ITA 2025 – 1st phase: Mathematics solved

Source: ITA – 1st phase exam and official answer key. Files on the ITA past exams page.

The 1st phase of ITA 2025 had 48 multiple-choice questions: 12 in Mathematics, 12 in Physics, 12 in Chemistry and 12 in English. This post covers the mathematics questions, numbered 1 to 12 as in the exam.

Exam conventions. The horizontal and vertical axes are $O_x$ and $O_y$, and $O$ is the origin; $\mathbb{N} = \{1, 2, 3, \dots\}$; $i^2 = -1$; $\overline{AB}$ is the segment and $AB$ the line through $A$ and $B$.

Answer key

Question 1: The triangle with vertices 1, z and z²

Statement

Let $z = a + 2i$ be a complex number, where $a \in \mathbb{R}$ is positive. The value of $a$ for which the representations of $1$, $z$ and $z^2$ in the Argand–Gauss plane form a triangle of area 200 is

• A) 14.
• B) 15.
• C) 16.
• D) 17.
• E) 18.

Solution

1. The points. $z^2 = (a + 2i)^2 = (a^2 - 4) + 4ai$. The vertices are $(1, 0)$, $(a, 2)$ and $(a^2 - 4, 4a)$.

2. Area from the determinant. With the vectors from $(1, 0)$: $\vec{u} = (a - 1, 2)$ and $\vec{v} = (a^2 - 5, 4a)$,

(the value is always positive, since $a^2 - 2a + 5 = (a - 1)^2 + 4$).

3. Solving. $a^2 - 2a + 5 = 200 \Rightarrow a^2 - 2a - 195 = 0 \Rightarrow a = \frac{2 \pm 28}{2}$. Since $a > 0$, $a = 15$.

Checking: $(15 - 1)^2 + 4 = 196 + 4 = 200$ ✔

Answer: option B (15).

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Question 2: Which polyhedra exist?

Statement

Consider the statements:

• I. There is a polyhedron formed by 3 triangular faces with all other faces quadrilaterals.
• II. There is a non-convex polyhedron with 5 vertices, 9 edges and 6 faces.
• III. There is a convex polyhedron with 7 vertices, 16 edges and 11 faces.

Which is (are) correct:

• A) I and II.
• B) I and III.
• C) II and III.
• D) only III.
• E) only II.

Solution

I. False. Adding up the sides of all the faces, each edge is counted twice: $3\cdot 3 + 4q = 2E$, where $q$ is the number of quadrilaterals. But $9 + 4q$ is odd, and $2E$ is even. Impossible.

II. True. Take a triangle $ABC$, a point $D$ above it and a point $E$ also above it, but lower, inside pyramid $DABC$. The solid between the two “roofs” (the upper one with faces $DAB$, $DBC$, $DCA$ and the lower one with faces $EAB$, $EBC$, $ECA$) is a polyhedron with a dent: non-convex. It has 5 vertices ($A$, $B$, $C$, $D$, $E$), 9 edges ($AB$, $BC$, $CA$ and the six from $D$ and $E$) and 6 faces.

III. False. In a convex polyhedron each face has at least 3 edges, so $2E \ge 3F$. Here, $32 \ge 33$ is false. (Equivalently, $E \le 3V - 6 = 15 < 16$.) Note that Euler’s relation ($7 - 16 + 11 = 2$) holds, but it is not enough.

Answer: option E (only II).

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Question 3: A constant term that does not appear

Statement

The constant term of the expansion of $\left(x + \dfrac{1}{x}\right)^6\left(x - \dfrac{2}{x}\right)^5$ is

• A) $-18$.
• B) $-12$.
• C) $0$.
• D) $12$.
• E) $18$.

Solution

1. The exponents. The terms of $\left(x + \frac{1}{x}\right)^6$ have the form $\binom{6}{j}x^{6 - 2j}$, and those of $\left(x - \frac{2}{x}\right)^5$, the form $\binom{5}{k}(-2)^k x^{5 - 2k}$.

2. Parity. In the product, the exponent is $(6 - 2j) + (5 - 2k) = 11 - 2(j + k)$, which is always odd. It is never 0.

So there is no $x^0$ term: the constant term is 0.

Answer: option C (0).

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Question 4: What the equality g(x²) = f(2x² − x + 1) forces

Statement

Let $f(x)$ and $g(x)$ be real functions defined for all $x \in \mathbb{R}$. If for all $x > 0$ the equality $g(x^2) = f(2x^2 - x + 1)$ holds, we can state that

• A) $f$ is not surjective.
• B) $f$ is not injective.
• C) $g$ is not surjective.
• D) $g$ is not injective.
• E) $f(a) \ne g(a)$ for all $a > 0$.

Solution

1. The idea. If two positive values $x_1 \ne x_2$ give the same $2x^2 - x + 1$, then $f$ receives the same number and

with $x_1^2 \ne x_2^2$. That shows $g$ is not injective.

2. Finding the values. The parabola $2x^2 - x + 1$ has its vertex at $x = \frac{1}{4}$, so values symmetric about $\frac{1}{4}$ give the same result: for example, $x_1 = \frac{1}{8}$ and $x_2 = \frac{3}{8}$:

Hence $g\left(\frac{1}{64}\right) = g\left(\frac{9}{64}\right)$.

3. The other options are not guaranteed.

• Take $f(t) = t$ and $g(u) = 2u - \sqrt{u} + 1$ for $u > 0$, $g(u) = u + \frac{7}{8}$ for $u \le 0$. The equality holds ($g(x^2) = 2x^2 - x + 1$), $f$ is bijective (A and B fail) and $g$ is surjective: for $u > 0$ it takes every value from $\frac{7}{8}$ upwards (minimum at $u = \frac{1}{16}$) and, for $u \le 0$, every value up to $\frac{7}{8}$ (C fails).
• With $f$ and $g$ both identically zero, the equality holds and $f(a) = g(a)$ for all $a$ (E fails).

Answer: option D ($g$ is not injective).

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Question 5: Partial sums as a difference

Statement

Let $(a_n)_{n \in \mathbb{N}}$ and $(b_n)_{n \in \mathbb{N}}$ be two numerical sequences such that

Consider the statements below:

• I. If $(b_n)_{n \in \mathbb{N}}$ is a geometric progression, then $(a_n)_{n \in \mathbb{N}}$ is also a geometric progression.
• II. If $(a_n)_{n \in \mathbb{N}}$ is a geometric progression, then $(b_n)_{n \in \mathbb{N}}$ is also a geometric progression.
• III. If $(a_n)_{n \in \mathbb{N}}$ is an arithmetic progression, then $(S_n)_{n \in \mathbb{N}}$ is also an arithmetic progression.

Which is (are) correct:

• A) I and II.
• B) II and III.
• C) I and III.
• D) only I.
• E) none.

Solution

Key fact. Subtracting $S_n - S_{n-1}$ (and using $S_1 = a_1 = b_2 - b_1$):

I. True. If $b_n = b_1 q^{n-1}$, then $a_n = b_1 q^{n-1}(q - 1)$: a geometric progression with the same ratio $q$. (If $q = 1$, $a_n$ is the zero sequence, the trivial geometric progression.)

II. False. $b_n = b_1 + S_{n-1}$ depends on the free value $b_1$. With $a_n = 1$ (ratio 1) and $b_1 = 1$: $b_n = n$, which is not a geometric progression.

III. False. If $a_n$ is an arithmetic progression with difference $r \ne 0$, $S_n$ is quadratic in $n$. For example, $a_n = n$ gives $S_n = 1, 3, 6, 10, \dots$, which is not an arithmetic progression.

Answer: option D (only I).

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Question 6: Two solutions for the exponential equation

Statement

The set of all values $a \in \mathbb{R}$ for which the equation $9^x - (3a + 4)3^x + 2a^2 + 9a - 5 = 0$ has two distinct real solutions is

• A) $(6, \infty)$.
• B) $\left(\frac{1}{2}, \infty\right) \setminus \{6\}$.
• C) $\left(\frac{3}{2}, \infty\right) \setminus \{6\}$.
• D) $\mathbb{R} \setminus \{6\}$.
• E) $\mathbb{R}$.

Solution

1. Change of variable. With $t = 3^x > 0$ (each $t > 0$ gives exactly one $x$):

2. Factoring. $2a^2 + 9a - 5 = (2a - 1)(a + 5)$ and $(2a - 1) + (a + 5) = 3a + 4$. So the roots are

3. Two solutions in $x$ require two positive and distinct roots $t$:

• $2a - 1 > 0 \Rightarrow a > \frac{1}{2}$;
• $a + 5 > 0 \Rightarrow a > -5$;
• $2a - 1 \ne a + 5 \Rightarrow a \ne 6$.

Answer: option B ($\left(\frac{1}{2}, \infty\right) \setminus \{6\}$).

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Question 7: Finding k for the systems to be equivalent

Statement

Let $m$, $n$ and $k$ be real numbers. Consider the systems below:

Given that the two systems are equivalent (that is, they have the same solution set), the value of $k$ is

• A) $\frac{1}{3}$.
• B) $\frac{1}{2}$.
• C) $1$.
• D) $2$.
• E) $3$.

Solution

1. Solving I. Subtracting the 1st equation from the 2nd: $y - 2z = 1$. Subtracting twice the 1st from the 3rd: $-y + 2z = -1$, the same equation. The system is consistent with infinitely many solutions: with $z = t$, $y = 1 + 2t$ and $x = 1 - y - z = -3t$.

2. Requiring these solutions to satisfy II.

• 2nd equation: $-3t + m(1 + 2t) + nt = 5$ for all $t$: $m = 5$ and $-3 + 2m + n = 0$, so $n = -7$.
• 3rd equation: $-3kt + (1 + 2t) - t = 3m + 2n = 15 - 14 = 1$, that is, $(1 - 3k)t = 0$ for all $t$: $k = \frac{1}{3}$.

3. Same solution set. With these values, the equations of II are $x + y + z = 1$, $x + 5y - 7z = 5$ and $\frac{1}{3}x + y - z = 1$. The 3rd is $\frac{1}{6}$ of the sum of the first two, and the first two are independent: II is also a line of solutions, the same as I ✔

Answer: option A ($k = \frac{1}{3}$).

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Question 8: The longest interval on which f is defined

Statement

Consider the function

Let $I = (a, b)$ be the longest interval contained in $[-\pi, 2\pi]$ such that $f(x)$ is defined for all $x \in I$. The value of $a + b$ is

• A) $0$.
• B) $\frac{\pi}{3}$.
• C) $\frac{2\pi}{3}$.
• D) $\frac{5\pi}{6}$.
• E) $\frac{13\pi}{6}$.

Solution

1. Factoring. With $\sin 2x = 2\sin x\cos x$:

and the denominator is $8(\cos 2x + 1) = 16\cos^2 x$.

2. Conditions. $f$ is defined when $\cos x \ne 0$ and $(2\sin x - 1)(2\cos x + 1) \ge 0$.

3. Signs over one turn $[0, 2\pi)$.

• $2\sin x - 1 \ge 0 \iff x \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$;
• $2\cos x + 1 \ge 0 \iff x \in \left[0, \frac{2\pi}{3}\right] \cup \left[\frac{4\pi}{3}, 2\pi\right)$.

The product is $\ge 0$ when both factors have the same sign:

• both $\ge 0$: $\left[\frac{\pi}{6}, \frac{2\pi}{3}\right]$, without $\frac{\pi}{2}$ (where $\cos x = 0$);
• both $\le 0$: $\left[\frac{5\pi}{6}, \frac{4\pi}{3}\right]$.

4. In $[-\pi, 2\pi]$. The pieces are $\left[-\pi, -\frac{2\pi}{3}\right]$ (the copy of $\left[\frac{5\pi}{6}, \frac{4\pi}{3}\right]$ shifted by $-2\pi$, cut at $-\pi$), $\left[\frac{\pi}{6}, \frac{\pi}{2}\right)$, $\left(\frac{\pi}{2}, \frac{2\pi}{3}\right]$ and $\left[\frac{5\pi}{6}, \frac{4\pi}{3}\right]$, with lengths $\frac{\pi}{3}$, $\frac{\pi}{3}$, $\frac{\pi}{6}$ and $\frac{\pi}{2}$. The longest is $I = \left(\frac{5\pi}{6}, \frac{4\pi}{3}\right)$.

(We checked by computer, testing the domain on a fine grid over $[-\pi, 2\pi]$: the longest continuous stretch goes from $\approx 0.8333\pi$ to $\approx 1.3333\pi$.)

Answer: option E ($\frac{13\pi}{6}$).

To review, see Sine, cosine and tangent inequalities.

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Question 9: Spinning triangle PO₁R

Question annulled by ITA. According to the official answer key, question 9 was annulled because none of the options is correct. Below is the computation, which shows the true value.

Statement

Two circles $C_1$ and $C_2$ with centres $O_1$ and $O_2$, respectively, have radius 5 cm and meet at the points $P$ and $Q$, determining a common chord $\overline{PQ}$ of length 6 cm. The intersections of each circle with the segment $\overline{O_1O_2}$ determine the points $R$ on $C_1$ and $S$ on $C_2$. The volume of the solid of revolution obtained by rotating triangle $PO_1R$ around line $O_1O_2$ is, in cm³,

• A) $\pi$.
• B) $3\pi$.
• C) $4\pi$.
• D) $6\pi$.
• E) $9\pi$.

Comment

1. The figure. The common chord is perpendicular to $O_1O_2$ and bisected by it at the point $M$. So $m(\overline{PM}) = 3$ and, in the right triangle $O_1MP$, $m(\overline{O_1M}) = \sqrt{25 - 9} = 4$.

2. The point $R$. $R$ is on $C_1$ and on segment $\overline{O_1O_2}$, so $m(\overline{O_1R}) = 5$, and $M$ lies between $O_1$ and $R$, with $m(\overline{MR}) = 1$.

3. The solid. Rotating triangle $PO_1R$ around $O_1R$, the point $P$ traces a circle of radius $3$ centred at $M$. The solid is two cones with that base, with heights $4$ and $1$:

The value $15\pi$ is not among the options, which confirms the annulment. (Option B, $3\pi$, is just the smaller cone.)

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Question 10: The common root of the two polynomials

Statement

A root common to the polynomials $p(x) = x^5 - x^3 + 12x - 18$ and $q(x) = x^4 - x^3 + 5x - 3$ is

• A) $-1 + \sqrt{2}\,i$.
• B) $\sqrt{2} + \sqrt{2}\,i$.
• C) $\sqrt{2} + i$.
• D) $1 + \sqrt{2}\,i$.
• E) $-\sqrt{2} + i$.

Solution

1. Euclid’s algorithm. A common root makes any combination $p - h\cdot q$ vanish. Computing:

So $p - (x + 1)q = -5(x^2 - 2x + 3)$.

2. The common roots. Every common root satisfies $x^2 - 2x + 3 = 0$, that is, $x = 1 \pm \sqrt{2}\,i$. And they are indeed common:

(check by multiplying).

Answer: option D ($1 + \sqrt{2}\,i$).

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Question 11: A hyperbola with the foci of the ellipse

Statement

Let $H$ be a hyperbola in the Cartesian plane whose foci are the foci of the ellipse $E: \dfrac{x^2}{4} + y^2 = 1$. Let $P$ be an intersection point of $E$ and $H$ in the first quadrant and $O$ the origin. Given that $\overline{OP}$ makes an angle $\theta$ with the horizontal axis $O_x$, with $\tan(\theta) = \dfrac{\sqrt{3}}{6}$, the eccentricity of $H$ is

• A) $\frac{2\sqrt{3}}{3}$.
• B) $\frac{\sqrt{6}}{2}$.
• C) $\frac{3\sqrt{3}}{4}$.
• D) $\frac{2\sqrt{6}}{3}$.
• E) $\frac{3\sqrt{3}}{2}$.

Solution

1. Foci. For the ellipse, $a = 2$, $b = 1$, $c = \sqrt{3}$: foci $F_1 = (-\sqrt{3}, 0)$ and $F_2 = (\sqrt{3}, 0)$.

2. The point $P$. On $\overline{OP}$, $y = \frac{\sqrt{3}}{6}x$. On the ellipse:

$P = \left(\sqrt{3}, \frac{1}{2}\right)$, exactly above the focus $F_2$.

3. The hyperbola. $|PF_1 - PF_2| = 2a_H$, with

So $2a_H = 3$, $a_H = \frac{3}{2}$, and with $c = \sqrt{3}$:

Checking with the ellipse: $PF_1 + PF_2 = 4 = 2\cdot 2$ ✔

Answer: option A ($\frac{2\sqrt{3}}{3}$).

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Question 12: 16m with 231 divisors

Statement

A number $m \in \mathbb{N}$, when multiplied by 16, gives a number with 231 positive integer divisors. Consider the statements:

• I. $m$ is an even number.
• II. $m$ is a perfect square.
• III. $16m$ has 3 distinct prime factors.

Which is (are) TRUE:

• A) only I.
• B) only II.
• C) I and II.
• D) I and III.
• E) II and III.

Solution

1. Counting divisors. Write $m = 2^e \cdot p_1^{f_1}\cdots p_r^{f_r}$ (odd primes). Then $16m = 2^{e + 4}p_1^{f_1}\cdots p_r^{f_r}$ and

2. Consequences.

• The factor $e + 5$ is a divisor of 231 that is at least 5: $7, 11, 21, 33, 77$ or $231$. So $e \in \{2, 6, 16, 28, 72, 226\}$, and always $e \ge 2$.
• All divisors of 231 are odd, so each $f_i + 1$ is odd and each $f_i$ is even; and $e$ is also even in every case.

I. True: $e \ge 2$, so $m$ is even.

II. True: all exponents of $m$ are even.

III. False: with $m = 2^{226}$, $16m = 2^{230}$ has $231$ divisors and a single prime factor.

Answer: option C (I and II).

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