Study questions platform-wide or filter by specific tests with correct answers revealed.
- They are hydrophobic due to alkyl groups.
- They are biocompatible and used in medical implants.
- They have high thermal stability and high dielectric strength (statement 'c' is incorrect because it says low dielectric strength).
- They are resistant to oxidation and chemicals.
PE Classification (ESC 2019):
| Category | Criteria | Treatment |
|---|---|---|
| Massive (High Risk) | Hemodynamic instability: SBP \(< 90 \text{ mmHg}\) or drop \(\geq 40 \text{ mmHg}\) \(> 15\) min | Systemic thrombolysis or surgical embolectomy |
| Submassive (Intermediate) | Hemodynamically stable + RV dysfunction or elevated troponin/BNP | Anticoagulation ± consider thrombolysis |
| Low Risk | Hemodynamically stable, no RV dysfunction | Anticoagulation (DOAC or heparin) |
Absolute contraindications to thrombolysis include: prior intracranial hemorrhage, active bleeding, recent surgery \(< 3\) weeks. If contraindicated → surgical embolectomy or catheter-directed thrombolysis.
In the $xy$-plane, line $k$ has equation $3x - 2y = 0$. Compare:
Quantity A: The $x$-intercept of line $k$
Quantity B: The $y$-intercept of line $k$
Find the intercepts of $3x - 2y = 0$:
x-intercept (set $y = 0$): $3x = 0 \Rightarrow x = 0$.
y-intercept (set $x = 0$): $-2y = 0 \Rightarrow y = 0$.
Both intercepts are $0$ (the line passes through the origin).
Quantity A $= 0 =$ Quantity B. The two quantities are equal.
For $x\in\mathbb{R}$, let $f(x)=|\log 2-\sin x|$ and $g(x)=f(f(x))$. Which of the following is correct?
Since $\log 2 \approx 0.693$ and $\sin 0=0$, we have $f(0)=\log 2 > 0$.
Near $x=0$: $f(x)=\log 2-\sin x$ (since $\log 2>\sin x$ locally). So $f$ is smooth near $x=0$.
$g(x) = f(f(x)) = |\log 2 - \sin(f(x))|$. At $x=0$: $f(0)=\log 2$, so $\sin(f(0))=\sin(\log 2)$. Since $\log 2 < 1 < \pi$, $\sin(\log 2)>0$.
$g(0) = \log 2 - \sin(\log 2) > 0$, so near $x=0$, $g(x)=\log 2 - \sin(f(x))$.
$g'(0)=-\cos(f(0))\cdot f'(0)=-\cos(\log 2)\cdot(-\cos 0)=\cos(\log 2)$
Two cars P and Q start simultaneously from the same point. Their positions are $x_P(t) = at + bt^2$ and $x_Q(t) = ft - t^2$. At what time do they have equal velocities?
$v_P = a + 2bt$, $\quad v_Q = f - 2t$
Setting equal: $a + 2bt = f - 2t \Rightarrow 2t(b+1) = f-a \Rightarrow t = \mathbf{\dfrac{f-a}{2(1+b)}}$
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