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JEE Main 202131 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Morning Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2021 · 31 August · Morning Shift · Mathematics · Continuity and Differentiability

The function f(x) = | x ^ 2 - 2x - 3| · e ^ 9 x ^ 2 - 12x + 4 is...

The function f(x)=∣x2−2x−3∣⋅e9x2−12x+4f\left(x\right) = \left|{x}^{2} - 2x - 3\right| \cdot {e}^{9{x}^{2} - 12x + 4} is not differentiable at exactly :

Options

  1. AFour points
  2. BTwo points
  3. Cthree points
  4. Done point

Correct answer

B. Two points

Step-by-step solution

Given thatf(x)=∣x2−2x−3∣e(9x2−12x+4)f\left(x\right) = \left|{x}^{2} - 2x - 3\right|{e}^{\left(9{x}^{2} - 12x + 4\right)}

Differentiating on both sides

f′(x)=∣x2−2x−3∣.e(9x2−12x+4).(18x−12)+∣x2−2x−3∣x2−2x−3.(2x−2).e(9x2−12x+4){f}^{'}\left(x\right) = \left|{x}^{2} - 2x - 3\right| . {e}^{\left(9{x}^{2} - 12x + 4\right)} . \left(18x - 12\right) + \dfrac{\left|{x}^{2} - 2x - 3\right|}{{x}^{2} - 2x - 3} . \left(2x - 2\right) . {e}^{\left(9{x}^{2} - 12x + 4\right)}

⇒f′(x)=∣x2−2x−3∣.e(9x2−12x+4){(18x−12)+(2x−2)x2−2x−3}⇒ f ' ( x ) = \left|{x}^{2} - 2x - 3\right| . {e}^{\left(9{x}^{2} - 12x + 4\right)}\left\{\left(18x - 12\right) + \dfrac{\left(2x - 2\right)}{{x}^{2} - 2x - 3}\right\}

⇒f′(x)=∣x2−2x−3∣.e(9x2−12x+4).{18x3−36x2−54x−12x2+24x+36+2x−2}x2−2x−3⇒ f ' \left(x\right) = \left|{x}^{2} - 2x - 3\right| . {e}^{\left(9{x}^{2} - 12x + 4\right)} . \dfrac{\left\{18{x}^{3} - 36{x}^{2} - 54x - 12{x}^{2} + 24x + 36 + 2x - 2\right\}}{{x}^{2} - 2x - 3}

⇒f′(x)=∣x2−2x−3∣.e(9x2−12x+4).{18x3−48x2−28x+34}(x−3)(x+1)⇒ f ' \left(x\right) = \left|{x}^{2} - 2x - 3\right| . {e}^{\left(9{x}^{2} - 12x + 4\right)} . \dfrac{\left\{18{x}^{3} - 48{x}^{2} - 28x + 34\right\}}{\left(x - 3\right)\left(x + 1\right)}

Here f′(x)f ' \left(x\right) is not defined at x=−1 and x=3x = - 1 \text{ and } x = 3.

f(x)f\left(x\right)is not differentiable at two points.

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