LimitsHard
Question
The function f(x) = (x2 - 1) | x2 - 3x + 2 | + cos (| x |) is not differentiable at
Options
A.-1
B.0
C.1
D.2
Solution
Function
f(x) = (x2 - 1) | x2 - 3x + 2| + cos (| x |) ....... (i)
Imp Note : In differentiable of | f(x) | we have to
consider critical points for which f(x) = 0
|x| is not differentiablity at x = 0
but cos |x| =
⇒ cos |x| =
Therefore it is differentiable at x = 0
Next, |x2 - 3x + 2| = |(x - 1) (x - 2)|
Therefore,
f(x) =
Now, x = 1, 2 are critical point for differentiability
Because f(x) is differentiable on other points in its domain
Differentiability at x = 1
L f′(1) =
= 0 - sin 1 = - sin 1
∴
(cos x)
at x = 1 - 0
= - sin x at x = 1 - 0
= - sin x at x = 1
= - sin1
and Rf′ (1) =
= 0 - sin 1 = - sin 1 (same approach)
∵ Lf′(1) = Rf′(1).
Therefore, function is differentiable
at x = 1.
Again Lf′(2) =
= - (4 - 1) (2 - 1) - sin 2 = - 3 - sin 2
and R f′(2) =
= (22 - 1) - sin 2 = 3 - sin 2
So L f′(2) ≠ R f′(2), f is not differentiable at x = 2
Therefore, (d) is the answer.
f(x) = (x2 - 1) | x2 - 3x + 2| + cos (| x |) ....... (i)
Imp Note : In differentiable of | f(x) | we have to
consider critical points for which f(x) = 0
|x| is not differentiablity at x = 0
but cos |x| =
⇒ cos |x| =
Therefore it is differentiable at x = 0
Next, |x2 - 3x + 2| = |(x - 1) (x - 2)|
Therefore,
f(x) =
Now, x = 1, 2 are critical point for differentiability
Because f(x) is differentiable on other points in its domain
Differentiability at x = 1
L f′(1) =
= 0 - sin 1 = - sin 1
∴
(cos x) at x = 1 - 0
= - sin x at x = 1 - 0
= - sin x at x = 1
= - sin1
and Rf′ (1) =
= 0 - sin 1 = - sin 1 (same approach)
∵ Lf′(1) = Rf′(1).
Therefore, function is differentiable
at x = 1.
Again Lf′(2) =
= - (4 - 1) (2 - 1) - sin 2 = - 3 - sin 2
and R f′(2) =
= (22 - 1) - sin 2 = 3 - sin 2
So L f′(2) ≠ R f′(2), f is not differentiable at x = 2
Therefore, (d) is the answer.
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