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

Answers (1)

Answer:

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.

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