Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
Q.1. (x+1)= 2*log(2^x+3)-2*log(1980-2^-x)
first log has base 2 and second one has base 4. so, question is sum of roots of the above equation
(a)3954 (b)log11(base=2) (c)log3954(base=2)
Q.2. lim ((x^(x^x))-x^x) =?
x tending to 0+
a.1) option b
2 log(4) (1980 - 2^-x) (4) means base it can be written as 2/2 log(2) (1980 - 2^-x ) by log properties
putting the value 2^x = y finally we get the equation as
log {(y + 3 )}^2 - log {(1980 y -1)/y} = x + 1
logs are all base 2 now applying the log properties again we get
log [{y (y + 3)^2}/(1980y - 1)] = x + 1
y (y + 3)^2}/(1980y - 1) = 2 ^(x+1) = 2^x . 2
as 2^x = y we get
y^2 - 3954 y + 11 = 0
the roots will be in form of 2^x1 and 2^x2
there product 2^x1*2^x2 = 2^(x1+x2) = 11
x1+x2 = log (2) 11
q.2) lim x tending to 0+ x ^ x = 1
now putting this result in the limits and applying the properties of limits we get
-1 as answer
We are all IITians and here to help you in your IIT JEE preparation.
Now you can win by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar : Click here to download the toolbar..
Thanks and Regards
Jitender
(askIITians expert)
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !