Board 2025 Class 12 Maths Q24 Solution: Where f(x) = 5x^(3/2) − 3x^(5/2) is increasing and decreasing

The question

Find the intervals in which function $f(x) = 5x^{\frac32} - 3x^{\frac52}$ is (i) increasing (ii) decreasing.

Exam answer

$f$ is defined for $x \ge 0$ ⁠.

Why?

$x^{\frac32} = (\sqrt x)^3$ and $x^{\frac52} = (\sqrt x)^5$ ⁠. Both need $\sqrt x$ ⁠, which is real only for $x \ge 0$ ⁠. So we look for the intervals inside $[0, \infty)$ ⁠.

\begin{aligned} f'(x) &= \frac{15}{2}x^{\frac12} - \frac{15}{2}x^{\frac32} \\ &= \frac{15}{2}\sqrt x\,(1 - x) \end{aligned}

Why?

Use the power rule $\frac{d}{dx}x^n = nx^{n - 1}$ on each term:

\begin{aligned} 5\cdot\frac32 = \frac{15}{2}&,\quad \frac32 - 1 = \frac12 \\ 3\cdot\frac52 = \frac{15}{2}&,\quad \frac52 - 1 = \frac32 \end{aligned}

\begin{aligned} f'(x) &= 5\cdot\frac32x^{\frac12} - 3\cdot\frac52x^{\frac32} \\ &= \frac{15}{2}x^{\frac12} - \frac{15}{2}x^{\frac32} \end{aligned}

Then take out the common factor $\frac{15}{2}x^{\frac12}$ ⁠, using $x^{\frac32} = x^{\frac12}\cdot x$ ⁠.

\begin{aligned} &\frac{15}{2}x^{\frac12} - \frac{15}{2}x^{\frac12}\cdot x \\ &= \frac{15}{2}x^{\frac12}(1 - x) \\ &= \frac{15}{2}\sqrt x\,(1 - x) \end{aligned}

f'(x) = 0 \Rightarrow x = 0, 1

Why?

A product is zero when a factor is zero: $\sqrt x = 0$ gives $x = 0$ ⁠, and $1 - x = 0$ gives $x = 1$ ⁠.

These points split $[0, \infty)$ into $[0, 1]$ and $[1, \infty)$ ⁠. We test the sign of $f'(x)$ on each piece.

For $0 < x < 1$ ⁠, $f'(x) > 0$ ⁠; for $x > 1$ ⁠, $f'(x) < 0$ ⁠.

Why?

For $x > 0$ ⁠, $\frac{15}{2}\sqrt x$ is positive, so $f'(x)$ has the same sign as $1 - x$ ⁠. For example, $x = \frac14$ gives $1 - x > 0$ and $x = 4$ gives $1 - x < 0$ ⁠.

\begin{aligned} x = \tfrac14&:\ 1 - \tfrac14 = \tfrac34 > 0 \\ x = 4&:\ 1 - 4 = -3 < 0 \end{aligned}

$f' > 0$ means increasing and $f' < 0$ means decreasing. The end points $0$ and $1$ may be included, because $f$ is continuous there.

(i) $f$ is increasing in $[0, 1]$ ⁠.

(ii) $f$ is decreasing in $[1, \infty)$ ⁠.

Where this comes from

Chapter 6, Application of Derivatives: Increasing and decreasing functions.

Question type: Find where a function increases and decreases

How to spot it: click to see

Type 5. Find where a function increases and decreases

Spot it: "Find the intervals in which $f$ is increasing or decreasing", or "strictly increasing or strictly decreasing".

Here: $f'(x) = \frac{15}{2}\sqrt x\,(1 - x)$ is 0 at $x = 0$ and $x = 1$ ⁠, which cut $[0, \infty)$ into $[0, 1]$ and $[1, \infty)$ ⁠. The sign of $1 - x$ decides each piece.

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Q24: Where f(x)=5x3/2−3x5/2f(x) = 5x^{3/2} - 3x^{5/2}f(x)=5x3/2−3x5/2 is increasing and decreasing

Q24: Where $f(x) = 5x^{3/2} - 3x^{5/2}$ is increasing and decreasing