Functions Question 744
Question: Let $ f:R\to R $ be a function such that $ f(x)=ax+3sinx+4cosx. $ Then f(x) is invertible if
Options:
A) $ a\in (-5,5) $
B) $ a\in (-\infty ,-5) $
C) $ a\in (0,+\infty ) $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
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For a function to be invertible, it must be both injective (one-to-one) and surjective (onto). In the case of a continuous function on R, it’s sufficient to prove that the function is strictly monotonic (always increasing or always decreasing).
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To check if f(x) is strictly monotonic, we need to examine its derivative. If f’(x) is always positive or always negative for all x, then f(x) is strictly monotonic.
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Let’s find f’(x): f’(x) = a + 3cos(x) - 4sin(x)
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Now, we need to find the conditions for a that make f’(x) always positive or always negative.
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We know that -1 ≤ sin(x) ≤ 1 and -1 ≤ cos(x) ≤ 1 for all x.
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The minimum value of f’(x) will occur when 3cos(x) is at its minimum (-3) and -4sin(x) is at its minimum (-4). So:
f’(x) ≥ a - 3 - 4 = a - 7
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The maximum value of f’(x) will occur when 3cos(x) is at its maximum (3) and -4sin(x) is at its maximum (4). So:
f’(x) ≤ a + 3 + 4 = a + 7
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For f(x) to be strictly increasing, we need f’(x) > 0 for all x. This means:
a - 7 > 0 a > 7
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For f(x) to be strictly decreasing, we need f’(x) < 0 for all x. This means:
a + 7 < 0 a < -7
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Therefore, f(x) is invertible if a > 7 or a < -7.
Looking at the given options, we can see that option B, a ∈ (-∞, -5), partially satisfies this condition (it includes all a < -7).
However, the correct range for a should be (-∞, -7) ∪ (7, ∞), which is not given as an option.
Therefore, the correct answer is D) None of these.
BETA
