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Quadratic Equation and Inequalities question

2005 · Shift 0 · Q109
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  5. /2005 · Shift 0 · Q109

Quadratic Equation and Inequalities question

2005 · Shift 0 · Q109

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
The value of aaa for which the sum of the squares of the roots of the equation x2−(a−2)x−a−1=0{x^2} - \left( {a - 2} \right)x - a - 1 = 0x2−(a−2)x−a−1=0 assume the least value is
  1. A
    111
  2. B
    000
  3. C
    333
  4. D
    222
View written solutionFree

Correct answer: A

  1. Given quadratic equation

    x2−(a−2)x−(a+1)=0x^2-(a-2)x-(a+1)=0x2−(a−2)x−(a+1)=0

    Let its roots be α\alphaα and β\betaβ.

  2. Use Vieta’s formulas

    For the quadratic x2−(a−2)x−(a+1)=0x^2-(a-2)x-(a+1)=0x2−(a−2)x−(a+1)=0,

    α+β=a−2\alpha+\beta=a-2α+β=a−2 αβ=−(a+1)\alpha\beta=-(a+1)αβ=−(a+1)

  3. Expression for sum of squares of roots

    We need to minimize

    α2+β2\alpha^2+\beta^2α2+β2

    Using the identity,

    α2+β2=(α+β)2−2αβ\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\betaα2+β2=(α+β)2−2αβ

    Substitute the Vieta values:

    α2+β2=(a−2)2−2(−(a+1))\alpha^2+\beta^2=(a-2)^2-2\bigl(-(a+1)\bigr)α2+β2=(a−2)2−2(−(a+1))

    =(a−2)2+2(a+1)=(a-2)^2+2(a+1)=(a−2)2+2(a+1)

    =a2−4a+4+2a+2=a^2-4a+4+2a+2=a2−4a+4+2a+2

    =a2−2a+6=a^2-2a+6=a2−2a+6

  4. Minimize this expression

    a2−2a+6=(a−1)2+5a^2-2a+6=(a-1)^2+5a2−2a+6=(a−1)2+5

    Since (a−1)2≥0(a-1)^2\ge 0(a−1)2≥0, the least value occurs when

    (a−1)2=0⇒a=1(a-1)^2=0 \Rightarrow a=1(a−1)2=0⇒a=1

  5. Check options

    • A: 111 ✅
    • B: 000
    • C: 333
    • D: 222

    Hence the correct option is A.

  6. Comparison with stored answer

    Stored correct answer: A

    Our derived answer: A

    They agree.

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