In the expansion of (√(x) + (1)/(x))^6, the rational (i.e. integer-power-of-x) terms count is:

The exponent (6 - 3k)/(2) is an integer when k is even: k ∈ {0, 2, 4, 6} — four values.

Daily Math Problems — June 7, 2026: Trigonometry, Combinatorics, Calculus

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Problems & worked solutions

RO M1

Problem 1 — Solving Triangles

In $△ A B C$ , $a = 5$ , $b = 6$ , $c = 7$ . The inradius $r$ is closest to:

Show answer & worked solution

A. $1.4$
B. $1.6$ ✓ correct
C. $2.4$
D. $3.0$

$s = 9$ . By Heron, Area $= \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = 6 \sqrt{6} \approx 14.70$ . So $r \approx \frac{14.70}{9} \approx 1.63$ .

RO M1

Problem 2 — Binomial Theorem

In the expansion of $⁣ {(\sqrt{x} + \frac{1}{x})}^{6}$ , the rational (i.e. integer-power-of- $x$ ) terms count is:

Show answer & worked solution

A. $1$
B. $2$
C. $3$ ✓ correct
D. $4$

The exponent $\frac{6 - 3 k}{2}$ is an integer when $k$ is even: $k \in {0, 2, 4, 6}$ — four values.

RO M1

Problem 3 — Rolle's Sign Method

By Rolle, the equation $x^{4} - 6 x^{2} + 8 = 0$ has how many distinct real solutions?

Show answer & worked solution

A. $0$
B. $1$
C. $2$
D. $4$ ✓ correct

$y^{2} - 6 y + 8 = 0 \Rightarrow y \in {2, 4}$ . Then $x^{2} = 2 \Rightarrow x = \pm \sqrt{2}$ and $x^{2} = 4 \Rightarrow x = \pm 2$ . Four distinct real solutions.

Problems & worked solutions

RO M1

Problem 1 — Solving Triangles

In $△ A B C$ , $a = 5$ , $b = 6$ , $c = 7$ . The inradius $r$ is closest to:

Show answer & worked solution

A. $1.4$
B. $1.6$ ✓ correct
C. $2.4$
D. $3.0$

$s = 9$ . By Heron, Area $= \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = 6 \sqrt{6} \approx 14.70$ . So $r \approx \frac{14.70}{9} \approx 1.63$ .

RO M1

Problem 2 — Binomial Theorem

In the expansion of $⁣ {(\sqrt{x} + \frac{1}{x})}^{6}$ , the rational (i.e. integer-power-of- $x$ ) terms count is:

Show answer & worked solution

A. $1$
B. $2$
C. $3$ ✓ correct
D. $4$

The exponent $\frac{6 - 3 k}{2}$ is an integer when $k$ is even: $k \in {0, 2, 4, 6}$ — four values.

RO M1

Problem 3 — Rolle's Sign Method

By Rolle, the equation $x^{4} - 6 x^{2} + 8 = 0$ has how many distinct real solutions?

Show answer & worked solution

A. $0$
B. $1$
C. $2$
D. $4$ ✓ correct

$y^{2} - 6 y + 8 = 0 \Rightarrow y \in {2, 4}$ . Then $x^{2} = 2 \Rightarrow x = \pm \sqrt{2}$ and $x^{2} = 4 \Rightarrow x = \pm 2$ . Four distinct real solutions.