∫ c dx = cx + C. So ∫ 5 dx = 5x + C.

Consider f: ℝ → ℝ, f(x) = x^2. Which statement is correct?

f(-1) = 1 = f(1), so f is not injective. The value -1 has no real preimage (since x^2 ≥ 0), so f is not surjective either.

f is neither injective nor surjective

f is injective but not surjective

f is surjective but not injective

Daily Math Problems — June 25, 2026: Calculus, Functions, Complex Numbers

Problems & worked solutions

RO M1

$\int 5 d x$ equals:

Show answer & worked solution

$\int c d x = c x + C$ . So $\int 5 d x = 5 x + C$ .

RO M1

Consider $f : R \to R$ , $f (x) = x^{2}$ . Which statement is correct?

Show answer & worked solution

$f (- 1) = 1 = f (1)$ , so $f$ is not injective. The value $- 1$ has no real preimage (since $x^{2} \geq 0$ ), so $f$ is not surjective either.

RO M1

The set of all complex solutions of $z^{2} = - 4$ is:

Show answer & worked solution

$z^{2} = - 4 \Rightarrow z^{2} + 4 = 0 \Rightarrow (z - 2 i) (z + 2 i) = 0$ . Solutions: $z = 2 i$ or $z = - 2 i$ .

Problems & worked solutions

RO M1

$\int 5 d x$ equals:

Show answer & worked solution

$\int c d x = c x + C$ . So $\int 5 d x = 5 x + C$ .

RO M1

Consider $f : R \to R$ , $f (x) = x^{2}$ . Which statement is correct?

Show answer & worked solution

$f (- 1) = 1 = f (1)$ , so $f$ is not injective. The value $- 1$ has no real preimage (since $x^{2} \geq 0$ ), so $f$ is not surjective either.

RO M1

The set of all complex solutions of $z^{2} = - 4$ is:

Show answer & worked solution

$z^{2} = - 4 \Rightarrow z^{2} + 4 = 0 \Rightarrow (z - 2 i) (z + 2 i) = 0$ . Solutions: $z = 2 i$ or $z = - 2 i$ .