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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Mathematics
Chapter
Matrices

JEE Advanced 2011 · Paper 1 · Mathematics · Matrices

Paragraph: Let a , b and c be three real numbers satisfying [ array...

Paragraph: Let aa, b and cc be three real numbers satisfying \left[abc\begin{array}{lll}a & b & c\end{array}\right]\left[197827737\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}$$\right]Question: Let b=6b=6, with aandca and c satisfying Eq. (E). If α\alpha and β\beta are the roots of the quadratic equation ax2+bx+c=0a x^2+b x+c=0, then ∑n=0∞(1α+1β)nis\sum_{n=0}^{\infty}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^n is

Options

  1. A6
  2. B7
  3. C67\frac{6}{7}
  4. D∞\infty

Correct answer

B. 7

Step-by-step solution

 Given, [abc]1×3[197827737]3×3=[000]⇒[a+8b+7c9a+2b+3c7a+7b+7c]=[000]⇒a+8b+7c=0⇒9a+2b+3c=0⇒a+b+c=0\begin{aligned}& \text{ Given, }[a b c]_{1 \times3}\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7 \end{array}\right]_{3 \times3}=\left[\begin{array}{lll}0 & 0 & 0 \end{array}\right] \\ & \Rightarrow \quad\left[\begin{array}{c}a+8 b+7 c \\ 9 a+2 b+3 c \\ 7 a+7 b+7 c \end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow \quad a+8 b+7 c=0 \\ & \Rightarrow 9 a+2 b+3 c=0 \\ & \Rightarrow \quad a+b+c=0 \\ & \end{aligned}On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get 7 a+c=0 Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get 6a−b=0∴b=6a and c=−7a\begin{array}{ll}& 6 a-b=0 \\ \therefore \quad & b=6 a \text{ and }c=-7 a \end{array}If b=6,a=1b=6, a=1 and c=−7c=-7∴ax2+bx+c=0⇒x2+6x−7=0⇒(x+7)(x−1)=0∴x=1,−7⇒∑n=0∞(11−17)n⇒∑n=0∞(67)n⇒1+67+(67)n+…∞=11−67=11/7=7\begin{array}{ll} \therefore& a x²+b x+c=0 \\ \Rightarrow & x²+6 x-7=0 \\ \Rightarrow & (x+7)(x-1)=0 \\ \therefore & x=1,-7 \\ \Rightarrow & \quad \sum_{n=0}^{\infty}\left(\frac{1}{1}-\frac{1}{7}\right)^n \Rightarrow \sum_{n=0}^{\infty}\left(\frac{6}{7}\right)^n \\ \Rightarrow & 1+\frac{6}{7}+\left(\frac{6}{7}\right)^n+\ldots \infty\\ & =\frac{1}{1-\frac{6}{7}}=\frac{1}{1 / 7}=7 \end{array}$
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