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Quadratic Equation and Inequalities question

2019 · 10 Jan · Shift 2 · Q30
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  5. /2019 · 10 Jan · Shift 2 · Q30

Quadratic Equation and Inequalities question

2019 · 10 Jan · Shift 2 · Q30

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
The value of λ\lambdaλ such that sum of the squares of the roots of the quadratic equation, x2 + (3 –λ\lambdaλ)x + 2 = λ\lambdaλ has the least value is -
  1. A
    1
  2. B
    2
  3. C
    158{{15} \over 8}815​
  4. D
    49{4 \over 9}94​
View written solutionFree

Correct answer: B

  1. Rewrite the quadratic equation in standard form

Given: x2+(3−λ)x+2=λx^2 + (3-\lambda)x + 2 = \lambdax2+(3−λ)x+2=λ

Bring all terms to one side: x2+(3−λ)x+(2−λ)=0x^2 + (3-\lambda)x + (2-\lambda)=0x2+(3−λ)x+(2−λ)=0

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

  1. Use relations between roots and coefficients

For x2+(3−λ)x+(2−λ)=0,x^2 + (3-\lambda)x + (2-\lambda)=0,x2+(3−λ)x+(2−λ)=0, we have: α+β=−(3−λ)=λ−3\alpha+\beta = -(3-\lambda)=\lambda-3α+β=−(3−λ)=λ−3 αβ=2−λ\alpha\beta = 2-\lambdaαβ=2−λ

  1. Form the sum of squares of the roots

We need: 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: α2+β2=(λ−3)2−2(2−λ)\alpha^2+\beta^2 = (\lambda-3)^2 - 2(2-\lambda)α2+β2=(λ−3)2−2(2−λ)

Now simplify: =λ2−6λ+9−4+2λ= \lambda^2 - 6\lambda + 9 - 4 + 2\lambda=λ2−6λ+9−4+2λ =λ2−4λ+5= \lambda^2 - 4\lambda + 5=λ2−4λ+5

  1. Minimize this expression

We must find the least value of f(λ)=λ2−4λ+5f(\lambda)=\lambda^2-4\lambda+5f(λ)=λ2−4λ+5

Complete the square: f(λ)=(λ−2)2+1f(\lambda)=(\lambda-2)^2+1f(λ)=(λ−2)2+1

This is minimum when (λ−2)2=0⇒λ=2(\lambda-2)^2=0 \Rightarrow \lambda=2(λ−2)2=0⇒λ=2

  1. Check the options

The required value is: 2\boxed{2}2​

So the correct option is B.

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