A comprehensive platform for Teachers to create standard question papers and Students to practice Case-Based, Assertion-Reason, and Critical Thinking questions.
Create professional PDF/Word papers with logo, instructions, and mixed question types in minutes.
Explore our repository by Class and Topic. Filter by "Knowledge" or "Competency" levels.
For Students. Take timed MCQ tests to check your understanding. Get instant feedback.
According to NEP 2020, rote learning is out. The focus has shifted to assessing a student's ability to apply concepts in real-life situations.
Questions derived from real-world passages to test analytical skills.
Testing the logic behind concepts, not just the definition.
Open-ended scenarios that require thinking beyond the textbook.
We provide complete AI-Powered Explanations for every question.
Given the function $f(x) = 5x^{\frac{3}{2}} - 3x^{\frac{5}{2}}$, we need to find its derivative $f'(x)$. $$f'(x) = \frac{d}{dx}(5x^{\frac{3}{2}} - 3x^{\frac{5}{2}})$$ $$f'(x) = 5 \cdot \frac{3}{2}x^{\frac{3}{2}-1} - 3 \cdot \frac{5}{2}x^{\frac{5}{2}-1}$$ $$f'(x) = \frac{15}{2}x^{\frac{1}{2}} - \frac{15}{2}x^{\frac{3}{2}}$$ $$f'(x) = \frac{15}{2}x^{\frac{1}{2}}(1 - x)$$
To find the intervals where the function is increasing or decreasing, we need to find the critical points by setting $f'(x) = 0$. $$\frac{15}{2}x^{\frac{1}{2}}(1 - x) = 0$$ This gives us $x^{\frac{1}{2}} = 0$ or $1 - x = 0$. So, $x = 0$ or $x = 1$. Also, we need to consider the domain of the function. Since we have $x^{\frac{3}{2}}$ and $x^{\frac{5}{2}}$, $x$ must be non-negative. Thus, $x \ge 0$.
We have the critical points $x = 0$ and $x = 1$. These points divide the domain $x \ge 0$ into the intervals $(0, 1)$ and $(1, \infty)$.
We will test the sign of $f'(x)$ in each interval. Interval $(0, 1)$: Choose a test point, say $x = 0.5$. $$f'(0.5) = \frac{15}{2}(0.5)^{\frac{1}{2}}(1 - 0.5) = \frac{15}{2}\sqrt{0.5}(0.5) > 0$$ Since $f'(x) > 0$ in $(0, 1)$, the function is increasing in this interval. Interval $(1, \infty)$: Choose a test point, say $x = 4$. $$f'(4) = \frac{15}{2}(4)^{\frac{1}{2}}(1 - 4) = \frac{15}{2}(2)(-3) = -45 < 0$$ Since $f'(x) < 0$ in $(1, \infty)$, the function is decreasing in this interval.
(i) Increasing: $(0, 1)$ (ii) Decreasing: $(1, \infty)$
Final Answer: Increasing: (0, 1), Decreasing: (1, ∞)