Sigma Percentile
JEE Main 2022 (29 July Shift 2)
LEVELJEE Main

Animated Solution for Mathematics - Differentiation: If the tangent to the curve at the point is also tangent to the curve at the point , then is equal to ________

Enter Numerical Value:

Visualized Solution

Visualize the Geometry

  • Given curves:
  • 1. at
  • 2. at
  • The tangent line is common to both points.

Slope of the Parabola at

  • Differentiate the parabola equation :

Calculate Numerical Slope

  • Substitute to find the slope :

Equation of the Tangent Line

  • Using point-slope form:
  • Substitute and :

Slope of the Cubic Curve at

  • Differentiate the cubic curve :
  • At point , the slope must be :

Solve for

  • Rearrange the quadratic equation:
  • Factorize by splitting the middle term:
  • Possible values: or

Verify Case 1:

  • Case 1:
  • Find from cubic curve:
  • Check if lies on the tangent :
  • RHS:
  • LHS = RHS. So, is valid.

Verify Case 2:

  • Case 2:
  • Find :
  • Check on line :
  • Case 2 is rejected.

Final Calculation

  • We have and .
  • Calculate :

Conclusion & Key Takeaway

  • Final Answer: 195
  • Key Takeaways:
  • 1. Tangency requires both slope equality () and point satisfaction ().
  • 2. Always verify multiple algebraic roots against the specific line equation.

The Sigma Insight: Tangents, Normals and Rate Measure

Solution Diagram

Analyzing the Setup

Welcome, future engineer. Today, we embark on a journey through the elegant world of calculus. We are not just solving a problem; we are uncovering the hidden geometry that connects two seemingly unrelated curves.
Imagine you are standing on a graph, looking at two distinct paths. One is a cubic curve, , and the other is a parabola, . The problem states that a single, straight line acts as a bridge, grazing both of these curves as a common tangent.

Phase 1

The Parabola (The Anchor)
We start with the parabola because we have a solid anchor point: . To find the slope of the tangent line here, we calculate the derivative:
At , the slope is . We have now locked down the steepness of our line.
Using the point-slope form , we substitute our values: . Expanding this, we obtain the equation of our common tangent line:

Phase 2

The Cubic Curve (The Challenge)
Now, we shift our focus to the cubic curve . We know this same line is tangent to the cubic curve at some unknown point .
If we differentiate the cubic equation, we get:
At the point where , this derivative must perfectly match the slope of our tangent line, which we know is . Setting these equal, we get , which simplifies to the quadratic equation:

Phase 3

The Trap of the Extraneous Root
Solving by splitting the middle term, we get . This yields two candidates: and .
For , the -coordinate is . We check if the point lies on our tangent line :
It matches perfectly. Now, for , the -coordinate is . Checking this on the line, we find it does not satisfy the equation, meaning the tangent at is parallel to our line but is not the same line. We reject this case.

The Final Celebration

With and confirmed, the final step is a simple calculation:
We have arrived at the final answer, 195. This problem demonstrates that in the JEE, the math is only half the battle; the other half is the geometric intuition and the rigor to verify every step.

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