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Limits and Derivatives · Question 6

JEE Main 2026 (Online) 28th January Morning ShiftMCQMedium📈 Topic in 32% papers+4−1
1

The value of x 0 e ( (ex) (e2x) (e10x) )e2 - e2 x is equal to

  1. (e10 - 1)2e2(e2 - 1)
  2. (e20 - 1)2e2(e2 - 1)
  3. (e10 - 1)2(e2 - 1)
  4. (e20 - 1)2(e2 - 1)

Answer

(4) (e20 - 1)2(e2 - 1)

Solution

Numerator N = k=110 e (ek x) = - k=110 e (ek x)

0 - e 2 = 0 1 - 2 = 12

x 0 Nx2 = 12 k=110 e2k = 12 e2(e20 - 1)e2 - 1

Denominator = e2 - e2 x = -e2 (e2 x - 2 - 1 )

2 x - 2 = -4 2 x2 0 and et - 1t 1

x 0 e2 - e2 xx2 = -e2 x 0 2 x - 2x2 = -e2(-1) = e2

Limit = e2(e20 - 1)2(e2 - 1) 1e2

= (e20 - 1)2(e2 - 1)

Answer: (4)

JEE Main 2026 (Online) 28th January Morning Shift

Keep practising — Limits and Derivatives

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If the function f(x) = ex (e x - x - 1 ) + e( x + x) - x x - x is continuous at x = 0, then the value of f(0) is equal to

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  3. 2
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3

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  2. 5
  3. 9
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4

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  2. 18
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  3. 25
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6

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  2. 1
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8

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  1. 20
  2. 19
  3. 18
  4. 17
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9

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  1. 3
  2. 1
  3. 2
  4. 0
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  1. -1
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