Sigma Percentile
JEE Main 2020 - 9 Jan (Morning)
LEVELJEE Main

Animated Solution for Mathematics - Quadratic Equations: Find number of real roots of equation is

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Visualized Solution

The Exponential Equation

  • Given equation:
  • We need to find the number of real roots, meaning values of where the function equals zero.

First Substitution:

  • Let
  • Since for all real , we must have .

Transforming to a Polynomial

  • Substituting into the equation:

Creating Symmetry

  • Divide the entire equation by (valid since ):

Grouping Terms

  • Rearrange and group the terms with equal powers:

Second Substitution

  • Let
  • Squaring both sides:
  • Therefore,

The Quadratic Equation in

  • Substitute and into the grouped equation:

Solving the Quadratic

  • Factorize the quadratic:
  • Possible values: or

Applying AM-GM Inequality

  • Recall and .
  • Apply the AM-GM inequality:
  • Reject . Accept .

Finding

  • Substitute back into :
  • Multiply by :

Finding the Root

  • Back-substitute :
  • Taking natural log on both sides:

Final Conclusion

  • The only real root is .
  • Number of real roots = 1
  • Key Takeaway: Always check domain constraints when substituting (e.g., and ).

The Sigma Insight: Transformation of Equations

Solution Diagram

Analyzing the Setup

The given equation is:
This is a quartic equation disguised in exponential form. To simplify the landscape, we let .
Since is always positive, we must respect the domain constraint . The equation transforms into the following polynomial:
This is a reciprocal equation because the coefficients mirror each other perfectly.

The Reduction Strategy

To exploit this symmetry, we divide the entire equation by . Since , we know $t^2 eq 0$, making this operation valid:
Next, we group the terms by their powers:

The Bridge of

Let . Squaring this expression yields:
Substituting this into our grouped equation, the quartic expression reduces to a quadratic:
Factoring the quadratic gives:
This provides two potential candidates: and .

Final Calculation and Verdict

We apply the AM-GM inequality to the substitution . For any :
The candidate fails this condition and is therefore rejected. We proceed with :
This yields . Back-substituting , we get , which implies:
We have successfully navigated the fortress to find the single, elegant solution.

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