JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 11)
If $$\alpha $$ and $$\beta $$ be two roots of the equation
x2 – 64x + 256 = 0. Then the value of
$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$ is :
x2 – 64x + 256 = 0. Then the value of
$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$ is :
1
3
2
4
Explanation
x2 – 64x + 256 = 0
$$\alpha $$ + $$\beta $$ = 64, $$\alpha $$$$\beta $$ = 256
$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$
= $${{{\alpha ^{{3 \over 8}}}} \over {{\beta ^{{5 \over 8}}}}} + {{{\beta ^{{3 \over 8}}}} \over {{\alpha ^{{5 \over 8}}}}}$$
= $${{\alpha + \beta } \over {{{\left( {\alpha \beta } \right)}^{{5 \over 8}}}}}$$
= $${{64} \over {{{\left( {256} \right)}^{{5 \over 8}}}}}$$
= 2
$$\alpha $$ + $$\beta $$ = 64, $$\alpha $$$$\beta $$ = 256
$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$
= $${{{\alpha ^{{3 \over 8}}}} \over {{\beta ^{{5 \over 8}}}}} + {{{\beta ^{{3 \over 8}}}} \over {{\alpha ^{{5 \over 8}}}}}$$
= $${{\alpha + \beta } \over {{{\left( {\alpha \beta } \right)}^{{5 \over 8}}}}}$$
= $${{64} \over {{{\left( {256} \right)}^{{5 \over 8}}}}}$$
= 2
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