Question

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

Which of the following is the only CORRECT combination?

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

Answer 1

The correct option is B $\text{(C)}\to \text{(P), (Q), (S)}$

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

$\text{(B)}$
Let $f\left(x\right)=a{x}^2+2bx+4c-16$
$f(-2)=4a-4b+4c-16=4(a-b+c-4)>0$
As the given equation has no real roots,
$\therefore f\left(x\right)>0\forall x\in R$
$\Rightarrow f\left(0\right)>0$
$\Rightarrow 4c-16>0\Rightarrow c>4$

$\text{(C)}$
Both roots are negative if, $D\ge 0$, the product of the roots must be positive and 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 $b\in (\frac{14}{9},2]\cup [7,\infty )$

$\text{(D)}$
$|x-|x-4||=2x+4$
$\left(i\right)x>4$
$4=2x+4\Rightarrow x=0$, 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.