Question

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If the quadratic equation}{x}^2-2\alpha x+2\alpha =0\text{has two real}& \text{(P)}& 0\\ & \text{and distinct roots}{x}_1\text{and}{x}_2\text{such that}|x_1–x_2|\le 2\sqrt{3},\text{then}& & \\ & \alpha \text{can be}& & \\ \text{(B)}& \text{If}y=\frac{x^2+4x}{x^2+4x+6},\text{then the integers in range of}y\text{is/are}& \text{(Q)}& 1\\ \text{(C)}& \text{If}(2-n)x^2-n<8x+4\forall x\in R,\text{then}\frac{n}{4}\text{can be}& \text{(R)}& 2\\ \text{(D)}& \text{If the equation}\cos 2x+2a\cos x+6a=17\text{has a solution,}& \text{(S)}& 3\\ & \text{then}a\text{can be}& & \\ & & \text{(T)}& -1\end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(A)}\to \text{(R)},\text{(S)}$
• B. $\text{(A)}\to \text{(S)},\text{(T)}$
• C. $\text{(B)}\to \text{(P)},\text{(Q)}$
• D. $\text{(B)}\to \text{(Q)},\text{(T)}$

Answer 1

The correct option is B $\text{(A)}\to \text{(S)},\text{(T)}$

$\left(A\right)$ If the quadratic equation $x^2-2\alpha x+2\alpha =0$ has two real and distinct roots, then
$D>0\Rightarrow 4{\alpha }^2-8\alpha >0\Rightarrow 4\alpha (\alpha -2)>0\Rightarrow \alpha \in (-\infty ,0)\cup (2,\infty )\dots \left(i\right)$

Also, given that
$|x_1-x_2|\le 2\sqrt{3}\Rightarrow (x_1+x_2{)}^2-4x_1{x}_2\le 12\Rightarrow 4{\alpha }^2-8\alpha \le 12\Rightarrow (\alpha -3\left)\right(\alpha +1)\le 0\Rightarrow \alpha \in [-1,3]\dots (ii)$

From $\left(i\right)$ and $\left(ii\right),$
$\alpha \in [-1,0)\cup (2,3]$
$\therefore \alpha$ can be $-1,3$
$\left(A\right)\to \left(S\right),\left(T\right)$

$\left(B\right)y=\frac{x^2+4x}{x^2+4x+6}\Rightarrow y=1-\frac{6}{(x+2{)}^2+2}\Rightarrow y\in [-2,1)$
$\therefore y=-2,-1,0$
$\left(B\right)\to \left(P\right),\left(T\right)$

$\left(C\right)(2-n)x^2-n<8x+4\forall x\in R$
$\Rightarrow (n-2)x^2+8x+(n+4)>0$
$\Rightarrow n-2>0\Rightarrow n>2\dots \left(i\right)$
and $D<0$
$\Rightarrow 64-4(n-2)(n+4)<0\Rightarrow n^2+2n-24>0\Rightarrow (n+6)(n-4)>0$
$\Rightarrow n\in (-\infty ,-6)\cup (4,\infty )\dots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right),$ we get
$\Rightarrow n>4$
$\Rightarrow \frac{n}{4}>1$
$\left(C\right)\to \left(R\right),\left(S\right)$

$\left(D\right)\cos 2x+2a\cos x+6a=17$
$\Rightarrow 2{\cos }^{2}x+2a\cos x+6a-18=0\Rightarrow 2(\cos x-3)(\cos x+3)+2a(\cos x+3)=0$
$\Rightarrow (\cos x+3)\left[2\right(\cos x-3)+2a]=0$
$\Rightarrow \cos x+3=0\Rightarrow \cos x=-3$ (not possible)
or $2(\cos x-3)+2a=0$
$\Rightarrow \cos x=3-a$

We know that,
$\Rightarrow -1\le \cos x\le 1\Rightarrow -1\le 3-a\le 1\Rightarrow -4\le -a\le -2\Rightarrow 2\le a\le 4$
$\left(D\right)\to \left(R\right),\left(S\right)$