Test: Functions Of One,Two Or Three Real Variables -3


20 Questions MCQ Test IIT JAM Mathematics | Test: Functions Of One,Two Or Three Real Variables -3


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QUESTION: 1

The function f(x) = - 2x3 - 9x2 - 12x + 1 is an increasing function in the interval

Solution:

Given that
f(x) = - 2x3 - 9x2 - 12x + 1
- + - 
- 2 - 1
implies f'(x) = - 6x2 - 18x - 12
                   = -6[x2 + 3x + 2]
                   ​= - 6[(x + l ) ( x + 2)]
implies f'(x) > 0, V x ∈ (-2, -1)
Hence, f(x) is increasing in the interval (-2, -1)

QUESTION: 2

If f(x) is real-valued function defined on [0, ∞] such that f(0) = 0 and then the function h(x)  is

Solution:

Given that,


Since, f''(x) > 0 and x > 0, therefore g'(x) > 0, 0 => g (x) is strictly increasing function in [0, ∞]
since, h'(x)
=> h'(x) > 0,
or h (x) is increasing in [0, ∞]

QUESTION: 3

 f(x) = x6 - x - 1, x ∈ [1, 2], consider the following statements
(i) f is increasing on [1, 2]
(ii) f has root in [1, 2]
(iii) f is decreasing in [1, 2]
(iv) f has no root in [1 ,2]
Q. which of the above is/are correct?

Solution:

Given that f(x) = x6 - x - 1
or f'(x) = 6x5- 1
or f'(x) > 0, 
Clearly, f(x) is continuous in the interval [1, 2]
Such that f(1) f(2) < 0, therefore , f(x ) has atleast one root between [1, 2]

QUESTION: 4

The value of ‘C’ of Lagrange’s mean value theorem for f( x ) = x (x - 1) in [1 ,2] is given by

Solution:

Clearly f(x) = x2 - x is continuous in the interval [1, 2] and differentiable in the open interval [1,2]. therefore,
or 2 = 2c - 1
or c = 3/2

QUESTION: 5

If f(x) =   then

Solution:

Given that,
f(x) =
 
Clearly the given function is not differentiable at x = 0. Therefore, Roll’s theorem does not apply to f in [-1,1]

QUESTION: 6

Let f(x, y) = then

Solution:

We are given that
f(x,y) = 
Let us take ε > 0 and (x, y) ≠ (0, 0). Consider
|f(x,y) - 0| =
Let x = r cos θ, y = r sin θ
Therefore,
Hence for given ε > 0, there exists δ > 0 such that
Therefore, f(x, y) is continuous at (0, 0). Hence option (a) is correct.
Next,

Hence, option (c) is incorrect.

QUESTION: 7

Let

Solution:

We are given that 
Let us take ε > 0 and x2 + y2 ≠ 0.
Now consider,

We know that, 
or equivalently

Thus,
| f(x ,y )-0 | ≤ | y | = 0- |x| + 1 • |y|
take 
Therefore, for given ε > 0, there exists δ > 0 such that
|f(x, y) - 0 | < ε. whenever 0<|x|<δ,
0< |y |< δ
Hence, 
Thus, f(x,y) is continuous at (0, 0).
Hence, option (a) is correct.
Now,
 f(0 + h ,0 + k) - f(0,0) = f(h , k) - f(*0,0)
= 0 • h + 0 • k + • 
So that A = 0, B = 0 which does not depends on h and k and

Now approaching along h = mk, we get 

which depends on m. Therefore does not exists. Hence, f(x, y) is not differentiable at origin. Hence, option (b) is incorrect.
Next,
Hence, option (c) is incorrect.
And,

Hence, option (d) is incorrect.

QUESTION: 8

Let

then

Solution:

ANSWER :- c

Solution :- Consider an approach along the line y=x; then f(x,y)=x^2/(x^2+x^2) = 1/2

for all x≠0. On the other hand, if we approach (0,0) along the line y=2x

f(x,y)=(2x)^2/[x^2+4x^2]

=2/5

So there are two different paths toward the origin, each giving a different limit. Hence, the limit does not exist.

Or more simply, approach along y=0

Therefore , f(x, y) is not differentiable at origin

QUESTION: 9

Let

Then, the directional derivative at c = (0, 0) in the direction u = (a1, a2) is

Solution:

We are given that

Then, the directional derivative at c = (0, 0) in the direction u(a1, a2) is

and 
Hence, option (c) is correct.

QUESTION: 10

Let

Then,

Solution:

We are given that 

Now,

Hence, option (b) is correct,

QUESTION: 11

The derivative of the function f(x) = x2m is

Solution:

Here f(x) = x2m is an even function f'(x) = 2mx2m-1 is an odd function.

QUESTION: 12

Let f(x) = | sin πx |, then

Solution:

The graph of the function f(x) = | sinπx | is y =f(x)

clearly, the graph have sharp edges at the points 0, ±1, ± 2, ±3, .. therefore function is continuous everywhere but not differentiable everywhere except at integral values of x.

QUESTION: 13

If f'(a) exists, then  is equal to

Solution:

Since, f'(a) exists, then, we have
Lf'(a) = f '(a) and Rf'(a) = f'(a)
Adding these two, we get

QUESTION: 14

A function is said to be ______________ if and only if f(a) = f(b) implies that a = b for all a and b in the domain of f. 

Solution:

A function is one-to-one if and only if f(a) ≠ f(b) whenever a ≠ b.

QUESTION: 15

The continuous function f : R -->R defined by f(x) = (x+ 1)2011 is

Solution:

Since  therefore is not onto and f(- 1) = f(1) , therefore, f(x) is not one-one.

QUESTION: 16

Let
f(x, y) = 2x2 - xy + 2y2 Then at (1, 2)

Solution:

We are given that
f(x, y) = 2x2 - xy + 2y2
then, 
Hence, = 4 x 1 - 2 = 2
and   = - x + 4y
Hence,  = - l + 4x 2 = 7
Thus,
Hence, option (d) is correct. 

QUESTION: 17

Let

then,

Solution:

We are given that

Now, 

and 

Hence, option (b) is correct.

QUESTION: 18

Given the function
 f(x ,y) = x- 2xy+y+ x3 - y3 + x5

Solution:

 

QUESTION: 19

Let

Then,

Solution:
QUESTION: 20

Let
f(x,y) = x3 + y3- 63 (x + y) + 12xy,
then

Solution: