Q79 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
$\text { Which of the following function is injective? }$
- A$f(x)=\dfrac{4 x^2+3 x-5}{4+3 x-5 x^2}, x \in(-\infty, \infty)$
- B$f(x)=(x-4)(x-5), x \in(-\infty, \infty)$
- Correct answerC$f(x)=|x+2|, x \in[-2, \infty)$
- D$f(x)=x^2+2, x \in(-\infty, \infty)$
Solution
A function $f(x)$ is injective if $f(x_1) = f(x_2)$ implies $x_1 = x_2$.
For option (LZ8Jm), the function $f(x) = \dfrac{4x^2 + 3x - 5}{4 + 3x - 5x^2}$ is a rational function. It is not injective as it is not monotonic over its entire domain.
For option (fCqBV), $f(x) = (x-4)(x-5) = x^2 - 9x + 20$. This is a parabola opening upwards, which is not injective on $(-\infty, \infty)$ because $f(4) = f(5) = 0$.
For option (bVs0N), $f(x) = |x+2|$ for $x \in [-2, \infty)$. Since $x \ge -2$, $x+2 \ge 0$, so $f(x) = x+2$. This is a linear function with a positive slope, which is strictly increasing. Therefore, it is injective.
For option (FeHkA), $f(x) = x^2 + 2$ is a parabola symmetric about the $y$-axis, so $f(x) = f(-x)$. Thus, it is not injective.
Answer: $f(x)=|x+2|, x \in[-2, \infty)$
For option (LZ8Jm), the function $f(x) = \dfrac{4x^2 + 3x - 5}{4 + 3x - 5x^2}$ is a rational function. It is not injective as it is not monotonic over its entire domain.
For option (fCqBV), $f(x) = (x-4)(x-5) = x^2 - 9x + 20$. This is a parabola opening upwards, which is not injective on $(-\infty, \infty)$ because $f(4) = f(5) = 0$.
For option (bVs0N), $f(x) = |x+2|$ for $x \in [-2, \infty)$. Since $x \ge -2$, $x+2 \ge 0$, so $f(x) = x+2$. This is a linear function with a positive slope, which is strictly increasing. Therefore, it is injective.
For option (FeHkA), $f(x) = x^2 + 2$ is a parabola symmetric about the $y$-axis, so $f(x) = f(-x)$. Thus, it is not injective.
Answer: $f(x)=|x+2|, x \in[-2, \infty)$
Hence, the correct answer is option: 'C'.