Question

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(I)}& \text{The value of}\underset{n\to \infty }{lim}\frac{\sqrt[3]{3n^2}\cdot \sin (n!)}{n+1}\text{is}& \text{(P)}& -1\\ \text{(II)}& \text{If}x\in [0,2\pi ]\text{and}{\log }_2\tan x+{\log }_2\tan 2x=0,& & \\ & \text{then the number of solutions is}& \text{(Q)}& 0\\ \text{(III)}& \text{An unbaised dice is thrown and the number}& & \\ & {\text{appear is put in the place of}}^{\prime }{p}^{\prime }\text{in equation}& & \\ & x^2+px+2=0.\text{If the probability of the}& & \\ & \text{equation having real roots is}\frac{a}{b},(a,b\in N)& & \\ & \text{then the least possible value of}(a+b)\text{is}& \left(R\right)& 1\\ \text{(IV)}& \text{Three lines through origin having direction}& & \\ & \text{ratios}(1,a,a^2);(1,b,b^2);(1,c,c^2)\text{are non-}& & \\ & \text{coplanar. But the lines with direction ratios}& & \\ & (a,a^2,1+a^3);(b,b^2,1+b^3);(c,c^2,1+c^3)\text{are}& & \\ & \text{coplanar, then the value of}{}^{\prime }ab{c}^{\prime }\text{is}& \left(S\right)& 2\\ & & \left(T\right)& 4\\ & & \left(U\right)& 5\end{array}$

Which of the following option is CORRECT ?

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

Answer 1

The correct option is D $\left(IV\right)\to \left(P\right)$

$\left(I\right)$
$L=\underset{n\to \infty }{lim}\frac{\sqrt[3]{3n^2}\cdot \sin (n!)}{n+1}=\underset{n\to \infty }{lim}\frac{\sin (n!)}{(1+\frac{1}{n})n^{1/3}}=0$
$\left(I\right)\to \left(Q\right)$

$\left(II\right)$
${\log }_2\tan x+{\log }_2\tan 2x=0\Rightarrow \frac{2{\tan }^{2}x}{1-{\tan }^{2}x}=1\Rightarrow \tan x=\frac{1}{\sqrt{3}}[\because \tan x>0\text{and}\tan 2x>0]\Rightarrow x=\frac{\pi }{6},\frac{7\pi }{6}$
Hence, the number of solutions is $2.$
$\left(II\right)\to \left(S\right)$

$\left(III\right)$
$x^2+px+2=0$
$\Rightarrow p^2-8\ge 0\Rightarrow p=3,4,5,6\text{Required probability}=\frac{4}{6}=\frac{2}{3}\Rightarrow a+b=5$
$\left(III\right)\to \left(U\right)$

$\left(IV\right)$
$\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|\ne 0\cdots \left(1\right)\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|+\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& c^3\end{array}\right|=0\Rightarrow (1+abc)\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=0\Rightarrow abc=-1$
$\left(IV\right)\to \left(P\right)$

Hence,
$\left(I\right)\to \left(Q\right)$
$\left(II\right)\to \left(S\right)$
$\left(III\right)\to \left(U\right)$
$\left(IV\right)\to \left(P\right)$