All Practice Exams
Free Practice Questions for Syrian Scientific Baccalaureate
Exam-style questions and explanations by OpenExamPrep.
✓ No registration✓ No credit card
94+ Questions
100% Free
Loading practice questions...
Same family resources
Explore More Syrian National Secondary and Basic Education Examinations
Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.
Sample Syrian Scientific Baccalaureate Practice Questions
Try these sample questions to review concepts for the Syrian Scientific Baccalaureate exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 94+ question experience with AI tutoring.
1What is the limit of the rational function $f(x) = \frac{2x^2 - 8}{x - 2}$ as $x$ approaches $2$?
A.8
B.4
C.0
D.The limit does not exist
Explanation: Factoring the numerator yields $2(x^2 - 4) = 2(x - 2)(x + 2)$. For $x \neq 2$, cancelling $(x - 2)$ gives $2(x + 2)$. Evaluating at $x = 2$ gives $2(2 + 2) = 8$.
2What is the derivative $f'(x)$ of the natural logarithmic function $f(x) = \ln(3x^2 + 5)$ for all $x \in \mathbb{R}$?
A.\frac{1}{3x^2 + 5}
B.\frac{6x}{3x^2 + 5}
C.\frac{6x}{(3x^2 + 5)^2}
D.6x \ln(3x^2 + 5)
Explanation: By the chain rule for logarithmic functions, $\frac{d}{dx}[\ln(u(x))] = \frac{u'(x)}{u(x)}$. Here $u(x) = 3x^2 + 5$ and $u'(x) = 6x$, so $f'(x) = \frac{6x}{3x^2 + 5}$.
3What is the value of the definite integral $\int_{0}^{1} (3x^2 - 2x + 1) \, dx$?
A.0
B.2
C.1
D.3
Explanation: The antiderivative is $F(x) = x^3 - x^2 + x$. Evaluating from 0 to 1 gives $F(1) - F(0) = (1^3 - 1^2 + 1) - 0 = 1$.
4For what value of $k$ is the function $f(x)$ continuous at $x = 1$, where $f(x) = \begin{cases} kx + 2 & x \le 1 \\ 3x^2 - 1 & x > 1 \end{cases}$?
A.1
B.2
C.-1
D.0
Explanation: For continuity at $x = 1$, the left-hand limit $\lim_{x \to 1^-} f(x) = k(1) + 2$ must equal the right-hand limit $\lim_{x \to 1^+} f(x) = 3(1)^2 - 1 = 2$. Setting $k + 2 = 2$ gives $k = 0$.
5What is the equation of the tangent line to the curve $y = e^{2x}$ at the point where $x = 0$?
A.y = 2x + 1
B.y = x + 1
C.y = 2x
D.y = e^2 x + 1
Explanation: At $x = 0$, $y = e^{0} = 1$. The derivative is $y' = 2e^{2x}$, which at $x = 0$ evaluates to $m = 2e^0 = 2$. The point-slope formula gives $y - 1 = 2(x - 0) \implies y = 2x + 1$.
6What are the inflection points of the polynomial function $f(x) = x^3 - 3x^2 + 4$?
A.(0, 4)
B.(1, 2)
C.(2, 0)
D.(1, 0)
Explanation: Compute the first derivative $f'(x) = 3x^2 - 6x$ and second derivative $f''(x) = 6x - 6$. Setting $f''(x) = 0$ gives $x = 1$. Evaluating $f(1) = 1 - 3 + 4 = 2$. Since $f''(x)$ changes sign at $x = 1$, $(1, 2)$ is an inflection point.
7What is the general solution of the first-order differential equation $y' - 3y = 0$?
A.y = 3x + C
B.y = C e^{-3x}
C.y = C e^{3x}
D.y = C e^{x/3}
Explanation: The differential equation is separable: $\frac{dy}{y} = 3 \, dx$. Integrating both sides yields $\ln|y| = 3x + c$, which gives $y = C e^{3x}$ where $C$ is an arbitrary constant.
8Using integration by parts, what is the value of $\int x \cos(x) \, dx$?
A.x \cos(x) - \sin(x) + C
B.x \sin(x) - \cos(x) + C
C.-x \sin(x) + \cos(x) + C
D.x \sin(x) + \cos(x) + C
Explanation: Let $u = x \implies du = dx$, and $dv = \cos(x) dx \implies v = \sin(x)$. Then $\int u \, dv = uv - \int v \, du = x \sin(x) - \int \sin(x) dx = x \sin(x) - (-\cos(x)) + C = x \sin(x) + \cos(x) + C$.
9In the complex plane, what is the modulus $r$ and principal argument $\theta$ of $z = -1 + i\sqrt{3}$?
A.r = 2, \theta = \frac{2\pi}{3}
B.r = 2, \theta = \frac{\pi}{3}
C.r = 4, \theta = \frac{2\pi}{3}
D.r = 2, \theta = \frac{5\pi}{6}
Explanation: The modulus is $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. Since $x = -1 < 0$ and $y = \sqrt{3} > 0$, $z$ lies in the second quadrant. $\tan(\alpha) = |\sqrt{3}/(-1)| = \sqrt{3} \implies \alpha = \pi/3$. In the second quadrant, $\theta = \pi - \pi/3 = \frac{2\pi}{3}$.
10What is the distance from the point $M(1, -2, 3)$ to the plane $(P): 2x - y + 2z - 5 = 0$ in 3D Euclidean space?
A.5
B.3
C.\frac{5}{3}
D.\frac{9}{5}
Explanation: The distance from a point $M(x_0, y_0, z_0)$ to a plane $ax + by + cz + d = 0$ is $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$. Substituting $M(1, -2, 3)$ and $2x - y + 2z - 5 = 0$: the numerator is $|2(1) - (-2) + 2(3) - 5| = |2 + 2 + 6 - 5| = 5$, and the denominator is $\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = 3$. Therefore, the distance is $\frac{5}{3}$.
About the Syrian Scientific Baccalaureate Practice Questions
Check the official exam sponsor for the exam format, time limit, passing standard, fees, and eligibility requirements. These details are not yet confirmed here.