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Let solution of inequality `(16^((1)/(x)))/2^(x+3)gt1` is `(-oo, a) cup (b, c)` thenA. `a^(2) + b^(2) + 9c^(2) = 25`B. `a + b + c gt 0`C. `a^(2) + b^(2) + c^(2) = 17`D. `a + b + c lt 0` |
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Answer» Correct Answer - A::C::D `(2^((4)/(x)))/(2^(x+3))gt1` `2^((4)/(x)-x-3)gt1` `(4)/(x)-x-3gt0` `(4-x^(2)-3x)/(x)gt0` `(x^(2)+3x-4)/(x)lt0` `((x+4)(x-1))/(x)lt0` `x in (-oo,-4)cup(0,1)` `a=-4, b=0, c=1` |
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