Question

If a real number $ a $ is such that $ \int_0^a x \, dx \leq a + 4 $, then what is the range of $ a $?

• A. \( -2 \leq a \leq 0 \)
• B. \( 0 \leq a \leq 4 \)
• C. \( -2 \leq a \leq 4 \)
• D. \( a \leq -2 \) or \( a \geq 4 \)

Hint:

For solving quadratic inequalities, use the sign method after factorizing the quadratic expression.

Answer 1

The Correct Option is C

The given inequality is \( \int_0^a x \, dx \leq a + 4 \). First, solve the integral: \[ \int_0^a x \, dx = \frac{a^2}{2} \] Now, substitute this into the inequality: \[ \frac{a^2}{2} \leq a + 4 \] Multiply through by 2 to eliminate the fraction: \[ a^2 \leq 2a + 8 \] Rearrange the terms: \[ a^2 - 2a - 8 \leq 0 \] Factorize the quadratic equation: \[ (a - 4)(a + 2) \leq 0 \] This inequality holds for \( -2 \leq a \leq 4 \). Thus, the solution is \( -2 \leq a \leq 4 \).