Application of Derivatives · Question 19

For the function f(x)=e |x|-|x|, x R, consider the following statements : In the light of the above statements, choose the correct answer from the options given below :

1. Both Statement I and Statement II are true
2. Both Statement I and Statement II are false
3. Statement I is true but Statement II is false
4. Statement I is false but Statement II is true

Let A= bmatrix1 & 3 & -1\ 2 & 1 & \ 0 & 1 & -1 bmatrix be a singular matrix. Let f(x)= 0x (t2+2t+3 )dt, x [1, ]. If M and m are respectively the maximum and the minimum values of f in [1, ], then 3(M-m) is equal to :

1. 64
2. 68
3. 72
4. 76

Let f : R R be a differentiable function such that f ( x+y3 )= f(x)+f(y)3 for all x, y R, and f'(0)=3. Then the minimum value of the function g(x)=3+ex f(x), is:

1. 3 ( e+1e )
2. 3 ( e-1e )
3. 3-ee
4. 3e

0 x (16 ( x2 ) 3 ( x2 ) ) is equal to :

1. 3 32
2. 3 3
3. 4 3
4. 6 3

The number of critical points of the function f(x)= cases | xx |, & x 0\ 1, & x=0 cases in the interval (-2 , 2 ) is equal to :

1. 1
2. 3
3. 5
4. 7

Let f(x) be a polynomial of degree 5, and have extrema at x=1 and x=-1. If x 0 ( f(x)x3 )=-5, then f(2)-f(-2) is equal to:

1. 0
2. 50
3. 92
4. 112

Let f be a differentiable function satisfying f(x) = 1 - 2x + 0x e(x-t) f(t) dt, x R and let g(x) = 0x (f(t) + 2)15(t - 4)6(t + 12)17 dt, x R. If p and q are respectively the points of local minima and local maxima of g, then the value of |p + q| is equal to .

Let (2 , ) be the largest interval in which the function f(t) = |t+1|t2, t < 0, is strictly decreasing. Then the local maximum value of the function g(x) = 2 e(x-2) + x2 + 4x - , x > 2, is

Consider the following three statements for the function f : (0, ) R defined by f(x) = | e x| - |x - 1| : (I) f is differentiable at all x > 0. (II) f is increasing in (0, 1). (III) f is decreasing in (1, ). Then

1. Only (I) is TRUE.
2. Only (I) and (III) are TRUE.
3. Only (II) and (III) are TRUE.
4. All (I), (II) and (III) are TRUE.

Let f(x) = x2025 - x2000, x [0, 1] and the minimum value of the function f(x) in the interval [0, 1] be (80)80(n)-81. Then n is equal to

1. -40
2. -41
3. -80
4. -81