If $\mathrm{a}$ and $\mathrm{b}$ are positive number such that $\mathrm{a}>\mathrm{b}$, then the minimum…

If $\mathrm{a}$ and $\mathrm{b}$ are positive number such that $\mathrm{a}>\mathrm{b}$, then the minimum value of $a \sec \theta-b \tan \theta\left(0 < \theta < \frac{\pi}{2}\right)$ is
  1. $\frac{1}{\sqrt{a^2-b^2}}$
  2. $\frac{1}{\sqrt{a^2+b^2}}$
  3. $\sqrt{a^2+b^2}$
  4. $\sqrt{a^2-b^2}$

Solution

$\begin{aligned} & \text { let } \mathrm{f}(\theta)=\mathrm{a} \sec \theta-b \tan \theta \\ & \therefore \quad \mathrm{f}^{\prime}(\theta)=\mathrm{a} \sec \theta \tan \theta-b \sec ^2 \theta \\ & \quad=\sec \theta(\mathrm{a} \tan \theta-\mathrm{b} \sec \theta) \\ & \therefore \quad \mathrm{f}^{\prime}(\theta)=0 \Rightarrow \sec \theta(\mathrm{a} \tan \theta-\mathrm{b} \sec \theta)=0 \\ & \Rightarrow \mathrm{a} \tan \theta-\mathrm{b} \sec \theta=0 \quad \ldots\left[\text { As } 0 < \pi < \frac{\pi}{2}, \sec \theta \neq 0\right] \\ & \\ & \Rightarrow \mathrm{a} \sin \theta-b=0 \quad \ldots\left[\text { As } 0 < \pi < \frac{\pi}{2}, \cos \theta \neq 0\right]\end{aligned}$ $\begin{aligned} & \Rightarrow \sin \theta=\frac{b}{a} \\ & \Rightarrow \sec \theta=\frac{a}{\sqrt{a^2-b^2}} \text { and } \tan \theta=\frac{b}{\sqrt{a^2-b^2}} \end{aligned}$ Now, $\begin{aligned} & \mathrm{f}^{\prime \prime}(\theta)=\sec \theta \tan \theta(\mathrm{a} \tan \theta-\mathrm{b} \sec \theta) \\ & +\sec \theta\left(a \sec ^2 \theta-b \sec \theta \tan \theta\right) \\ & =\mathrm{a} \tan ^2 \theta \sec \theta-\mathrm{b} \sec ^2 \theta \tan \theta \\ & +a \sec ^3 \theta-b \sec ^2 \theta \tan \theta \\ & =a \sec \theta\left(\tan ^2 \theta+\sec ^2 \theta\right) \\ & =a \sec \theta\left(1+2 \tan ^2 \theta\right) \\ & >0 \quad \ldots\left[\because \text { a is positive and } 0 < \theta < \frac{\pi}{2}\right] \\ & \end{aligned}$ $\therefore \\f(\theta)$ is minimum when $\sin \theta=\frac{b}{a}$. $\begin{aligned}\therefore \ & \text { Minimum value of } f(\theta) \\ & =a\left(\frac{a}{\sqrt{a^2-b^2}}\right)-b\left(\frac{b}{\sqrt{a^2-b^2}}\right) \quad \ldots[\text { From (i) }] \\ & =\frac{a^2-b^2}{\sqrt{a^2-b^2}} \\ & =\sqrt{a^2-b^2} \end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 2)

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