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 $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}x=\sqrt{6+\sqrt{6+\sqrt{6+\cdots \text{up to}\infty }}},& & \\ & \text{then}x\text{is equal to}& \text{(P)}& 4\\ \text{(B)}& \text{If}a\text{and}x\text{are positive integers such}& & \\ & \text{that}x<a\text{and}\sqrt{a-x},\sqrt{x},\sqrt{a+x}& \text{(Q)}& 5\\ & \text{are in A.P., then least possible value of}a\text{is}& & \\ \text{(C)}& \text{If}3a+2b+4c=0,a,b,c\in R\text{and the}& & \\ & \text{line}ax+by+c=0\text{always passes}& & \\ & \text{through a fixed point}(p,q)\text{, then the}& & \\ & \text{value of}2p+q\text{is}& \text{(R)}& 2\\ \text{(D)}& \text{If}k(\sin {18}^{\circ }+\cos {36}^{\circ }{)}^2=5,\text{then the}& & \\ & \text{value of}k\text{is}& \text{(S)}& 3\\ & & \text{(T)}& 6\end{array}$

Which of the following is the only CORRECT combination?

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

Answer 1

The correct option is D $C\to R;D\to P$

$\left(A\right)x=\sqrt{6+\sqrt{6+\sqrt{6+\cdots \text{up to}\infty }}}$
$\Rightarrow x=\sqrt{6+x}\Rightarrow x^2-x-6=0\Rightarrow x=-2,3$
As $x>0$, so $x=3$
$A\to S$

$\left(B\right)\sqrt{a-x},\sqrt{x},\sqrt{a+x}$ are in A.P., then
$2\sqrt{x}=\sqrt{a-x}+\sqrt{a+x}$
Squaring both the sides, we get
$4x^2=2a+2\sqrt{(a^2-x^2)}\Rightarrow 4x=2a+2\sqrt{a^2-x^2}\Rightarrow 2x-a=\sqrt{a^2-x^2}$
Again squaring both the sides, we get
$4x^2+a^2-4xa=a^2-x^2\Rightarrow 5x^2=4xa\Rightarrow x(5x-4a)=0$
As $x$ and $a$ are positive integers, so $x=\frac{4a}{5}$
Hence, the minimum value of $a=5$
$B\to S$

$\left(C\right)3a+2b+4c=0$
$\Rightarrow b=-\frac{(3a+4c)}{2}$
Now, $ax+by+c=0$
$\Rightarrow ax-\frac{(3a+4c)y}{2}+c=0\Rightarrow a(x-\frac{3y}{2})+c(1-2y)=0\Rightarrow (x-\frac{3y}{2})+\frac{c}{a}(1-2y)=0$
The fixed point is $(p,q)=(\frac{3}{4},\frac{1}{2})$
Therefore, $2p+q=2$
$C\to R$

$\left(D\right)$
$k(\sin {18}^{\circ }+\cos {36}^{\circ }{)}^2=5\Rightarrow k{(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{5}+1}{4})}^2=5\therefore k=4$
$D\to P$