-
Pages
-
Categories
- algebra
- applications of maths
- Basic Set Theory and Logic
- calculus
- careers in mathematics
- Cnennai Math Institute Entrance Exam
- co-ordinate geometry
- combinatorics or permutations and combinations
- Complex Numbers
- Fun with Mathematics
- geometry
- IITJEE Advanced
- IITJEE Advanced Mathematics
- IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha)
- IITJEE Foundation mathematics
- IITJEE Mains
- IMO International Mathematical Olympiad IMU
- Inequalities
- Information about IITJEE Examinations
- INMO
- ISI Kolkatta Entrance Exam
- KVPY
- mathematicians
- memory power concentration retention
- miscellaneous
- motivational stuff
- physicisrs
- Pre-RMO
- probability theory
- pure mathematics
- RMO
- RMO Number Theory
- Statistics
- time management
- Trigonometry
-
Archives
- March 2021
- January 2021
- November 2020
- October 2020
- September 2020
- August 2020
- July 2020
- June 2020
- May 2020
- April 2020
- March 2020
- February 2020
- January 2020
- December 2019
- November 2019
- October 2019
- September 2019
- August 2019
- June 2019
- May 2019
- March 2019
- February 2019
- January 2019
- November 2018
- October 2018
- September 2018
- August 2018
- July 2018
- June 2018
- May 2018
- April 2018
- March 2018
- February 2018
- January 2018
- December 2017
- November 2017
- October 2017
- September 2017
- August 2017
- July 2017
- June 2017
- May 2017
- April 2017
- March 2017
- February 2017
- January 2017
- November 2016
- October 2016
- September 2016
- August 2016
- July 2016
- June 2016
- May 2016
- April 2016
- March 2016
- February 2016
- January 2016
- December 2015
- November 2015
- October 2015
- September 2015
- August 2015
- July 2015
- June 2015
- May 2015
- April 2015
- March 2015
- February 2015
- January 2015
- December 2014
- November 2014
- October 2014
- September 2014
- August 2014
- July 2014
- June 2014
Author Archives: Nalin Pithwa
Annual Essay Contest: Mentoris Project: Aug 2021
Match making, STEM, math, algorithm by 18 year old
Derivatives: part 11: IITJEE maths tutorial problems for practice
Problem 1: Find .
Choose (a) (b)
(c)
(d)
Solution 1:
Let . Hence,
. Differentiating both sides w.r.t. x, we get the following:
But,
Hence, the answer is . Option c.
Problem 2: Find if
Choose (a) (b)
(c)
(d)
Solution 2:
The given equation is . Differentiating both sides wrt x,
is the answer. Option D.
Problem 3: If then
is
choose (a) (b)
(c)
(d)
Solution 3:
Given that so that we have
so now differentiating both sides w.r.t. x,
Now, we also know that
But, note that by laws of logarithms, on simplification, we get
and
so that on squaring, we get
so that now we get
, which all put together simplifies to
so that the answer is option C.
Problem 4: Find
Choose option (a) (b)
(c)
(d)
Solution 4:
Let us consider the first differential. Let us substitute . Hence,
and so we
, and so also, we get
so we get
required derivative
. Answer is option C.
Problem 5: Find
Choose option (a) zero (b) 26 (c) 26! (d) does not exist
Solution 5: the expression also includes a term so that the final answer is zero only.
Problem 6: Find .
Solution 6: Let
so
so so that differentiating both sides w.r.t. x, we get
we get
we get
we get
so the answer is option B.
Choose option (a): (b)
(c)
(d) none of these
Problem 7:
Find
Choose option (a) (b)
(c)
(d)
Solution 7: Let so that taking logarithm of both sides
so that
. Differentiating both sides w.r.t.x we get:
so that we get now
$latex\frac{1}{y(\log{y})} \times \frac{dy}{dx} = 1 + \log{x} $
so we get option a as the answer.
Problem 8:
Find
Choose option (a): (b)
(c)
(d) none of these.
Solution 8:
let taking logarithm of both sides we get
and now differentiating both sides w.r.t.x, we get
and now let
and again take logarithm of both sides so that we get (this is quite a classic example…worth memorizing and applying wherever it arises):
The answer is option C.
Problem 9:
Find .
Choose option (a): (b)
(c)
(d) none of these
Solution 9:
Given that
Remark: Simply multpilying out thinking the symmetry will simplify itself is going to lead to a mess…because there will be no cancellation of terms …:-) The way out is a simple algebra observation…this is why we should never ever forget the fundamentals of our foundation math:-)
note that the above can be re written as follows:
Now, we are happy like little children because many terms cancel out 🙂 hahaha…lol 🙂
and now differentiating both sides w.r.t.x we get
The answer is option A.
Problem 10:
If and
then find the value of
at
Choose option (a): (b)
(c)
(d)
Solution 10:
Answer is option D.
Regards,
Nalin Pithwa.
A problem of log, GP and HP…
Question: If and
, pyrove that:
Solution: This is same as proving: y is Harmonic Mean (HM) of x and z;
That is, to prove that is the same as the proof for :
Now, it is given that —– I
and —– II
Let say. By definition of logarithm,
;
;
;
;
.
Now let us see what happens to the following two algebraic entities, namely, and
;
Now, …call this III
Now,
Hence, ….equation IV
but it is also given that …see equation II
Hence,
Take log of above both sides w.r.t. base N:
So, above is equivalent to
But now see relations III and IV:
Hence,
Hence,
Hence, as desired.
Regards,
Nalin Pithwa
Express a given integral number in any scale (radix)
Several scales (radix) have been used by mathematicians. Binary (2), Ternary (3), Quaternary (4), Quinary (5), Senary (6), Septenary (7), Octenary(8), Nonary (9), Denary (10/Decimal), Undenary(11), Duodenary (12) and of course, hexadecimal (16). Note that in any scale the base radix is “10”. Thus, “10” stands for 2 in “binary”, “ten” in “decimal”, 8 for “octal” radix respectively, etc.
Let the digits used in a proposed scale(radix r) be . Let us express an integer in this scale. Let
be unit’s digits. Analagous to the place value system (in decimal):
Now, let us say we want to express this number N in terms of these digits s.
Dividing N by , we get the unit’s digit
as the remainder; and the quotient is:
.
Dividing the above quotient by r, we get as the remainder and the quotient as:
, and so on.
Example: Express the denary number 5213 in the scale of seven.
Solution: gives 5 as remainder and
as quotient.
gives 2 as remainder and
as remainder.
Continuing this way, we are able to express:
. That is
. You can check the equivalence by converting both to decimal values.
Cheers,
Nalin Pithwa.
STEM for Australia
Derivatives: Part 10: IITJEE maths tutorial problems for practice
Problem 1: If , and
, then
is equal to:
(a) (b)
(c)
(d)
Problem 2: If , and
, then
is equal to:
(a) (b)
(c)
(d)
Problem 3: is equal to:
(a) (b)
(c) cosec(x) (d)
Problem 4: , then
is:
(a) (b)
(c)
(d)
Problem 5: is equal to:
(a) (b)
(c)
(d)
Problem 6: then
is equal to :
(a) (b)
(c)
(d)
Problem 7: If then
is
(a) (b)
(c)
(d)
Problem 8: is:
(a) (b)
(c) (d)
Problem 9: If then
is:
(a) (b) 0 (c) 1 (d)
Problem 10: is:
(a) (b)
(c)
(d)
Regards,
Nalin Pithwa.
Derivatives: Part 9: IITJEE maths tutorial problems practice
Problem 1: is equal to:
(a) (b)
(c) (d)
Problem 2: is equal to:
(a) (b)
(c)
(d)
Problem 3: If where
, then
is given by :
(a) (b)
(c)
(d)
Problem 4: is equal to:
(a) (b)
(c)
(d)
Problem 5:
If , and
, then
is equal to:
(a) (b)
(c)
(d)
Problem 6: is equal to:
(a) (b)
(c)
(d)
Problem 7: is equal to
(a) 0 (b) (c)
(d)
Problem 8: If , then
is equal to:
(a) (b)
(c)
(d)
Problem 9: is equal to:
(a) (b)
(c)
9d)
Problem 10: If then
is equal to:
(a) (b)
(c)
(d)
Cheers,
Nalin Pithwa
Derivatives: part 8: IITJEE mains tutorial problems practice
Problem 1: If , then
is equal to:
(a) (b) -1 (c) 0 (d) b
Problem 2: If , then
is:
(a) (b)
(c)
(d)
Problem 3: is equal to:
(a) (b)
(c)
(d)
Problem 4: If , then
is equal to:
(a) (b)
(c)
(d)
Problem 5: is equal to:
(a) (b)
(c)
(d)
Problem 6: If then the value of
is
(a) (b)
(c)
(d)
Problem 7: is equal to:
(a) (b)
(c)
(d) none
Problem 8: is equal to:
(a) (b)
(c) (d)
Problem 9: If and
, then
is :
(a) (b)
(c)
(d)
Problem 10: is equal to:
(a) (b)
(c)
(d)
Cheers,
Nalin Pithwa.