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1.Relation and Function
medium
આપેલ પૈકી . . . . યુગ્મ વિધેય છે.
A
$f(x) = \frac{{{a^x} + 1}}{{{a^x} - 1}}$
B
$f(x) = x\left( {\frac{{{a^x} - 1}}{{{a^x} + 1}}} \right)$
C
$f(x) = \frac{{{a^x} - {a^{ - x}}}}{{{a^x} + {a^{ - x}}}}$
D
$f(x) = \sin x$
Solution
(b) In $(a)$, $f( – x) = \frac{{{a^{ – x}} + 1}}{{{a^{ – x}} – 1}} = \frac{{1 + {a^x}}}{{1 – {a^x}}} = – \frac{{{a^x} + 1}}{{{a^x} – 1}} = – f(x)$
So, it is an odd function.
In $(b)$, $f( – x) = ( – x)\frac{{{a^{ – x}} – 1}}{{{a^{ – x}} + 1}} = – x\frac{{1 – {a^x}}}{{1 + {a^x}}} = x\frac{{{a^x} – 1}}{{{a^x} + 1}} = f(x)$
So, it is an even function.
In $(c)$, $f( – x) = – \sin \left[ {\log (x + \sqrt {1 + {x^2}} )} \right]$
So, it is an odd function.
In $(d)$, $f( – x) = \sin ( – x) = – \sin x = – f(x)$
So, it is an odd function.
Standard 12
Mathematics
