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1.Relation and Function
hard
વિધેય $f(x) = {\sin ^{ - 1}}(1 + 3x + 2{x^2})$ નો પ્રદેશ મેળવો.
A
$( - \infty ,\;\infty )$
B
$( - 1,\;1)$
C
$\left[ { - \frac{3}{2},\;0} \right]$
D
$\left( { - \infty ,\;\frac{{ - 1}}{2}} \right) \cup (2,\;\infty )$
Solution
(c) $ – 1 \le 1 + 3x + 2{x^2} \le 1$
$Case I :$ $2{x^2} + 3x + 1 \ge – 1$; $2{x^2} + 3x + 2 \ge 0$
$x = \frac{{ – 3 \pm \sqrt {9 – 16} }}{6}$ $ = \frac{{ – 3 \pm i\sqrt 7 }}{6}$ (imaginary).
$Case II :$ $2{x^2} + 3x + 1 \le 1$
==> $2{x^2} + 3x \le 0$ ==> $2x\,\left( {x + \frac{3}{2}} \right) \le 0$
==> $\frac{{ – 3}}{2} \le x \le 0$ ==> $x \in \left[ { – \frac{3}{2},\,\,0} \right]$
In $case I$, we get imaginary value hence, rejected
$\therefore$ Domain of function = $\left[ {\frac{{ – 3}}{2},\,0} \right]$.
Standard 12
Mathematics
