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Test: Increasing And Decreasing Functions - JEE MCQ


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5 Questions MCQ Test - Test: Increasing And Decreasing Functions

Test: Increasing And Decreasing Functions for JEE 2024 is part of JEE preparation. The Test: Increasing And Decreasing Functions questions and answers have been prepared according to the JEE exam syllabus.The Test: Increasing And Decreasing Functions MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Increasing And Decreasing Functions below.
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Test: Increasing And Decreasing Functions - Question 1
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Separate the interval  into sub-intervals in which f (x) = sin4 x + cos4 x is increasing or decreasing

Detailed Solution for Test: Increasing And Decreasing Functions - Question 1

f’(x) = 4sin3x cosx - 4cos3x sinx
= - sin4x
As x is in the interval [0, ∏/2] hence 4x is in the interval [0, 2∏]

Test: Increasing And Decreasing Functions - Question 2
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The function f(x) = ax, 0 < a < 1 is​

Detailed Solution for Test: Increasing And Decreasing Functions - Question 2

 f(x) = ax
Taking log bth the sides, log f(x) = xloga
f’(x)/ax = loga
f’(x) = ax loga   {ax > 0 for all x implies R, 
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.

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Test: Increasing And Decreasing Functions - Question 3
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The function  increases in​

Detailed Solution for Test: Increasing And Decreasing Functions - Question 3

 f'(x) = [(logx).2−2x.1/x](logx)2
= 2(logx−1)/(logx)^2
∴f'(x)>0
⇔logx−1>0
⇔logx>1
⇔logx>loge
⇔x>e
∴f(x) is increasing in (e,∞)

Test: Increasing And Decreasing Functions - Question 4
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The function f (x) = -3x + 12 on R.is​
Detailed Solution for Test: Increasing And Decreasing Functions - Question 4

f(x) = -3x + 12
f(0) = -3(0) + 12 = 0 - 12 = 0
f(1) = 9
f(2)= 6
f(3) = 3
f(4) = 0
f(5) = -3

Test: Increasing And Decreasing Functions - Question 5
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What is the nature of function f(x) = x3 – 3x2 + 4x on R?
Detailed Solution for Test: Increasing And Decreasing Functions - Question 5

  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. 

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