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Let f (x) = 4 and f ' (x) = 4. Then given by
Apply L H Rule
denotes greatest integer less than or equal to x)
Since does not exist, hence the required limit
does not exist.
If f(1) = 1, f ^{1} (1) = 2, then
form using L’ Hospital’s rule
f is defined in [5, 5] as f(x) = x if x is rational = –x if x is irrational. Then
Let a is a rational number other than 0, in [–5, 5], then
[As in the immediate neighbourhood of a rational number, we find irrational numbers]
∴ f (x) is not continuous at any rational number
If a is irrational number, then
∴ f (x) is not continuous at any irrational number clearly
∴ f (x) is continuous at x = 0
f(x) and g(x) are two differentiable functions on [0, 2] such that f ''( x)  g '' (x)= 0, f ' (1) =2g'(1)=4 f(2) =3g(2) = 9 then f(x)–g(x) at x = 3/2 is
and f (5) = 2,f '(0)= 3 , then f ' (5) is
f (x + y) = f (x) * f (y)
Differentiate with respect to x, treating y as constant
f ' (x + y) = f ' (x) f (y)
Putting x = 0 and y = x, we get f '(x)= f '(0) f (x) ;
⇒ f ' (5) = 3 f (5) = 3 × 2 = 6.
The given expression can be written as
the value of k is
The value of
Let f (a) = g (a) = k and their nth derivatives
f^{n} (a) , g^{n} (a) exist and are not equal for some n. Further if
then the value of k is
then the values of a and b, are
is continuous
equals
⇒ Given limit is equal to value of integral
Let a and b be the distinct roots of ax^{2} + bx +c = 0 , then
Suppose f(x) is differentiable at x = 1 and
As function is differentiable so it is continuous as it is given that and hence f (1) = 0
Let f be differentiable for all x. If f (1) = – 2 and f '(x) > 2 for x ∈ [1, 6], then
Applying Lagrange’s mean value theorem
If f is a real valued differentiable function satisfying
f (x) – f (y)  < ( x y)^{2} , x, y ∈ R and f (0) = 0, then f (1) equals
Let f : R → R be a function defined by f (x) = min {x + 1,x+ 1} ,Then which of the following is true ?
Hence, f (x) is differentiable everywhere for all x ∈ R.
The function f : R /{0} → R given by
can be made continuous at x = 0 by defining f (0) as
∴ using, L'Hospital rule
Then which one of the following is true?
Let this finite number be l
∴ f is not differentiable at x = 1
Let f : R →R be a positive incr easing function with
f(x) is a positive increasing function
By Sandwich Theorem.
The values of p and q for which the function
Let f : R → [0, ∞) be such that exists and
If f : R → R is a function defined by f (x) = [x] where [x] denotes the greatest integer function, then f is .
Consider the function, f (x) = x – 2+ x – 5, x ∈ R.
Statement1 : f '(4) = 0
Statement2 : f is continuous in [2,5], differentiable in (2,5) and f (2) = f (5).
∴ statement2 is also true and a correct explanation for statement 1.
Multiply and divide by x in the given expression, we get
Multiply and divide by x in the given expression, we get
If the function
is differentiable, then the value of k + m is :
Since g (x) is differentiable, it will be continuous at x = 3
g (x) = f (f (x))
In the neighbourhood of x = 0,
(x) =  log2 – sin x = (log 2 – sin x)
∴ g (x) = log 2 – sin log 2 – sin x  = (log 2 – sin(log 2 – sin x))
∴ g (x) is differentiable and g'(x) = – cos(log 2 – sin x) (– cos x)
⇒ g'(0) = cos (log 2)
then log p is equal to :
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