1. The table above shows some values for the function f. If f(x)=ka^x for some constants k and a, what is the value of a?
A. 1\2
B. 1\4
C. 2
D. 4 Correct
Explanation
To find the value of a, we can use the points provided in the table and the given function f(x)=kax. First, use the point (0,1/2) since any base raised to the power of 0 is 1. Substituting these values into the function gives 1/2=ka^0, which simplifies to 1/2=k(1), so k=1/2. Now that we have the value of k, we can substitute it back into the function to get f(x)=1/2ax. Next, use another point from the table, like (1,2), to solve for a. Substituting the values into the equation gives 2=1/2a^1. To isolate a, multiply both sides by 2, resulting in a=4. This value can be confirmed with the remaining point (−1,1/8), where 1/2(4)^-1=1/2(1/4)=1/8, which is consistent. The other options are incorrect as they do not satisfy the function with the given points.
2. If f(x)=f(−x) for all x, then f(x) could NOT be
A. x^2-2
B. x^2+2
C. x^4-x^2
D. (x^3-1)^2 Correct
E. (x^3-x)^2
Explanation
The condition f(x)=f(−x) defines an even function, which is a function that is symmetric with respect to the y-axis. To check if a function is even, one must substitute −x for x and see if the resulting expression is the same as the original function. When we test option D, f(x)=(x^3−1)^3, and substitute −x, we get f(−x)=((−x)^3−1)^2=(−x^3−1)^2. Expanding this results in (x^3+1)^2, which is not equivalent to the original function (x^3−1)^2. In contrast, options A, B, C, and E all satisfy the condition because any term with an even exponent remains unchanged when −x is substituted for x, making them even functions
3. sin 45 deg * cos 45 deg * tan 45 deg =
A. 01-Apr
B. (sqrt(2))/4
C. 01-Feb Correct
D. (sqrt(2))/2
E. sqrt(2)
Explanation
To solve the expression, first replace each trigonometric function with its value at 45 degrees. The sine of 45 degrees is the square root of 2 divided by 2. The cosine of 45 degrees is also the square root of 2 divided by 2. The tangent of 45 degrees is 1. When you multiply the values, you get (the square root of 2 / 2) multiplied by (the square root of 2 / 2) multiplied by 1. The product of the two square roots of 2 is 2, and the product of the two denominators is 4. This results in the fraction 2/4, which simplifies to 1/2.
4. Which expression can be factored into the form (ax+by)^2 where a and b are real number constants?
A. 36x^2-49y^2
B. 81x^2+64y^2
C. 4x^2+20xy+25y^2 Correct
D. 16x^2-24xy-9y^2
Explanation
An expression that can be factored into the form of a squared binomial is called a perfect square trinomial. This type of expression always has three terms: the square of a first term, a middle term that is twice the product of the two terms in the binomial, and the square of a second term. The expression 4x^2+20xy+25y^2 fits this exact pattern. The first term, 4x^2, is the square of 2x, and the last term, 25y^2, is the square of 5y. The middle term, 20xy, is exactly twice the product of 2x and 5y. The other expressions do not fit this pattern; the first is a difference of two squares, the second is a sum of two squares, and the last has a negative term where a positive term is required for it to be a perfect square trinomial.
5. The graph of y=5x^2-20x+17 in the xy-plane is shown above. If k is a constant such that the graph of y=5x^2-20x+(17+k) does not intersect the x-axis, which of the following could be the value of k?
A. -3
B. 2
C. 3
D. 4 Correct
Explanation
The original graph of y=5x^2−20x+17 is a parabola that opens upwards and has its vertex at a y-value of -3. For the new parabola, y=5x^2−20x+(17+k), to not intersect the x-axis, its vertex must be shifted upwards so that its lowest point is above the x-axis. The term plus k represents a vertical shift. Therefore, the y-value of the new vertex, which is the original y-value plus k, must be greater than 0. The original y-value is -3, so we need to find a value for k such that −3+k>0. By adding 3 to both sides of the inequality, we find that k must be greater than 3. The only answer choice that satisfies this condition is 4.