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Question: Answered & Verified by Expert
Let
$A=\left|\begin{array}{cc}2 & e^{i \pi} \\ -1 & i^{2012}\end{array}\right|, C=\frac{d}{d x}\left(\frac{1}{x}\right)_{x=1}$,
$D=\int_{e^2}^1 \frac{d x}{x}$
If the sum of two roots of the equation $A x^3+B x^2+C x-D=0$ is equal to zero, then $B$ is equal to
MathematicsQuadratic EquationTS EAMCETTS EAMCET 2012
Options:
  • A $-1$
  • B $0$
  • C $1$
  • D $2$
Solution:
2692 Upvotes Verified Answer
The correct answer is: $2$
Given, $\begin{aligned} A & =\left|\begin{array}{cc}2 & e^{i \pi} \\ -1 & i^{2012}\end{array}\right| \\ & =\left|\begin{array}{cc}2 & \cos \pi+i \sin \pi \\ -1 & \left(1^4\right)^{503}\end{array}\right|\end{aligned}$
$\begin{aligned} & =\left|\begin{array}{cc}2 & -1 \\ -1 & 1\end{array}\right|=2-1=1 \\ C & =\frac{d}{d x}\left(\frac{1}{x}\right)_{x=1}=\left(-\frac{1}{x^2}\right)_{x=1}=-1\end{aligned}$
and $\begin{aligned} D & =\int_{e^2}^1 \frac{d x}{x} \\ & =[\log x]_{e^2}^1 \\ & =\log 1-\log e^2=0-2 \\ & =-2\end{aligned}$
Let $\alpha, \beta$ and $\gamma$ are the roots of the equation
$\begin{aligned} & A x^3+B x^2+C x+D=0 \\ & x^3+B x^2-x+2=0 \\ & \alpha+\beta+\gamma=-B \\ & \gamma=-B \quad(\because \alpha+\beta=0 \text { given })\end{aligned}$
$\therefore \quad-B$ satisfies the Eq. (ii),
$\begin{array}{ll}\therefore & (-B)^3+B^2+B+2=0 \\ \Rightarrow & B=2\end{array}$

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