Step 1: Find the derivative of the function
The given function is:
f(x)=2sinx+sin2x
To find the critical points, we differentiate f(x) with respect to x:
f′(x)=dxd(2sinx+sin2x)
f′(x)=2cosx+2cos2x
Step 2: Find the critical points
Set f′(x)=0 to look for critical values in the given interval [0,23π]:
2cosx+2cos2x=0
cosx+cos2x=0
Using the trigonometric identity cos2x=2cos2x−1:
2cos2x+cosx−1=0
This is a quadratic equation in terms of cosx. Factoring it gives:
2cos2x+2cosx−cosx−1=0
2cosx(cosx+1)−1(cosx+1)=0
(2cosx−1)(cosx+1)=0
This yields two cases:
2cosx−1=0⟹cosx=21
cosx+1=0⟹cosx=−1
Now, find the values of x that lie within the interval [0,23π]:
For cosx=21⟹x=3π
For cosx=−1⟹x=π
Step 3: Evaluate the function at critical points and boundary points
To find the absolute maximum (greatest value), we evaluate f(x) at the critical points (x=3π,π) and the endpoints (x=0,23π).
At x=0:
f(0)=2sin(0)+sin(0)=0
At x=3π:
f(3π)=2sin(3π)+sin(32π)
f(3π)=2(23)+23=3+23=233
At x=π:
f(π)=2sin(π)+sin(2π)=2(0)+0=0
At x=2π⋅3=23π:
f(23π)=2sin(23π)+sin(3π)
f(23π)=2(−1)+0=−2
Step 4: Compare the values
Comparing the computed values:
f(0)=0
f(3π)=233≈2.598
f(π)=0
f(23π)=−2
The greatest value among these is 233.
Correct Answer
The correct option is (d).