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Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsFunctions1 Mark
MCQ
The function $\text{f}:\Big[\frac{-1}{2},\frac{1}{2},\frac{1}{2}\Big]\rightarrow\ \Big[\frac{-\pi}{2},\frac{\pi}{2}\Big],$ defined by $\text{f(x)}=\sin^{-1}(3\text{x}-4\text{x}^3),$ is:
  • ✓
    Bijection.
  • B
    Injection but not a surjection.
  • C
    Surjection but not an injection.
  • D
    Neither an injection nor a surjection.
✓

Answer

Correct option: A.
Bijection.
$\text{f(x)}=\sin^{-1}(3\text{x}-4\text{x}^3)$
$\Rightarrow\ \text{f(x)}=3\sin^{-1}\text{x}$
Injectivity: Let $x$ and $y$ be two elements in the domain $\Big[\frac{-1}{2},\frac{1}{2}\Big],$ such that
$f(x) = f(y)$
$\Rightarrow\ 3\sin^{-1}\text{x}=3\sin^{-1}\text{y}$
$\Rightarrow\ \sin^{-1}\text{x}=\sin^{-1}\text{y}$
$\Rightarrow\ \text{x}=\text{y}$
So, $f$ is one$-$one.
Surjectivity: Let $y$ be any element in the co$-$domain, such that
$f(x) = y$
$\Rightarrow\ 3\sin^{-1}\text{x}=\text{y}$
$\Rightarrow\ \sin^{-1}\text{x}=\frac{\text{y}}{3}$
$\Rightarrow\ \text{x}=\sin\frac{\text{y}}{3}\in\Big[\frac{-1}{2},\frac{1}{2}\Big]$
$\Rightarrow f$ is onto.
$\Rightarrow f$ is a bijection.

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