Question

Let f: $X\to Y$ be a function defined by $f\left(x\right)=a\sin (x+\frac{\pi }{4})+b\cos x+c$. If f is both one-one and onto, then find the sets $X$ and $Y$

• A. $X\in [-\frac{\pi }{2}-\alpha ,\frac{\pi }{2}-\alpha ]$ and $Y\in [c-r,c+r]$ where $\alpha ={\tan }^{-1}\left(\frac{a+b\sqrt{2}}{a}\right)$ and $r=\sqrt{a^2+\sqrt{2}ab+b^2}$
• B. $X\in [-\frac{\pi }{1}-\alpha ,\frac{\pi }{1}-\alpha ]$ and $Y\in [c-r,c+r]$ where $\alpha ={\tan }^{-1}\left(\frac{a+b\sqrt{2}}{a}\right)$ and $r=\sqrt{a^2+\sqrt{2}ab+b^2}$
• C. $X\in [-\frac{\pi }{4}-\alpha ,\frac{\pi }{4}-\alpha ]$ and $Y\in [c-r,c+r]$ where $\alpha ={\tan }^{-1}\left(\frac{a+b\sqrt{2}}{a}\right)$ and $r=\sqrt{a^2+\sqrt{2}ab+b^2}$
• D. $X\in [-\frac{\pi }{8}-\alpha ,\frac{\pi }{8}-\alpha ]$ and $Y\in [c-r,c+r]$ where $\alpha ={\tan }^{-1}\left(\frac{a+b\sqrt{2}}{a}\right)$ and $r=\sqrt{a^2+\sqrt{2}ab+b^2}$

Answer 1

The correct option is A $X\in [-\frac{\pi }{2}-\alpha ,\frac{\pi }{2}-\alpha ]$ and $Y\in [c-r,c+r]$ where $\alpha ={\tan }^{-1}\left(\frac{a+b\sqrt{2}}{a}\right)$ and $r=\sqrt{a^2+\sqrt{2}ab+b^2}$

$f\left(x\right)=a\sin (x+\frac{\pi }{4})+b\cos x+c=\frac{a}{\sqrt{2}}\sin x+\frac{a}{\sqrt{2}}\cos x+b\cos x+c=\frac{a}{\sqrt{2}}\sin x+\frac{a+b\sqrt{2}}{\sqrt{2}}\cos x+c=r\cos \alpha \sin x+r\sin \alpha \cos x+c(\because r=\sqrt{a^2+b^2+\sqrt{2}ab}and\tan \alpha =\frac{a+b\sqrt{2}}{a})=r\sin (x+\alpha )+c$

Since $f\left(x\right)$ is a one-one onto function, $\sin (x+\alpha )$ values must not be repetitive.

$\therefore -\frac{\pi }{2}\le x+\alpha \le \frac{\pi }{2}$ and $c-r\le f\left(x\right)\le c+r$