MCQ
Domain of the function $f(x) = {\sin ^{ - 1}}(1 + 3x + 2{x^2})$ is
- A$( - \infty ,\;\infty )$
- B$( - 1,\;1)$
- ✓$\left[ { - \frac{3}{2},\;0} \right]$
- D$\left( { - \infty ,\;\frac{{ - 1}}{2}} \right) \cup (2,\;\infty )$
$Case I :$ $2{x^2} + 3x + 1 \ge - 1$; $2{x^2} + 3x + 2 \ge 0$
$x = \frac{{ - 3 \pm \sqrt {9 - 16} }}{6}$ $ = \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 = $\left[ {\frac{{ - 3}}{2},\,0} \right]$.
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$1.$ The real number $s$ lies in the interval
$(A)$ $\left(-\frac{1}{4}, 0\right)$ $(B)$ $\left(-11,-\frac{3}{4}\right)$
$(C)$ $\left(-\frac{3}{4},-\frac{1}{2}\right)$ $(D)$ $\left(0, \frac{1}{4}\right)$
$2.$ The area bounded by the curve $y=f(x)$ and the lines $x=0, y=0$ and $x=t$, lies in the interval
$(A)$ $\left(\frac{3}{4}, 3\right)$ $(B)$ $\left(\frac{21}{64}, \frac{11}{16}\right)$
$(C)$ $(9,10)$ $(D)$ $\left(0, \frac{21}{64}\right)$
$3.$ The function $f^{\prime}(x)$ is
$(A)$ increasing in $\left(-t,-\frac{1}{4}\right)$ and decreasing in $\left(-\frac{1}{4}, t\right)$
$(B)$ decreasing in $\left(-t,-\frac{1}{4}\right)$ and increasing in $\left(-\frac{1}{4}, t\right)$
$(C)$ increasing in (-t, t) $(D)$ decreasing in ( $-\mathrm{t}, \mathrm{t})$
Give the answer question $1,2$ and $3.$