QJEEM19Ma2b41: JEE Main 2019 Mathematics - Class 11 JEEM19Ma2b41 Solution: If the fourth term in the binomial expansion of is equal to…

Fourth term (T₄)

= ${}^6{C_3}{\left( {\sqrt {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} } \right)^3}{\left( {{x^{{1 \over {12}}}}} \right)^3}$

= $20{\left( {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} \right)^{{3 \over 2}}}\left( {{x^{{1 \over 4}}}} \right)$

= $20 \times {x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right){3 \over 2}}} \times {x^{{1 \over 4}}}$

= $20 \times {x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$

Given, T₄ = 200

$\therefore$ $20 \times {x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$ = 200

$\Rightarrow$ ${x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$ = 10

Taking log₁₀ on both sides

$\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right){\log _{10}}x$ = 1

put log₁₀ x = t

$\left( {{3 \over {2\left( {1 + t} \right)}} + {1 \over 4}} \right)t$ = 1

$\Rightarrow$ $\left( {{{\left( {1 + t} \right) + 6} \over {4\left( {1 + t} \right)}}} \right) \times t$ = 1

$\Rightarrow$ t² + 7t = 4 + 4t

$\Rightarrow$ t² + 3t - 4 = 0

$\Rightarrow$ (t + 4)(t - 1) = 0

$\Rightarrow$ t = 1 or t = - 4

$\therefore$ log₁₀ x = 1

$\Rightarrow$ x = 10

or log₁₀ x = - 4

$\Rightarrow$ x = 10^-4

But as x > 1 so x $\ne$ 10^-4

$\therefore$ x = 10