MCQ
$\int_{\,0}^{\,\pi } {\sqrt {\frac{{1 + \cos 2x}}{2}} \,dx} $ is equal to
- A$0$
- ✓$2$
- C$1$
- D$ - 1$
$I = \int_{\,0}^{\,\pi /2} {\cos x\,dx} - \int_{\,\pi /2}^{\,\pi } {\cos x\,dx} $
$= [\sin x]_0^{\pi /2} - [\sin x]_{\pi /2}^\pi $
$I = \left[ {\sin \frac{\pi }{2} - \sin 0} \right] - \left[ {\sin \pi - \sin \frac{\pi }{2}} \right] $
$=1+ 1 = 2.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$PROPERTY 1$ if $\lim _{ h \rightarrow 0} \frac{ f ( h )- f (0)}{\sqrt{| h |}}$ exists and is finite, and $PROPERTY 2$ if $\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h^2}$ exists and is finite.
Then which of the following options is/are correct ?
$(1)$ $f(x)=x|x|$ has $PROPERTY$ $2$ $(2)$ $f(x)=x^{2 / 3}$ has $PROPERTY$ $1$ $(3)$ $f(x)=\sin x$ has $PROPERTY$ $2$ $(4)$ $f(x)=|x|$ has $PROPERTY$ $1$
$T_p=\left\{A=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{a}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1, \ldots ., \mathrm{p}-1\}\right\}$
$1.$ The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and $\operatorname{det}(\mathrm{A})$ divisible by $\mathrm{p}$ is
$(A)$ $(\mathrm{p}-1)^2$ $(B)$ $2(\mathrm{p}-1)$
$(C)$ $(\mathrm{p}-1)^2+1$ $(D)$ $2 \mathrm{p}-1$
$2.$ The number of $A$ in $T_p$ such that the trace of $A$ is not divisible by $p$ but det $(A)$ is divisible by $p$ is [Note: The trace of a matrix is the sum of its diagonal entries.]
$(A)$ $(\mathrm{p}-1)\left(\mathrm{p}^2-\mathrm{p}+1\right)$ $(B)$ $\mathrm{p}^3-(\mathrm{p}-1)^2$
$(C)$ $(\mathrm{p}-1)^2$ $(D)$ $(p-1)\left(p^2-2\right)$
$3.$ The number of $A$ in $T_p$ such that det $(A)$ is not divisible by $p$ is
$(A)$ $2 \mathrm{p}^2$ $(B)$ $p^3-5 p$ $(C)$ $p^3-3 p$ $(D)$ $p^3-p^2$
Give the answer question $1,2$ and $3.$