Mathematics Questions and Answers – Derivatives Application – Increasing and Decreasing Functions

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

1. What is the nature of function f(x) = 7x-4 on R?
a) Increasing
b) Decreasing
c) Strictly Increasing
d) Increasing and Decreasing
View Answer

Answer: c
Explanation: Let x1 and x2 be any two numbers in R.
Then x1<x2 => 7×1 < 7×2.
=> 7×1 – 4 < 7×2 – 4.
As f(x1) < f(x2), thus the function f is strictly increasing on R.
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2. What is the nature of function f(x) = x3 – 3x2 + 4x on R?
a) Increasing
b) Decreasing
c) Constant
d) Increasing and Decreasing
View Answer

Answer: a
Explanation: f(x) = x3 – 3x2 + 4x
f’(x) = 3x2 – 6x + 4.
f’(x) = 3(x2 – 2x + 1) + 1.
=> 3(x-1)2 + 1>0, in every interval of R. Therefore the function f is increasing on R.

3. Find the interval in which function f(x) = x2 – 4x + 5 is increasing.
a) (2, ∞)
b) (-∞, 2)
c) (3, ∞)
d) (-∞, ∞)
View Answer

Answer: a
Explanation: f(x) = x2 – 4x + 5.
f’(x) = 2x – 4. Therefore f’(x) = 0 gives x = 2.
Now this point x=2 divides the line into two disjoint intervals and the interval namely (2, ∞) is increasing on f(x).
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4. Find the interval in which function f(x) = x2 – 4x + 5 is decreasing.
a) (2, ∞)
b) (-∞, 2)
c) (3, ∞)
d) (-∞, ∞)
View Answer

Answer: b
Explanation: f(x) = x2 – 4x + 5.
f’(x) = 2x – 4. Therefore f’(x) = 0 gives x = 2.
Now this point x=2 divides the line into two disjoint intervals and the interval namely (-∞, 2) is decreasing on f(x).

5. Find the interval in which function f(x) = sinx+cosx, 0 ≤ x ≤ 2π is decreasing.
a) (π/4, 5π/4)
b) (-π/4, 5π/4)
c) (π/4, -5π/4)
d) (-π/4, π/4)
View Answer

Answer: a
Explanation: f(x) = sinx+cosx.
f’(x) = cosx – sinx. Now f’(x) = 0 gives sinx = cosx which gives that x=π/4, 5π/4 as 0 ≤ x ≤ 2π.
Therefore on checking the values we get f is decreasing in (π/4, 5π/4).
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6. Find the interval in which function f(x) = sinx+cosx is increasing.
a) (5π/4, 2π)
b) [0, π/4) and (5π/4, 2π]
c) (π/4, -5π/4)
d) (-π/4, π/4)
View Answer

Answer: b
Explanation: f(x) = sinx+cosx.
f’(x) = cosx – sinx. Now f’(x) = 0 gives sinx = cosx which gives that x= π/4, 5π/4 as 0 ≤ x ≤ 2π.
The points x = π/4 and x = 5π/4 divide the interval [0, 2π] into three disjoint intervals which are
[0, π/4), (π/4, 5π/4) and (5π/4, 2π].
Therefore on checking the values we get f is increasing in [0, π/4) and (5π/4, 2π].

7. Is the function f(x) = 3x+10 is increasing on R?
a) True
b) False
View Answer

Answer: a
Explanation: f(x) = 3x+10.
f’(x) = 3, which shows 3 > 0 for all x ∈ R.
Thus function f(x) is increasing.
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8. Find the intervals in which f(x) = 2x2 – 3x is increasing.
a) (-1/4, ∞)
b) (-3/4, ∞)
c) (1/4, ∞)
d) (3/4, ∞)
View Answer

Answer: d
Explanation: f(x) = 2x2-3x.
f’(x) = 4x – 3.
As we know f’(x) = 0, x=-3/4. This shows that function f is increasing in interval (-3/4, ∞) for all x ∈ R.

9. Find the intervals in which f(x) = x2 + 2x – 5 is strictly increasing.
a) x>1
b) x<-1
c) x>-1
d) x>2
View Answer

Answer: c
Explanation: f(x) = x2+2x -5.
f’(x) = 4x – 3.
As we know f’(x) = 0, x=-3/4, which shows that function f is increasing in interval (-3/4, ∞) for all x ∈ R.
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10. Nature of the function f(x) = e2x is _______
a) increasing
b) decreasing
c) constant
d) increasing and decreasing
View Answer

Answer: a
Explanation: f(x) = e2x.
f’(x) = 2e2x.
As we know 2e2x > 0, so it always has a value greater than zero.
Which shows that function f is increasing for all x ∈ R.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter