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Factoring both the numerator and the denominator gives $\frac{x^2 - 4}{x^2 - 3x + 2} = \frac{(x - 2)(x + 2)}{(x - 2)(x - 1)}$. For $x \neq 2$, we can cancel the common factor $(x - 2)$ to obtain $\frac{x + 2}{x - 1}$. Taking the limit as $x \to 2$ yields $\frac{2 + 2}{2 - 1} = \frac{4}{1} = 4$.
Recall the half-angle identity $1 - \cos(4x) = 2\sin^2(2x)$. Substituting this into the limit gives $\lim_{x \to 0} \frac{2\sin^2(2x)}{x\sin(2x)} = \lim_{x \to 0} \frac{2\sin(2x)}{x}$. Since $\lim_{x \to 0} \frac{\sin(2x)}{2x} = 1$, we have $\lim_{x \to 0} 2 \cdot 2 \cdot \frac{\sin(2x)}{2x} = 4 \cdot 1 = 4$. Alternatively, applying L'Hôpital's rule: $\frac{4\sin(4x)}{\sin(2x) + 2x\cos(2x)} \to \frac{16\cos(4x)}{2\cos(2x) + 2\cos(2x) - 4x\sin(2x)} = \frac{16}{4} = 4$.
Multiplying and dividing by the conjugate expression $\sqrt{x^2 + 6x + 5} + x$ yields $\frac{(x^2 + 6x + 5) - x^2}{\sqrt{x^2 + 6x + 5} + x} = \frac{6x + 5}{\sqrt{x^2(1 + 6/x + 5/x^2)} + x}$. For $x > 0$, $\sqrt{x^2} = x$, so we can factor $x$ from the denominator: $\frac{x(6 + 5/x)}{x(\sqrt{1 + 6/x + 5/x^2} + 1)} = \frac{6 + 5/x}{\sqrt{1 + 6/x + 5/x^2} + 1}$. As $x \to +\infty$, this approaches $\frac{6 + 0}{\sqrt{1 + 0 + 0} + 1} = \frac{6}{2} = 3$.
Using the product rule and the chain rule: $f'(x) = \frac{d}{dx}[x^2] \cdot e^{3x} + x^2 \cdot \frac{d}{dx}[e^{3x}] = 2x e^{3x} + x^2 (3e^{3x}) = (2x + 3x^2)e^{3x}$. Evaluating at $x = 1$: $f'(1) = (2(1) + 3(1)^2)e^{3(1)} = (2 + 3)e^3 = 5e^3$.
First, find the $y$-coordinate: $y(1) = \ln(2(1) - 1) = \ln(1) = 0$, so the point of tangency is $(1, 0)$. Next, compute the derivative using the chain rule: $y' = \frac{1}{2x - 1} \cdot \frac{d}{dx}(2x - 1) = \frac{2}{2x - 1}$. At $x = 1$, the slope is $m = y'(1) = \frac{2}{2(1) - 1} = 2$. The equation of the tangent line is $y - y_0 = m(x - x_0) \implies y - 0 = 2(x - 1) \implies y = 2x - 2$.
By the inverse function differentiation theorem, if $y_0 = f(x_0)$ and $f'(x_0) \neq 0$, then $(f^{-1})'(y_0) = \frac{1}{f'(x_0)}$. Here $x_0 = 2$ corresponds to $y_0 = f(2) = 5$. Since $f'(2) = 4$, we have $(f^{-1})'(5) = \frac{1}{f'(2)} = \frac{1}{4} = 0.25$.
Differentiating both sides with respect to $x$: $\frac{d}{dx}(x^2 + 4y^2) = \frac{d}{dx}(8) \implies 2x + 8y \frac{dy}{dx} = 0$. Solving for $\frac{dy}{dx}$ gives $\frac{dy}{dx} = -\frac{2x}{8y} = -\frac{x}{4y}$. Evaluating at the point $(2, 1)$: $\frac{dy}{dx}\Big|_{(2, 1)} = -\frac{2}{4(1)} = -\frac{1}{2}$.
Compute the first derivative: $f'(x) = 1 \cdot e^{-x} + x(-e^{-x}) = (1 - x)e^{-x}$. Compute the second derivative: $f''(x) = -1 \cdot e^{-x} + (1 - x)(-e^{-x}) = (-1 - 1 + x)e^{-x} = (x - 2)e^{-x}$. Since $e^{-x} > 0$ for all $x$, $f''(x) = 0 \iff x = 2$. For $x < 2$, $f''(x) < 0$ (concave downward); for $x > 2$, $f''(x) > 0$ (concave upward). Because the concavity changes at $x = 2$, $(2, 2e^{-2})$ is an inflection point.
Find the antiderivative: $\int (3x^2 - 4x + 1) \, dx = x^3 - 2x^2 + x$. Applying the Fundamental Theorem of Calculus: $\left[x^3 - 2x^2 + x\right]_0^2 = (2^3 - 2(2^2) + 2) - (0^3 - 2(0^2) + 0) = (8 - 8 + 2) - 0 = 2$.
Use substitution: let $u = x^2 + 1$, then $du = 2x \, dx$. When $x = 0$, $u = 0^2 + 1 = 1$. When $x = 1$, $u = 1^2 + 1 = 2$. The integral becomes $\int_{1}^{2} \frac{1}{u} \, du = \left[\ln |u|\right]_1^2 = \ln 2 - \ln 1 = \ln 2 - 0 = \ln 2$.
Use integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \ln x \implies du = \frac{1}{x} \, dx$, and $dv = x \, dx \implies v = \frac{x^2}{2}$. Then $\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \frac{1}{2}\int x \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4}$. Evaluating from $1$ to $e$: at $x = e$, $\frac{e^2}{2}(1) - \frac{e^2}{4} = \frac{e^2}{4}$. At $x = 1$, $\frac{1}{2}(0) - \frac{1}{4} = -\frac{1}{4}$. Subtracting the lower limit: $\frac{e^2}{4} - \left(-\frac{1}{4}\right) = \frac{e^2 + 1}{4}$.
By the Fundamental Theorem of Calculus combined with the chain rule: $\frac{d}{dx} \left(\int_{a}^{u(x)} f(t) \, dt\right) = f(u(x)) \cdot u'(x)$. Here $u(x) = x^2$ and $f(t) = \sqrt{1 + t^2}$, so $u'(x) = 2x$. Therefore, $F'(x) = \sqrt{1 + (x^2)^2} \cdot (2x) = 2x \sqrt{1 + x^4}$. Evaluating at $x = 2$: $F'(2) = 2(2) \sqrt{1 + 2^4} = 4 \sqrt{1 + 16} = 4\sqrt{17}$.
We evaluate the limits of the bounding functions as $x \to 0$: $\lim_{x \to 0} (2 - x^2) = 2 - 0 = 2$ and $\lim_{x \to 0} (2 + x^2) = 2 + 0 = 2$. Since both bounds converge to the same value 2, by the Squeeze Theorem (Κριτήριο Παρεμβολής), $\lim_{x \to 0} f(x) = 2$.
For the finite limit to exist while the denominator approaches $1 - 1 = 0$, the numerator must also vanish at $x = 1$: $1^2 + a(1) + b = 0 \implies b = -a - 1$. Substituting $b$: $x^2 + ax - (a + 1) = (x - 1)(x + a + 1)$. The limit becomes $\lim_{x \to 1} \frac{(x - 1)(x + a + 1)}{x - 1} = \lim_{x \to 1} (x + a + 1) = 1 + a + 1 = a + 2$. Setting $a + 2 = 5$ gives $a = 3$. Then $b = -3 - 1 = -4$.
For $f$ to be continuous at $x = 0$, we must have $\lim_{x \to 0} f(x) = f(0)$. The limit is $\lim_{x \to 0} \frac{\sin(3x)}{x} = 3 \lim_{x \to 0} \frac{\sin(3x)}{3x} = 3(1) = 3$. Therefore, $k^2 - 1 = 3 \implies k^2 = 4$. Since $k > 0$, we have $k = 2$.
$P(x)$ is continuous on $[0, 1]$ as a polynomial function. We evaluate the endpoints: $P(0) = 0^3 - 3(0) + 1 = 1 > 0$ and $P(1) = 1^3 - 3(1) + 1 = -1 < 0$. Since $P(0)P(1) = 1(-1) = -1 < 0$, by Bolzano's Theorem there exists at least one $x_0 \in (0, 1)$ such that $P(x_0) = 0$.
Define the auxiliary function $g(x) = f(x) - x$ on $[a, b]$. Since $f$ and $x$ are continuous on $[a, b]$, $g$ is continuous on $[a, b]$. Because $f([a, b]) \subseteq [a, b]$, we have $f(a) \ge a \implies g(a) = f(a) - a \ge 0$, and $f(b) \le b \implies g(b) = f(b) - b \le 0$. If $g(a) = 0$, then $c = a$; if $g(b) = 0$, then $c = b$. If $g(a) > 0$ and $g(b) < 0$, then $g(a)g(b) < 0$, and by Bolzano's Theorem there exists $c \in (a, b)$ such that $g(c) = 0 \iff f(c) = c$.
The Intermediate Value Theorem (Θεώρημα Ενδιάμεσων Τιμών) states that if $f$ is continuous on $[a, b]$, it takes every value between $f(a)$ and $f(b)$ at least once. Here $f(1) = 2$ and $f(5) = 10$, so $f$ must attain every value $\eta \in (2, 10)$. Among the choices, only 6 lies strictly between 2 and 10.
The volume of a sphere of radius $r$ is $V = \frac{4}{3}\pi r^3$. Differentiating with respect to time $t$: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. Given $\frac{dV}{dt} = 36\pi\text{ cm}^3/\text{s}$ and $r = 3\text{ cm}$: $36\pi = 4\pi (3)^2 \frac{dr}{dt} = 36\pi \frac{dr}{dt}$. Dividing both sides by $36\pi$ yields $\frac{dr}{dt} = 1\text{ cm/s}$.
A particle is momentarily at rest when its instantaneous velocity $v(t) = s'(t) = 0$. Differentiating $s(t)$: $s'(t) = 3t^2 - 12t + 9$. Setting $s'(t) = 0$: $3(t^2 - 4t + 3) = 0 \implies 3(t - 1)(t - 3) = 0$. Thus, $t = 1\text{ s}$ and $t = 3\text{ s}$.
First, evaluate the point on the curve: $f(1) = 1^3 - 2(1) + 1 = 0$, so the point is $(1, 0)$. Next, find the derivative: $f'(x) = 3x^2 - 2$. At $x = 1$, the tangent slope is $m_{\text{tan}} = f'(1) = 3(1)^2 - 2 = 1$. The slope of the normal line is the negative reciprocal: $m_{\text{norm}} = -\frac{1}{m_{\text{tan}}} = -\frac{1}{1} = -1$. The normal line equation is $y - 0 = -1(x - 1) \implies y = -x + 1$.
Rolle's Theorem requires three specific hypotheses: (1) $f$ is continuous on $[a, b]$, (2) $f$ is differentiable on $(a, b)$, and (3) $f(a) = f(b)$. It does NOT require $f'$ to be continuous on $[a, b]$ or even defined at the endpoints $a$ and $b$.
$f(x)$ is a polynomial, hence continuous on $[0, 2]$ and differentiable on $(0, 2)$. Endpoints: $f(0) = 0$ and $f(2) = 2^3 - 3(2)^2 + 2(2) = 8 - 12 + 4 = 0$. Since $f(0) = f(2) = 0$, Rolle's Theorem applies. The derivative is $f'(x) = 3x^2 - 6x + 2$. Setting $f'(x) = 0$: $x = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = 1 \pm \frac{2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}$. Since $\frac{\sqrt{3}}{3} \approx 0.577$, both $1 - 0.577 \approx 0.423$ and $1 + 0.577 \approx 1.577$ lie strictly within $(0, 2)$.
$f(x) = \sqrt{x}$ is continuous on $[1, 4]$ and differentiable on $(1, 4)$. The average rate of change is $\frac{f(4) - f(1)}{4 - 1} = \frac{\sqrt{4} - \sqrt{1}}{3} = \frac{2 - 1}{3} = \frac{1}{3}$. The derivative is $f'(x) = \frac{1}{2\sqrt{x}}$. Setting $f'(\xi) = \frac{1}{3}$: $\frac{1}{2\sqrt{\xi}} = \frac{1}{3} \implies 2\sqrt{\xi} = 3 \implies \sqrt{\xi} = \frac{3}{2} \implies \xi = \frac{9}{4} = 2.25$. Note that $2.25 \in (1, 4)$.
Applying the Mean Value Theorem to $f$ on the interval $[1, 4]$: there exists $\xi \in (1, 4)$ such that $\frac{f(4) - f(1)}{4 - 1} = f'(\xi)$. Since $f'(x) \le 3$ for all $x$, we have $\frac{f(4) - 2}{3} \le 3 \implies f(4) - 2 \le 9 \implies f(4) \le 11$. Thus, the maximum possible value is 11.
Fermat's Theorem states that if $f$ has a local extremum at an interior point $x_0$ and $f$ is differentiable at $x_0$, then its derivative must be zero: $f'(x_0) = 0$.
Compute the first derivative: $f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2)$. The derivative vanishes at $x = 1$ and $x = 2$. For $x \in (1, 2)$, $(x - 1) > 0$ and $(x - 2) < 0$, so $f'(x) < 0$. Since $f$ is continuous on $[1, 2]$ and $f'(x) < 0$ on $(1, 2)$, $f$ is strictly decreasing on $[1, 2]$.
Compute the first derivative using the quotient rule: $f'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$. Setting $f'(x) = 0$ gives $1 - \ln x = 0 \implies \ln x = 1 \implies x = e$. For $0 < x < e$, $f'(x) > 0$ (increasing); for $x > e$, $f'(x) < 0$ (decreasing). Thus $x = e$ is a global and local maximum. The maximum value is $f(e) = \frac{\ln e}{e} = \frac{1}{e}$.
Performing polynomial long division of $2x^2 + 5x - 3$ by $x + 1$: $2x^2 + 5x - 3 = (x + 1)(2x + 3) - 6$. Thus, $f(x) = 2x + 3 - \frac{6}{x + 1}$. As $x \to +\infty$, $\lim_{x \to +\infty} [f(x) - (2x + 3)] = \lim_{x \to +\infty} \left(-\frac{6}{x + 1}\right) = 0$. Therefore, the line $y = 2x + 3$ is the oblique asymptote.
Factor the numerator and denominator: $f(x) = \frac{(x - 3)(x + 3)}{(x - 2)(x - 3)}$. The domain is $\mathbb{R} \setminus \{2, 3\}$. For $x \neq 3$, $f(x) = \frac{x + 3}{x - 2}$. At $x = 3$: $\lim_{x \to 3} f(x) = \frac{3 + 3}{3 - 2} = 6$. Because the limit is finite, $x = 3$ is a removable discontinuity (a hole), NOT a vertical asymptote. At $x = 2$: $\lim_{x \to 2^+} f(x) = +\infty$ and $\lim_{x \to 2^-} f(x) = -\infty$. Thus, the only vertical asymptote is $x = 2$.
Let $x$ be the length of each of the two sides perpendicular to the river, and let $y$ be the length of the side parallel to the river. The total perimeter fence used is $2x + y = 800 \implies y = 800 - 2x$. The area function is $A(x) = x y = x(800 - 2x) = 800x - 2x^2$. Differentiating to find the maximum: $A'(x) = 800 - 4x = 0 \implies x = 200\text{ m}$. Since $A''(x) = -4 < 0$, this is a maximum. The optimal side parallel to the river is $y = 800 - 2(200) = 400\text{ m}$. The maximum area is $A = 200 \times 400 = 80,000\text{ m}^2$.
The square of the distance from any point $(x, x^2)$ on the parabola to $(0, \frac{3}{2})$ is $D(x) = (x - 0)^2 + \left(x^2 - \frac{3}{2}\right)^2 = x^2 + x^4 - 3x^2 + \frac{9}{4} = x^4 - 2x^2 + \frac{9}{4}$. Differentiating with respect to $x$: $D'(x) = 4x^3 - 4x = 4x(x^2 - 1) = 4x(x - 1)(x + 1)$. Critical points are $x = 0$ and $x = \pm 1$. Evaluating $D(x)$: $D(0) = \frac{9}{4} = 2.25$, while $D(\pm 1) = 1 - 2 + 2.25 = 1.25$. Since $1.25 < 2.25$, the minimum distance occurs at $x = \pm 1$ (the points are $(1, 1)$ and $(-1, 1)$).
The volume of a cylinder is $V = \pi r^2 h = 27\pi \implies h = \frac{27}{r^2}$. Because the top is open, the surface area consists of one circular base plus the cylindrical side: $S(r) = \pi r^2 + 2\pi r h = \pi r^2 + 2\pi r \left(\frac{27}{r^2}\right) = \pi r^2 + \frac{54\pi}{r}$. Differentiating with respect to $r$: $S'(r) = 2\pi r - \frac{54\pi}{r^2}$. Setting $S'(r) = 0$: $2\pi r = \frac{54\pi}{r^2} \implies r^3 = 27 \implies r = 3\text{ cm}$. Since $S''(r) = 2\pi + \frac{108\pi}{r^3} > 0$, this guarantees a minimum.
Total revenue is $R(x) = x \cdot p(x) = 50x - 0.1x^2$. Total profit is $P(x) = R(x) - C(x) = (50x - 0.1x^2) - (200 + 10x + 0.1x^2) = 40x - 0.2x^2 - 200$. Differentiating to maximize profit: $P'(x) = 40 - 0.4x = 0 \implies 0.4x = 40 \implies x = 100$. Since $P''(x) = -0.4 < 0$, $x = 100$ is the profit-maximizing output.
The points of intersection with the $x$-axis are given by $4 - x^2 = 0 \implies x = \pm 2$. Since $4 - x^2 \ge 0$ on $[-2, 2]$, the enclosed area is $A = \int_{-2}^{2} (4 - x^2) \, dx$. Utilizing symmetry about the $y$-axis: $A = 2 \int_{0}^{2} (4 - x^2) \, dx = 2 \left[4x - \frac{x^3}{3}\right]_0^2 = 2 \left(4(2) - \frac{8}{3}\right) = 2 \left(8 - \frac{8}{3}\right) = 2 \left(\frac{16}{3}\right) = \frac{32}{3}$.
Since e²ˣ > 0 on [0, ln 2], the area is given by the definite integral A = ∫₀^(ln 2) e²ˣ dx = [(1/2) e²ˣ]₀^(ln 2) = (1/2) [e^(2 ln 2) − e⁰]. Since e^(2 ln 2) = e^(ln 4) = 4 and e⁰ = 1, A = (1/2)(4 − 1) = 3/2 area units.
The transformation has the form z' = az + b with a = -i = e^(-i π/2). Since |a| = 1 and a ≠ 1, it is a rotation of angle arg(a) = -π/2. The invariant center Ω satisfies ω = -i ω + 1 + 2i ⇒ ω(1 + i) = 1 + 2i ⇒ ω = (1 + 2i)/(1 + i) = (1 + 2i)(1 − i)/2 = (3 + i)/2 = 3/2 + (1/2)i.
Let A be the fixed point with affix z_A = 2 − i, which has Cartesian coordinates (2, -1). The condition |z − z_A| = 3 expresses that the distance AM is constant and equal to 3. Therefore, the set of points M is a circle of center A(2, -1) and radius R = 3.
Since the common ratio q = 2/3 satisfies |q| < 1, lim(n→+∞) (2/3)ⁿ = 0, so lim(n→+∞) vₙ = 0. The infinite sum of a convergent geometric series is S = v₀ / (1 − q) = 5 / (1 − 2/3) = 5 / (1/3) = 15.
The fixed point of the recurrence is L = (1/2)L + 3 ⇒ (1/2)L = 3 ⇒ L = 6. Then v_{n+1} = u_{n+1} − 6 = (1/2)uₙ + 3 − 6 = (1/2)(uₙ − 6) = (1/2)vₙ. Thus (vₙ) is geometric with ratio 1/2 and initial term v₀ = u₀ − 6 = 10 − 6 = 4. Hence vₙ = 4(1/2)ⁿ, which gives uₙ = vₙ + 6 = 4(1/2)ⁿ + 6.
A point M(x, y, z) lies on line (D) if and only if there exists a scalar t ∈ ℝ such that AM = t u, which translates to (x − x_A, y − y_A, z − z_A) = t(u_x, u_y, u_z). Substituting A(1, -2, 3) and u(2, 1, -1) yields x = 1 + 2t, y = -2 + t, and z = 3 − t.
A point M(x, y, z) belongs to plane (P) if and only if AM · n = 0. Expanding gives 3(x − 2) − 1(y − 0) + 2(z − 1) = 0 ⇒ 3x − 6 − y + 2z − 2 = 0 ⇒ 3x − y + 2z − 8 = 0.
The distance from point M(x₀, y₀, z₀) to plane ax + by + cz + d = 0 is d(M, P) = |a x₀ + b y₀ + c z₀ + d| / √(a² + b² + c²). Substituting gives |2(1) − 1(2) + 2(3) − 1| / √(2² + (-1)² + 2²) = |2 − 2 + 6 − 1| / √(4 + 1 + 4) = |5| / √9 = 5/3.
The normal vectors to the two planes are n₁ = (1, 2, -1) and n₂ = (2, -1, 1). The intersection line (Δ) is perpendicular to both normal vectors, so its direction vector is given by the cross product n₁ × n₂ = (2(1) − (-1)(-1), (-1)(2) − (1)(1), 1(-1) − 2(2)) = (2 − 1, -2 − 1, -1 − 4) = (1, -3, -5).
Cross-multiplying yields z(z + 1) = z − 2i ⇒ z² + z = z − 2i ⇒ z² = -2i. Writing -2i in polar form: -2i = 2e^(-i π/2). The two square roots are z = √2 e^(-i π/4) = √2 [cos(-π/4) + i sin(-π/4)] = √2 [√2/2 − i √2/2] = 1 − i, and its opposite -(1 − i) = -1 + i. The root with positive real part is 1 − i.
Admittance is the sum of conductance G and susceptance B: Y = 1/R + 1/ZC = 1/20 + 1/(-j10) = 0.05 + j/10 = 0.05 + j 0.10 S (Siemens).
Tolerance width is USL - LSL = 50.15 - 49.85 = 0.30 mm. Process spread is 6 · σ = 6 · 0.025 mm = 0.15 mm. The capability index is Cp = 0.30 / 0.15 = 2.00 (indicating a capable 6-sigma process).
Total number of ways to draw 3 balls from 9 is C(9, 3) = (9 · 8 · 7) / (3 · 2 · 1) = 84. The number of favorable outcomes (2 red from 5 and 1 green from 4) is C(5, 2) · C(4, 1) = 10 · 4 = 40. The probability is 40/84 = 10/21.
Using the binomial probability formula P(X = k) = C(n, k) p^k (1 − p)^(n−k) with n = 5, k = 2, p = 0.2, and 1 − p = 0.8: P(X = 2) = C(5, 2) · (0.2)² · (0.8)³ = 10 · 0.04 · 0.512 = 0.4 · 0.512 = 0.2048.
The expectation E(X) is the probability-weighted mean of the values of X, the value expected on average over many trials. Probability and random variables are part of the common maths programme.