Question:medium
The function $f(x)=2x^{3}-15x^{2}+36x-24$ is strictly decreasing in the interval is
The function $f(x)=2x^{3}-15x^{2}+36x-24$ is strictly decreasing in the interval is
Show Hint
Logic Tip: For any upward-opening quadratic inequality $ax^2 + bx + c<0$ (where $a>0$) with real roots $r_1<r_2$, the solution is always the bounded interval $(r_1, r_2)$. You don't need to test numbers if you remember this geometric rule.
Updated On: Apr 27, 2026
- (2,3)
- (1,3)
- (2,4)
- (1,4)
- (2,5)
Show Solution
The Correct Option is A
Solution and Explanation
Was this answer helpful?
0
Top Questions on Application of derivatives
- If the function \( f(x) \), defined below, is continuous on the interval \([0,8]\), then: \[ f(x) = \begin{cases} x^2 + ax + b, & 0 \leq x < 2 \\ 3x + 2, & 2 \leq x \leq 4 \\ 2ax + 5b, & 4 < x \leq 8 \end{cases} \]
- BITSAT - 2024
- Mathematics
- Application of derivatives
- If \( x\sqrt{1 + y} + y\sqrt{1 + x} = 0 \), then find \( \frac{dy}{dx} \).
- BITSAT - 2024
- Mathematics
- Application of derivatives
- If \( y = \tan^{-1}\left( \frac{\sqrt{x} - x}{1 + x^{3/2}} \right) \), then \( y'(1) \) is equal to:
- BITSAT - 2024
- Mathematics
- Application of derivatives
- At \( x = \frac{\pi^2}{4} \), \( \frac{d}{dx} \left( \tan^{-1}(\cos\sqrt{x}) + \sec^{-1}(e^x) \right) = \)
- BITSAT - 2024
- Mathematics
- Application of derivatives
- Want to practice more? Try solving extra ecology questions todayView All Questions
