JEE Main 2021 (Online) 26th August Morning ShiftQuestion 1
The sum of all integral values of $k$ ($k\neq0$) for which the equation $\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$ in $x$ has no real roots, is $\_\_\_\_\_\_$ .
Quadratic Equation · Nature of Roots and Discriminant
Answer$66$
Solution
Concept. Clear the denominators to get a quadratic; it has no real roots when $D<0$ (and $x=1,2$ never sneak in for $k\ne0$). Formulas used. (1) $\dfrac{2}{x-1}-\dfrac{1}{x-2}=\dfrac{x-3}{(x-1)(x-2)}$Step 1 - Quadratic. By (1), $k(x-3)=2(x-1)(x-2)$: $$2x^2-(6+k)x+(4+3k)=0.$$ At $x=1$ and $x=2$ the left side equals $2k$ and $k$, so for $k\ne0$ these are never roots. Step 2 - No real roots. $$(6+k)^2-8(4+3k)=k^2-12k+4<0\ \Longrightarrow\ 6-4\sqrt2<k<6+4\sqrt2.$$ Step 3 - Integers. $k=1,2,\dots,11$; sum $66$. Answer: 66