A full teaching guide covering every topic on the M1 (Mate-Info) Bacalaureat mathematics syllabus \u2014 definitions, formulas, worked technique, and the reasoning behind each one, from 9th grade foundations through 12th grade calculus. This is built to actually teach the material, not just list it.<\/p>\n
Click to read the full guide \u2192<\/a><\/p>\n Everything builds on N \u2282 Z \u2282 Q \u2282 R. A number is rational if it can be written as a fraction of integers; irrational numbers (like \u221a2 or \u03c0) cannot. When a radical sits in a denominator, rationalize it by multiplying top and bottom by the conjugate. For a single radical: multiply by \u221aa\/\u221aa. For a binomial like 1\/(\u221aa + \u221ab), multiply by (\u221aa \u2212 \u221ab)\/(\u221aa \u2212 \u221ab), using (x+y)(x\u2212y) = x\u00b2 \u2212 y\u00b2 to eliminate the radicals.<\/p>\n Worked example:<\/strong> 1\/(\u221a5 + \u221a3) \u2192 multiply by (\u221a5 \u2212 \u221a3)\/(\u221a5 \u2212 \u221a3) \u2192 (\u221a5 \u2212 \u221a3)\/(5 \u2212 3) = (\u221a5 \u2212 \u221a3)\/2.<\/p>\n |x| = x if x \u2265 0, and |x| = \u2212x if x < 0. Geometrically, |x \u2212 a| is the distance between x and a on the number line. To solve |x| = k (k > 0): x = k or x = \u2212k. To solve |x| < k: this means \u2212k < x < k. To solve |x| > k: this means x < \u2212k or x > k. Mixing up the AND\/OR logic here is one of the most common early errors.<\/p>\n f(x) = ax + b is a line with slope a and y-intercept b. If a > 0 the function is increasing; if a < 0 it’s decreasing. The root is x = \u2212b\/a. To find the equation of a line through two points (x\u2081,y\u2081) and (x\u2082,y\u2082): compute slope a = (y\u2082\u2212y\u2081)\/(x\u2082\u2212x\u2081), then substitute into y \u2212 y\u2081 = a(x \u2212 x\u2081).<\/p>\n f(x) = ax\u00b2 + bx + c, with a \u2260 0. The discriminant \u0394 = b\u00b2 \u2212 4ac tells you everything: \u0394 > 0 gives two distinct real roots, \u0394 = 0 gives one repeated root, \u0394 < 0 gives no real roots. Roots (when they exist): x\u2081,\u2082 = (\u2212b \u00b1 \u221a\u0394)\/2a. The vertex is at (\u2212b\/2a, \u2212\u0394\/4a) \u2014 a minimum if a > 0, a maximum if a < 0.<\/p>\n Vi\u00e8te’s relations:<\/strong> for roots x\u2081 and x\u2082, sum S = x\u2081+x\u2082 = \u2212b\/a and product P = x\u2081\u00b7x\u2082 = c\/a. These let you answer questions about the roots (like x\u2081\u00b2+x\u2082\u00b2 = S\u00b2\u22122P) without solving for them individually.<\/p>\n Worked example \u2014 find m so both roots are positive, for x\u00b2 \u2212 (m+1)x + m = 0:<\/strong> Need \u0394 \u2265 0 (always true here since it simplifies to (m\u22121)\u00b2 \u2265 0), sum > 0 \u2192 m+1 > 0 \u2192 m > \u22121, and product > 0 \u2192 m > 0. Combined: m > 0<\/strong>. This is the standard pattern for “both roots satisfy X” problems \u2014 turn the condition into a system of inequalities on \u0394, S, and P.<\/p>\n In a right triangle: sin = opposite\/hypotenuse, cos = adjacent\/hypotenuse, tan = opposite\/adjacent. Know exact values for 30\u00b0, 45\u00b0, 60\u00b0, 90\u00b0 by heart. Fundamental identity: sin\u00b2x + cos\u00b2x = 1. Law of Sines: a\/sin(A) = b\/sin(B) = c\/sin(C). Law of Cosines: c\u00b2 = a\u00b2 + b\u00b2 \u2212 2ab\u00b7cos(C).<\/p>\n Isolate the radical, then square both sides. Squaring can introduce extraneous solutions, so always verify your answer in the original equation. Before squaring, note the domain restriction from the radical (expression under the root \u2265 0) and, if the radical is set equal to an expression, that expression must be \u2265 0 too.<\/p>\n Worked example:<\/strong> \u221a(x+3) = x\u22121. First, x\u22121 \u2265 0 is required (a root can’t equal a negative), so x \u2265 1. Squaring: x+3 = x\u00b2\u22122x+1 \u2192 x\u00b2\u22123x\u22122 = 0 \u2192 x = (3\u00b1\u221a17)\/2. Only the root satisfying x \u2265 1 survives \u2014 check both against that condition before finalizing.<\/p>\n f(x) = a\u02e3, a > 0, a \u2260 1. Increasing if a > 1, decreasing if 0 < a < 1. Rules: a\u1d50\u00b7a\u207f = a\u1d50\u207a\u207f, a\u1d50\/a\u207f = a\u1d50\u207b\u207f, (a\u1d50)\u207f = a\u1d50\u207f, a\u207b\u207f = 1\/a\u207f. To solve equations, match bases and equate exponents, or substitute t = a\u02e3 (t > 0) to reduce to a polynomial in t.<\/p>\n Worked example:<\/strong> 4\u02e3 \u2212 3\u00b72\u02e3 \u2212 4 = 0. Since 4\u02e3 = (2\u02e3)\u00b2, let t = 2\u02e3: t\u00b2 \u2212 3t \u2212 4 = 0 \u2192 (t\u22124)(t+1) = 0 \u2192 t = 4 or t = \u22121. Reject t = \u22121 (must be positive). So 2\u02e3 = 4 \u2192 x = 2<\/strong>.<\/p>\n log\u2090(x) = y means a\u02b8 = x, defined only for x > 0, a > 0, a \u2260 1. Properties: log\u2090(mn) = log\u2090m + log\u2090n, log\u2090(m\/n) = log\u2090m \u2212 log\u2090n, log\u2090(m\u207f) = n\u00b7log\u2090m, and change of base log\u2090b = logc(b)\/logc(a). Always establish the domain (every log argument > 0) before solving, and check your solution against it afterward \u2014 this is the step most often skipped.<\/p>\n Permutations: P\u2099 = n! \u2014 arrangements of all n objects. Arrangements: A(n,k) = n!\/(n\u2212k)! \u2014 choosing and ordering k of n objects. Combinations: C(n,k) = n!\/(k!(n\u2212k)!) \u2014 choosing k of n objects, order irrelevant. The deciding question: does order matter?<\/p>\n Binomial theorem:<\/strong> (a+b)\u207f = \u03a3 C(n,k)\u00b7a\u207f\u207b\u1d4f\u00b7b\u1d4f, summed k=0 to n. To find a specific term (the (k+1)-th), use T\u2096\u208a\u2081 = C(n,k)\u00b7a\u207f\u207b\u1d4f\u00b7b\u1d4f; to find a term with a given power, set the exponent equal to the target and solve for k.<\/p>\n P(event) = favorable outcomes \/ total outcomes, assuming equally likely outcomes. Complement: P(not A) = 1 \u2212 P(A). Union: P(A or B) = P(A) + P(B) \u2212 P(A and B). Most exam problems reduce to correctly counting outcomes with combinatorics first \u2014 the probability formula itself is the easy part.<\/p>\n Distance: d = \u221a[(x\u2082\u2212x\u2081)\u00b2 + (y\u2082\u2212y\u2081)\u00b2]. Midpoint: ((x\u2081+x\u2082)\/2, (y\u2081+y\u2082)\/2). Line through two points: slope m = (y\u2082\u2212y\u2081)\/(x\u2082\u2212x\u2081), then y \u2212 y\u2081 = m(x \u2212 x\u2081). Parallel lines have equal slopes; perpendicular lines have slopes whose product is \u22121.<\/p>\n Area of triangle A(x\u2081,y\u2081), B(x\u2082,y\u2082), C(x\u2083,y\u2083): \u00bd\u00b7|x\u2081(y\u2082\u2212y\u2083) + x\u2082(y\u2083\u2212y\u2081) + x\u2083(y\u2081\u2212y\u2082)|.<\/p>\n Add matrices by adding corresponding entries (same size required). Multiply by a scalar by multiplying every entry. Matrix multiplication A\u00b7B is defined only when A’s column count equals B’s row count, computed row-by-column via dot products. Matrix multiplication is generally not commutative<\/strong>: A\u00b7B \u2260 B\u00b7A. The transpose A\u1d40 flips rows and columns; the identity matrix I satisfies A\u00b7I = I\u00b7A = A.<\/p>\n For [[a,b],[c,d]]: determinant = ad \u2212 bc. For 3\u00d73, use Sarrus’ rule \u2014 repeat the first two columns to the right, sum the three down-right diagonal products, subtract the three down-left diagonal products. Key properties: swapping two rows flips the sign; a scalar multiple of a row scales the whole determinant; identical or proportional rows give determinant 0; det(A\u00b7B) = det(A)\u00b7det(B).<\/p>\n Form the coefficient matrix, compute its determinant \u0394. If \u0394 \u2260 0, the system has a unique solution: each unknown x\u1d62 = \u0394\u1d62\/\u0394, where \u0394\u1d62 replaces the i-th column with the constants column. If \u0394 = 0, check whether the \u0394\u1d62 are also all zero to distinguish “infinitely many solutions” from “no solution.”<\/p>\n A is invertible only if det(A) \u2260 0. A\u207b\u00b9 = (1\/det(A))\u00b7adj(A). For a 2\u00d72 matrix [[a,b],[c,d]]: A\u207b\u00b9 = (1\/(ad\u2212bc))\u00b7[[d,\u2212b],[\u2212c,a]] \u2014 swap the diagonal entries, negate the off-diagonal ones, divide by the determinant.<\/p>\n A set G with operation * is a group<\/strong> if the operation is associative, there’s an identity element e with a*e = e*a = a, and every element has an inverse. If also commutative, it’s abelian. A ring<\/strong> has two operations where the set is an abelian group under addition, multiplication is associative, and distributes over addition. A field<\/strong> is a ring where every nonzero element also has a multiplicative inverse. Exam problems typically hand you a specific (set, operation) pair and ask you to verify each axiom in sequence \u2014 mechanical once you know the checklist.<\/p>\n Remainder theorem: dividing P(x) by (x\u2212a) gives remainder P(a). Factor theorem: (x\u2212a) divides P(x) exactly when P(a) = 0. Generalized Vi\u00e8te’s relations connect a degree-n polynomial’s roots to its coefficients (for degree 3: x\u2081+x\u2082+x\u2083 = \u2212b\/a, x\u2081x\u2082+x\u2081x\u2083+x\u2082x\u2083 = c\/a, x\u2081x\u2082x\u2083 = \u2212d\/a).<\/p>\n Arithmetic: constant difference r, a\u2099 = a\u2081+(n\u22121)r, sum S\u2099 = n(a\u2081+a\u2099)\/2. Geometric: constant ratio q, a\u2099 = a\u2081\u00b7q\u207f\u207b\u00b9, sum (q\u22601) S\u2099 = a\u2081(q\u207f\u22121)\/(q\u22121). Identify the type first by checking whether consecutive differences (arithmetic) or consecutive ratios (geometric) are constant.<\/p>\n Key known limits: lim(1\/n) = 0, lim(q\u207f) = 0 if |q|<1, lim(q\u207f) = \u221e if q>1, lim(1+1\/n)\u207f = e. For rational expressions in n, divide numerator and denominator by the highest power of n present to resolve \u221e\/\u221e.<\/p>\n Worked example:<\/strong> lim (3n\u00b2+n)\/(n\u00b2+5) as n\u2192\u221e. Divide every term by n\u00b2: (3+1\/n)\/(1+5\/n\u00b2) \u2192 3\/1 = 3<\/strong>.<\/p>\n The remarkable limit sin(x)\/x \u2192 1 as x\u21920 underlies most trigonometric limit problems. For 0\/0 forms with polynomials, factor and cancel the shared root. For \u221e\u2212\u221e forms with radicals, multiply by the conjugate to reach a simplifiable form.<\/p>\n f is continuous at a if f(a) is defined, the limit as x\u2192a exists, and that limit equals f(a). For piecewise functions, compute the left-hand and right-hand limits at the boundary and set both equal to each other and to the function’s value there \u2014 exactly how “find m such that f is continuous” problems are solved.<\/p>\n f'(a) = limit of [f(x)\u2212f(a)]\/(x\u2212a) as x\u2192a \u2014 the slope of the tangent at that point. Rules: (f+g)’ = f’+g’; (fg)’ = f’g+fg’ (product rule); (f\/g)’ = (f’g\u2212fg’)\/g\u00b2 (quotient rule); (f(g(x)))’ = f'(g(x))\u00b7g'(x) (chain rule). Know cold: (x\u207f)’ = nx\u207f\u207b\u00b9, (e\u02e3)’ = e\u02e3, (ln x)’ = 1\/x, (sin x)’ = cos x, (cos x)’ = \u2212sin x. Tangent line at (a, f(a)): y = f(a) + f'(a)(x\u2212a).<\/p>\n Where f'(x) > 0, f is increasing; where f'(x) < 0, decreasing. Critical points (f’=0 or undefined) are candidates for local extrema \u2014 sign change positive-to-negative gives a local max, negative-to-positive a local min. The second derivative refines this: f”(a) > 0 at a critical point signals a local minimum, f”(a) < 0 a local maximum. Where f” changes sign, you have an inflection point.<\/p>\n Worked example \u2014 study f(x) = x\u00b7e\u02e3 on \u211d:<\/strong> f'(x) = e\u02e3(x+1). Since e\u02e3 > 0 always, sign of f’ matches sign of (x+1): decreasing for x<\u22121, increasing for x>\u22121 \u2192 local minimum at x=\u22121<\/strong>, value \u22121\/e. f”(x) = e\u02e3(x+2), changing sign at x=\u22122 \u2192 inflection point at x=\u22122<\/strong>, value \u22122\/e\u00b2. Note the full chain here: compute f’, build the sign table, state the conclusion, then repeat for f” \u2014 every step written out, not skipped.<\/p>\n Always follow this order: (1) domain, (2) limits at boundaries and infinity to find asymptotes, (3) f’ and a sign table for monotony\/extrema, (4) f” for concavity\/inflection points, (5) the full variation table combining everything, (6) the graph sketch built from that table. This exact sequence is what “studiul func\u021biei” problems are asking for, and skipping a step (even an “obvious” one) is where points get lost.<\/p>\n F'(x) = f(x) defines an antiderivative. Core formulas: \u222bx\u207fdx = x\u207f\u207a\u00b9\/(n+1)+C (n\u2260\u22121), \u222b(1\/x)dx = ln|x|+C, \u222be\u02e3dx = e\u02e3+C, \u222bsin(x)dx = \u2212cos(x)+C, \u222bcos(x)dx = sin(x)+C.<\/p>\n Substitution:<\/strong> when an inner function and its derivative both appear in the integral, substitute u = (inner function) and rewrite entirely in terms of u before integrating.<\/p>\n Integration by parts:<\/strong> \u222bu dv = uv \u2212 \u222bv du. Choose u as the part that simplifies under differentiation (often a polynomial), dv as the part that’s easy to integrate (often exponential or trigonometric).<\/p>\n Fundamental theorem of calculus: \u222b\u2090\u1d47 f(x)dx = F(b) \u2212 F(a). All antiderivative techniques (substitution, by parts) carry over \u2014 just remember to either change your bounds when substituting, or substitute back before plugging in the original bounds.<\/p>\n Worked example:<\/strong> \u222b\u2080\u00b2 (2x+3)dx. F(x) = x\u00b2+3x. F(2)\u2212F(0) = (4+6)\u22120 = 10<\/strong>.<\/p>\n Area between a curve and the x-axis over [a,b]: \u222b\u2090\u1d47 |f(x)| dx \u2014 take the absolute value seriously, since a portion of the curve below the axis must count as positive area, not be subtracted out. Area between two curves f and g: \u222b\u2090\u1d47 |f(x)\u2212g(x)| dx. Volume of a solid formed by rotating f(x) around the x-axis over [a,b]: V = \u03c0\u00b7\u222b\u2090\u1d47 [f(x)]\u00b2 dx.<\/p>\n To prove P(n) for all n \u2265 n\u2080: (1) verify the base case P(n\u2080) directly, (2) assume P(k) is true (induction hypothesis), (3) prove that assumption forces P(k+1) to be true. Both steps matter \u2014 skipping the base case, or not actually using the hypothesis in step 3, are the two most common ways points are lost here.<\/p>\n Across irrational, logarithmic, and rational expressions: determine the domain before solving, and check your final answer against it before submitting. An algebraically correct solution that falls outside the valid domain is not a correct solution \u2014 this single habit prevents a large share of avoidable point losses across the entire exam.<\/p>\n Subjects II and III specifically reward visible reasoning \u2014 writing out the sign table, stating which theorem justifies a step, showing the substitution explicitly \u2014 rather than jumping straight to a final numeric answer. A correct method with a small arithmetic slip typically scores far better than a correct-looking final answer with no visible work, because partial credit is built around seeing the intermediate steps.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":" How to Ace the Romanian Bacalaureat M1: The Complete Mathematics Guide A full teaching guide covering every topic on the M1 (Mate-Info) Bacalaureat mathematics syllabus \u2014 definitions, formulas, worked technique, and the reasoning behind each one, from 9th grade foundations through 12th grade calculus. This is built to actually teach the material, not just list … Continue reading “10 at Mathematics Baccalaureate Romania M1”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-38","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/danielmarcu.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/38","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/danielmarcu.com\/blog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/danielmarcu.com\/blog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/danielmarcu.com\/blog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/danielmarcu.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=38"}],"version-history":[{"count":2,"href":"https:\/\/danielmarcu.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/38\/revisions"}],"predecessor-version":[{"id":40,"href":"https:\/\/danielmarcu.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/38\/revisions\/40"}],"wp:attachment":[{"href":"https:\/\/danielmarcu.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=38"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/danielmarcu.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=38"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/danielmarcu.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=38"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Class IX: Foundations<\/h2>\n
Number Sets and Radicals<\/h3>\n
Absolute Value<\/h3>\n
The Linear Function<\/h3>\n
The Quadratic Function<\/h3>\n
Basic Trigonometry<\/h3>\n
Class X: Extending the Toolkit<\/h2>\n
Irrational Equations<\/h3>\n
Exponential Functions and Equations<\/h3>\n
Logarithms and Logarithmic Equations<\/h3>\n
Combinatorics<\/h3>\n
Probability<\/h3>\n
Analytic Geometry<\/h3>\n
Class XI: Algebra Deepens, Analysis Begins<\/h2>\n
Matrices<\/h3>\n
Determinants<\/h3>\n
Systems of Linear Equations and Cramer’s Rule<\/h3>\n
Inverse of a Matrix<\/h3>\n
Algebraic Structures<\/h3>\n
Polynomials<\/h3>\n
Sequences (Progressions)<\/h3>\n
Limits of Sequences<\/h3>\n
Limits of Functions<\/h3>\n
Continuity<\/h3>\n
Class XII: Calculus and Synthesis<\/h2>\n
Derivatives<\/h3>\n
Monotony and Extrema<\/h3>\n
Full Function Study<\/h3>\n
Antiderivatives (Primitives)<\/h3>\n
Definite Integrals<\/h3>\n
Applications of Integrals<\/h3>\n
Threads That Run Through Everything<\/h2>\n
Mathematical Induction<\/h3>\n
Domain Discipline<\/h3>\n
Showing Justification<\/h3>\n