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QJEEM21Maa8a5: JEE Main 2021 Mathematics - Class 11 JEEM21Maa8a5 Solution: The sum of all integral values of k (k 0) for which the…

The sum of all integral values of k (k ≠\ne 0) for which the equation 2x−1−1x−2=2k{2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k} in x has no real roots, is ____________.

Step-by-Step Verified Solution

2x−1−1x−2=2k{2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k}

x∈R−{1,2}x \in R - \{ 1,2\}

⇒k(2x−4−x+1)=2(x2−3x+2)\Rightarrow k(2x - 4 - x + 1) = 2({x^2} - 3x + 2)

⇒k(x−3)=2(x2−3x+2)\Rightarrow k(x - 3) = 2({x^2} - 3x + 2)

for x ≠\ne 3, k=2(x−3+2x−3+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

⇒2(x−3+2x−3+3)∈(−∞,6−42]∪[6+42,∞)\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∈(6−42,6+42)−{0}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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