#### Need solution for RD Sharma maths class 12 chapter Determinants exercise 5 multiple choise question 15

Correct option (c)

Hint:

Simply solve this determinant, then conclude answer

Given:

If $a$ > 0  and discriminant of $ax^{2}+2bx+c$ is negative.

Here we have to find $\Delta$.

Solution:

If $a$ > 0  and discriminant of $ax^{2}+2bx+c$ is negative.

$\Rightarrow (2b)^{2}-4ac< 0$

$\Rightarrow 4b^{2}-4ac< 0$

$\Rightarrow b^{2}-ac< 0\: but\: \: a> 0$                                            .......(i)

$\Delta =\begin{vmatrix} a &b &ax+b \\ b &c &bx+c \\ ax+b &bx+c &0 \end{vmatrix}$

R1→R1+R2

$\Rightarrow \Delta =\begin{vmatrix} ax+b &bx+c &ax^{2}+2bx+c \\ b &c &bx+c \\ ax+b &bx+c &0 \end{vmatrix}$

R1→R1-R3

$\Rightarrow \Delta =\begin{vmatrix} 0 &0 &ax^{2}+2bx+c \\ b &c &bx+c \\ ax+b &bx+c &0 \end{vmatrix}$

Expanding along R1

$\Rightarrow \Delta =(ax^{2}+2bx+c)[b(bx+c)-c(ax+b)]$

$\Rightarrow \Delta =(ax^{2}+2bx+c)[b^{2}x+bc-acx-bc]$

$\Rightarrow \Delta =(ax^{2}+2bx+c)[b^{2}x-acx]$

$\Rightarrow \Delta =(ax^{2}+2bx+c)x(b^{2}-ac)$

$As\: \: b^{2}-ac< 0,\: \Delta\: \: must\: \: be\: \: negative$.