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MCQMediumJEE 2026Nature of Roots & Formation of Equations

JEE Mathematics 2026 Question with Solution

If α,β\alpha,\beta where α<β\alpha<\beta, are the roots of the equation λx2−(λ+3)x+3=0\lambda x^2-(\lambda+3)x+3=0 such that 1α−1β=13\frac{1}{\alpha}-\frac{1}{\beta}=\frac{1}{3}, then the sum of all possible values of λ\lambda is:

  • A

    88

  • B

    66

  • C

    44

  • D

    22

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: α,β\alpha,\beta are the roots of λx2−(λ+3)x+3=0\lambda x^2-(\lambda+3)x+3=0 with α<β\alpha<\beta and 1α−1β=13\frac{1}{\alpha}-\frac{1}{\beta}=\frac{1}{3}.

Find: The sum of all possible values of λ\lambda.

Using Vieta’s formulas,

α+β=λ+3λ,αβ=3λ\alpha+\beta=\frac{\lambda+3}{\lambda},\quad \alpha\beta=\frac{3}{\lambda}

From the given condition,

1α−1β=β−ααβ=13\frac{1}{\alpha}-\frac{1}{\beta}=\frac{\beta-\alpha}{\alpha\beta}=\frac{1}{3}

So,

β−α=αβ3=1λ\beta-\alpha=\frac{\alpha\beta}{3}=\frac{1}{\lambda}

Now use

(β−α)2=(α+β)2−4αβ(\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta

Hence,

1λ2=(λ+3λ)2−12λ\frac{1}{\lambda^2}=\left(\frac{\lambda+3}{\lambda}\right)^2-\frac{12}{\lambda}

Multiplying through by λ2\lambda^2,

1=(λ+3)2−12λ1=(\lambda+3)^2-12\lambda

Therefore,

1=λ2−6λ+91=\lambda^2-6\lambda+9 (λ−3)2=1+8=13(\lambda-3)^2=1+8=13

So the possible values are

λ=3±13\lambda=3\pm\sqrt{13}

Their sum is

(3+13)+(3−13)=6(3+\sqrt{13})+(3-\sqrt{13})=6

Therefore, the correct option is B.

Using reciprocal condition carefully

Given: 1α−1β=13\frac{1}{\alpha}-\frac{1}{\beta}=\frac{1}{3} and α,β\alpha,\beta are roots of λx2−(λ+3)x+3=0\lambda x^2-(\lambda+3)x+3=0.

Find: The sum of all possible values of λ\lambda.

First rewrite the reciprocal condition:

1α−1β=β−ααβ\frac{1}{\alpha}-\frac{1}{\beta}=\frac{\beta-\alpha}{\alpha\beta}

Thus,

β−ααβ=13\frac{\beta-\alpha}{\alpha\beta}=\frac{1}{3}

From Vieta’s formulas for λx2−(λ+3)x+3=0\lambda x^2-(\lambda+3)x+3=0,

α+β=λ+3λ,αβ=3λ\alpha+\beta=\frac{\lambda+3}{\lambda},\quad \alpha\beta=\frac{3}{\lambda}

Substituting αβ=3λ\alpha\beta=\frac{3}{\lambda} into the reciprocal condition gives

β−α=13⋅3λ=1λ\beta-\alpha=\frac{1}{3}\cdot\frac{3}{\lambda}=\frac{1}{\lambda}

Now apply the identity

(β−α)2=(α+β)2−4αβ(\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta

So,

(1λ)2=(λ+3λ)2−4(3λ)\left(\frac{1}{\lambda}\right)^2=\left(\frac{\lambda+3}{\lambda}\right)^2-4\left(\frac{3}{\lambda}\right)

That is,

1λ2=(λ+3)2λ2−12λ\frac{1}{\lambda^2}=\frac{(\lambda+3)^2}{\lambda^2}-\frac{12}{\lambda}

Multiplying by λ2\lambda^2,

1=(λ+3)2−12λ1=(\lambda+3)^2-12\lambda

Expand:

1=λ2+6λ+9−12λ1=\lambda^2+6\lambda+9-12\lambda 1=λ2−6λ+91=\lambda^2-6\lambda+9 λ2−6λ+8=0\lambda^2-6\lambda+8=0

Equivalently,

(λ−3)2=13(\lambda-3)^2=13

Hence,

λ=3+13orλ=3−13\lambda=3+\sqrt{13} \quad \text{or} \quad \lambda=3-\sqrt{13}

Therefore, the required sum is

66

So the correct option is B.

Common mistakes

  • Using 1α−1β=α−βαβ\frac{1}{\alpha}-\frac{1}{\beta}=\frac{\alpha-\beta}{\alpha\beta} instead of β−ααβ\frac{\beta-\alpha}{\alpha\beta}. This reverses the sign and leads to the wrong equation for λ\lambda. Always combine the fractions carefully before substituting Vieta’s relations.

  • Applying Vieta’s formulas incorrectly by taking α+β=λ+31\alpha+\beta=\frac{\lambda+3}{1} or αβ=3\alpha\beta=3. For ax2+bx+c=0ax^2+bx+c=0, use α+β=−ba\alpha+\beta=-\frac{b}{a} and αβ=ca\alpha\beta=\frac{c}{a}, so here they are λ+3λ\frac{\lambda+3}{\lambda} and 3λ\frac{3}{\lambda}.

  • Forgetting to use the identity (β−α)2=(α+β)2−4αβ(\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta. Trying to relate β−α\beta-\alpha directly to α+β\alpha+\beta without squaring misses the standard connection between sum, product, and difference of roots.

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