# Mathematics Questions and Answers – Calculus Application – Increasing and Decreasing Functions

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Calculus Application – Increasing and Decreasing Functions”.

1. If x > 0, then which one is correct?
a) x > log(x + 1)
b) x < log (x + 1)
c) x = log(x + 1)
d) x ≥ log(x + 1)

Explanation: Assume, f(x) = x – log(1 + x)
Therefore, f’(x) = 1 – (1/(1 + x))
= x/(x + 1) > 0, when x > 0
Hence, we must have, f(x) > f(0), when x > 0
But f(0) = 0 – log1 = 0
Thus, f(x) > 0, when x > 0 i.e.,
x – log(1 + x) > 0, when x > 0
Or x > (1 + x), when x > 0.

2. If x > 0, then which one is correct?
a) log(1 + x) > x – (x2/2)
b) log(1 + x) < x – (x2/2)
c) log(1 + x) ≥ x – (x2/2)
d) log(1 + x) ≤ x– (x2/2)

Explanation: Assume, g(x) = log(1 + x) – (x – (x2/2))
Therefore, g’(x) = 1/(1 + x) – (1 – ½(2x))
= x2/(1 + x) > 0, when x > 0
Hence, we must have g(x) > g(0), when x > 0
But g(0) = log1 – 0 = 0
Therefore, g(x) > 0, when x > 0 i.e., log(1 + x) – (x – (x2/2)) > 0, when x > 0
Or log(1 + x) > x – (x2/2), when x > 0.

3. What will be the range of the function f(x) = 2x3 – 9x2 – 24x + 5 which increases with x?
a) x > 4
b) x > 4 or x < -1
c) x < -1
d) Can’t be determined

Explanation: Since f(x) = 2x3 – 9x2 – 24x + 5
Therefore, f’(x) = 6x2 – 18x + 24
= 6(x – 4)(x + 1)
If x > 4, then, x – 4 > 0 and x + 1 > 0
Thus, (x – 4)(x + 1) > 0 i.e., f’(x) > 0, when x > 4
Again, if x < -1, then, x – 4 < 0 and x + 1 < 0
So, from here,
(x – 4)(x + 1) > 0 i.e., f’(x) > 0, when x < -1
Hence, f’(x) > 0, when x > 4 Or x < -1
Therefore, f(x) increases with x when, x > 4 or x < -1
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4. What will be the range of the function f(x) = 2x3 – 9x2 – 24x + 5 which decreases with x?
a) -1 < x < 4
b) 1 < x < 4
c) -1 ≤ x < 4
d) -1 < x ≤ 4

Explanation: Since f(x) = 2x3 – 9x2 – 24x + 5
Therefore, f’(x) = 6x2 – 18x + 24
= 6(x – 4)(x + 1)
If -1 < x < 4, then x – 4 < 0 and x + 1 > 0
Thus, (x – 4)(x + 1) < 0 i.e.,
f’(x) < 0, when -1 < x < 4
Therefore, f(x) decreases with x, when, -1 < x < 4.

5. What will be nature of the f(x) = 10 – 9x + 6x2 – x3 for x > 3?
a) Decreases
b) Increases
c) Cannot be determined for x > 3
d) A constant function

Explanation: f(x) = 10 – 9x + 6x2 – x3
Thus, f’(x) = – 9 + 12x –3 x2
= -3(x2 – 4x + 3)
Or f’(x) = -3(x – 1)(x – 3) ……….(1)
If x > 3, then x – 1 > 0 and x – 3 > 0
Hence, (x – 1)(x – 3) > 0
Thus, from (1) it readily follows that, f’(x) < 0, when x > 3
So, f(x) decreases for values of x > 3.

6. What will be nature of the f(x) = 10 – 9x + 6x2 – x3 for x < 1?
a) Decreases
b) Increases
c) Cannot be determined for x < 1
d) A constant function

Explanation: f(x) = 10 – 9x + 6x2 – x3
Thus, f’(x) = – 9 + 12x –3 x2
= -3(x2 – 4x + 3)
Or f’(x) = -3(x – 1)(x – 3) ……….(1)
If x < 1, then x – 1 < 0 and x – 3 < 0
Hence, (x – 1)(x – 3) > 0
Thus, from (1) it readily follows that, f’(x) < 0, when x < 1
So, f(x) decreases for values of x < 1.

7. What will be nature of the f(x) = 10 – 9x + 6x2 – x3 for 1 < x < 3?
a) Decreases
b) Increases
c) Cannot be determined for 1 < x < 3
d) A constant function

Explanation: f(x) = 10 – 9x + 6x2 – x3
Thus, f’(x) = – 9 + 12x –3 x2
= -3(x2 – 4x + 3)
Or f’(x) = -3(x – 1)(x – 3) ……….(1)
If 1 < x < 3, then x – 1 > 0 and x – 3 < 0
Hence, (x – 1)(x – 3) < 0
Thus, from (1) it readily follows that, f’(x) > 0, when 1 < x < 3
So, f(x) increases for values of 1 < x < 3.

8. What will be the nature of the equation (sinθ)/θ for 0 < θ < π/2 if θ increases continuously?
a) Decreases
b) Increases
c) Cannot be determined for 0 < θ < π/2
d) A constant function

Explanation: Let, f(θ) = (sinθ)/θ
Differentiating both sides of (1) with respect to θ we get,
f’(x) = (θcosθ – sinθ)/θ2 ……….(1)
Further, assume that F(θ) = θcosθ – sinθ
Then, F’(x) = -θsinθ – cosθ + cosθ
= -θsinθ
Clearly, F’(x) < 0, when 0 < θ < π/2
Thus, F(θ) < F(0), when 0 < θ < π/2
But F(0) = 0*cos0 – sin0 = 0
Thus, F(θ) < 0, when 0 < θ < π/2
Therefore, from (1) it follows that,
f’(θ) < 0 in 0 < θ < π/2
Hence, f(θ) = (sinθ)/θ is a decreasing function for 0 < θ < π/2
i.e., for 0 < θ < π/2, f(θ) = (sinθ)/θ steadily decreases as θ continuously increases.

9. If y = 3x((x + a)/(x + b)) + 5 where, a and b are constants and a > b, be the total cost for x unit of output of a commodity. What will be the nature of marginal cost as the output increases continuously?
a) Does not change
b) Increases continuously
c) Falls continuously
d) Changes as the interval of y changes

Explanation: We have y = 3x((x + a)/(x + b)) + 5
From the definition we get,
Marginal cost = dy/dx = d/dx[3*((x2 + xa)/(x+b))] + 0
Solving it further, we get,
= [((x2 + xa)/(x+b))]d/dx+ 3 d[((x2 + xa)/(x+b))]/dx
= 0 + 3 [(x+b)d(x2 + xa)/dx – (x2 + xa)d(x+b)/dx]/(x+b)2
= 3[1+(b(a – b)/(x+b)2)]
By problem, a > 0, b > 0 and a > b; hence from the expression of dy/dx, it is evident that dy/dx decreases as x increases.
Hence, we conclude the marginal cost falls continuously as the output increases.

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