin 2x = 2sin x cos x yields [ f(x) = 2sqrt{3}sin^2 x + 2(2sin x cos x) - sqrt{3}. ] Combining like terms results in [ f(x) = 2sqrt{3}sin^2 x + 4sin x cos x - sqrt{3} = 2sqrt{3}left(frac{1 - cos 2x}{2}right) + 2sin 2x - sqrt{3}, ] and further simplification gives [ f(x) = sqrt{3} - sqrt{3}cos 2x + 2sin 2x. ] We can rewrite this using the sum-to-product formulas as [ f(x) = 2sinleft(2x - frac{pi}{3}right). ] Since 0 leq x leq frac{pi}{2}, it follows that -frac{pi}{3} leq 2x - frac{pi}{3} leq frac{pi}{3}. Therefore, the range of the sine function within this interval is [ - frac{sqrt{3}}{2} leq sinleft(2x - frac{pi}{3}right) leq 1. ] Multiplying this inequality by 2 gives [ - sqrt{3} leq 2sinleft(2x - frac{pi}{3}right) leq 2. ] Thus, the range of f(x) is boxed{[-sqrt{3}, 2]}.

answer:Let's denote the number of oranges as O and the number of apples as A. According to the problem, there are twice as many apples as oranges, so we can write: A = 2O We also know that there are 5 bananas, so let's denote the number of bananas as B = 5. The total number of fruits on the display is 35, so we can write the following equation: A + O + B = 35 Substituting A = 2O and B = 5 into the equation, we get: 2O + O + 5 = 35 Combining like terms, we have: 3O + 5 = 35 Subtracting 5 from both sides, we get: 3O = 30 Dividing both sides by 3, we find the number of oranges: O = 10 Now that we know there are 10 oranges and 5 bananas, we can find the ratio of the number of oranges to the number of bananas: Ratio of oranges to bananas = O : B = 10 : 5 To simplify the ratio, we divide both numbers by the greatest common divisor, which is 5: 10 ÷ 5 : 5 ÷ 5 = 2 : 1 So, the ratio of the number of oranges to the number of bananas on the display is boxed{2:1} .