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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
The function $f(x) = ({x^2} - 1)|{x^2} - 3x + 2| + \cos (|x|)$ is not differentiable at
  • A
    $-1$
  • B
    $0$
  • C
    $1$
  • ✓
    $2$
✓

Answer

Correct option: D.
$2$
d
(d) Since function $|x|$ is not differentiable at $x = 0$

$\therefore \,|{x^2} - 3x + 2| = |(x - 1)(x - 2)|$

Hence is not differentiable at $x = 1$ and $2$

Now $f(x) = ({x^2} - 1)|{x^2} - 3x + 2|\cos (|x|)$ is not differentiable at $x = 2$

For $1 < x < 2$, $f(x) = - ({x^2} - 1)({x^2} - 3x + 2) + \cos x$

For $2 < x < 3$, $f(x) = + ({x^2} - 1)({x^2} - 3x + 2) + \cos x$

$Lf'(x) = - ({x^2} - 1)(2x - 3) - 2x({x^2} - 3x + 2) - \sin x$

$Lf'(2) = - 3 - \sin 2$

$Rf'(x) = ({x^2} - 1)(2x - 3) + 2x({x^2} - 3x + 2) - \sin x$

$Rf'(2) = (4 - 1)(4 - 3) + 0 - \sin 2 = 3 - \sin 2$

Hence $Lf'(2) \ne Rf'(2)$.

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