Q89 / 155 · WBJEE 2025 · Mathematics

If the sum of the squares of the roots of the equation $x^2-(a-2) x-(a+1)=0$ is least for an appropriate value of the variable parameter $a$, then the value of ' $a$ ' will be

  • A
    $3$
  • B
    $2$
  • Correct answer
    C
    $1$
  • D
    $0$

Solution

$\begin{aligned} & \text { } \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\ & =(a-2)^2+2(a+1) \\ & =a^2-2 a+6=(a-1)^2+5\end{aligned}$

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

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