- A$f$ is one-one onto
- ✓$f$ is one-one into
- C$f$ is many one onto
- D$f$ is many one into
$f(x) = f(y) \Rightarrow \frac{{x - m}}{{x - n}} = \frac{{y - m}}{{y - n}} \Rightarrow x = y$
$\therefore$ $f$ is one-one.
Let $\alpha$ $\in$ $R$ such that $f(x) = \alpha \Rightarrow \frac{{x - m}}{{x - n}} = \alpha $
==> $x = \frac{{m - n\alpha }}{{1 - \alpha }}$
Clearly $x \notin R$ for $\alpha = 1$. So, $f$ is not onto.
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$g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1}$
where
$f(\theta)=\frac{1}{2}\left|\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}\sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _e\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _e\left(\frac{\pi}{4}\right) & \tan \pi\end{array}\right|$.
Let $p (x)$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g(\theta)$, and $p(2)=2-\sqrt{2}$. Then, which of the following is/are TRUE ?
$(A)$ $p \left(\frac{3+\sqrt{2}}{4}\right)<0$
$(B)$ $p \left(\frac{1+3 \sqrt{2}}{4}\right)>0$
$(C)$ $p \left(\frac{5 \sqrt{2}-1}{4}\right)>0$
$(D)$ $p \left(\frac{5-\sqrt{2}}{4}\right)<0$