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Quadratic Equation and Inequalities question

2005 · Shift 0 · Q69
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  5. /2005 · Shift 0 · Q69

Quadratic Equation and Inequalities question

2005 · Shift 0 · Q69

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. Let the roots of x2−(a−2)x−(a+1)=0x^2-(a-2)x-(a+1)=0x2−(a−2)x−(a+1)=0 be α\alphaα and β\betaβ.

  2. Using Vieta’s formulas for x2−(a−2)x−(a+1)=0,x^2-(a-2)x-(a+1)=0,x2−(a−2)x−(a+1)=0, we get α+β=a−2,αβ=−(a+1).\alpha+\beta=a-2, \qquad \alpha\beta=-(a+1).α+β=a−2,αβ=−(a+1).

  3. The sum of squares of the roots is α2+β2=(α+β)2−2αβ.\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta.α2+β2=(α+β)2−2αβ. Substitute the 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)). So, α2+β2=(a−2)2+2(a+1).\alpha^2+\beta^2=(a-2)^2+2(a+1).α2+β2=(a−2)2+2(a+1).

  4. Simplify: (a−2)2+2(a+1)=a2−4a+4+2a+2=a2−2a+6.(a-2)^2+2(a+1)=a^2-4a+4+2a+2=a^2-2a+6.(a−2)2+2(a+1)=a2−4a+4+2a+2=a2−2a+6. Thus we need the least value of f(a)=a2−2a+6.f(a)=a^2-2a+6.f(a)=a2−2a+6.

  5. Complete the square: f(a)=a2−2a+6=(a−1)2+5.f(a)=a^2-2a+6=(a-1)^2+5.f(a)=a2−2a+6=(a−1)2+5. This is minimum when (a−1)2=0⇒a=1.(a-1)^2=0 \Rightarrow a=1.(a−1)2=0⇒a=1.

  6. Checking options:

  • A: 111 ✅
  • B: 000 gives f(0)=6f(0)=6f(0)=6
  • C: 333 gives f(3)=9f(3)=9f(3)=9
  • D: 222 gives f(2)=6f(2)=6f(2)=6

Hence the required value of aaa is 1.\boxed{1}.1​.

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