The function is:
\[ f(x) = \cos x - x + 1. \]
The derivative is:
\[ f'(x) = -\sin x - 1. \]
Since \( -\sin x - 1 < 0 \) for all \( x \in \mathbb{R} \), \( f(x) \) is strictly decreasing in \([0, \pi]\).
At \( x = 0 \),
\[ f(0) = \cos(0) - 0 + 1 = 2. \]
At \( x = \pi \),
\[ f(\pi) = \cos(\pi) - \pi + 1 = -\pi < 0. \]
By the intermediate value theorem, \( f(x) = 0 \) has exactly one root in \([0, \pi]\). Thus, (S1) is correct.
(S2) is incorrect because \( f(x) \) is strictly decreasing in \([0, \pi]\).
Final Answer: Only (S1) is correct.
The value of \[ \lim_{x \to \infty} \left( x - \sqrt{x^2 + x} \right) \] is equal to:
Integration of \(\ln(x)\) with \(x\), i.e. \(\int \ln(x)dx =\) __________.
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: