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  • If the  term in the expansion of the binomial <img alt="\left[\sqrt{2^{\log \left(10-3^x\right)}}+\sqrt[5

If the 6^{th} term in the expansion of the binomial \left[\sqrt{2^{\log \left(10-3^x\right)}}+\sqrt[5]{2^{(x-2) \log 3}}\right]^m is equal to 21 and it is known that the binomial coefficients of the 2nd, 3rd and 4th terms in the expansion represent respectively the first, third and fifth terms of an A.P. (the symbol log stands for logarithm to the base 10), then value of x will be 

Option: 1

4


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

Since coefficients { }^m C_1,{ }^m C_2 and { }^m C_3 of T_2, T_3, T_4 i.e. are the first, third and fifth terms of an A. P., which will also be in A. P. of common difference 2d.

Hence 2^m C_2={ }^m C_1+{ }^m C_3 \Rightarrow(m-2)(m-7)=0

Since 6^{th} term is 21, m = 2 is ruled out and we have m = 7 and 

\begin{aligned} & T_6=21={ }^7 C_5\left[\sqrt{2^{\log \left(10-3^x\right)}}\right]^{7-5} \times\left[\sqrt[5]{2^{(x-2)} \log 3}\right]^5 \\ & \Rightarrow 21=21.2^{\log \left(10-3^x\right)+\log 3^{x-2}} \\ & \Rightarrow 2^{\log \left[\left(10-3^x\right) 3^{x-2}\right]}=1=2^0 \end{aligned}

Thus, \log \left[\left(10-3^x\right) 3^{x-2}\right] = 0

Both x = 0 and x=2 will satisfy the equation

 

Posted by

Kshitij

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