Which of the following functions is always increasing?

x + sin 2 x

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x - sin 2 x

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2 x + sin 3 x

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2 x - sin x

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Solution

The correct option is D $2 x - sin x$
A function is said to be always increasing if its derivative is always greater than zero. i.e, $\frac{d y}{d x} > 0$
$(a) y = x + s i n 2 x$
$\frac{d y}{d x} = 1 + 2 c o s 2 x$
$\Rightarrow$ $- 1 \leq (1 + 2 c o s 2 x) \leq 3 ∵ (- 1 \leq c o s 2 x \leq 1)$

$(b) y = x - s i n 2 x$
$\frac{d y}{d x} = 1 - 2 c o s 2 x$
$\Rightarrow$ $- 1 \leq (1 - 2 c o s 2 x) \leq 3 ∵ (- 1 \leq c o s 2 x \leq 1)$

$(c) y = 2 x + s i n 3 x$
$\frac{d y}{d x} = 3 + 3 c o s 3 x$
$\Rightarrow$ $0 \leq (3 + 3 c o s 3 x) \leq 6 ∵ (- 1 \leq c o s 3 x \leq 1)$

$(d) y = 2 x - s i n x$
$\frac{d y}{d x} = 2 - c o s x$
$\Rightarrow$ $1 \leq (2 - c o s x) \leq 3 ∵ (- 1 \leq c o s x \leq 1)$

Clearly in the case of (d) dy/dx > 0
$∴ 2 x - sin x$ is always increasing.

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