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Test: Logarithmic Functions - JEE MCQ


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10 Questions MCQ Test Mathematics (Maths) Class 12 - Test: Logarithmic Functions

Test: Logarithmic Functions for JEE 2024 is part of Mathematics (Maths) Class 12 preparation. The Test: Logarithmic Functions questions and answers have been prepared according to the JEE exam syllabus.The Test: Logarithmic Functions MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Logarithmic Functions below.
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Test: Logarithmic Functions - Question 1

The differential coefficient  dy/dx  of the function yx = xy

Detailed Solution for Test: Logarithmic Functions - Question 1

 d(xy)=d(e(y⋅log(x)))
=e^(y⋅log(x))d(y⋅log(x))
=(xy)(dy⋅log(x) + y⋅d(log(x))
=(xy
d(yx) = (yx)(log(y)dx + x/ydy). 
Since  d(xy) = d(yx) , and simplifying by  xy = yx , we get
log(y)dx + x/ydy = log(x)dy + y/xdx. 
Removing the denominators leads to:
xylog(y)dx + x2dy = xylog(x)dy + y2dx 
(xylog(y) − y2)dx = (xylog(x) − x2)dy 
dy/dx = (xylog(y)−y2)/(xy⋅log(x)−x2)

Test: Logarithmic Functions - Question 2

The derivative of 2x tan x is​

Detailed Solution for Test: Logarithmic Functions - Question 2

dy/dx = 2x(tanx)' + tanx (2x)'
dy/dx = 2x (secx)2 + 2x tanx log2 , {since , (ax)' = ax loga (x)'}
dy/dx = 2x[(Secx)2 + log2 tanx]

Test: Logarithmic Functions - Question 3

The differential coefficient of (log x)tanx is:

Detailed Solution for Test: Logarithmic Functions - Question 3

y = (lnx)tanx
ln(y) = tanx(ln(lnx))
d(lny)/dx = d(tanx(ln(lnx)))/dx
Using product rule -
(1/y)dy/dx = (secx)2(ln(lnx)) + (1÷xlnx)tanx
dy/dx = [(secx)2(ln(lnx)) + (1÷xlnx)tanx ]×y
dy/dx = [(secx)2(ln(lnx)) + (1÷xlnx)tanx ]×[(lnx)tanx]

Test: Logarithmic Functions - Question 4

The differential coefficient of the function f(x) = asin x, where a is positive constant is:​

Test: Logarithmic Functions - Question 5

Detailed Solution for Test: Logarithmic Functions - Question 5

Test: Logarithmic Functions - Question 6

Detailed Solution for Test: Logarithmic Functions - Question 6

 y = (1 - log x)/(1 + log x)
Applying the Quoitent rule and differential for ln x
dy / dx = [(1 - ln x)(1 / x) - (1+ ln x)( -1 / x)]/(1 - ln x)2
= [(1 - ln x + 1 + ln x)]/x(1 + ln x)2
= 2 / x(1+ln x)2

Test: Logarithmic Functions - Question 7

If x = a cos φ and y = b sin φ then dy/dx = 

Test: Logarithmic Functions - Question 8

If sin(x + y) = log(x + y),  :

Detailed Solution for Test: Logarithmic Functions - Question 8

Given equation,

Test: Logarithmic Functions - Question 9

The differential coffcient   of the equation yx = e(x - y) is :

Detailed Solution for Test: Logarithmic Functions - Question 9

Taking log both the sides
xlogy = (x-y)loge
Differentiate it with respect to x, we get
x/y dy/dx = logy = loge - xloge dy/dx 
dy/dx = (loge - logy)/(x/y + loge)  
= y(1 - loge)/(x + y)

Test: Logarithmic Functions - Question 10

Detailed Solution for Test: Logarithmic Functions - Question 10


On differentiating both sides with respect to x

 

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