Sets, Relations & Functions
Functions
nta_abhyas_2025
Grade 11
Question:
Let $f: \mathbb{R} \to (2, \infty)$ be a function defined as $f(x) = x^3 - 12ax + 15 - 2a + 3b(x)$. If $f(x)$ is surjective on $\mathbb{R}$, then the value of $a$ is equal to
$-\frac{1}{4}$
$\frac{1}{2}$
$\frac{11}{2}$
$\frac{33}{2}$
Step-by-Step Solution
Key Concept: A quadratic function is surjective on $\mathbb{R}$ only when specific conditions on its parameters are met.
Rewriting the given function: $f(x) = (x - 6a)^2 + 15 - 2a$. Since $f(x)$ is surjective on $\mathbb{R}$, the range must be $\mathbb{R}$, which is impossible for a quadratic with minimum value. The condition requires $15 - 2a = 2$ to minimize constraints, giving $2a = 13$, so $a = \frac{13}{2}$.
Correct Answer: 2