Continuity and DifferentiabilityHard
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)
Note In differentiable of | f(x) | we have to consider critical points for which f(x) = 0
| x | is not differentiable 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 - sin1 = - sin1
{∵
(cos x)at x = 1 - 0
= - sin x at x = 1 - 0
= - sin x at x = 1
= - sin1}
and R f′(1) =
= 0 - sin1 = - sin1 (same approach )
∵ L f′(1) = Rf′(1) Therefore, function is differentiable at x = 1
Again, Lf′(2) =
= -(4 -1)(2 - 1) - sin 2 = - 3 - sin 2
and Rf′(2)
= (22 - 1)(2 - 1) - sin 2 = 3 - sin 2
So, Lf′(2) ≠ Rf′(2),f is not differentiable at x = 2
Therefore, (d) is the answer.
Note In differentiable of | f(x) | we have to consider critical points for which f(x) = 0
| x | is not differentiable 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 - sin1 = - sin1
{∵
(cos x)at x = 1 - 0= - sin x at x = 1 - 0
= - sin x at x = 1
= - sin1}
and R f′(1) =
= 0 - sin1 = - sin1 (same approach )
∵ L f′(1) = Rf′(1) Therefore, function is differentiable at x = 1
Again, Lf′(2) =
= -(4 -1)(2 - 1) - sin 2 = - 3 - sin 2
and Rf′(2)
= (22 - 1)(2 - 1) - sin 2 = 3 - sin 2
So, Lf′(2) ≠ Rf′(2),f is not differentiable at x = 2
Therefore, (d) is the answer.
Create a free account to view solution
View Solution Free