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Let f be an odd function defined on the set of real number such that for x greater than or equal to 0, f(x)=3sinx 4cosx then f(x) at x = -11pi÷6. (1. 3÷2 2root3 2. -3÷2 2root3 3.3÷2-2root2 4. -3÷2-2root3)which of these option is correct and explain it?
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Let f be an odd function defined on the set of real number such that f...
Solution:

Given, f(x) is an odd function defined on the set of real numbers such that for x greater than or equal to 0, f(x) = 3sinx 4cosx.

Since f(x) is an odd function, we can say that f(-x) = -f(x) for all x in the domain of f.

Therefore, for x greater than or equal to 0, we have

f(-x) = -f(x)
=> f(-11pi/6) = -f(11pi/6)
=> f(-11pi/6) = -3sin(11pi/6) + 4cos(11pi/6)

Now, we know that sin(11pi/6) = -1/2 and cos(11pi/6) = -sqrt(3)/2.

Substituting these values in the above equation, we get

f(-11pi/6) = -3(-1/2) + 4(-sqrt(3)/2)
=> f(-11pi/6) = 3/2 - 2sqrt(3)

Therefore, the correct option is 3. 3/2 - 2sqrt(3).

Explanation:

We are given an odd function f(x) which is defined on the set of real numbers such that for x greater than or equal to 0, f(x) = 3sinx 4cosx.

We need to find the value of f(x) at x = -11pi/6.

Since f(x) is an odd function, we can use the property f(-x) = -f(x) to find the value of f(-11pi/6).

Substituting the values of sin(11pi/6) and cos(11pi/6) in the expression for f(-x), we get the value of f(-11pi/6) as 3/2 - 2sqrt(3).

Therefore, the correct option is 3. 3/2 - 2sqrt(3).
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Let f be an odd function defined on the set of real number such that f...
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Let f be an odd function defined on the set of real number such that for x greater than or equal to 0, f(x)=3sinx 4cosx then f(x) at x = -11pi÷6. (1. 3÷2 2root3 2. -3÷2 2root3 3.3÷2-2root2 4. -3÷2-2root3)which of these option is correct and explain it?
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