Test: Functions Of One,Two Or Three Real Variables - 4 - Mathematics MCQ

2 and 3 are the divisors of 72, which are prime. So, the roster form of set x is (2, 3}.

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 

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

At x =1/2

Since 
therefore, f(x) is not continuous x =1/2

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.

Concept: Tabular form / Roaster form: In this method, a set is described by listing all the elements, separated by commas, within the braces {}.

Example: A = {2, 3, 5} is a set of first three prime numbers. Set –builder form: In this method, all the elements of the set possess a single common property, which is being enlisted. Example: B = {x : 6 ≤ x ∈ N ≤ 12}

Calculation: Given: A = {x : x2 – 5x + 6 = 0} x2 – 5x + 6 = 0 ⇒ x2 – 2x -3x + 6 = 0 ⇒ x(x - 2) – 3(x - 2) = 0 ⇒ (x - 3)(x - 2) = 0 ⇒ x = 3 or 2 As we know that, roaster form of a set is described by listing all the elements, separated by commas, within the braces {}.

Hence the required roaster form of set B = {2, 3}

Given function f(x) = sin x



the value of a₂, a₇ a₄, a₃ in (i) are

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, 

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 .

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 xⁿ + a₁.x^n-1 + . . . + a_n-1 + a_n = 0 
 

We have

Now f(0 + 0)= f(0).f(0)

[Since f(x)  0 ,]
from eq. (ii) we get
Gives