JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 13)
The sum of all integral values of k (k $$\ne$$ 0) for which the equation $${2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k}$$ in x has no real roots, is ____________.
Answer
66
Explanation
$${2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k}$$
$$x \in R - \{ 1,2\} $$
$$ \Rightarrow k(2x - 4 - x + 1) = 2({x^2} - 3x + 2)$$
$$ \Rightarrow k(x - 3) = 2({x^2} - 3x + 2)$$
for x $$\ne$$ 3, $$k = 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right)$$
$$x - 3 + {2 \over {x - 3}} \ge 2\sqrt 2 ,\forall x > 3$$
& $$x - 3 + {2 \over {x - 3}} \le - 2\sqrt 2 ,\forall x < - 3$$
$$ \Rightarrow 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right) \in \left( { - \infty ,6 - 4\sqrt 2 } \right] \cup \left[ {6 + 4\sqrt 2 ,\infty } \right)$$
for no real roots
$$k \in (6 - 4\sqrt 2 ,6 + 4\sqrt 2 ) - \{ 0\} $$
Integral k$$\in$${1, 2 ..... 11}
Sum of k = 66
$$x \in R - \{ 1,2\} $$
$$ \Rightarrow k(2x - 4 - x + 1) = 2({x^2} - 3x + 2)$$
$$ \Rightarrow k(x - 3) = 2({x^2} - 3x + 2)$$
for x $$\ne$$ 3, $$k = 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right)$$
$$x - 3 + {2 \over {x - 3}} \ge 2\sqrt 2 ,\forall x > 3$$
& $$x - 3 + {2 \over {x - 3}} \le - 2\sqrt 2 ,\forall x < - 3$$
$$ \Rightarrow 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right) \in \left( { - \infty ,6 - 4\sqrt 2 } \right] \cup \left[ {6 + 4\sqrt 2 ,\infty } \right)$$
for no real roots
$$k \in (6 - 4\sqrt 2 ,6 + 4\sqrt 2 ) - \{ 0\} $$
Integral k$$\in$${1, 2 ..... 11}
Sum of k = 66
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