Q45 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

If fx=e2x2+x: x>0ax+b: x≤0 is differentiable at x=0, then

  • A
    a=1,b=-1
  • B
    a=-1,b=1
  • Correct answer
    C
    a=1,b=1
  • D
    a=-1,b=-1

Solution

f ′ 0 − = lim h→ 0 + f 0−h −f 0 −h ⇒ lim h→ 0 + f −ah+b − b −h =a

f'0+=limh→0+f0+h−f0h⇒limh→0+e2h2+h−bh  ... 1

For limit to exists, it must be 00 form

∴e20+0-b=0⇒b=1 ... 2

From 1, f'0+=limh→0+e2h2+h−1h

=limh→0+e2h2+h−12h2+h×2h2+hh=1×1=1 ... 3

∵ fx is differentiable at x=0

∴ f'0−=f'0+

⇒a=1

Hence, the correct answer is option: 'C'.

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