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The set of values of the parameter $a$ for which the function $f(x) = 8ax – a \sin 6x – 7x – \sin 5x$ increases and has no critical points in $\mathbb R$, is

(A) $[–1, 1]$

(B) $(–\infty, –6)$

(C) $(6, +\infty)$

(D) $[6, +\infty)$

For the function to be increasing its derivative should be greater then zero. Proceeding in the same path I get $$a=\frac{5\cos5x+7}{8-6\cos6x}.$$

What does this suggest? How to proceed with this problem?

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  • $\begingroup$ When you solved for $a$, did you throw in an inequality, i.e. $0<f(x)$, or did you solve $f(x)=0$? $\endgroup$ – Ian Coley Sep 13 '13 at 17:00
  • $\begingroup$ I actually solved for f'(x)=0. I was actually trying to eliminate the values of a for which the function would have a critical point as it is given in the question that 'the value of a for which there should not be any critical points.'How to proceed with this please help. $\endgroup$ – Rajath Krishna R Sep 14 '13 at 3:50
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As Ian Coley indicated, your problem is that you stuck in an $=$ where you should have gotten an inequality. After that, think about how to make a fraction big.

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