Q112 / 180 · COMEDK 2025 (10 May Shift 2) · Mathematics
In the interval $(0,1)$ the function $f(x)=x^2-x+1$ is
- AIncreasing
- Correct answerBNeither increasing nor decreasing
- CStrictly decreasing
- DDecreasing
Solution
The function is given by $f(x) = x^2 - x + 1$.
To determine the monotonicity of the function, we find its derivative with respect to $x$:
$f'(x) = \dfrac{d}{dx}(x^2 - x + 1) = 2x - 1$.
We analyze the sign of $f'(x)$ in the interval $(0, 1)$.
For $0 < x < \dfrac{1}{2}$, $2x < 1$, which implies $2x - 1 < 0$. Thus, $f'(x) < 0$ in the interval $(0, \dfrac{1}{2})$, meaning the function is strictly decreasing in this sub-interval.
For $\dfrac{1}{2} < x < 1$, $2x > 1$, which implies $2x - 1 > 0$. Thus, $f'(x) > 0$ in the interval $(\dfrac{1}{2}, 1)$, meaning the function is strictly increasing in this sub-interval.
Since the derivative changes sign within the interval $(0, 1)$, the function is neither increasing nor decreasing on the entire interval $(0, 1)$.
Answer: Neither increasing nor decreasing
To determine the monotonicity of the function, we find its derivative with respect to $x$:
$f'(x) = \dfrac{d}{dx}(x^2 - x + 1) = 2x - 1$.
We analyze the sign of $f'(x)$ in the interval $(0, 1)$.
For $0 < x < \dfrac{1}{2}$, $2x < 1$, which implies $2x - 1 < 0$. Thus, $f'(x) < 0$ in the interval $(0, \dfrac{1}{2})$, meaning the function is strictly decreasing in this sub-interval.
For $\dfrac{1}{2} < x < 1$, $2x > 1$, which implies $2x - 1 > 0$. Thus, $f'(x) > 0$ in the interval $(\dfrac{1}{2}, 1)$, meaning the function is strictly increasing in this sub-interval.
Since the derivative changes sign within the interval $(0, 1)$, the function is neither increasing nor decreasing on the entire interval $(0, 1)$.
Answer: Neither increasing nor decreasing
Hence, the correct answer is option: 'B'.