JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Quadratic Equations and Inequalities

The sum of all integral values of

k (k \neq = 0)

for which the equation

\frac{2}{x - 1} - \frac{1}{x - 2} = \frac{2}{k}

x

has no real roots, is ______.

Answer: 66

Solution

Combining the left side,

\frac{2 ( x - 2 ) - ( x - 1 )}{( x - 1 ) ( x - 2 )} = \frac{x - 3}{( x - 1 ) ( x - 2 )} = \frac{2}{k},

k (x - 3) = 2 (x - 1) (x - 2)

, that is

2 x^{2} - (6 + k) x + (4 + 3 k) = 0.

For

k \neq = 0

neither

x = 1

nor

x = 2

satisfies this quadratic, so every real root of the quadratic is a genuine root of the equation, and the equation has no real root exactly when the discriminant is negative:

(6 + k)^{2} - 8 (4 + 3 k) < 0 \Rightarrow k^{2} - 12 k + 4 < 0 \Rightarrow 6 - 42 < k < 6 + 42 .

Since

6 - 42 \approx 0.34

and

6 + 42 \approx 11.66

, the admissible non-zero integers are

1, 2, \dots, 11

, and their sum is

\frac{11 \cdot 12}{2} = 66

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Numerical answer

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JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Quadratic Equations and Inequalities

Solution

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