Question

$\begin{array}{cc}\text{(I) If}{x}^2+x-a=0\text{has integral roots}& \left(P\right)2\\ \text{and}a\in N,\text{then}a\text{can be equal to}& \\ \text{(II) If the equation}a{x}^2+2bx+4c=16& \left(Q\right)12\\ \text{has no real roots and}a+c>b+4,& \\ \text{then the integral value of}c\text{can be}& \\ \text{(III) If equation}{x}^2+2bx+9b-14=0& \left(R\right)1\\ \text{has only negative roots, then the integral}& \\ \text{values of}b\text{can be}& \\ \text{(IV) If}N\text{be the number of solutions of}& \left(S\right)30\\ \text{the equation}|x-|4-x||-2x=4,\text{then}& \\ \text{the value of}N\text{is}& \end{array}$

Which of the following is the only CORRECT combination?

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

Answer 1

The correct option is C $\left(I\right)\to \left(P\right),\left(Q\right),\left(S\right)$

(I) $D=\sqrt{1+4a}$
$1+4a=(2\lambda +1{)}^2$
$a=\lambda (\lambda +1),\lambda \in I^{+}$

(II) $Letf\left(x\right)=a{x}^2+2bx+4c-16$
$f(-2)=4a-4b+4c-16=4(a-b+c-4)>0$

$\text{As the given equation has no real roots.}$
$\therefore f\left(x\right)>0\forall x\in R$

$\Rightarrow f\left(0\right)>0\Rightarrow 4c-16>0\Rightarrow c>4$

(III) Both roots are negative if, $D\ge 0$, the products of the roots must be positive & the sum of the roots must be negative

$\{\begin{array}{c}-2b<0\Rightarrow b>0\\ 4b^2-4(9b-14)\ge 0\\ 9b-14>0\end{array}$

we get $b\in (\frac{14}{9},2]\cup [7,\infty )$

(IV) $|x-|x-4||=2x+4$

(i) $x>4$
$4=2x+4\Rightarrow x=0$, which is not possible.

$\left(ii\right)x\le 4$

$|x+x-4|=2x+42|x-2|=2x+4$

$\left(a\right)2\le x\le 4$
$2x-4=2x+4$, which is not possible.

$\left(b\right)x\le 2$
$-2x+4=2x+4\Rightarrow x=0$

So, the equation has only one solution.