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


20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Test: Functions Of One,Two Or Three Real Variables - 4


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QUESTION: 1

Convert the set x in roster form if set x contains the positive prime number, which divides 72. 

Solution:

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

QUESTION: 2

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)?

Solution:

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 

QUESTION: 3

Let l  = { 1 } ∪ { 2 }  For x ∈ R, let φ(x) = dis (x, l) = ln f{| x —y | : y ∈ l} then

Solution:

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

QUESTION: 4

Let f(x) = 

then f is

Solution:


At x =1/2

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

QUESTION: 5

If the function defined as

Then, the function f (x) has

Solution:


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.

QUESTION: 6

Let

then,

Solution:
QUESTION: 7

Let

then,

Solution:
QUESTION: 8

Let

then,

Solution:
QUESTION: 9

Let

then,

Solution:
QUESTION: 10

Let

then,

Solution:
QUESTION: 11

Consider

then,

Solution:
QUESTION: 12

If we expand sinx by Taylor’s series about  then a1, a7, a4, a3 are

Solution:

Given function f(x) = sin x



the value of a2, a7 a4, a3 in (i) are

QUESTION: 13

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?

Solution:

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, 

QUESTION: 14

The function f(x) = | x + 2 | is not differentiable at a point

Solution:

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 .

QUESTION: 15

Using Rolle’s theorem, the equation. = 0 has atleast one root between 0 and 1, if

Solution:

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.xn-1 + . . . + an-1 + a= 0 
 

QUESTION: 16

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

Solution:

We have

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

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

 

QUESTION: 17

Which one of the following is true?

Solution:
QUESTION: 18

Let

then,

Solution:
QUESTION: 19

Let

then,

Solution:
QUESTION: 20

Let

then,

Solution:

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