Courses

# 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

Description
This mock test of Test: Functions Of One,Two Or Three Real Variables -3 for IIT JAM helps you for every IIT JAM entrance exam. This contains 20 Multiple Choice Questions for IIT JAM Test: Functions Of One,Two Or Three Real Variables -3 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Functions Of One,Two Or Three Real Variables -3 quiz give you a good mix of easy questions and tough questions. IIT JAM students definitely take this Test: Functions Of One,Two Or Three Real Variables -3 exercise for a better result in the exam. You can find other Test: Functions Of One,Two Or Three Real Variables -3 extra questions, long questions & short questions for IIT JAM on EduRev as well by searching above.
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:

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: