Question

A function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \begin{cases} 2x+3, & x \le 4/3 \\ -3x^2+8x, & x>4/3 \end{cases}$ is

• A. One-one function
• B. not onto
• C. a bijective function
• D. constant function

Hint:

To check if a function is onto, always compare its range with the codomain. If even a single value in the codomain is not achieved, the function is not surjective.

Answer 1

The Correct Option is B

Step 1: Checking injectivity (one-one nature):
For $x \le \frac{4}{3}$, the function is $f(x) = 2x + 3$. Since its derivative $f'(x) = 2>0$, the function is strictly increasing in this interval.
For $x>\frac{4}{3}$, the function becomes $f(x) = -3x^2 + 8x$. Its derivative is $f'(x) = -6x + 8$. For all $x>\frac{4}{3}$, we have $f'(x)<0$, so the function is strictly decreasing in this region.
Since the function increases up to $x = \frac{4}{3}$ and decreases afterward, it fails the one-one test.
For instance, \[ f(1) = 2(1) + 3 = 5 \] Now solving $-3x^2 + 8x = 5$, \[ 3x^2 - 8x + 5 = 0 \Rightarrow (3x - 5)(x - 1) = 0 \] This gives $x = 1$ or $x = \frac{5}{3}$. Since $\frac{5}{3}>\frac{4}{3}$, \[ f\left(\frac{5}{3}\right) = 5 \] Thus, $f(1) = f\left(\frac{5}{3}\right)$, proving that the function is not one-one.

Step 2: Checking surjectivity (onto nature):
The codomain is $\mathbb{R}$, so we determine the range.
For $x \le \frac{4}{3}$, the function is linear and increasing, attaining its maximum at $x = \frac{4}{3}$: \[ f\left(\frac{4}{3}\right) = \frac{17}{3} \] Hence, the range here is $(-\infty, \frac{17}{3}]$.
For $x>\frac{4}{3}$, the function $f(x) = -3x^2 + 8x$ is a downward-opening parabola. Its maximum occurs at $x = \frac{4}{3}$, but since this point is excluded, the range becomes $(-\infty, \frac{16}{3})$.
Combining both parts, the overall range is \[ (-\infty, \frac{17}{3}] \]
Step 3: Comparing range and codomain:
The range is $(-\infty, \frac{17}{3}]$, whereas the codomain is $\mathbb{R}$.
Since values greater than $\frac{17}{3}$ are not attained, the function does not cover the entire codomain.
Therefore, the function is not onto.