If $f(x) = sin\, (sin\, x)$ and $f'' (x) + tan \,x\,f'(x)+ g(x) = 0,$ then $g(x) $ is :

Question

Let $f(x)=|\log_e x|-|x-1|+5$. Statement 1: $f(x)$ is differentiable for all $x\in(0,\infty)$
Statement 2: $f(x)$ is increasing in $(1,\infty)$
Statement 3: $f(x)$ is decreasing in $(0,1)$
Which of the following is correct?

Answer 1

To solve the problem, we need to find the function $ g(x) $ given the function $ f(x) = \sin (\sin x) $ and the differential equation:

f''(x) + \tan x \, f'(x) + g(x) = 0

Let's differentiate $ f(x) $:

First derivative of $ f(x) = \sin(\sin x) $:
\[ f'(x) = \frac{d}{dx}[\sin(\sin x)] = \cos(\sin x) \cdot \cos x \]
Second derivative of $ f(x) $:
\[ f''(x) = \frac{d}{dx}[\cos(\sin x) \cdot \cos x] \]

Using the product rule: \[ f''(x) = \frac{d}{dx}[\cos(\sin x)] \cdot \cos x + \cos(\sin x) \cdot \frac{d}{dx}[\cos x] \]

Calculating each part: \[ \frac{d}{dx}[\cos(\sin x)] = -\sin(\sin x) \cdot \cos x \] \[ \frac{d}{dx}[\cos x] = -\sin x \]

Thus, \[ f''(x) = [-\sin(\sin x) \cdot \cos x] \cdot \cos x + \cos(\sin x) \cdot [-\sin x] \]

Which simplifies to: \[ f''(x) = -\sin(\sin x) \cdot \cos^2 x - \cos(\sin x) \cdot \sin x \]

Substitute $ f'(x) $ and $ f''(x) $ into the differential equation:

f''(x) + \tan x \, f'(x) + g(x) = 0\

Substitute the expressions: \[ -\sin(\sin x) \cdot \cos^2 x - \cos(\sin x) \cdot \sin x + \tan x \cdot [\cos(\sin x) \cdot \cos x] + g(x) = 0 \]

Simplifying the middle term: \[ \tan x \cdot [\cos(\sin x) \cdot \cos x] = \frac{\sin x}{\cos x} \cdot [\cos(\sin x) \cdot \cos x] = \cos(\sin x) \cdot \sin x \]

Substitute and cancel terms: \[ -\sin(\sin x) \cdot \cos^2 x - \cos(\sin x) \cdot \sin x + \cos(\sin x) \cdot \sin x + g(x) = 0 \]

The $\cos(\sin x) \cdot \sin x$ terms cancel out, leaving: \[ -\sin(\sin x) \cdot \cos^2 x + g(x) = 0 \]

This implies: \[ g(x) = \sin(\sin x) \cdot \cos^2 x \]

So, the function $ g(x) $ is: \cos^2 x \sin(\sin x)\

Thus, the correct answer is: \cos^2 x \sin(\sin x)\

Answer 2