Sample 2026-27 Class 12 Maths Q9 Solution: Interval where y = a{x + 2a sin(x/a)} decreases

The question

If $y = a\left\{x + 2a\sin\left(\frac{x}{a}\right)\right\}$ ⁠, $a > 0$ ⁠, $x \in [0, \pi a]$ ⁠; is a decreasing function, then the interval in which $x$ lies is

(A) $\left(\frac{5\pi}{6}, \pi\right]$

(B) $\left(\frac{2\pi}{3}, \pi\right]$

(C) $\left(\frac{5\pi a}{-6}, \pi a\right]$

(D) $\left(\frac{2\pi a}{3}, \pi a\right]$

Exam answer

\begin{aligned} \frac{dy}{dx} &= a\left\{1 + 2a\cos\left(\frac{x}{a}\right) \cdot \frac{1}{a}\right\} \\ &= a\left\{1 + 2\cos\left(\frac{x}{a}\right)\right\} \end{aligned}

Why?

$a$ is a constant. By the chain rule, the inner function $\frac{x}{a}$ has derivative $\frac{1}{a}$ ⁠:

\frac{d}{dx}\sin\left(\frac{x}{a}\right) = \frac{1}{a}\cos\left(\frac{x}{a}\right)

So the $2a$ in front becomes $2a \cdot \frac{1}{a} = 2$ ⁠. In the next step, dividing by $a > 0$ keeps the sign of the inequality.

$y$ is decreasing, so $\frac{dy}{dx} < 0$ ⁠:

\begin{aligned} &a\left\{1 + 2\cos\left(\frac{x}{a}\right)\right\} < 0 \\ \Rightarrow\ &\cos\left(\frac{x}{a}\right) < -\frac{1}{2} \quad [a > 0] \end{aligned}

Why?

One move per line. Divide by $a > 0$ ⁠, take 1 from both sides, then divide by 2:

\begin{aligned} 1 + 2\cos\left(\tfrac{x}{a}\right) &< 0 \\ 2\cos\left(\tfrac{x}{a}\right) &< -1 \\ \cos\left(\tfrac{x}{a}\right) &< -\frac{1}{2} \end{aligned}

As $x \in [0, \pi a]$ ⁠, $\frac{x}{a} \in [0, \pi]$ ⁠. On $[0, \pi]$ ⁠, $\cos\theta = -\frac{1}{2}$ at $\theta = \frac{2\pi}{3}$ ⁠, and $\cos\theta$ decreases. So

\begin{aligned} &\frac{x}{a} \in \left(\frac{2\pi}{3}, \pi\right] \\ \Rightarrow\ &x \in \left(\frac{2\pi a}{3}, \pi a\right] \end{aligned}

Why?

$\cos\theta$ falls from $1$ at $\theta = 0$ to $-1$ at $\theta = \pi$ ⁠, passing $-\frac{1}{2}$ at $\frac{2\pi}{3}$ ⁠, since

\cos\frac{2\pi}{3} = -\cos\frac{\pi}{3} = -\frac{1}{2}

So $\cos\theta < -\frac{1}{2}$ exactly for $\theta$ beyond $\frac{2\pi}{3}$ ⁠. Multiplying by $a > 0$ gives the interval for $x$ ⁠.

\begin{aligned} \frac{2\pi}{3} < \frac{x}{a} &\le \pi \\ \frac{2\pi}{3} \times a < x &\le \pi \times a \\ \frac{2\pi a}{3} < x &\le \pi a \end{aligned}

Watch out: (B) is the interval for $\frac{x}{a}$ ⁠, not for $x$ ⁠.

Answer: (D) $\left(\dfrac{2\pi a}{3}, \pi a\right]$

Where this comes from

Chapter 6, Application of Derivatives: 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: $\frac{dy}{dx} = a\left\{1 + 2\cos\left(\frac{x}{a}\right)\right\}$ is negative where $\cos\left(\frac{x}{a}\right)$ is below $-\frac12$ ⁠. With $\frac{x}{a}$ in $[0, \pi]$ ⁠, that is $\frac{x}{a}$ in $\left(\frac{2\pi}{3}, \pi\right]$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Q9: Interval where y=a{x+2asin⁡(xa)}y = a\left\{x + 2a\sin\left(\frac{x}{a}\right)\right\}y=a{x+2asin(ax​)} decreases

Q9: Interval where $y = a\left\{x + 2a\sin\left(\frac{x}{a}\right)\right\}$ decreases