Verified Solution • 2025–26 Session moderate

QJEEM21Mae536: JEE Main 2021 Mathematics - Class 12 JEEM21Mae536 Solution: If f ( x ) = { ∫ 0 x ( 5 +

If f(x)={∫0x(5+|1−t|)dt,x>25x+1,x≤2, then

Multiple Choice Options

  • A f(x) is not continuous at x = 2
  • B f(x) is everywhere differentiable
  • C f(x) is continuous but not differentiable at x = 2
  • D f(x) is not differentiable at x = 1

Step-by-Step Verified Solution

f(x)=∫01(5+(1−t))dt+∫1x(5+(t−1))dtf(x) = \int\limits_0^1 {(5 + (1 - t))dt + \int\limits_1^x {(5 + (t - 1))dt} }

=6−12+(4t+t22)∣1x= \left. {6 - {1 \over 2} + \left( {4t + {{{t^2}} \over 2}} \right)} \right|_1^x

=112+4x+x22−4−12= {{11} \over 2} + 4x + {{{x^2}} \over 2} - 4 - {1 \over 2}

=x22+4x−1= {{{x^2}} \over 2} + 4x - 1

f(2+)=2+8+1=11f({2^ + }) = 2 + 8 + 1 = 11

f(2)=f(2−)=5×2+1=11f(2) = f({2^ - }) = 5 \times 2 + 1 = 11

⇒\Rightarrow continuous at x = 2

Clearly differentiable at x = 1

Lf' (2) = 5

Rf' (2) = 6

⇒\Rightarrow Not differentiable at x = 2

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