Basics & LogarithmsmediumPYQ · JEE Main · 24 Jun 2022 · Shift 2 (Afternoon)Free

Basics & Logarithms: Sum Real Roots Equation (JEE Main 2022)

JEE Maths question with a full step-by-step solution.

Question
The sum of all the real roots of the equation (e2x−4)(6e2x−5ex+1)=0(e^{2x}-4)(6e^{2x}-5e^{x}+1)=0 is
Alog⁡e3\log_{e}3
B−log⁡e3-\log_{e}3correct
Clog⁡e6\log_{e}6
D−log⁡e6-\log_{e}6
Solution
Step 1: A product is zero when a factor is zero. So either e2x−4=0e^{2x}-4=0 or 6e2x−5ex+1=06e^{2x}-5e^{x}+1=0. Step 2: First factor: e2x=4⇒ex=2e^{2x}=4\Rightarrow e^{x}=2 (taking positive root since ex>0e^{x}>0). So x=ln⁡2x=\ln 2. Step 3: Second factor: Let u=ex>0u=e^{x}>0:
6u2−5u+1=06u^{2}-5u+1=0
 Step 4: Factor:
(3u−1)(2u−1)=0  ⇒  u=13 or u=12(3u-1)(2u-1)=0\;\Rightarrow\;u=\dfrac{1}{3}\text{ or }u=\dfrac{1}{2}
 Step 5: Solve for xx:
ex=13  ⇒  x=−ln⁡3e^{x}=\dfrac{1}{3}\;\Rightarrow\;x=-\ln 3
ex=12  ⇒  x=−ln⁡2e^{x}=\dfrac{1}{2}\;\Rightarrow\;x=-\ln 2
 Step 6: Sum of all real roots:
ln⁡2+(−ln⁡3)+(−ln⁡2)=−ln⁡3\ln 2+(-\ln 3)+(-\ln 2)=-\ln 3
 Answer: (2) −log⁡e3-\log_{e}3
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