Question

​​​​​​$\begin{array}{cc}List-I& List-II\\ \left(I\right)\text{If}\alpha ,\beta ,\gamma ,\delta \text{are real numbers such that}& \\ {\alpha }^2+{\beta }^2=9,{\gamma }^2+{\delta }^2=4\&\alpha \delta -\beta \gamma =6& \\ \text{then maximum value of}\alpha \gamma \text{is}& \left(P\right)5\\ \left(II\right)\text{If range of the function}& \\ f\left(x\right)={\sin }^{-1}x+2{\tan }^{-1}x+x^5-5x^4+5x^3+1& \\ \text{is}[a,b]\text{then}\left[b\right],\text{where}[.]\text{is the greatest integer}& \\ \text{function, is}& \left(Q\right)3\\ \left(III\right)\text{If for each}r,P\left(r\right)\text{be the number of points}(x,y),& \\ \text{where}x,y\in I\text{which lie inside or on the curve}& \\ |xy|+r^2=r|x|+r|y|,r\in N\text{and}& \\ \underset{n\to \infty }{lim}\frac{\sum _{r=1}^n\sqrt{P\left(r\right)}}{a{n}^3+b{n}^2+c}=\frac{1}{2}\text{then}|a+b|\text{is}& \left(R\right)2\\ \left(IV\right)\text{Let a continuous function}f\left(x\right)& \\ \text{satisfying the equation}& \\ \underset{0}{\overset{2}{\int }}f\left(x\right)(6x-f(x\left)\right)dx=24\text{then}f\left(2\right)\text{is}& \left(S\right)6\end{array}$

Which of the following is only "CORRECT" combination?

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

Answer 1

The correct option is C $\left(IV\right)\left(S\right)$

Let
$\left(I\right)$
$\alpha =3\cos \theta ,\beta =3\sin \theta \gamma =2\cos \varphi ,\delta =2\sin \varphi$
$\alpha \delta -\beta \gamma =6\Rightarrow \sin (\varphi -\theta )=1\Rightarrow \varphi =\frac{\pi }{2}+\theta$
So,
$\alpha \gamma =3\cos \theta \cdot 2\cos \varphi \Rightarrow \alpha \gamma =-3\sin 2\theta$
Hence the maximum value of $\alpha \gamma$ is $3$
$\left(I\right)\to \left(Q\right)$

$\left(II\right)$
Domain of the function will be $[-1,1]$
$f^{\prime }\left(x\right)=\frac{1}{\sqrt{1-x^2}}+\frac{2}{1+x^2}+5x^4-20x^3+15x^2\Rightarrow f^{\prime }\left(x\right)=\frac{1}{\sqrt{1-x^2}}+\frac{2}{1+x^2}+5x^2(x-1)(x-3)>0$
in $(-1,1)$

$f\left(1\right)=\pi +2$ and $f(-1)=-\pi -10$
Hence $\left[b\right]=[\pi +2]=5$
$\left(II\right)\to \left(P\right)$

$\left(III\right)$
When $r=1$
$|xy|+1=|x|+|y|\Rightarrow (|x|-1)(|y|-1)=0$
The above equation represents a square having length $1$ and center at Origin.

The number of integer points on or inside the curve will be $9$,
$(0,0),(0,1),(0,-1),(1,0),(-1,0),(1,1),(1,-1),(-1,-1),(-1,1)$

$\to$
When $r=2$ then
$|xy|+4=2|x|+2|y|$

So now the number of integral points will be $25$

For any $r$ the number of integral points will be $(2r+1{)}^2$
$P\left(r\right)=(2r+1{)}^2$
Now,
$\sum _{r=1}^n\sqrt{P\left(r\right)}=\sum _{r=1}^n(2r+1)=n(n+1)+n=n^2+2n$
The limit,
$\underset{n\to \infty }{lim}\frac{\sum _{r=1}^n\sqrt{P\left(r\right)}}{a{n}^3+b{n}^2+c}=\frac{1}{2}\Rightarrow \underset{n\to \infty }{lim}\frac{n^2+2n}{a{n}^3+b{n}^2+c}=\frac{1}{2}$
For the limit to exist $a=0$
$\underset{n\to \infty }{lim}\frac{n^2+2n}{b{n}^2+c}=\frac{1}{2}\Rightarrow \underset{n\to \infty }{lim}\frac{1+\frac{2}{n}}{b+\frac{c}{n}}=\frac{1}{2}\Rightarrow b=2$
Hence $|a+b|=2$
$\left(III\right)\to \left(R\right)$

$\left(IV\right)$
$\underset{0}{\overset{2}{\int }}f\left(x\right)(6x-f(x\left)\right)dx=24\Rightarrow \underset{0}{\overset{2}{\int }}-\left(f^2\right(x)-6xf(x)+(3x{)}^2)+(3x{)}^{2}dx=24\Rightarrow \underset{0}{\overset{2}{\int }}-\left(f\right(x)-3x{)}^{2}dx+[\frac{9x^3}{3}{]}_0^2=24\Rightarrow \underset{0}{\overset{2}{\int }}-\left(f\right(x)-3x{)}^{2}dx=0\Rightarrow f\left(x\right)=3x\Rightarrow f\left(2\right)=6$
$\left(IV\right)\to \left(S\right)$