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Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsFunctions5 Marks
Question
If $\text{f(x)}=\sqrt{1-\text{x}}$ and $\text{g(x)}=\log_\text{e}\text{x}$ are two real functions, then describe, functions fog and gof.
✓

Answer

$\text{f(x)}={1-\text{x}}$For domain,
$1-\text{x}\geq0$
$\text{x}\leq1$
⇒ domain of g
$\text{f}:(-\infty,1]\rightarrow0,\infty=\log_\text{e}\text{x}$
Clearly, $\text{g}:0,\infty\rightarrow\text{R}$
Computation of fog: Clearly, the range of g is not a subset of the domain of f.
Therefore,
We need to compute the domain of fog.
⇒ Domain fog = x : x $\in$ Domain g and $\text{g}(\text{x})\in$ Domain of f
⇒ Domain fog = x : x : x $\in0,\ \infty$ and $\log_\text{e}\text{x}\in(-\infty,1]$
⇒ Domain fog = x : x : x $\in0,\ \infty$ and $\text{x}\in(0,\text{e}]$
⇒ Domain fog = x : x $\in(0,\text{e}]$
⇒ Domain fog = (0, e]
⇒ fog : 0, e → R
Therefore,
(fog)(x) = f(g(x))
$=\text{f}(\log_\text{e}\text{x})$
$=\sqrt{1-\log_\text{e}\text{x}}$
Computation of gof: Clearly, the range of f is a subset of the domain of g.
$\Rightarrow\ \text{gof}:(-\infty,1]\rightarrow\text{R}$
$(\text{gof})(\text{x})=\text{g(f(x))}$
$=\text{g}(\sqrt{1-\text{x}})$
$\Rightarrow\ \log_\text{e}\sqrt{1-\text{x}}$
$=\log_\text{e}(1-\text{x})^\frac{1}{2}$
$=\frac{1}{2}\log_\text{e}(1-\text{x})$

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