Question

Which of the following functions is always increasing?
A.$x + \sin 2x$
B. $x - \sin 2x$
C. $2x + \sin 3x$
D. $x + \sin 2x$

Answer 1

Hint: Differentiate every given option with respect to x and when whether the result is always greater than zero or not. If the result is always greater than zero then the function is increasing.

Formula Used:
A function $f(x)$ is increasing if $f'(x) > 0,\forall x.$
$\dfrac{d}{{dx}}(x) = 1$
$\dfrac{d}{{dx}}(\sin nx) = n\cos nx$
The value of cosine always lies between -1 to 1, that is $ - 1 \le \cos \theta \le 1$.

Complete step by step solution:
A. Differentiate $x + \sin 2x$ with respect to x.
$\dfrac{d}{{dx}}(x + \sin 2x) = 1 + 2\cos 2x$
Now,
$ - 1 \le \cos 2x \le 1$
Multiply both sides of the inequality by 2.
$ - 2 \le 2\cos x \le 2$
Add 1 to both sides of the inequality.
 $1 - 2 \le 1 + 2\cos x \le 1 + 2$
$ - 1 \le 1 + 2\cos x \le 3$
B. Differentiate $x - \sin 2x$ with respect to x.
$\dfrac{d}{{dx}}(x - \sin 2x) = 1 - 2\cos 2x$
Now,
$ - 1 \le \cos 2x \le 1$
Multiply both sides of the inequality by -2.
$2 \ge 2\cos x \ge - 2$
$ - 2 \le 2\cos x \le 2$
Add 1 to both sides of the inequality.
 $1 - 2 \le 1 - 2\cos x \le 1 + 2$
$ - 1 \le 1 - 2\cos x \le 3$
C. Differentiate $2x + \sin 3x$ with respect to x.
$\dfrac{d}{{dx}}(2x + \sin 3x) = 2 + 3\cos 3x$
Now,
$ - 1 \le \cos 3x \le 1$
Multiply both sides of the inequality by 3.
$ - 3 \le 3\cos x \le 3$
Add 2 to both sides of the inequality.
 $2 - 3 \le 1 + 2\cos x \le 2 + 3$
$ - 1 \le 1 + 2\cos x \le 5$
D. Differentiate $2x - \sin x$ with respect to x.
$\dfrac{d}{{dx}}(2x - \sin x) = 2 - \cos x$
Now,
$ - 1 \le \cos x \le 1$
Multiply both sides of the inequality by -1.
$ - 1 \le - \cos x \le 1$
Add 2 to both sides of the inequality.
 $2 - 1 \le 2 - \cos x \le 1 + 2$
$1 \le 2 - \cos x \le 3$

Option ‘D’ is correct

Note: One has to be careful when doing the calculations with the inequality, as the inequality signs get reversed when multiplied by negative signs.