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Relations and Functions — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSRelations and Functions1 Mark
Question
The function $f: R \rightarrow R$ defined by $f(x)=4+3 \cos x$ is
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Answer

We have, $f(x)=4+3 \cos x, \forall x \in R$
At $x=\frac{\pi}{2}, f\left(\frac{\pi}{2}\right)=4+3 \cos \frac{\pi}{2}=4 \Rightarrow f\left(-\frac{\pi}{2}\right)=4+3 \cos \left(-\frac{\pi}{2}\right)=4$
Since, $f\left(\frac{\pi}{2}\right)=f\left(-\frac{\pi}{2}\right)$, But $\frac{\pi}{2} \neq-\frac{\pi}{2}$
Therefore, $f$ is not one-one.
As $-1 \leq \cos x \leq 1, \forall x \in R \Rightarrow 1 \leq 4+3 \cos x \leq 7, \forall x \in R$
$\Rightarrow f(x) \in[1,7]$, where $[1,7]$ is subset of $R . \therefore f$ is not onto.

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