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For a real number y, let [y] denotes the greatest integer less than or equal to y : Then the function
By def. [x – π] is an integer whatever be the value of x.
And so p[x – π] is an integral multiple of p.
Consequently tan
And since 1 + [x]2 ≠ 0 for any x, we conclude that f (x) = 0.
Thus f (x) is constant function and so, it is continuous and differentiable any no. of times, that is f '(x), f ''(x), f '''(x),... all exist for every x, their value being 0 at every pt. x. Hence, out of all the alternatives only (d) is correct.
There exist a function f (x), satisfying f (0) = 1, f '(0) = –1, f (x) > 0 for all x, and
f (x) = e^{–x} is one such function.
has the value
If f (a) = 2, f ' (a) = 1 , g (a) = 1 , g ' (a) = 2 , then the value of
The function is not defined at x = 0. The value which should be assigned to f at x = 0 so that it is continuous at x = 0, is
For f (x) to be continuous at x = 0
= a + b
Where [x] denotes the greatest integer less than or equal to x. then
The given function can be restated as
Let f : R → R be a differentiable function and f (1) = 4. Then the value of
We have f : R → R, a differentiable function and f (1) = 4
NOTE THIS STEP
Let [.] den ote th e greatest integer function and f (x) = [tan ^{2}x], then:
We have f (x) = [tan^{2} x]
tan x is an increasing function for
The function denotes the greatest integer function, is discontinuous at
When x is not an integer, both the functions [x] and cos are continuous.
∴f (x) is continuous on all non integral points.
∴ f is continuous at all integral pts as well.
Thus, f is continuous everywhere.
The function f(x) = [x]^{2} – [x^{2}] (where [y] is the greatest integer less than or equal to y), is discontinuous at
We have f (x) = [x]^{2} – [x^{2}]
At x = 0,
f (1) = [1]^{2} – [1^{2}] = 1 – 1 = 0
∴ L.H.L. = R.H.L. = f (1)
∴ f (x) is continuous at x = 1.
Clearly at other integral pts f (x) is not continuous.
The function f (x) = (x^{2}  1)  x^{2}  3x + 2 +cos (x) is NOT differentiable at
∴ Given function can be written as
This function is differentiable at all points except possibly at x = 1 and x = 2.
Lf (2) ≠ Rf '(2)
∴ f is not differentiable at x = 2.
For x ∈ R
The lefthand derivative of f(x) = [x] sin(p x) at x = k, k an integer, is
Let f : R → R be a function defined by f (x) = max {x, x^{3}}. The set of all points where f (x) is NOT differentiable is
KEY CONCEPT
A continuous function f (x) is not differentiable at x = a if graphically it takes a sharp turn at x = a.
Graph of f (x) = max {x, x^{3}} is as shown with solid lines.
From graph of f (x) at x = – 1, 0, 1, we have sharp turns.
∴ f (x) is not differentiable at x = – 1, 0, 1.
Which of the following functions is differentiable at x = 0?
Let us test each of four options.
The domain of the derivative of the function
The given function is
∴ f (x) is discontinuous at x = – 1
Also we can prove in the same way, that f (x) is discontinuous at x = 1
∴ f ' (x) can not be found for x = ± 1 or domain of f ' (x) = R – {– 1, 1}
The integer n for which is a finite nonzero number is
Given that,
For this limit to be finite, n – 3 = 0 ⇒ n = 3
Let f : R → R be such that f (1) = 3 and f '(1) = 6. Then
Given that f ; R → R such that
f (1) = 3 and f ' (1) = 6
where n is nonzero real number, then a is equal to
We are given that
given that f ' (2) = 6 and f '(1) = 4
∴ Applying L Hospital's rule, we get
If (x) is differentiable and strictly increasing function, then the value of
Using L.H. Rule, we get
The function given by y = x – 1 is differentiable for all real numbers except th e points
Graph of y =   x  – 1 is as follows :
The graph has sharp turnings at x = – 1, 0 and 1; and hence not differentiable at x = – 1, 0, 1.
If f (x) is continuous an d differ en tiable function and
Given that f (x) is a contin uous and differ entiable function and
The value of
Let f (x) be differentiable on the interval (0, ∞) such that for each x > 0. Then f (x) is
Given that f (x) is differentiable on (0, ∞) with
[Linear differential equation]
Integrating factor
NOTE THIS STEP
KEY CONCEPT
On applying L' Hospital's rule, we get
m and n are integers, m ≠ 0,n > 0 , and let p be the left hand derivative of x – 1 at x = 1.
As per question,
p = left hand derivative of x –1 at x = 1 ⇒ p = –1
then the value of θ is
For this limit to be finite 1  a = 0 ⇒ a = 1 then given limit reduces to
i s not differ entiable at x = 2.
Let α(a) and β (a) be the roots of the equation where
The given equation is
Dividing both sides by y – 1, we get
Taking limit as y → 1 i.e. a → 0 on both sides we get
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