JEE Main202628 January 2026Morning ShiftMathematicsLimitsActual
The value of _ x 0 _ e ( (e x) (e² x ) (e¹⁰ x ) ) e²-e^ 2 x is equal to
Options
- A(e²⁰-1 ) 2 (e²-1 )
- B(e¹⁰-1 ) 2 (e²-1 )
- C(e¹⁰-1 ) 2 e² (e²-1 )
- D(e²⁰-1 ) 2 e² (e²-1 )
Correct answer
A. (e²⁰-1 ) 2 (e²-1 )
Step-by-step solution
Numerator: _e( (ex) (e^2x) (e¹⁰x)) = _ k=1 ¹⁰ _e( (e^kx)) . Using (u) 1 + u^2 2 for small u , we get _e( (e^kx)) e^ 2k x^2 2 . Thus the numerator is x^2 2 _ k=1 ¹⁰ e^ 2k = x^2 e^2(e²⁰-1) 2(e^2-1) . Denominator: Using x 1 - x^2 2 , we have e^ 2 x e^2(1-x^2) , so e^2 - e^ 2 x e^2 x^2 . Therefore: _ x 0 = x^2 e^2(e²⁰-1) 2(e^2-1) e^2 x^2 = e²⁰-1 2(e^2-1) .