Sample 2020-21 Class 12 Maths Q32 Solution: Where f(x) = tan x − 4x increases and decreases on (0, π/2)

The question

Find the intervals in which the function $f$ given by $f(x) = \tan x - 4x$ ⁠, $x \in \left(0, \frac{\pi}{2}\right)$ is

a) strictly increasing

b) strictly decreasing

Exam answer

\begin{aligned} f(x) &= \tan x - 4x \\ f'(x) &= \sec^2 x - 4 \end{aligned}

Why?

Standard derivatives: $\frac{d}{dx}\tan x = \sec^2 x$ and $\frac{d}{dx}(4x) = 4$ ⁠.

a) For $f(x)$ to be strictly increasing, $f'(x) > 0$ ⁠.

Why?

The sign of the derivative decides: if $f'(x) > 0$ on an interval, $f$ is strictly increasing there; if $f'(x) < 0$ ⁠, strictly decreasing. So both parts come down to solving an inequality.

\begin{aligned} &\Rightarrow \sec^2 x - 4 > 0 \Rightarrow \sec^2 x > 4 \\ &\Rightarrow \cos^2 x < \frac14 = \left(\frac12\right)^2 \\ &\Rightarrow -\frac12 < \cos x < \frac12 \\ &\Rightarrow \frac{\pi}{3} < x < \frac{\pi}{2} \quad [\because x \in (0, \tfrac{\pi}{2})] \end{aligned}

Why?

$\sec^2 x = \frac{1}{\cos^2 x}$ ⁠. Both sides of $\frac{1}{\cos^2 x} > 4$ are positive, so taking reciprocals reverses the inequality: $\cos^2 x < \frac14$ ⁠.

\begin{aligned} \sec^2 x - 4 &> 0 \\ \sec^2 x &> 4 \quad (+4) \\ \frac{1}{\cos^2 x} &> 4 \\ 1 &> 4\cos^2 x \quad (\times \cos^2 x) \\ \frac14 &> \cos^2 x \quad (\div 4) \end{aligned}

Then $\cos^2 x < \left(\frac12\right)^2$ means $\cos x$ lies between $-\frac12$ and $\frac12$ ⁠.

In $\left(0, \frac{\pi}{2}\right)$ ⁠, $\cos x$ is positive and decreasing, and $\cos\frac{\pi}{3} = \frac12$ ⁠. So $\cos x < \frac12$ exactly when $x > \frac{\pi}{3}$ ⁠.

$\therefore f$ is strictly increasing in $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ ⁠.

b) For $f(x)$ to be strictly decreasing, $f'(x) < 0$ ⁠.

\begin{aligned} &\Rightarrow \sec^2 x - 4 < 0 \Rightarrow \sec^2 x < 4 \\ &\Rightarrow \cos^2 x > \frac14 = \left(\frac12\right)^2 \\ &\Rightarrow \cos x > \frac12 \quad [\because x \in (0, \tfrac{\pi}{2})] \\ &\Rightarrow 0 < x < \frac{\pi}{3} \end{aligned}

Why?

The same steps with the inequality the other way. $\cos^2 x > \frac14$ would allow $\cos x < -\frac12$ as well, but $\cos x > 0$ in $\left(0, \frac{\pi}{2}\right)$ ⁠, so only $\cos x > \frac12$ is possible.

Since $\cos x$ is decreasing there, $\cos x > \cos\frac{\pi}{3}$ means $x < \frac{\pi}{3}$ ⁠. At $x = \frac{\pi}{3}$ itself $f'(x) = 0$ ⁠: that is where $f$ changes from decreasing to increasing.

$\therefore f$ is strictly decreasing in $\left(0, \dfrac{\pi}{3}\right)$ ⁠.

Where this comes from

Chapter 6, Application of Derivatives: Intervals of increase and decrease.

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: $f'(x) = \sec^2 x - 4$ is 0 where $\cos x = \frac12$ ⁠, at $x = \frac{\pi}{3}$ ⁠. That cuts $\left(0, \frac{\pi}{2}\right)$ in two: $f'$ is negative on $\left(0, \frac{\pi}{3}\right)$ and positive on $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ ⁠.

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Q32: Where f(x)=tan⁡x−4xf(x) = \tan x - 4xf(x)=tanx−4x increases and decreases on (0,π2)\left(0, \frac{\pi}{2}\right)(0,2π​)

Q32: Where $f(x) = \tan x - 4x$ increases and decreases on $\left(0, \frac{\pi}{2}\right)$