The area bounded by lines y=x and x = 2 in first quadrant is

$T_p=\left\{A=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{a}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1, \ldots ., \mathrm{p}-1\}\right\}$

$1.$ 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 $3.$

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$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2}}}\left[ {\cos \frac{1}{{{n^2}}} + 2\cos \frac{4}{{{n^2}}} + 3\cos \frac{9}{{{n^2}}} + .... + 2n\,\cos 4} \right]$ is equal to-

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If the tangent to the curve $x = at^2, y = 2$ at is perpendicular to $x-$ axis, then its point of contact is :

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If $\int \frac{1}{ x } \sqrt{\frac{1- x }{1+ x }} dx = g ( x )+ c , g (1)=0$, then $g \left(\frac{1}{2}\right)$ is equal to

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