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If $\log_{1/2}(x^2-5x+7)>0$ then exhaustive range of values of x is |
(−∞, 2)∪(3, ∞) (2, 3) (−∞, 1)∪(2, ∞) none of these |
(2, 3) |
$\log_{1/2}(x^2-5x+7)>0⇒x^2-5x+7≤(\frac{1}{2})0$ as $\frac{1}{2}<1$ So $x^2-5x+7<1⇒x^2-5x+6<0$ So $(x-2)(x-3)<0$ $⇒x∈(2, 3)$
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