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QJEEM23Maedc7: JEE Main 2023 Mathematics - Class 11 JEEM23Maedc7 Solution: Let the sixth term in the binomial expansion of ( 2 log 2 ( 10…

Let the sixth term in the binomial expansion of (2log2(10−3x)+2(x−2)log235)m in the increasing powers of 2(x−2)log⁡232^{(x-2) \log _{2} 3} , be 21 . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of xx is __________.

Step-by-Step Verified Solution

mC1,mC2,mC3{ }^m C_1,{ }^m C_2,{ }^m C_3 are first, third and fifth term of APA P

∴amp;a=mC1amp;a+2d=mC2amp;a+4d=mC3∴amp;2mC2−mC3=m⇒amp;m=7 or m=2∵amp;m=2 is not possible ∴amp;m=7\begin{aligned}\therefore \quad & a={ }^m C_1 \\\\& a+2 d={ }^m C_2 \\\\& a+4 d={ }^m C_3 \\\\\therefore \quad & 2{ }^m C_2-{ }^m C_3=m \\\\\Rightarrow & m=7 \text { or } m=2 \\\\\because & m=2 \text { is not possible } \\\\\therefore & m=7\end{aligned}

T6=mC5(10−3x)m−52⋅(3x−2)=21\mathrm{T}_6={ }^{\mathrm{m}} \mathrm{C}_5\left(10-3^{\mathrm{x}}\right)^{\frac{\mathrm{m}-5}{2}} \cdot\left(3^{\mathrm{x}-2}\right)=21

Putting value of m = 7, we get

amp;T5+1=7C5(10−3x)7−523x−2=21amp;⇒10.3x−(3x)232=1\begin{aligned} & T_{5+1}={ }^7 C_5\left(10-3^x\right)^{\frac{7-5}{2}} 3^{x-2}=21 \\\\ & \Rightarrow \frac{10.3^x-\left(3^x\right)^2}{3^2}=1\end{aligned}

amp;⇒(3x)2−10⋅3x+9=0amp;⇒3x=9,1amp;⇒x=0,2\begin{aligned}& \Rightarrow \left(3^x\right)^2-10 \cdot 3^x+9=0 \\\\& \Rightarrow 3^x=9,1 \\\\& \Rightarrow x=0,2\end{aligned}

Sum of squares of values of x = 02 + 22 = 4

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