Convert the set x in roster form if set x contains the positive prime number, which divides 72.
2 and 3 are the divisors of 72, which are prime. So, the roster form of set x is (2, 3}.
Let f(x + y) = f(x)f(y) for all x and y and f(5) =  2 and f'(0) = 3. What is the value of f'(5)?
f'(0) = 3
implies
or
[Since f(5 + 0) = f(5) • f(0)
implies f(5) = f(5)  f(0)
implies f(0) = 1]
Now, we have
[by eq. (1)]
Hence, f'(5) = 6
Let l = { 1 } ∪ { 2 } For x ∈ R, let φ(x) = dis (x, l) = ln f{ x —y  : y ∈ l} then
Given that,
= ln f{x .y : y ∈ 1}
where l = { l } ∪ { 2 }
= ln f{ x l , x 2  }
the graph of the φ(x) is given by.
Clearly, the graph of the function have sharp edges at x = 1,3/2 and 2. Therefore, f(x) is not differentiable at x
= 1, 3/2 and 2.
Hence, φ(x) is continuous on R but not
differentiable at x = 1, 3/2 and 2
Let f(x) =
then f is
At x =1/2
Since
therefore, f(x) is not continuous x =1/2
If the function defined as
Then, the function f (x) has
So, the function is not continuous at the point x = 1. The discontinuity is of the first kind and can be removed by defining function as f (x) = – 1.
Let
then,
Let
then,
Let
then,
Let
then,
Let
then,
Consider
then,
If we expand sinx by Taylor’s series about then a_{1, }a_{7, }a_{4, }a_{3} are
Given function f(x) = sin x
the value of a_{2}, a_{7} a_{4}, a_{3} in (i) are
If f(x) is twice differentiable and  f(x)  < α, f'(x) < β, in the range x > α, then which of the following is correct?
Let x > a and n a positive number then
or
Now, for maxima or minima of f(h), we have
or
and
Hence, the least value of φ(h)
= 2√AB
thus,
The function f(x) =  x + 2  is not differentiable at a point
Given that
Clearly, the rule of the function is changing at x = 0, so we shall test the differentiability of f(x) only at the point x = 2 obviously, being polynomial, at all other points the function is differentiable.
Since Lf '( 2) ≠ R'f( 2) , therefore , f(x) is not differentiable at x =  2 .
Using Rolle’s theorem, the equation. = 0 has atleast one root between 0 and 1, if
Consider the function f defined by
Since , f(x) is a polynomial, it is continuous and differentiablex. Consequently f(x) is continuous in the closed interval [0, 1] and differentiable in the open interval [0,1] Also , f(0) = 0 and
i.e. , f(0) = f(1)
Thus, all the three conditions of Rolle’s theorem are satisfied
Hence, there is atleast one value of x in the open interval [0, 1]
where f'(x) = 0
i.e., cos x^{n} + a_{1}.x^{n1} + . . . + a_{n1 }+ a_{n }= 0
A function f : R>R satisfies the e q .f(x +y) = f(x).f(y), and f(x) ≠ 0, . If f(x) is differentiate at 0 a n d f'(0 ) = 2 then f'(x) is equal to
We have
Now f(0 + 0)= f(0).f(0)
[Since f(x) 0 ,]
from eq. (ii) we get
Gives
Which one of the following is true?
Let
then,
Let
then,
Let
then,








