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Quadratic Equation and Inequalities question

2022 · 24 Jun · Shift 2 · Q23
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  5. /2022 · 24 Jun · Shift 2 · Q23

Quadratic Equation and Inequalities question

2022 · 24 Jun · Shift 2 · Q23

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
The sum of all the real roots of the equation (e2x−4)(6e2x−5ex+1)=0({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0(e2x−4)(6e2x−5ex+1)=0 is
  1. A
    log⁡e3{\log _e}3loge​3
  2. B
    −log⁡e3- {\log _e}3−loge​3
  3. C
    log⁡e6{\log _e}6loge​6
  4. D
    −log⁡e6- {\log _e}6−loge​6
View written solutionFree

Correct answer: B

  1. Given equation
(e2x−4)(6e2x−5ex+1)=0(e^{2x}-4)(6e^{2x}-5e^x+1)=0(e2x−4)(6e2x−5ex+1)=0

We need the sum of all real roots.

  1. Use zero product property

So either

e2x−4=0e^{2x}-4=0e2x−4=0

or

6e2x−5ex+1=06e^{2x}-5e^x+1=06e2x−5ex+1=0
  1. Solve the first factor
e2x=4e^{2x}=4e2x=4

Taking natural log:

2x= rac{}{}\ln 4=2\ln 2

Hence,

x=ln⁡2x=\ln 2x=ln2
  1. Solve the second factor

Let

y=exy=e^xy=ex

Since ex>0e^x>0ex>0, we must have y>0y>0y>0.

Then the equation becomes

6y2−5y+1=06y^2-5y+1=06y2−5y+1=0

Factorizing:

6y2−5y+1=(3y−1)(2y−1)=06y^2-5y+1=(3y-1)(2y-1)=06y2−5y+1=(3y−1)(2y−1)=0

So,

y=13ory=12y=\frac13 \quad \text{or} \quad y=\frac12y=31​ory=21​

Since y=exy=e^xy=ex, we get

x=ln⁡(13)=−ln⁡3x=\ln\left(\frac13\right)=-\ln 3x=ln(31​)=−ln3

and

x=ln⁡(12)=−ln⁡2x=\ln\left(\frac12\right)=-\ln 2x=ln(21​)=−ln2
  1. List all real roots

The real roots are:

x=ln⁡2,−ln⁡3,−ln⁡2x=\ln 2,\quad -\ln 3,\quad -\ln 2x=ln2,−ln3,−ln2
  1. Find their sum
ln⁡2+(−ln⁡3)+(−ln⁡2)=−ln⁡3\ln 2+(-\ln 3)+(-\ln 2)=-\ln 3ln2+(−ln3)+(−ln2)=−ln3

So the sum of all real roots is

−log⁡e3-\log_e 3−loge​3
  1. Match with options

This corresponds to Option B.

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