JEE Main 2021 — Mathematics Algebra

If three numbers $a,b,c$a,b,c are in arithmetic progression, then $2b=a+c.$2b=a+c.

Given $1,{\mathrm{log}}_{10}({4}^{x}-2),{\mathrm{log}}_{10}({4}^{x}+\frac{18}{5})$1,log10​(4x−2),log10​(4x+518​) are in arithmetic progression, then we have

$$2{\mathrm{log}}_{10}({4}^{x}-2)=1+{\mathrm{log}}_{10}({4}^{x}+\frac{18}{5})$$2log10​(4x−2)=1+log10​(4x+518​)

Using ${\mathrm{log}}_{10}{m}^{n}=n{\mathrm{log}}_{10}m,$log10​mn=nlog10​m, we get

$${\mathrm{log}}_{10}{({4}^{x}-2)}^{2}={\mathrm{log}}_{10}(10)+{\mathrm{log}}_{10}({4}^{x}+\frac{18}{5})$$log10​(4x−2)2=log10​(10)+log10​(4x+518​)

Using ${\mathrm{log}}_{10}(m)+{\mathrm{log}}_{10}(n)={\mathrm{log}}_{10}(mn),$log10​(m)+log10​(n)=log10​(mn), we get

$${\mathrm{log}}_{10}{({4}^{x}-2)}^{2}={\mathrm{log}}_{10}(10({4}^{x}+\frac{18}{5}))$$log10​(4x−2)2=log10​(10(4x+518​))

$$\Rightarrow {({4}^{x}-2)}^{2}=10({4}^{x}+\frac{18}{5})$$⇒(4x−2)2=10(4x+518​)

$$\Rightarrow {({4}^{x})}^{2}+4-4\cdot {4}^{x}=10\cdot {4}^{x}+36$$⇒(4x)2+4−4⋅4x=10⋅4x+36

$$\Rightarrow {({4}^{x})}^{2}-14({4}^{x})-32=0$$⇒(4x)2−14(4x)−32=0

$$\Rightarrow ({4}^{x}-16)({4}^{x}+2)=0$$⇒(4x−16)(4x+2)=0

$\Rightarrow {4}^{x}=16$⇒4x=16 as ${4}^{x}+2\neq 0$4x+2=0

$$\Rightarrow x=2$$⇒x=2

Now, $|\begin{matrix}2(x-\frac{1}{2}) & x-1 & {x}^{2} \\ 1 & 0 & x \\ x & 1 & 0\end{matrix}|$∣2(x−21​)1x​x−101​x2x0​∣

$$=|\begin{matrix}3 & 1 & 4 \\ 1 & 0 & 2 \\ 2 & 1 & 0\end{matrix}|$$=∣312​101​420​∣

$$=3(0-2)-1(0-4)+4(1-0)$$=3(0−2)−1(0−4)+4(1−0)

$$=-6+4+4=2.$$=−6+4+4=2.