Question

Let the equation $x^2$ + 2(a - 1)x + a + 5 = 0, where 'a' is a parameter, match the real value of 'a' so that the given equation has :-

Column - I	Column - II
(A) Imaginary roots	(P) $(-\infty ,-\frac{8}{7})$
(B) One root less than 3 other root is greater than 3	(Q) (-1, 4)
(C) One root less than 1 & other root is greater than 3	(R) $(-\infty ,-\frac{4}{3})$

• A. (A) $\to$ (Q), (B) $\to$ (P), (C) $\to$ (R)
• B. (A) $\to$ (P), (B) $\to$ (Q), (C) $\to$ (R)
• C. (A) $\to$ (Q), (B) $\to$ (R), (C) $\to$ (P)
• D. (A) $\to$ (R), (B) $\to$ (P), (C) $\to$ (Q)

Answer 1

The correct option is A (A) $\to$ (Q), (B) $\to$ (P), (C) $\to$ (R)

Given equation
$x^2+2(a-1)x+a+5=0$
$D=4(a-1{)}^2-4a-20=4a^2-12a-12$
(A) For imaginary roots
$D<0$
$\Rightarrow 4(a^2-3a-3)<0$
$\Rightarrow (a-4)(a+1)<0$
$\Rightarrow a\in (-1,4)$
(B) Condition for roots to lie on opposite side of a number k is $af\left(k\right)<0$
So, here $f\left(3\right)<0$
$\Rightarrow 9+6(a-1+a+5)<0$
$\Rightarrow 7a+8<0$
$\Rightarrow a<-\frac{8}{7}$
So, $a\in (-\infty ,-\frac{8}{7})$
(C)For this , f(1) < 0
$\Rightarrow 3a+4<0$
$\Rightarrow a<\frac{4}{3}$ ...(1)
Also, $f\left(3\right)<0$
$\Rightarrow 9+6(a-1+a+5)<0$
$\Rightarrow 7a+8<0$
$\Rightarrow a<-\frac{8}{7}$
So, $a\in (-\infty ,-\frac{8}{7})$ ....(2)
From (1) and (2)
$a\in (-\infty ,-\frac{4}{3})$