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

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)

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, ∞]

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]

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

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]

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.

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.

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

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.

We are given that 

Now,

Hence, option (b) is correct,

Here f(x) = x^2m is an even function f'(x) = 2mx^2m-1 is an odd function.

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.

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

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

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

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. 

We are given that

Now, 

and 

Hence, option (b) is correct.