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)

View Answer

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 – (x^{2}/2)

b) log(1 + x) < x – (x^{2}/2)

c) log(1 + x) ≥ x – (x^{2}/2)

d) log(1 + x) ≤ x– (x^{2}/2)

View Answer

Explanation: Assume, g(x) = log(1 + x) – (x – (x

^{2}/2))

Therefore, g’(x) = 1/(1 + x) – (1 – ½(2x))

= x

^{2}/(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 – (x

^{2}/2)) > 0, when x > 0

Or log(1 + x) > x – (x

^{2}/2), when x > 0.

3. What will be the range of the function f(x) = 2x^{3} – 9x^{2} – 24x + 5 which increases with x?

a) x > 4

b) x > 4 or x < -1

c) x < -1

d) Can’t be determined

View Answer

Explanation: Since f(x) = 2x

^{3}– 9x

^{2}– 24x + 5

Therefore, f’(x) = 6x

^{2}– 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

4. What will be the range of the function f(x) = 2x^{3} – 9x^{2} – 24x + 5 which decreases with x?

a) -1 < x < 4

b) 1 < x < 4

c) -1 ≤ x < 4

d) -1 < x ≤ 4

View Answer

Explanation: Since f(x) = 2x

^{3}– 9x

^{2}– 24x + 5

Therefore, f’(x) = 6x

^{2}– 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 + 6x^{2} – x^{3} for x > 3?

a) Decreases

b) Increases

c) Cannot be determined for x > 3

d) A constant function

View Answer

Explanation: f(x) = 10 – 9x + 6x

^{2}– x

^{3}

Thus, f’(x) = – 9 + 12x –3 x

^{2}

= -3(x

^{2}– 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 + 6x^{2} – x^{3} for x < 1?

a) Decreases

b) Increases

c) Cannot be determined for x < 1

d) A constant function

View Answer

Explanation: f(x) = 10 – 9x + 6x

^{2}– x

^{3}

Thus, f’(x) = – 9 + 12x –3 x

^{2}

= -3(x

^{2}– 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 + 6x^{2} – x^{3} for 1 < x < 3?

a) Decreases

b) Increases

c) Cannot be determined for 1 < x < 3

d) A constant function

View Answer

Explanation: f(x) = 10 – 9x + 6x

^{2}– x

^{3}

Thus, f’(x) = – 9 + 12x –3 x

^{2}

= -3(x

^{2}– 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

View Answer

Explanation: Let, f(θ) = (sinθ)/θ

Differentiating both sides of (1) with respect to θ we get,

f’(x) = (θcosθ – sinθ)/θ

^{2}……….(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

View Answer

Explanation: We have y = 3x((x + a)/(x + b)) + 5

From the definition we get,

Marginal cost = dy/dx = d/dx[3*((x

^{2}+ xa)/(x+b))] + 0

Solving it further, we get,

= [((x

^{2}+ xa)/(x+b))]d[3]/dx+ 3 d[((x

^{2}+ xa)/(x+b))]/dx

= 0 + 3 [(x+b)d(x

^{2}+ xa)/dx – (x

^{2}+ 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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