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TS EAMCET 10 May 2024 Shift 1 Solved Paper

Section: Mathematics

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Question : 6 of 160

Marks: +1, -0

If

a \neq b \neq c, \Delta_1 = \begin{vmatrix} 1 & a^2 & bc \\ 1 & b^2 & ca \\ 1 & c^2 & ab \end{vmatrix}, \Delta_2 = \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}

\; \frac{\Delta_1}{\Delta_2} = \frac{6}{11}

11(a+b+c)=

[TG EAPCET 10 May 2024 Shift 1]

0
1
$ab+bc+ca$
$6(ab+bc+ca)$

Solution:

Given,

\; \Delta_1 = \begin{vmatrix} 1 & a^2 & bc \\ 1 & b^2 & ca \\ 1 & c^2 & ab \end{vmatrix}

\; R_2 \rightarrow R_2 - R_1 \; \text{ and } \; R_3 \rightarrow R_3 - R_1

\; \Delta_1 = \begin{vmatrix} 1 & a^2 & bc \\ 0 & b^2 - a^2 & c(a-b) \\ 0 & c^2 - a^2 & b(a-c) \end{vmatrix}

= (a-b)(c-a) \begin{vmatrix} 1 & a^2 \\ 0 & -(a+b) \\ c & 0 \\ c+a & \text{---} \end{vmatrix}

= (a-b)(c-a)(ab+b^2-c^2-ac)

= (a-b)(c-a)(a(b-c)+(b-c)(b+c))

\; \Delta_1 = (a-b)(b-c)(c-a)(a+b+c) \ldots

and

\Delta_2 = \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}

C_2 \rightarrow C_2 - C_1

and

C_3 \rightarrow C_3 - C_1

= \begin{vmatrix} 1 & 0 & 0 \\ a^2 & b^2-a^2 & c^2-a^2 \\ a^3 & b^3-a^3 & c^3-a^3 \end{vmatrix}

= -(a-b)(c-a)

\begin{vmatrix} 1 & 0 & 0 \\ a^2 & b-c & c+a \\ a^3 & (b^2-c^2)+a(b-c) & c^2+a^2+ac \end{vmatrix}

= -(a-b)(b-c)(c-a)

\; \begin{vmatrix} 1 & 0 & 0 \\ a^2 & 1 & c+a \\ a^3 & b+c+a & c^2+a^2+ac \end{vmatrix}

= -(a-b)(b-c)(c-a)

\; [c^2 + a^2 + ac - bc - c^2 - ac - ab - ac - a^2]

= -(a-b)(b-c)(c-a)[-bc - ab - ac]

\Delta_2 = (a-b)(b-c)(c-a)[ab+bc+ca]

From Ist and 2 nd equation replace

\Delta_1

and

\Delta_2

\; \frac{\Delta_1}{\Delta_2} = \frac{6}{11} \Rightarrow \; \frac{a+b+c}{ab+bc+ca} = \frac{6}{11}

\therefore 11(a+b+c) = 6(ab+bc+ca)

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