Global Minima
Trending Questions
If , then points at which are
None of these
- one-one and onto
- one-one and into
- many-one and onto
- many-one and into
- 12
- 13(1+1√2)
- 16
- 13(1−1√2)
If the points and lie on the graph, of the line , then the sum of value of ‘’ and ‘’ is
The function in the interval has
A point of maximum
A point of minimum
Point of maximum as well as of a minimum
Neither a point of maximum nor minimum
The function is increasing in the set
(where [.] denotes the greatest integer function)
- [5, 7)
- (5, 7)
- [5, 6)
- (5, 6)
Evaluate:
State, whether the following set is finite or infinite?
(v) {x ∈ Z, x<5}
Absolute value of
If f(x) = |x| + |x - 1| + |x - 2|, then
f (x) has maxima at x = 0
f (x) has maxima or minima at x = 3
None of these
f (x) has minima at x = 1
- [0, 1]
(0, 1]
- [0, 12]
- [12, 1]
- 19
- 20
- 39
- 40
(i) −x
(ii) (−x)−1
(iii) sin(x+1)
(iv) cos(x−π8)
N characters of information are held on magnetic tape, in batches of x character each; the batch processing time is α+βx2 seconds, α, β are constants. The optimum value of x for fast processing is
αβ
βα
√αβ
√βα
f(x)=⎧⎨⎩[x]2+sin[x][x]for [x]≠00for [x]=0
where [x] denotes the greatest integer less than or equal to x, then limx→0f(x) equals
- 1
- 0
- −1
- limit does not exist.
- 13
- −13
- 23
- 2
f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩sin(a+1)x+sin2x2x, if x<0 b, if x=0√x+bx3−√xbx5/2, if x>0
If f is continuous at x=0, then the value of a+b is equal
- −2
- −52
- −32
- −3
Let f(x)={xnsin1x, x≠00, x=0 , then f(x) is continuous but not differentiable at x=0 if
- n∈(0, 1]
- n=0
- n∈[1, ∞)
- n∈(−∞, 0)
- R
- (0, ∞)
- [0, ∞)
- R−{0}
f(x) = ⎧⎨⎩sinxx, ifx<0x+1, ifx≥0
- {1, 2, 3, 4, 5}
- {1, 2, 3}
- {1, 2, 3, 4}
- {1, 2, 3, 4, 5, 6}
Column-I | Column-II |
(A) Let fn(x)=ex∙ex2∙ex3….exn, n ϵ N and g(x)=limn→∞ fn(x) then g′(12)=λ⋅e, then λ is greater than | (p) 0 |
(B) a, b are distinct real numbers satisfying |a–1¬|+|b–1|=|a|+|b|=|a+1|+|b+1|. If the minimum value of |a–b| is μ, then μ is greater than | (q) 1 |
(C) Let n ϵN. If the value of c prescribed in Rolle’s theorem for the function f(x)=2x(x–3)n on [0, 3] is 34, then n is equal to | (r) 2 |
(D) If x1, x2 are abscissae of two points on the curve f(x)=x–x2 in the interval [0, 1], then the maximum value of expression(x1+x2)−(x21+x22)is less than |
(s) 3 |
(t) 4 |
- A(P, Q, R, S); B(P, Q); C(S); D(Q, R, S, T)
- A(P, Q, R, S); B(P, Q); C(S); D(Q, R, T, S)
- 6
- 9
- None of these
- 5