Question

The most general values of $\theta$ for which $\sin \theta -\cos \theta =\underset{a\in R}{min}(1,a^2-6a+10)$ are given by

• A. $n\pi +{(-1)}^n\frac{\pi }{4}-\frac{\pi }{4}$
• B. $n\pi +{(-1)}^n\frac{\pi }{4}+\frac{\pi }{4}$
• C. $2n\pi +\frac{\pi }{4}$
• D. None of these

Answer 1

The correct option is B $n\pi +{(-1)}^n\frac{\pi }{4}+\frac{\pi }{4}$

We have $\sin \theta -\cos \theta ={min}_{a\in R}(1,a^2-6a+10)$

Since $a^2-6a+11={(a-3)}^2+2>2$ for all $a$

$⟹a^2-6a+10>1$

$\therefore \sin \theta -\cos \theta =1$ ............ min value

$\Rightarrow \sin (\theta -\frac{\pi }{4})=\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}$

$\Rightarrow \theta -\frac{\pi }{4}=n\pi +{(-1)}^n\frac{\pi }{4}$

$\therefore \theta =n\pi +{(-1)}^n\frac{\pi }{4}+\frac{\pi }{4}$, where $n\in Z$