Final answer:
The value of k for which the line y = 3x + k is tangent to the curve y = x³ is 2 or -2, corresponding to choice (a). This is determined by setting the derivative of the curve equal to the slope of the line and solving for x, then finding the corresponding y values on the curve and solving for k in the line equation.
Step-by-step explanation:
To find the value of k for which the line y = 3x + k is tangent to the curve y = x³, we first need to determine the point at which they touch by equating their derivatives. The derivative of the curve, which gives us the slope at any point, is found by differentiating y = x³ with respect to x, yielding dy/dx = 3x². As the slope of the line is constantly 3 (as given by the coefficient of x in the line equation), we set 3x² = 3 to find the point where the slopes agree, resulting in x² = 1.
Then, we solve for x, which gives us x = 1 or x = -1. Next, we substitute these values back into the original curve equation to get the corresponding y values: for x = 1, y = 1³ = 1, and for x = -1, y = (-1)³ = -1. Lastly, we use these points to solve for k in the linear equation y = 3x + k by plugging in the values of x and y. For x = 1 and y = 1, we get 1 = 3(1) + k, which simplifies to k = -2. For x = -1 and y = -1, we get -1 = 3(-1) + k, which simplifies to k = 2.Therefore, the value of k for the line to be tangent to the curve is 2 or -2, which corresponds to the choice (a) in the given options.he line y = 3x + k is tangent to the curve y = x³ when k is equal to a.2 or -2To determine when the line is tangent to the curve, we need to find the value of k that makes the slope of the line equal to the slope of the curve at the point of tangency.The slope of the line y = 3x + k is always 3, so we need to find the value of x where the slope of the curve y = x³ is also 3. By finding the derivative of y = x³ and setting it equal to 3, we can solve for x and then find the corresponding value of k.After solving for x, we get x = 1, which means k can be either 2 or -2.