First Derivative Test for Local Minimum
Trending Questions
S2: If f(x) is decreasing function with upwards concavity, then concavity of f−1(x) is also upwards.
- S1 and S2 both are true
- S1 is true and S2 is false
- S1 is false and S2 is true
- S1 and S2 both are false
- q=2
- p=2
- p=3
- q=1
A differentiable function f(x) will have a local minimum at x = b if -
f’(b) = 0 , f’(b-h) < 0 & f’(b+h) > 0
f’(b) = 0 , f’(b-h) > 0 & f’(b+h) < 0
f’(b) = 0
None of the above
- None of these
- (-4, ∞)
- (-3, 3)
- (3, ∞)
Match the entries of col. I with those of col. II.
Column−IColumn−II(a)f(x)=1−x+x21+x−x2 on [0, 1](p)Greatest value of f=1(b)f(x)=2tanx−tan2x on [0, π2](q)Least value of f=35(c)f(x)=2π(sin2x−x) on [−π2, π2](r)Least value of f=−1(d)f(x)=12, (x3−3x2+6x−2) on (−1, 1)(s)Least value of f=−6
(a)→(p), (b)→(q), (c)→(r), (d)→(s)
(a)→(p, q), (b)→(p), (c)→(p, r), (d)→(p, s)
(a)→(q), (b)→(p), (c)→(p), (d)→(s)
(a)→(s), (b)→(q), (c)→(p, r), (d)→(r, s)
- neither a maxima nor a minima
- only one maxima
- both maxima and minima
- only one minima
- g(x) is decreasing in (0, 43)
- g(x) is decreasing in (43, 2)
- g(110)<g(1110)
- g(x) has a local minimum at x=43
Then which of the following options is/are correct?
- F has two local maxima and one local minimum in (0, ∞)
- F(x)≠0 for all x∈(0, 5)
- F has a local maximum at x=2
- F has a local minimum at x=1
- Neither a maximum nor a minimum
- Only one absolute minima
- Only one absolute maxima & only one absolute minima.
- Only one absolute maxima
- (−4, ∞)
- (−3, 3)
- (3, ∞)
- R
- g(x) is decreasing in (0, 43)
- g(x) is decreasing in (43, 2)
- g(110)<g(1110)
- g(x) has a local minimum at x=43
- only one maximum
- neither a maximum nor a minimum
- only one minimum
- none
- Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
- Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
- Assertion is correct but Reason is incorrect
- Assertion is incorrect but Reason is correct
- −1
- Has no real value.
- 2
- 4
Then which of the following options is/are correct?
- F has a local minimum at x=1
- F has a local maximum at x=2
- F has two local maxima and one local minimum in (0, ∞)
- F(x)≠0 for all x∈(0, 5)
- None of these
- (3, ∞)
- (-3, 3)
- (-4, ∞)
A differentiable function f(x) will have a local minimum at x = b if -
f’(b) = 0 , f’(b-h) < 0 & f’(b+h) > 0
f’(b) = 0 , f’(b-h) > 0 & f’(b+h) < 0
f’(b) = 0
None of the above
Match the entries of col. I with those of col. II.
Column−IColumn−II(a)f(x)=1−x+x21+x−x2 on [0, 1](p)Greatest value of f=1(b)f(x)=2tanx−tan2x on [0, π2](q)Least value of f=35(c)f(x)=2π(sin2x−x) on [−π2, π2](r)Least value of f=−1(d)f(x)=12, (x3−3x2+6x−2) on (−1, 1)(s)Least value of f=−6
(a)→(p, q), (b)→(p), (c)→(p, r), (d)→(p, s)
(a)→(q), (b)→(p), (c)→(p), (d)→(s)
(a)→(s), (b)→(q), (c)→(p, r), (d)→(r, s)
(a)→(p), (b)→(q), (c)→(r), (d)→(s)
A differentiable function f(x) will have a local minimum at x = b if -
f’(b) = 0 , f’(b-h) < 0 & f’(b+h) > 0
f’(b) = 0 , f’(b-h) > 0 & f’(b+h) < 0
f’(b) = 0
None of the above