If domain of function 3 y + 2 x 4 = 2 4 x 2 − 1 can be expressed as :
( − 2 b 2 a + 1 , − 2 c − 3 d ) U ( e − f , h g + 1 )
Find a × b × c × d × e × f × g × h
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:( I was so close just didnt multiply h. I pressed 108 . I uaed wavy curve method :P and was a bit astonished to see why there were 8 variables XD. I know that you intended it right?
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The Question is easy , but i made it tough :P by adding 8 variables to it
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we have 3 y + 2 x 4 = 2 4 x 2 − 1 ⟹ 3 y = 2 4 x 2 − 1 − 2 x 4 , since 3 y is always positive 2 4 x 2 − 1 − 2 x 4 > 0
\color{#333333}{= 2^{{4x^2-1}}>2^{x^4} \implies 4x^2-1>x^4\\ = x^4-4x^2+1<0 \\ = (x^2-[2+\sqrt{3}\text{ ] })(x^2-[2-\sqrt{3}\text{ ] })<0 \\ (x-\sqrt{2+\sqrt{3}})(x+\sqrt{2+\sqrt{3}})(x-\sqrt{2-\sqrt{3}} )(x+\sqrt{2-\sqrt{3}})<0 \\ \text { Domain }\implies \bigg{(}-\sqrt{2+\sqrt{3}}\text{ , }-\sqrt{2-\sqrt{3}}\bigg{)} \text { U }\bigg{(}\sqrt{2-\sqrt{3}} \text{ , }\sqrt{2+\sqrt{3}}\bigg{)} \\ \text{We can simplify } \sqrt{2+\sqrt{3}} \\ \implies \sqrt{ \dfrac{4+2\sqrt{3}}{2} } \implies \sqrt{ \dfrac{3+1+2\sqrt{3}}{2} }\implies \dfrac{\sqrt{3}+1}{\sqrt{2}} \\\text{In comparison }a=6,b=1,c=1,d=1,e=2,f=3,g=3,h=2 \\ a\times{b}\times{c}\times{d}\times{e}\times{f}\times{g}\times{h}=\boxed{216} }