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The function f(x) =  2x^{3}  9x^{2}  12x + 1 is an increasing function in the interval
Given that
f(x) =  2x^{3}  9x^{2}  12x + 1
 + 
 2  1
implies f'(x) =  6x^{2}  18x  12
= 6[x^{2} + 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)
If f(x) is realvalued function defined on [0, ∞] such that f(0) = 0 and then the function h(x) is
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, ∞]
f(x) = x^{6}  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?
Given that f(x) = x^{6}  x  1
or f'(x) = 6x^{5} 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]
The value of ‘C’ of Lagrange’s mean value theorem for f( x ) = x (x  1) in [1 ,2] is given by
Clearly f(x) = x^{2}  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 x^{2} + y^{2} ≠ 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
Let
Then, the directional derivative at c = (0, 0) in the direction u = (a_{1}, a_{2}) is
We are given that
Then, the directional derivative at c = (0, 0) in the direction u(a_{1}, a_{2}) is
and
Hence, option (c) is correct.
We are given that
Now,
Hence, option (b) is correct,
The derivative of the function f(x) = x^{2m }is
Here f(x) = x^{2m} is an even function f'(x) = 2mx^{2m1} is an odd function.
Let f(x) =  sin πx , then
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.
If f'(a) exists, then is equal to
Since, f'(a) exists, then, we have
Lf'(a) = f '(a) and Rf'(a) = f'(a)
Adding these two, we get
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.
A function is onetoone if and only if f(a) ≠ f(b) whenever a ≠ b.
The continuous function f : R >R defined by f(x) = (x^{2 }+ 1)^{2011} is
Since therefore is not onto and f( 1) = f(1) , therefore, f(x) is not oneone.
Let
f(x, y) = 2x^{2}  xy + 2y^{2} Then at (1, 2)
We are given that
f(x, y) = 2x^{2}  xy + 2y^{2}
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.
If any four numbers are selected and they are multiplied, then the probability that the last digit will be 1, 3, 5 or 7 is
The total number of digits in any number at the units place is 10.
Therefore, n (S) = 10
If the last digit is 1, 3, 5 or 7, then it is necessary that the last digit in each number must be 1, 3, 5 or 7.
Therefore, n (A) = 4
P (A) = 4 / 10 = 2 / 5
Hence, the required probability is (2 / 5)^{4} = 16 / 625.
Let
f(x,y) = x^{3} + y^{3} 63 (x + y) + 12xy,
then
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