Question

​​​​​​$\begin{array}{cc}Column-I& Column-II\\ \left(I\right)\text{A circle}{x}^2+y^2-6x-10y+k=0& \\ \text{doesn't touch or intersect the coordinate axes}& \\ \text{and the point}(2,4)\text{lies inside the circle}& \\ \text{then the number of integral value of}k\text{is}& \left(P\right)5\\ \left(II\right)\text{If}\overrightarrow{V_1}=\hat{i}-2\hat{j}+3\hat{k},\overrightarrow{V_2}=a\hat{i}+b\hat{j}+c\hat{k}& \\ V_2\text{is non-zero vectors}\&a,b,c\in \{-1,0,1,2,3\},& \\ a,b,c\text{are choosen such that}\overrightarrow{V_1}\cdot \overrightarrow{V_2}=0& \\ \text{then the maximum of}(a+b+c)\text{is}& \left(Q\right)2\\ \left(III\right)\text{If a circle having radius}r\text{described on}& \\ \text{a normal chord of}{y}^2=4x\text{as diameter and}& \\ \text{passes through the vertex of the parabola,}& \\ \text{then}\left[r\right]\text{is}\left(\right[\left]\text{denotes G.I.F.}\right)& \left(R\right)6\\ \left(IV\right)\text{Consider}f\left(x\right)={\sin }^{-1}(\frac{x+3}{2x+5})\text{and}& \\ g\left(x\right)={\sin }^{-1}(\frac{a{x}^2+b}{x^2+5})\&\underset{x\to \infty }{lim}\left(f\right(x)-g(x\left)\right)=0,& \\ \underset{x\to 0}{lim}\left(f\right(x)+g(x\left)\right)=\frac{\pi }{4}\text{, then}10b^2\text{is}& \left(S\right)7\end{array}$

Which of the following is only "INCORRECT" combination?

• A. $\left(I\right)\left(R\right)$
• B. $\left(II\right)\left(S\right)$
• C. $\left(III\right)\left(Q\right)$
• D. $\left(IV\right)\left(P\right)$

Answer 1

The correct option is C $\left(III\right)\left(Q\right)$

$\left(I\right)$
$x^2+y^2-6x-10y+k=0\Rightarrow (x-3{)}^2+(y-5{)}^2=34-k$
For the circle to exist
$34-k>0\Rightarrow k<34$
Now we know that it will not intersect or touch the coordinate axis so,
$\sqrt{34-k}<3\Rightarrow 34-k<9\Rightarrow k>25$
We know that $(1,4)$ is inside the circle so,
$r>\sqrt{2}\Rightarrow 34-k>2\Rightarrow k<32$
Hence $k\in (25,32)$ so the number of integral values of $k$ will be $6$
$\left(I\right)\to \left(R\right)$

$\left(II\right)$
$\overrightarrow{V_1}\cdot \overrightarrow{V_2}=0\Rightarrow a-2b+3c=0$
Case 1:
$a=-1,b=1,c=1\Rightarrow a+b+c=1$
Case 2:
$a=2,b=1,c=0\Rightarrow a+b+c=3$
Case 3:
$a=1,b=-1,c=-1\Rightarrow a+b+c=-1$
Case 4:
$a=1,b=2,c=1\Rightarrow a+b+c=4$
Case 5:
$a=3,b=0,c=-1\Rightarrow a+b+c=2$
Case 6:
$a=3,b=3,c=1\Rightarrow a+b+c=7$
Hence the maximum of $a+b+c$ is $7$
$\left(II\right)\to \left(S\right)$

$\left(III\right)$
Let the coordinates of
$P=(t_1^2,2t_1)$ and $Q=(t_2^2,2t_2)$
We know that,
$t_2=-t_1-\frac{2}{t_1}\cdots \left(i\right)$
$PQ$ is the diameter so it will subtends ${90}^{\circ }$ at the vertex,
$\frac{2t_1-0}{t_1^2-0}\times \frac{2t_2-0}{t_2^2-0}=-1\Rightarrow t_1{t}_2=-4\Rightarrow -t_1^2-2=-4\Rightarrow t_1=\pm \sqrt{2}\Rightarrow t_2=\pm 2\sqrt{2}$
$P=(2,2\sqrt{2}),Q=(8,-4\sqrt{2}),PQ=6\sqrt{3}$
Hence the radius will be $3\sqrt{3}$
$\left[r\right]=5$
$\left(III\right)\to \left(P\right)$

$\left(IV\right)$
$\underset{x\to \infty }{lim}{\sin }^{-1}(\frac{x+3}{2x+5})=\frac{\pi }{6}\underset{x\to \infty }{lim}{\sin }^{-1}(\frac{a{x}^2+b}{x^2+5})=\frac{\pi }{6}\Rightarrow a=\frac{1}{2}$
$\underset{x\to 0}{lim}\left(f\right(x)+g(x\left)\right)=\frac{\pi }{4}\Rightarrow {\sin }^{-1}(\frac{3}{5})+{\sin }^{-1}(\frac{b}{5})=\frac{\pi }{4}\Rightarrow \frac{b}{5}=\sin ({\sin }^{-1}(\frac{3}{5})+\frac{\pi }{4})\Rightarrow \frac{b}{5}=\frac{1}{\sqrt{2}}\times \frac{4}{5}-\frac{1}{\sqrt{2}}\times \frac{3}{5}\Rightarrow b=\frac{1}{\sqrt{2}}\Rightarrow 10b^2=5$
$\left(IV\right)\to \left(P\right)$