JEE MAIN - Mathematics Hindi (2020 - 6th September Morning Slot - No. 11)
यदि $$\alpha $$ और $$\beta $$ समीकरण के दो मूल हो
x2 – 64x + 256 = 0. तब
$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$ का मान है :
x2 – 64x + 256 = 0. तब
$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$ का मान है :
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