Q235 / 270 · NDA 2026 (Phase 1) · Mathematics
Passage: Let $S$ and $T$ be the sets where $f(x)=\dfrac{x^3}{3}-\dfrac{5x^2}{2}+6x+7$ decreases and increases respectively.
Question: What is $T$ equal to ?
Question: What is $T$ equal to ?
- Correct answerA$\{x\leq2\}\cup\{x\geq3\}$
- B$\{x<2\}\cup\{x>3\}$
- C$(2,3)$
- D$[2,3]$
Solution
Given $f(x) = \dfrac{x^3}{3} - \dfrac{5x^2}{2} + 6x + 7$
Differentiating with respect to $x$, we get:
$f'(x) = x^2 - 5x + 6$
$f'(x) = (x - 2)(x - 3)$
For the function to be increasing, $f'(x) \geq 0$
$(x - 2)(x - 3) \geq 0$
This inequality holds true for $x \leq 2$ or $x \geq 3$.
Therefore, the set $T$ where the function increases is $\{x \leq 2\} \cup \{x \geq 3\}$.
Answer: $\{x\leq2\}\cup\{x\geq3\}$
Differentiating with respect to $x$, we get:
$f'(x) = x^2 - 5x + 6$
$f'(x) = (x - 2)(x - 3)$
For the function to be increasing, $f'(x) \geq 0$
$(x - 2)(x - 3) \geq 0$
This inequality holds true for $x \leq 2$ or $x \geq 3$.
Therefore, the set $T$ where the function increases is $\{x \leq 2\} \cup \{x \geq 3\}$.
Answer: $\{x\leq2\}\cup\{x\geq3\}$
Hence, the correct answer is option: 'A'.