JEE Main 2026 (Online) 28th January Morning ShiftQuestion 1
If $\alpha,\beta$, where $\alpha<\beta$, are the roots of the equation $\lambda x^2-(\lambda+3)x+3=0$ such that $\frac1\alpha-\frac1\beta=\frac13$, then the sum of all possible values of $\lambda$ is :
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. The quadratic factorises, so the roots are $1$ and $\frac3\lambda$; which one is smaller depends on $\lambda$. Formulas used. (1) $\lambda x^2-(\lambda+3)x+3=(\lambda x-3)(x-1)$Step 1 - Roots. By (1), the roots are $1$ and $\frac3\lambda$; their reciprocals are $1$ and $\frac\lambda3$. Step 2 - Case $\lambda>3$. $\frac3\lambda<1$, so $\alpha=\frac3\lambda$, $\beta=1$: $\frac\lambda3-1=\frac13$ gives $\lambda=4$. Step 3 - Case $0<\lambda<3$. $\alpha=1$, $\beta=\frac3\lambda$: $1-\frac\lambda3=\frac13$ gives $\lambda=2$. Step 4 - Case $\lambda<0$. $\alpha=\frac3\lambda<0<\beta=1$: $\frac\lambda3-1=\frac13$ gives $\lambda=4$, not negative. No solution. Step 5 - Sum. $2+4=6$. Answer: (C)