The maximum value of f(x) = sin x + cos x over ℝ is:

sin x + cos x = √(2) (sin x · (√(2))/(2) + cos x · (√(2))/(2)) = √(2)sin (x + (π)/(4)). Since sin has maximum 1, the maximum of f is √(2), attained at x = (π)/(4).

Daily Math Problems — June 1, 2026: Polynomials, Combinatorics, Trigonometry

dailymath

Problems & worked solutions

RO M1

Problem 1 — Polynomials in ℂ

For $P (X) = X^{3} - 2 X + 5$ , the value of $P (2)$ is:

Show answer & worked solution

A. $5$
B. $7$
C. $9$ ✓ correct
D. $13$

$P (2) = 8 - 4 + 5 = 9$ .

RO M1

Problem 2 — Binomial Theorem

The coefficient of $x^{3}$ in $(x + 2)^{5}$ is:

Show answer & worked solution

A. $20$
B. $32$
C. $40$ ✓ correct
D. $80$

$5 - k = 3 \Rightarrow k = 2$ . Coefficient: $(\binom{5}{2}) \cdot 2^{2} = 10 \cdot 4 = 40$ .

RO M1

Problem 3 — Trigonometric Identities

The maximum value of $f (x) = \sin x + \cos x$ over $R$ is:

Show answer & worked solution

A. $1$
B. $\sqrt{2} - 1$
C. $\sqrt{2}$ ✓ correct
D. $2$

$\sin x + \cos x = \sqrt{2} ⁣ (\sin x \cdot \frac{\sqrt{2}}{2} + \cos x \cdot \frac{\sqrt{2}}{2}) = \sqrt{2} \sin ⁣ (x + \frac{π}{4})$ .

Since $\sin$ has maximum $1$ , the maximum of $f$ is $\sqrt{2}$ , attained at $x = \frac{π}{4}$ .

Problems & worked solutions

RO M1

Problem 1 — Polynomials in ℂ

For $P (X) = X^{3} - 2 X + 5$ , the value of $P (2)$ is:

Show answer & worked solution

A. $5$
B. $7$
C. $9$ ✓ correct
D. $13$

$P (2) = 8 - 4 + 5 = 9$ .

RO M1

Problem 2 — Binomial Theorem

The coefficient of $x^{3}$ in $(x + 2)^{5}$ is:

Show answer & worked solution

A. $20$
B. $32$
C. $40$ ✓ correct
D. $80$

$5 - k = 3 \Rightarrow k = 2$ . Coefficient: $(\binom{5}{2}) \cdot 2^{2} = 10 \cdot 4 = 40$ .

RO M1

Problem 3 — Trigonometric Identities

The maximum value of $f (x) = \sin x + \cos x$ over $R$ is:

Show answer & worked solution

A. $1$
B. $\sqrt{2} - 1$
C. $\sqrt{2}$ ✓ correct
D. $2$

$\sin x + \cos x = \sqrt{2} ⁣ (\sin x \cdot \frac{\sqrt{2}}{2} + \cos x \cdot \frac{\sqrt{2}}{2}) = \sqrt{2} \sin ⁣ (x + \frac{π}{4})$ .

Since $\sin$ has maximum $1$ , the maximum of $f$ is $\sqrt{2}$ , attained at $x = \frac{π}{4}$ .