MCQ
Let $f:R \to R$ be a function defined by $f(x) = \frac{{x - m}}{{x - n}}$, where $m \ne n$. Then
- 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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The value of objective function is maximum under linear constraints
($S1$) $f(x)=0$ for only one value of $x$ is $[0, \pi]$.
($S2$) $\mathrm{f}(\mathrm{x})$ is decreasing in $\left[0, \frac{\pi}{2}\right]$ and increasing in $\left[\frac{\pi}{2}, \pi\right] .$