Question

$\text{List I}$ has four entries and $\text{List II}$ has five entries. Each entry of $\text{List I}$ is to be matched with one entry of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{The minimum value of}ab\text{if roots of}& & \\ & x^3-a{x}^2+bx-2=0\text{are positive, is}& \text{(P)}& 36\\ \text{(B)}& \text{The number of quadrilateral formed}& & \\ & \text{in an octagon having two sides common}& \text{(Q)}& 24\\ & \text{with the octagon, is}& & \\ \text{(C)}& \text{If}{}^{2n}{C}_4,{}^{2n}{C}_5\text{and}{}^{2n}{C}_6\text{are in}A.P.,& & \\ & \text{then the value of}2n\text{is}& \text{(R)}& 18\\ \text{(D)}& \text{The value of}72\sin \frac{\pi }{18}\sin \frac{5\pi }{18}\sin \frac{7\pi }{18}\text{is}& \text{(S)}& 14\\ & & \text{(T)}& 9\end{array}$

Which of the following is the only CORRECT combination?

• A. $A\to S;B\to T$
• B. $A\to R;B\to P$
• C. $A\to T;B\to Q$
• D. $A\to P;B\to S$

Answer 1

The correct option is B $A\to R;B\to P$

$\left(A\right)$ Let $x_1,x_2,x_3$ be the roots of the equation $x^3-a{x}^2+bx-2=0,$ then
$A.M.\ge H.M.\Rightarrow \frac{x_1+x_2+x_3}{3}\ge \frac{3x_1{x}_2{x}_3}{x_1{x}_2+x_2{x}_3+x_3{x}_1}\Rightarrow \frac{a}{3}\ge \frac{3\times 2}{b}\therefore ab\ge 18$
Hence, the minimum value of $ab$ is $18$.
$A\to R$

$\left(B\right)$ Number of quadrilateral having two adjacent sides common
$={}^8{C}_1\times {}^3{C}_1=8\times 3=24$

Number of quadrilateral not having two adjacent sides common
$=\frac{1}{2}({}^8{C}_1\times {}^3{C}_1)=12$

Hence, total number of quadrilateral $=24+12=36$

$\left(C\right){}^{2n}{C}_4,{}^{2n}{C}_5$ and ${}^{2n}{C}_6$ are in $A.P.$, so
$2\times {}^{2n}{C}_5={}^{2n}{C}_4+{}^{2n}{C}_6\Rightarrow 2=\frac{{}^{2n}{C}_4}{{}^{2n}{C}_5}+\frac{{}^{2n}{C}_6}{{}^{2n}{C}_5}\Rightarrow 2=\frac{5}{2n-4}+\frac{2n-5}{6}\Rightarrow 2n^2-21n+49=0\Rightarrow (n-7)(2n-7)=0\therefore n=7(\because n\in N)$
Hence, $2n=14$
$C\to S$

$\left(D\right)72\sin \frac{\pi }{18}\sin \frac{5\pi }{18}\sin \frac{7\pi }{18}=72\sin 10{}^{\circ }\sin 50{}^{\circ }\sin 70{}^{\circ }=72\sin (60{}^{\circ }-10{}^{\circ })\sin \left(10{}^{\circ }\right)\sin (60{}^{\circ }+10{}^{\circ })=72\times \frac{1}{4}\sin (3\times 10{}^{\circ })=9$
$D\to T$