If the following function is continuous, what is the value of a + b?ℎ(𝑡)={3𝑡2−2𝑡+1,if 𝑡<0𝑎cos(𝑡)+𝑏,if 0≤𝑡≤𝜋34sin2𝑡,if 𝑡>𝜋3h(t)= ⎩⎨⎧ 3t 2 −2t+1,acos(t)+b,4sin 2 t, if t<0if 0≤t≤ 3π if t> 3π A.1B.0C.4D.3E.2
Question
If the following function is continuous, what is the value of a + b?ℎ(𝑡)={3𝑡2−2𝑡+1,if 𝑡<0𝑎cos(𝑡)+𝑏,if 0≤𝑡≤𝜋34sin2𝑡,if 𝑡>𝜋3h(t)= ⎩⎨⎧ 3t 2 −2t+1,acos(t)+b,4sin 2 t, if t<0if 0≤t≤ 3π if t> 3π A.1B.0C.4D.3E.2
Solution
For a function to be continuous, the function values must be the same at the point where the function changes form. In this case, the function changes form at t=0 and t=π/3.
At t=0, the two parts of the function are 3t^2 - 2t + 1 and acos(t) + b. Setting these equal to each other gives:
3(0)^2 - 2(0) + 1 = a*cos(0) + b 1 = a + b ----(1)
At t=π/3, the two parts of the function are acos(t) + b and 4sin^2(t). Setting these equal to each other gives:
acos(π/3) + b = 4sin^2(π/3) a*(1/2) + b = 4*(3/4) a/2 + b = 3 ----(2)
Subtracting equation (1) from equation (2) gives:
a/2 = 2 a = 4
Substituting a = 4 into equation (1) gives:
4 + b = 1 b = -3
Therefore, a + b = 4 - 3 = 1. So, the answer is A.1.
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