Question

The number of real roots of the equation \[{5+}\left| {{2}^{x}}-1 \right|={{2}^{x}}\left( {{2}^{x}}-2 \right)\] is
A) 2
B) 3
C) 4
D) 1

Answer 1

Hint: In this we will find the roots of \[\text{5+}\left| {{2}^{x}}-1 \right|={{2}^{x}}\left( {{2}^{x}}-2 \right).\] By consider first \[\left| {{2}^{x}}-1 \right|\] positive and then converting the given equation it into quadratic equation and also by taking \[\left| {{2}^{x}}-1 \right|\] negative and then converting the given equation into quadratic equation. By find the roots of converted quadratic we will find the roots of equation \[\text{5+}\left| {{2}^{x}}-1 \right|={{2}^{x}}\left( {{2}^{x}}-2 \right).\]
Consider a quadratic equation \[a{{x}^{2}}+bx+c=0\]. To find the roots of a quadratic equation we will split the coefficient of x i.e. b its sum will be the product a x c the by taking the common term out we will get factorisation of quadratic equation i.e. we will get roots of quadratic equation.

Complete step-by-step answer:
The given equation is,
\[\text{5+}\left| {{2}^{x}}-1 \right|={{2}^{x}}\left( {{2}^{x}}-2 \right).....(1)\]
When \[\left| {{2}^{x}}-1 \right|\] positive
\[\left| {{2}^{x}}-1 \right|={{2}^{x}}-1\]
Equation (1) becomes
\[\text{5+}{{2}^{x}}-1={{2}^{x}}\left( {{2}^{x}}-2 \right)\]
\[\text{4+}{{2}^{x}}={{2}^{2x}}-2\cdot {{2}^{x}}\]
\[{{2}^{2x}}-2\cdot {{2}^{x}}-{{2}^{x}}-4=0\]
\[{{2}^{2x}}-3\cdot {{2}^{x}}-4=0\]
Put \[{{2}^{x}}=t\]
\[{{t}^{2}}-3t-4=0\]
Which is the quadratic equation we can solve by factorisation method i.e. we split the coefficient of t such that their product is the last constant term.
\[\begin{align}
  & {{t}^{2}}-4t+t-4=0 \\
 & t\left( t-4 \right)+1\left( t-4 \right)=0 \\
 & \left( t+1 \right)\left( t-4 \right)=0 \\
\end{align}\]
\[\left( t+1 \right)=0\text{ or }\left( t-4 \right)=0\]
t = -1 or 4
Since, \[t={{2}^{x}}\]
\[{{2}^{x}}=-1\text{ or }{{2}^{x}}=4\]
By taking logarithm, we get
\[\log {{2}^{x}}=\log \left( -1 \right)\text{ or log}{{2}^{x}}=\log 4\]
Since logarithm of negative numbers is not possible.
\[\log {{2}^{x}}\ne \log \left( -1 \right)\]
\[\text{log}{{2}^{x}}=\log 4=\log {{2}^{2}}\]
\[\text{xlog}2=2\log 2\]
By cancelling log2 on both sides, we get x=2.
When \[\left| {{2}^{x}}-1 \right|\] negative
\[\left| {{2}^{x}}-1 \right|=-\left( {{2}^{x}}-1 \right)=-{{2}^{x}}+1\]
Equation (1) becomes
\[\text{5-}{{2}^{x}}+1={{2}^{x}}\left( {{2}^{x}}-2 \right)\]
\[\text{6-}{{2}^{x}}={{2}^{2x}}-2\cdot {{2}^{x}}\]
\[{{2}^{2x}}-2\cdot {{2}^{x}}+{{2}^{x}}-6=0\]
\[{{2}^{2x}}-{{2}^{x}}-6=0\]
Put \[{{2}^{x}}=t\]
\[{{t}^{2}}-t-6=0\]
Which is the quadratic equation we can solve by factorisation method i.e. we split the coefficient of t such that their product is the last constant term.

\[\begin{align}
  & {{t}^{2}}-3t+2t-6=0 \\
 & t\left( t-3 \right)+2\left( t-3 \right)=0 \\
 & \left( t+2 \right)\left( t-3 \right)=0 \\
\end{align}\]
\[\left( t+2 \right)=0\text{ or }\left( t-3 \right)=0\]
t = -2 or 3
Since, \[t={{2}^{x}}\]
\[{{2}^{x}}=-2\text{ or }{{2}^{x}}=3\]
By taking logarithm, we get
\[\log {{2}^{x}}=\log \left( -2 \right)\text{ or log}{{2}^{x}}=\log 3\]
Since logarithm of negative numbers is not possible.
\[\log {{2}^{x}}\ne \log \left( -2 \right)\]
\[\text{log}{{2}^{x}}=\log 3\]
\[\text{xlog}2=\log 3\]
\[\text{x}=\dfrac{\log 3}{\text{log}2}\]
Hence there are two real root of equation \[\text{5+}\left| {{2}^{x}}-1 \right|={{2}^{x}}\left( {{2}^{x}}-2 \right).\]

So, the correct answer is “Option A”.

Note: In this problem, students should note that logarithm of negative numbers is not possible. Alternative method to find roots of quadratic equations. Consider the quadratic equation \[a{{x}^{2}}+bx+c=0\]. Then roots of the equation can be find by following formula:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.