A line L : $y=\mathit{\text{m}}x+3$ meets $y-\mathrm{a}\mathrm{x}\mathrm{\text{is}}$ at E(0,3) and the arc of the parabola $y^2=16x\text{,}0\le y\le 6$ at the point $F\left(x_0,y_0\right)$.The tangent to the parabola at $F\left(x_0,y_0\right)$ intersects the y-axis at $G\left(0,y_1\right)$.The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum.
Match List I with List II and select the correct answer using the code given below the lists :
 

	List I		List II
A.	m =	P.	$\frac{1}{2}$
B.	Maximum area of ΔEFG is	Q.	4
C.	y0​ =	R.	2
D.	y1 =	S.	1

Tangents drawn from the point \(P(1,8)\) to the circle \(x^2+y^2-6 x-4 y-11=0\) touch the circle at the points \(A\) and \(B\). The equation of the circumcircle of \(\triangle P A B\) is

Let \(f:[1, \infty) \rightarrow[2, \infty)\) be a differentiable function such that \(f(1)=2\). If \(6 \int_1^x f(t) d t=3 x f(x)-x^3\) for all \(x \geq 1\), then the value of \(f(2)\) is

Let $f:\mathrm{ℝ}\to \mathrm{ℝ}$ be a differentiable function with $f\left(0\right)=0$. If $y=f\left(x\right)$ satisfies the differential equation $\frac{dy}{dx}=\left(2+5y\right)\left(5y-2\right)$, then the value of  $\underset{x\to -\infty }{\lim }f\left(x\right)$ is 

In a $∆PQR$, $P$ is the largest angle and $\cos P=\frac{1}{3}$. Further the incircle of the triangle touches the sides $PQ, QR$ and $RP$ at $N, L$ and $M$ respectively, such that the lengths of $PN, QL$ and $RM$ are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)

Let \(f\) be a real-valued function defined on the interval \((-1,1)\) such that \(e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t, \quad\) for all \(x \in(-1,1)\) and let \(f^{-1}\) be the inverse function of \(f\). Then \(\left(f^{-1}\right)^{\prime}(2)\) is equal to

The integral \(\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{\frac{9}{2}}} d x\) equals (for some arbitrary constant \(\mathrm{K}\) )

Paragraph:
Read the following passage and answer the questions.
Question:
The value of \(\left[\begin{array}{lll}3 & 2 & 0\end{array}\right]\left[\begin{array}{l}3 \\ 2 \\ 0\end{array}\right]\) is

Let ${\mathrm{E}}_1\left(\mathrm{r}\right),\mathrm{ }{\mathrm{E}}_2\left(\mathrm{r}\right)$ and ${\mathrm{E}}_3\left(\mathrm{r}\right)$ be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density $\mathrm{\lambda }$ , and an infinite plane with uniform surface charge density $\mathrm{\sigma }$ . If ${\mathrm{E}}_1\left({\mathrm{r}}_0\right)={\mathrm{E}}_2\left({\mathrm{r}}_0\right)={\mathrm{E}}_3\left({\mathrm{r}}_0\right)$ at a given distance ${\mathrm{r}}_0$ , then

A glass beaker has a solid, plano-convex base of refractive index 1.60, as shown in the figure. The radius of curvature of the convex surface (SPU) is \(9 \mathrm{~cm}\), while the planar surface (STU) acts as a mirror. This beaker is filled with a liquid of refractive index \(n\) up to the level QPR. If the image of a point object \(\mathrm{O}\) at a height of \(h\) (OT in the figure) is formed onto itself, then, which of the following option(s) is(are) correct?

The center of a disk of radius \(r\) and mass \(m\) is attached to a spring of spring constant \(k\), inside a ring of radius \(R>r\) as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke's law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as \(T=\frac{2 \pi}{\omega}\). The correct expression for \(\omega\) is ( \(g\) is the acceleration due to gravity):

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1	Column 2	Column 3
(I) $x^2+y^2=a^2$	(i) $my=m^{2}x+a$	(P) $\left(\frac{a}{m^2},\frac{2a}{m}\right)$
(II) $x^2+a^2{y}^2=a^2$	(ii) $y=mx+a \sqrt{m^2+1}$	(Q) $\left(\frac{-ma}{\sqrt{m^2+1}},\frac{a}{\sqrt{m^2+1}}\right)$
(III) $y^2=4ax$	(iii) $y=mx+ \sqrt{a^2{m}^2-1}$	(R) $\left(\frac{-a^{2}m}{\sqrt{a^2{m}^2+1}},\frac{1}{\sqrt{a^2{m}^2+1}}\right)$
(IV) $x^2-a^2{y}^2=a^2$	(iv) $y=mx+\sqrt{a^2{m}^2+1}$	(S) $\left(\frac{-a^{2}m}{\sqrt{a^2{m}^2-1}},\frac{-1}{\sqrt{a^2{m}^2-1}}\right)$

The tangent to a suitable conic (Column 1) at $\left(\sqrt{3},\frac{1}{2}\right)$ is found to be $\sqrt{3}x+2y=4$ , then which of the following options is the only Correct combination?

Let the function \(g:(-\infty, \infty) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) be given by \(g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}\). Then, \(g\) is