Q71 / 160 · TS EAMCET 2015 · Mathematics

Let $\left|\begin{array}{ccc}x^2+x+1 & x+1 & 2 x-3 \\ 3 x^2-1 & x+2 & x-1 \\ x^2+5 x+1 & 2 x+3 & x+4\end{array}\right|$ $=a x^4+b x^3+c x^2+d x+e$ be an identity in $x$. If $a, b, c, d$ are known, then the value of $e$ is

  • Correct answer
    A
    29
  • B
    24
  • C
    16
  • D
    9

Solution

We have, $$ \begin{array}{r} \left|\begin{array}{ccc} x^2+x+1 & x+1 & 2 x-3 \\ 3 x^2-1 & x+2 & x-1 \\ x^2+5 x+1 & 2 x+3 & x+4 \end{array}\right| \\ =a x^4+b x^3+c x^2+d x+e \end{array} $$ Put $x=0$ on both sides, we get $$ \begin{aligned} & \left|\begin{array}{ccc} 1 & 1 & -3 \\ -1 & 2 & -1 \\ 1 & 3 & 4 \end{array}\right|=e \\ \Rightarrow \quad & =1(8+3)-1(-4+1)-3(-3-2) \\ & =11+3+15=29 \end{aligned} $$

Hence, the correct answer is option: 'A'.

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