MCQ
Primitive of $(x^2 + 4)^{-1/2}$ $w.r.t$ $x^2 + 2$ is equal to :-
- A$\frac{2}{{\sqrt {{x^2} + 4} }} + C$
- B$\sqrt {{x^2} + 4} +\frac{1}{{\sqrt {{x^2} + 4} }} + C$
- ✓$2\sqrt {{x^2} + 4} + C$
- DNone
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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 $3.$
$f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2} & , x<0 \\ \alpha & , x=0, \text { where } \alpha, \beta \in R \text {. If } \\ \frac{\beta \sqrt{1-\cos x}}{x} & , x>0\end{cases}$
$f$ is continuous at $\mathrm{x}=0$, then $\alpha^2+\beta^2$ is equal to :