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JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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The range of values of aa for which the function f(x)=(a2−7a+12)cos⁡x+2(a−4)x+3e5f\left( x \right) = \left( {{a^2} - 7a + 12} \right)\cos x + 2\left( {a - 4} \right)x + 3{e^5} does not possess critical points, is
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Correct option: B
(1,  4)∪(4,  5)\left( {1,\;4} \right) \cup \left( {4,\;5} \right)
Option B ((1,  4)∪(4,  5)\left( {1,\;4} \right) \cup \left( {4,\;5} \right)) is correct because it represents the values of 'a' for which the function has no critical points, derived from the condition that the derivative has no real solutions.

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