Question

$\begin{array}{cccc}& \text{List- I}& & \text{List-II}\\ \text{(I)}& \frac{3}{2}\underset{0}{\overset{3}{\int }}\frac{1+x+\sqrt{x^2+x^3}}{x+\sqrt{1+x}}dx& \text{(P)}& 1\\ \text{(II)}& \text{Number of integer(s) in the range of the}& \text{(Q)}& 2\\ & \text{function}f\left(x\right)=\frac{28}{5+x^2+x^4}\text{is}& & \\ \text{(III)}& \text{If}\underset{x\to \frac{\pi }{2}}{lim}{\tan }^{2}x(\sqrt{4{\sin }^{2}x+\sin x+3}& \text{(R)}& 3\\ & -\sqrt{3{\sin }^{2}x+4\sin x+1})=\frac{1}{\sqrt{32k}},& & \\ & \text{then}k=& & \\ \text{(IV)}& \overrightarrow{a},\overrightarrow{b}\text{and}\overrightarrow{c}\text{are three vectors such}& \text{(S)}& 4\\ & \text{that there magnitudes are in the ratio}& & \\ & 1:2:3.\text{The angle between each of}& & \\ & \text{them is}\frac{\pi }{3}\text{and}|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=10,& & \\ & \text{then the magnitude of}\overrightarrow{a}\text{is}& & \\ & & \text{(T)}& 5\\ & & \text{(U)}& 7\end{array}$


Which of the following is the only CORRECT combination?

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

Answer 1

The correct option is B $\left(II\right)\to \left(T\right)$, $\left(III\right)\to \left(S\right)$

$\frac{3}{2}\underset{0}{\overset{3}{\int }}\frac{1+x+\sqrt{x^2+x^3}}{x+\sqrt{1+x}}dx$
$=\frac{3}{2}\underset{0}{\overset{3}{\int }}\frac{\sqrt{1+x}(\sqrt{1+x}+x)}{x+\sqrt{1+x}}dx$
$=\frac{3}{2}{\left[\frac{2}{3}\right(1+x{)}^{\frac{3}{2}}]}_0^3$
$=7$


The denominator of $f\left(x\right)$ will always be positive and will lie in $[5,\infty )$
At $x=0,$ we get $f\left(x\right)=\frac{28}{5}$
As $x\to \infty ,$ $f\left(x\right)\to 0$.
So, the range of $f\left(x\right)$ is $(0,\frac{28}{5}]$
Number of integers will be $5$ i.e., $\{1,2,3,4,5\}$

$\underset{x\to \frac{\pi }{2}}{lim}{\tan }^{2}x(\sqrt{4{\sin }^{2}x+\sin x+3}-\sqrt{3{\sin }^{2}x+4\sin x+1})$
$=\underset{x\to \frac{\pi }{2}}{lim}\frac{{\sin }^{2}x}{{\cos }^{2}x}\left(\frac{4{\sin }^{2}x+\sin x+3-3{\sin }^{2}x-4\sin x-1}{\sqrt{4{\sin }^{2}x+\sin x+3}+\sqrt{3{\sin }^{2}x+4\sin x+1}}\right)$
$=\underset{x\to \frac{\pi }{2}}{lim}\frac{{\sin }^{2}x}{(1-\sin x)(1+\sin x)}\left(\frac{(\sin x-1)(\sin x-2)}{\sqrt{4{\sin }^{2}x+\sin x+3}+\sqrt{3{\sin }^{2}x+4\sin x+1}}\right)$
$=\underset{x\to \frac{\pi }{2}}{lim}\frac{{\sin }^{2}x}{(1+\sin x)}\left(\frac{(2-\sin x)}{\sqrt{4{\sin }^{2}x+\sin x+3}+\sqrt{3{\sin }^{2}x+4\sin x+1}}\right)$
$=\frac{1\times (2-1)}{(1+1)(\sqrt{8}+\sqrt{8})}$
$=\frac{1}{8\sqrt{2}}=\frac{1}{\sqrt{128}}=\frac{1}{\sqrt{32k}}$
$\therefore k=4$


$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=10$
$\Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2=100$
$\Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})$

$\Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2=|\overrightarrow{a}{|}^2+4|\overrightarrow{a}{|}^2+9|\overrightarrow{a}{|}^2+2(2|\overrightarrow{a}{|}^2\cos \left(\frac{\pi }{3}\right)+6|\overrightarrow{a}{|}^2\cos \left(\frac{\pi }{3}\right)+3|\overrightarrow{a}{|}^2\cos \left(\frac{\pi }{3}\right))$

$\Rightarrow 25|\overrightarrow{a}{|}^2=100$
$\Rightarrow |\overrightarrow{a}|=2$