If the following function is continuous, then what is the value of a?ℎ(𝑡)={2𝑡+𝑏𝑖𝑓 𝑡<02cos⁡(𝑡)−3𝑖𝑓 0≤𝑡≤𝜋2𝑎sin⁡(𝑡)+5𝑏𝑖𝑓 𝑡>𝜋2h(t)= ⎩⎪⎨⎪⎧​ 2t+b2cos(t)−3asin(t)+5b​ if t<0if 0≤t≤ 2π​ if t> 2π​ ​ A.1B.ℎ(𝑡)={2𝑡+𝑏𝑖𝑓 𝑡<02cos⁡(𝑡)−3𝑖𝑓 0≤𝑡≤𝜋2𝑎sin⁡(𝑡)+5𝑏𝑖𝑓 𝑡>𝜋2h(t)= ⎩⎪⎨⎪⎧​ 2t+b2cos(t)−3asin(t)+5b​ if t<0if 0≤t≤ 2π​ if t> 2π​ ​ C.4D.0E.2

The question seems to be a bit jumbled, but it appears to be asking for the value of 'a' that would make the piecewise function h(t) continuous.

For a function to be continuous, the limit as t approaches a certain value from the left (t-) must equal the limit as t approaches that value from the right (t+), and both must equal the value of the function at that point.

Given the function:

h(t) = { 2t+b if t<0, 2cos(t)-3 if 0≤t≤π/2, asin(t)+5b if t>π/2 }

We need to find the value of 'a' that makes this function continuous.

First, we need to find the value of the function at t=0 and t=π/2, and set the left and right limits equal to these values.

At t=0, h(t) = 2cos(0)-3 = 2-3 = -1.

At t=π/2, h(t) = 2cos(π/2)-3 = 0-3 = -3.

So, we need to find 'a' such that:

1. The limit as t approaches 0 from the left (2t+b) equals -1, and
2. The limit as t approaches π/2 from the right (asin(t)+5b) equals -3.

From the first condition, we can see that b must be -1 (since 2*0 - 1 = -1).

Substituting b=-1 into the second condition gives us:

asin(π/2) - 5 = -3 a1 - 5 = -3 a = -3 + 5 a = 2

So, the value of 'a' that makes the function continuous is 2 (Option E).