Question 1
A monotonic function on [a,b] is either a monotonically increasing or monotonically decreasing function.
A. FALSE
B. TRUE
View Answer
Answer: Option B
Explanation:
Maximum and minimum values of a function are represented by monotonicity. A monotonic function on [a,b] is either a monotonically increasing or monotonically decreasing on [a,b].
Question 2
A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?
A. 1
B. 2
C. 3
D. 4
View Answer
Answer: Option C
Explanation:
Assume that the velocity of the particle at time t second is vcm/sec.Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1So, v = dx/dt = t3/3 – 2t2/ + 3t + 1Thus, dv/dt = t2 – 4t + 3And d2v/dt2 = 2t – 4For maximum and minimum value of v we have,dv/dt = 0Or t2 – 4t + 3 = 0Or (t – 1)(t – 3) = 0Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0Thus, v is minimum at t = 3.
Question 3
A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. What is the minimum velocity?
A. 1 cm/sec
B. 2 cm/sec
C. 3 cm/sec
D. 4 cm/sec
View Answer
Answer: Option A
Explanation:
Assume that the velocity of the particle at time t second is vcm/sec.Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1So, v = dx/dt = t3/3 – 2t2/ + 3t + 1Thus, dv/dt = t2 – 4t + 3And d2v/dt2 = 2t – 4For maximum and minimum value of v we have,dv/dt = 0Or t2 – 4t + 3 = 0Or (t – 1)(t – 3) = 0Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0Thus, v is minimum at t = 3.Putting t = 3 in (1) we get,33/3 – 2(3)2/ + 3(3) + 1= 1 cm/sec.
Question 4
At which point does f(x) = |x – 1| has itslocal minimum?
A. They are unequal
B. They are equal
C. Depend on the numbers
D. Can’t be predicted
View Answer
Question 5
For which value of x will (x – 1)(3 – x) have its maximum?
A. 0
B. 1
C. 2
D. -2
View Answer
Answer: Option C
Explanation:
Let, y = (x – 1)(3 – x) = 4x – x2 – 3Then, dy/dx = 0Or 4 – 2x = 0Or 2x = 4Or x = 2Now, [d2y/dx2] = -2 which is negative.Therefore, (x – 1)(3 – x) will have its maximum at x = 2.
Question 6
Given, f(x) = x3 – 12x2 + 45x + 8. At which point does f(x) has its maximum?
A. 1
B. 2
C. 3
D. 4
View Answer
Answer: Option C
Explanation:
We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)Differentiating both sides of (1) with respect to x wef’(x) = 3x2 – 24x + 453x2 – 24x + 45 = 0Or x2 – 8x + 15 = 0Or (x – 3)(x – 5) = 0Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5Therefore, f’(x) = 0 for x = 3 and x = 5.If h be a positive quantity, however small, then,f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.So, f(x) has maximum at 3.
Question 7
Given, f(x) = x3 – 12x2 + 45x + 8. At which point does f(x) has its minimum?
A. 1
B. 7
C. 3
D. 5
View Answer
Answer: Option D
Explanation:
We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)Differentiating both sides of (1) with respect to x wef’(x) = 3x2 – 24x + 453x2 – 24x + 45 = 0Or x2 – 8x + 15 = 0Or (x – 3)(x – 5) = 0Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5Therefore, f’(x) = 0 for x = 3 and x = 5.If h be a positive quantity, however small, then,f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.So, f(x) has minimum at 5.
Question 8
Given, f(x) = x3 – 12x2 + 45x + 8. What is the maximum value of f(x)?
A. 61
B. 62
C. 63
D. 54
View Answer
Answer: Option B
Explanation:
We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)Differentiating both sides of (1) with respect to x wef’(x) = 3x2 – 24x + 453x2 – 24x + 45 = 0Or x2 – 8x + 15 = 0Or (x – 3)(x – 5) = 0Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5Therefore, f’(x) = 0 for x = 3 and x = 5.If h be a positive quantity, however small, then,f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.So, f(x) has maximum at 3.Putting, x = 3 in (1)Thus, its maximum value is,f(3) = 33 – 12*32 + 45*3 + 8 = 62.
Question 9
Given, f(x) = x3 – 12x2 + 45x + 8. What is the minimum value of f(x)?
A. -1
B. 0
C. 1
D. Value does not exist
View Answer
Answer: Option C
Explanation:
We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)Differentiating both sides of (1) with respect to x wef’(x) = 3x2 – 24x + 453x2 – 24x + 45 = 0Or x2 – 8x + 15 = 0Or(x – 3)(x – 5) = 0Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5Therefore, f’(x) = 0 for x = 3 and x = 5.If h be a positive quantity, however small, then,f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.So, f(x) has minimum at 5.Putting, x = 5 in (1)Thus, its maximum value is,f(3) = 53 – 12*52 + 45*5 + 8 = 58.
Question 10
What is a monotonically increasing function?
A. x1 > x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b) ∀ c ∈ a
B. x1 < x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b)
C. x1 < x2 ⇒ f(x1) = f(x2) ∀ x1, x2 ∈ (a,b)
D. x1 = x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b)
View Answer
Answer: Option B
Explanation:
A function f : (a,b) → R is said to be monotonically increasing on (a,b) if x1 < x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b). A monotonically increasing function can also be called as non-decreasing function.
Question 11
What is the condition for a function f to be constant if f be continuous and differentiable on (a,b)?
A. f’(x) > 0 ∀ x1, x2 ∈ (a,b)
B. f’(x) < 0 ∀ x1, x2 ∈ (a,b)
C. f’(x) = 0 ∀ x1, x2 ∈ (a,b)
D. f’(x) ≤ 0 ∀ x1, x2 ∈ (a,b)
View Answer
Answer: Option C
Explanation:
One of the properties of a function is to be constant. A function is said to be constant when it satisfies the condition f’(x) = 0 ∀ x1, x2 ∈ (a,b) where the function ‘f’ should be continuous and differentiable on (a,b).
Question 12
What is the condition for a function f to be decreasing if f be continuous and differentiable on (a,b)?
A. f’(x) > 0 ∀ x1, x2 ∈ (a,b)
B. f’(x) < 0 ∀ x1, x2 ∈ (a,b)
C. f’(x) = 0 ∀ x1, x2 ∈ (a,b)
D. f’(x) ≤ 0 ∀ x1, x2 ∈ (a,b)
View Answer
Answer: Option D
Explanation:
If a function f be continuous and differentiable on (a,b), then f’(x) ≤ 0 ∀ x1, x2 ∈ (a,b) is the condition for the function f(x) to be decreasing on (a,b).
Question 13
What is the condition for a function f to be increasing if f be continuous and differentiable on (a,b)?
A. f’(x) < 0 ∀ x1, x2 ∈ (a,b)
B. f’(x) > 0 ∀ x1, x2 ∈ (a,b)
C. f’(x) = 0 ∀ x1, x2 ∈ (a,b)
D. f’(x) ≥ 0 ∀ x1, x2 ∈ (a,b)
View Answer
Answer: Option D
Explanation:
If a function f be continuous and differentiable on (a,b), then f’(x) ≥ 0 ∀ x1, x2 ∈ (a,b) is the condition for the function f(x) to be increasing on (a,b).
Question 14
What is the condition for a function f to be strictly decreasing if f be continuous and differentiable on (a,b)?
A. f’(x) > 0 ∀ x1, x2 ∈ (a,b)
B. f’(x) < 0 ∀ x1, x2 ∈ (a,b)
C. f’(x) = 0 ∀ x1, x2 ∈ (a,b)
D. f’(x) ≤ 0 ∀ x1, x2 ∈ (a,b)
View Answer
Answer: Option B
Explanation:
The mathematical expression for strictly decreasing function is f’(x) < 0 ∀ x1, x2 ∈ (a,b). This is the condition for strictly decreasing function and only possible when function ‘f’ is continuous and differentiable on (a,b).
Question 15
What is the condition for a function f to be strictly increasing if f be continuous and differentiable on (a,b)?
A. f’(x) > 0 ∀ x1, x2 ∈ (a,b)
B. f’(x) < 0 ∀ x1, x2 ∈ (a,b)
C. f’(x) ≤ 0 ∀ x1, x2 ∈ (a,b)
D. f’(x) = 0 ∀ x1, x2 ∈ (a,b)
View Answer
Answer: Option A
Explanation:
The condition for a function ‘f’ to be strictly increasing is f’(x) > 0 ∀ x1, x2 ∈ (a,b) where the the function ‘f’ should be continuous and differentiable on (a,b).
Question 16
What is the mathematical expression for a function to be strictly decreasing on (a,b)?
A. x1 = x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b)
B. x1 < x2 ⇒ f(x1) > f(x2) ∀ x1, x2 ∈ (a,b)
C. x1 < x2 ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ (a,b)
D. x1 < x2 ⇒ f(x1) = f(x2) ∀ x1, x2 ∈ (a,b)
View Answer
Answer: Option B
Explanation:
There are two types of decreasing functions in maxima and minima. They are strictly decreasing and monotonically decreasing. The mathematical expression for strictly decreasing function is x1 < x2 ⇒ f(x1) > f(x2) ∀ x1, x2 ∈ (a,b).
Question 17
What is the mathematical expression for a function to be strictly increasing on (a,b)?
A. x1 < x2 ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ (a,b)
B. x1 < x2 ⇒ f(x1) ≥ f(x2) ∀ x1, x2 ∈ (a,b)
C. x1 = x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b)
D. x1 = x2 ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ (a,b)
View Answer
Answer: Option A
Explanation:
A function f : (a,b) → R is said to be strictly increasing on (a,b) if x1< x2 ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ (a,b). (a,b) may be replaced by [a,b] or any interval in the definition.
Question 18
What is the mathematical expression for monotonically decreasing function?
A. x1 < x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b)
B. x1 < x2 ⇒ f(x1) ≥ f(x2) ∀ x1, x2 ∈ (a,b)
C. x1 = x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b)
D. x1 < x2 ⇒ f(x1) = f(x2) ∀ x1, x2 ∈ (a,b)
View Answer
Answer: Option B
Explanation:
The definition of monotonically decreasing function is if a function f : (a,b) → R is said to be monotonically decreasing on (a,b) if x1 < x2 ⇒ f(x1) ≥ f(x2) ∀ x1, x2 ∈ (a,b). Hence, the mathematical expression is x1< x2 ⇒ f(x1) ≥ f(x2) ∀ x1, x2 ∈ (a,b).
Question 19
What is the mathematical expression for monotonically non-increasing function?
A. x1 < x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b)
B. x1 < x2 ⇒ f(x1) ≥ f(x2) ∀ x1, x2 ∈ (a,b)
C. x1 = x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b)
D. x1 < x2 ⇒ f(x1) = f(x2) ∀ x1, x2 ∈ (a,b)
View Answer
Answer: Option B
Explanation:
The meaning of a monotonic function is it either never decreases or never increases. The condition for a function to be monotonically non-increasing is x1 < x2 ⇒ f(x1) ≥ f(x2) ∀ x1, x2 ∈ (a,b).
Question 20
What is the mathematical expression of non-decreasing function?
A. x1 > x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b) ∀ c ∈ a
B. x1 < x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b)
C. x1 < x2 ⇒ f(x1) = f(x2) ∀ x1, x2 ∈ (a,b)
D. x1 = x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b)
View Answer
Answer: Option B
Explanation:
A function f : (a,b) → R is said to be monotonically increasing on (a,b) if x1 < x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b). A monotonically increasing function can also be called as non-decreasing function.
Question 21
What is the nature of the function f(x) = 2/3(x3) – 6x2 + 20x – 5?
A. Possess only minimum value
B. Possess only maximum value
C. Does not possess a maximum or minimum value
D. Datainadequate
View Answer
Answer: Option C
Explanation:
We have, f(x) = 2/3(x3) – 6x2 + 20x – 5 ……….(1)Differentiating both side of (1) with respect to x, we get,f’(x) = 2x2 – 12x + 20Now, for a maximum and minimum value of f(x) we have,f’(x) = 0Or 2x2 – 12x + 20 = 0Or x2 – 6x + 10= 0So, x = [6 ± √(36 – 4*10)]/2x = (6 ± √-4)/2, which is imaginary.Hence, f’(x) does not vanishes at any point of x.Thus, f(x) does not possess a maximum or minimum value.
Question 22
What is the relation between f(x) and ℓ when the maximum value or greatest value function f is defined on a set A and ℓ ∈ f(A)?
A. f(x) < ℓ ∀ x ∈ A
B. f(x) ≤ ℓ ∀ x ∈ A
C. f(x) = ℓ ∀ x ∈ A
D. f(x) > ℓ ∀ x ∈ A
View Answer
Answer: Option B
Explanation:
A function f defined on a set A and ℓ ∈ f(A), then ℓ is the maximum or the greatest value of f in A if f(x) ≤ ℓ ∀ x ∈ A and the minimum or the least value of f in A if f(x) ≥ ℓ ∀ x ∈ A.
Question 23
What is the relation between f(x) and ℓ when the minimum value or least value function f is defined on a set A and ℓ ∈ f(A)?
A. f(x) < ℓ ∀ x ∈ A
B. f(x) ≤ ℓ ∀ x ∈ A
C. f(x) ≥ ℓ ∀ x ∈ A
D. f(x) > ℓ ∀ x ∈ A
View Answer
Answer: Option C
Explanation:
The relation between f(x) and ℓ when the minimum value or least value function f is f(x) ≥ (ℓ) ∀ x ∈ A where the function is defined on a set A and ℓ ∈ f(A).
Question 24
What will be the maxima for the function f(x) = x4 –8x3 + 22x2 –24x + 8?
A. 0
B. 1
C. 2
D. 3
View Answer
Answer: Option C
Explanation:
We have, x4 –8x3 + 22x2 –24x + 8 ……….(1)Differentiating both sides of (1) with respect to x, we get,f’(x) = 4x3 – 24x2 + 44x – 24 and f”(x) = 12x2 – 48x + 44 ……….(2)At an extremum of f(x), we have f’(x) = 0Or 4x3 – 24x2 + 44x – 24 = 0Or x2(x – 1) – 5x(x – 1) + 6(x – 1) = 0Or (x – 1)(x2 – 5x + 6) = 0Or (x – 1)(x – 2)(x – 3) = 0So, x = 1, 2, 3Now, f”(x) = 12x2 – 48x + 44f”(1) = 8 > 0f”(2) = -4 < 0f”(3) = 8 < 0So, f(x) has maximum at x = 2.
Question 25
What will be the maximum value of the function 2x3 + 3x2 – 36x + 10?
A. 71
B. 81
C. 91
D. 0
View Answer
Answer: Option C
Explanation:
Let y = 2x3 + 3x2 – 36x + 10 ……….(1)Differentiating both sides of (1) with respect to x we get,dy/dx = 6x2 + 6x – 36And d2y/dx2 = 12x + 6For maxima or minima value of y, we have,dy/dx = 0Or 6x2 + 6x – 36 = 0Or x2 + x – 6 = 0Or (x + 3)(x – 2) = 0Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2Now, d2y/dx2 = 12x + 6 = 12(-3) + 6 = -30 < 0Putting x = -3 in (1) we get its maximum value as,2x3 + 3x2 – 36x + 10 = 2(-3)3 + 3(-3)2 – 36(-3) + 10= 91
Question 26
What will be the minima for the function f(x) = x4 – 8x3 + 22x2 – 24x + 8?
A. -1
B. 0
C. 2
D. 3
View Answer
Answer: Option D
Explanation:
We have, x4 – 8x3 + 22x2 – 24x + 8 ……….(1)Differentiating both sides of (1) with respect to x, we get,f’(x) = 4x3 – 24x2 + 44x – 24 and f”(x) = 12x2 – 48x + 44 ……….(2)At an extremum of f(x), we have f’(x) = 0Or 4x3 – 24x2 + 44x – 24 = 0Or x2(x – 1) – 5x(x – 1) + 6(x – 1) = 0Or (x – 1)(x2 – 5x + 6) = 0Or (x – 1)(x – 2)(x – 3) = 0So, x = 1, 2, 3Now, f”(x) = 12x2 – 48x + 44f”(1) = 8 > 0f”(2) = -4 < 0f”(3) = 8 < 0So, f(x) has minimum at x = 1 and 3.
Question 27
What will be the minimum value of the function 2x3 + 3x2 – 36x + 10?
A. -31
B. 31
C. -34
D. 34
View Answer
Answer: Option C
Explanation:
Let y = 2x3 + 3x2 – 36x + 10 ……….(1)Differentiating both sides of (1) with respect to x we get,dy/dx = 6x2 + 6x – 36And d2y/dx2 = 12x + 6For maxima or minima value of y, we have,dy/dx = 0Or 6x2 + 6x – 36 = 0Or x2 + x – 6 = 0Or (x + 3)(x – 2) = 0Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2Now, d2y/dx2 = 12x + 6 = 12(2) + 6 = 30 > 0Putting x = 2 in (1) we get its minimum value as,2x3 + 3x2 – 36x + 10 = 2(2)3 + 3(2)2 – 36(2) + 10= -34
Question 28
What will be the nature of the equation sin(x + α)/sin(x + β)?
A. Possess only minimum value
B. Possess only maximum value
C. Does not possess a maximum or minimum value
D. Data inadequate
View Answer
Answer: Option C
Explanation:
Let, y = sin(x + α)/sin(x + β)Then,dy/dx = [cos(x + α)sin(x + β) – sin(x + α)cos(x + β)]/sin2(x + β)= sin(x+β – x-α)/sin2(x + β)Or sin(β – α)/sin2(x + β)So, for minimum or maximum value of x we have,dy/dx = 0Or sin(β – α)/sin2(x + β) = 0Or sin(β – α) = 0 ……….(1)Clearly, equation (1) is independent of x; hence, we cannot have a real value of x as root of equation (1).Therefore, y has neither a maximum or minimum value.
Question 29
What will be the point of maximum of the function 2x3 + 3x2 – 36x + 10?
A. -1
B. -2
C. -3
D. -4
View Answer
Answer: Option C
Explanation:
Let y = 2x3 + 3x2 – 36x + 10 ……….(1)Differentiating both sides of (1) with respect to x we get,dy/dx = 6x2 + 6x – 36And d2y/dx2 = 12x + 6For maxima or minima value of y, we have,dy/dx = 0Or 6x2 + 6x – 36 = 0Or x2 + x – 6 = 0Or (x + 3)(x – 2) = 0Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2Now, d2y/dx2 = 12x + 6 = 12(-3) + 6 = -30, which is < 0.
Question 30
What will be the point of minimum of the function 2x3 + 3x2 – 36x + 10?
A. 1
B. 2
C. 3
D. 4
View Answer
Answer: Option B
Explanation:
Let y = 2x3 + 3x2 – 36x + 10 ……….(1)Differentiating both sides of (1) with respect to x we get,dy/dx = 6x2 + 6x – 36And d2y/dx2 = 12x + 6For maxima or minima value of y, we have,dy/dx = 0Or 6x2 + 6x – 36 = 0Or x2 + x – 6 = 0Or (x + 3)(x – 2) = 0Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2Now, d2y/dx2 = 12x + 6 = 12(2) + 6 = 30, which is > 0.
Question 31
What will be the value of x for which the value of cosx is minimum?
A. 0
B. -1
C. 1
D. Cannot be determined
View Answer
Answer: Option B
Explanation:
Let, f(x) = cosxThen, f’(x) = -sinx and f”(x) = -cosxAt an extreme point of f(x) we must have,f’(x) = 0Or -sinx = 0Or x = nπ where, n – any integer.If n is an odd integer i.e., n = 2m + 1 where m is any integer, then at,x = (2m + 1)π we have, f”(x) = [(2m + 1)π] = -cos(2mπ + π) = -cosπ = -1(-1) = 1So, f”(x) is positive at x = (2m + 1)πHence, f(x) = cosx is minimum at x = (2m + 1)π.So, the minimum value of cosx is cos(2mπ + π) = cosπ = -1.
Question 32
What will be the values of x for which the value of cosx is minimum?
A. (2m + 1)π
B. (2m)π
C. (2m + 1)π/2
D. (2m – 1)π
View Answer
Answer: Option A
Explanation:
Let, f(x) = cosxThen, f’(x) = -sinx and f”(x) = -cosxAt an extreme point of f(x), we must have,f’(x) = 0Or -sinx = 0Or x = nπ where, n is any integer.If n is an odd integer i.e., n = 2m + 1 where m is any integer, then at,x = (2m + 1)π we have, f”(x) = [(2m + 1)π] = -cos(2mπ + π) = -cosπ = -1(-1) = 1So, f”(x) is positive at x = (2m + 1)πHence, f(x) = cosx is minimum at x = (2m + 1)π.