JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 22)
If the fourth term in the binomial expansion of $${\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}$$
(x > 0) is 20 × 87, then a value of
x is :
8–2
82
83
8
Explanation
$${\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}$$
Given T4 = 20 × 87
$$ \Rightarrow $$ $${}^6{C_3}{\left( {{2 \over x}} \right)^3}{\left( {{x^{{{\log }_8}x}}} \right)^3}$$ = 20 × 87
$$ \Rightarrow $$ 20$$ \times $$$${8 \over {{x^3}}} \times {x^{3{{\log }_8}x}}$$ = 20 × 87
$$ \Rightarrow $$ $${x^{3{{\log }_8}x - 3}}$$ = 86
Taking $${{{\log }_8}}$$ both side
$$ \Rightarrow $$ ($${3{{\log }_8}x - 3}$$) $$ \times $$ $${{{\log }_8}x}$$ = 6
$$ \Rightarrow $$ $$3{\left( {{{\log }_8}x} \right)^2}$$ - 3$${{{\log }_8}x}$$ = 6
$$ \Rightarrow $$ $${\left( {{{\log }_8}x} \right)^2}$$ - $${{{\log }_8}x}$$ = 2
$$ \Rightarrow $$ ($${{{\log }_8}x}$$ - 2)($${{{\log }_8}x}$$ + 1) = 0
$$ \Rightarrow $$ $${{{\log }_8}x}$$ = 2 or $${{{\log }_8}x}$$ = -1
$$ \Rightarrow $$ x = 82 or x = $${1 \over 8}$$
Given T4 = 20 × 87
$$ \Rightarrow $$ $${}^6{C_3}{\left( {{2 \over x}} \right)^3}{\left( {{x^{{{\log }_8}x}}} \right)^3}$$ = 20 × 87
$$ \Rightarrow $$ 20$$ \times $$$${8 \over {{x^3}}} \times {x^{3{{\log }_8}x}}$$ = 20 × 87
$$ \Rightarrow $$ $${x^{3{{\log }_8}x - 3}}$$ = 86
Taking $${{{\log }_8}}$$ both side
$$ \Rightarrow $$ ($${3{{\log }_8}x - 3}$$) $$ \times $$ $${{{\log }_8}x}$$ = 6
$$ \Rightarrow $$ $$3{\left( {{{\log }_8}x} \right)^2}$$ - 3$${{{\log }_8}x}$$ = 6
$$ \Rightarrow $$ $${\left( {{{\log }_8}x} \right)^2}$$ - $${{{\log }_8}x}$$ = 2
$$ \Rightarrow $$ ($${{{\log }_8}x}$$ - 2)($${{{\log }_8}x}$$ + 1) = 0
$$ \Rightarrow $$ $${{{\log }_8}x}$$ = 2 or $${{{\log }_8}x}$$ = -1
$$ \Rightarrow $$ x = 82 or x = $${1 \over 8}$$
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