Find all points of discontinuity of f, where f is defined as follows:

OR
Find
, if y = (cos x)x + (sin x)1/x.
The given function is defined at all points of real line. So, let d be any point on the number line.
Case I: When d < -3
For this case, we can define function as,
f(x) = -x + 3, x ≤ -3
So,
At d < -3, f(x) = -x + 3
And f(d) = -d + 3
That is,
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Therefore, the function is continuous at all points x < -3.
Case II: When d = -3
We have the function f(x) = -x + 3, x = -3
So,
At d = -3, f(x) = -x + 3
And f(-3) = -(-3) + 3
⇒ f(-3) = 3 + 3
⇒ f(-3) = 6
For the function to be continuous at x = -3,
Left Hand Limit (LHL) = Right Hand Limit (RHL) = f(-3)
That is, ![]()
For LHL,
We have the function f(x) = -x + 3, x < -3
So,
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For RHL,
We have the function f(x) = -2x, x > -3
So,
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Since, ![]()
Therefore, the function is continuous at x = -3.
Case III: When -3 < d < 3
For this case, we have the function f(x) = -2x, -3 < x < 3
At -3 < d < 3, f(x) = -2x
And f(d) = -2d
That is,
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Therefore, the function is continuous at all points -3 < x < 3.
Case IV: When d = 3
We have the function f(x) = 6x + 2, x = 3
So,
At d = 3, f(x) = 6x + 2
And f(3) = 6(3) + 2
⇒ f(3) = 18 + 2
⇒ f(3) = 20
For the function to be continuous at x = 3,
Left Hand Limit (LHL) = Right Hand Limit (RHL) = f(3)
That is, ![]()
For LHL,
We have the function f(x) = -2x, x < 3
So,
![]()
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For RHL,
We have the function f(x) = 6x + 2, x > 3
So,
![]()
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Since, ![]()
Therefore, the function is discontinuous at x = 3.
Case V: When d > 3
For this case, we have the function f(x) = 6x + 2, x > 3
So,
At d > 3, f(x) = 6x + 2
And f(d) = 6d + 2
That is,
![]()
![]()
Therefore, the function is continuous at all points x > 3.
Thus, the function is discontinuous at x = 3.
OR
We need to find the differentiation of y with respect to x.
Let u = (cos x)x …(i)
Let v = (sin x)1/x …(ii)
For u = (cos x)x
Taking log on both sides, we get
log u = log (cos x)x
⇒ log u = x log (cos x)
Differentiate u with respect to x,
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Substitute value of u from eq (i),
…(iii)
For v = (sin x)1/x
Taking log on both sides, we get
log v = log (sin x)1/x
![]()
Differentiate v with respect to x,
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Substitute value of v from eq (ii),
…(iv)
We know,
y = u + v
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Substituting the value of
and
from eq (iii) and (iv),
![]()
Rearranging it,
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Couldn't generate an explanation.
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