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3sinx + 4cosx + r is always greater than or equal to 0. What is the smallest value ‘r’ can to take?
- a)5
- b)-5
- c)4
- d)3
Correct answer is option 'A'. Can you explain this answer?
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3sinx + 4cosx + r is always greater than or equal to 0. What is the sm...
Therefore, the answer is Option A.
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3sinx + 4cosx + r is always greater than or equal to 0. What is the sm...
To find the smallest value, we need to find the minimum value of the expression 3sinx + 4cosx.
We can rewrite this expression using the identity sin^2(x) + cos^2(x) = 1:
3sinx + 4cosx = sqrt(3^2 + 4^2) * (3/5 * sinx + 4/5 * cosx)
= 5 * (3/5 * sinx + 4/5 * cosx)
= 5 * cos(x - α)
where α is the angle whose cosine is 3/5 and sine is 4/5.
Since cos(x - α) is always between -1 and 1, the minimum value of 5 * cos(x - α) is -5.
Therefore, the smallest value of the expression 3sinx + 4cosx is -5.
We can rewrite this expression using the identity sin^2(x) + cos^2(x) = 1:
3sinx + 4cosx = sqrt(3^2 + 4^2) * (3/5 * sinx + 4/5 * cosx)
= 5 * (3/5 * sinx + 4/5 * cosx)
= 5 * cos(x - α)
where α is the angle whose cosine is 3/5 and sine is 4/5.
Since cos(x - α) is always between -1 and 1, the minimum value of 5 * cos(x - α) is -5.
Therefore, the smallest value of the expression 3sinx + 4cosx is -5.
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3sinx + 4cosx + r is always greater than or equal to 0. What is the smallest value ‘r’ can to take?a)5b)-5c)4d)3Correct answer is option 'A'. Can you explain this answer?
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