Question

The expression $y=a{x}^2+bx+c(a,b,c\in R$ and $a\ne 0)$ represents a parabola which cuts the $x$-axis at the points which are roots of the equation $a{x}^2+bx+c=0$. Column-II contains values which correspond to the nature of roots mentioned in column-I.

Column I		Column II
A	For $a=1,c=4$, if both roots are greater than 2 then $b$ can be equal to	P	4
B	For $a=-1,b=5$, if roots lie on either side of -1 then $c$ can be equal to	Q	8
C	For $b=6,c=1$, if one root is less than -1 and the other root greater than (	R	10$\frac{-1}{2}$ then a can be equal to
		S	no real value

• A. (A) S (B) Q, R (C) P
• B. (A) S (B) Q, R (C) Q
• C. (A) S (B) Q, R (C) R
• D. (A) S (B) Q, R (C) S

Answer 1

Answer: (A)

(A) We have $f(x)=x^2+bx+4$
$\therefore D\ge 0\Rightarrow b^2-16>0\Rightarrow b\in (-\infty ,-4]\cup [4,\infty )$
$\text{ and }f(2)>0\Rightarrow 4+2b+4>0\Rightarrow b>-4$
Also $2<\frac{-b}{2}\Rightarrow 4<-b\Rightarrow b<-4$.

Hence $b\in \varphi$.
(B) We have $f(x)=-x^2+5x+c$
Clearly $f(-1)>0\Rightarrow -1-5+c>0$
Hence $c>6$.

(C) We have $f(x)=a{x}^2+6x+1$
Now $D>0\Rightarrow 36-4a>0\Rightarrow a<9$
Case-I : If $a>0,f(-1)<0\Rightarrow a-6+1<0\Rightarrow a<5$.
and $f\left(\frac{-1}{2}\right)<0\Rightarrow \frac{a}{4}-3+1<0\Rightarrow a<8$.

Hence $a\in (0,5)$
Case-II : If $a<0,f(-1)>0$ and $f\left(\frac{-1}{2}\right)>0$

$\Rightarrow a-6+1>0\text{ and }\frac{a}{4}-3+1>0\Rightarrow a>5\text{ and }\frac{a}{4}>2$
$\therefore a>8\text{ (Rejected) }$
Hence we conclude from case-I and case-II that $a\in (0,5)]$