iit jee maths
1. A particle moves according to a law of motion
s(t) = t^{3 } 12t^{2} + 36t , t ≥ 0
where, t is measured in seconds and s in meters.
a) Find the acceleration at time t and after 3 seconds.
b) Graph the position, velocity and acceleration function for 0≤t≤8
c) When is the particle speeding up? When is it slowing down?
SCRA 2013 SET D, QUES. 10
( [x]+[2x]+[3x]+.....[nx] ) / n^{2}
SCRA 2013
f(x) = x^{2}2 ; 3<x<3
QUES. Consider The Following Statements:
1. The Absolute Maxmimum Value of The Function is 2.
2. The Absolute Minimum Value of The Fuction is 0.
WHICH OF THE ABOVE STATEMNTS IS/ARE CORRECT?
why a funtion is always continue when it is differential but not always differential when it is continue?
d/dx((1+x^2+x^4)/(1+x+x^2))=ax+b
then a=?, b=?
1+x^{2}+x^{4} = (1+x+x^{2})(1x+x^{2})
now solve
HOW TO INTEGRATE "GREATEST INTEGER FUNCTION"?
Find a formula for a function g(x) satisfying the following conditions
a) domain of g is (∞ , ∞ ) b) range of g is [2 , 8] c) g has a period π d) g(2) = 3
g(x) = 35sin(2x4)
Let f(x) = x^{135} + x^{125}  x^{115 }+ x^{5} +1. If f(x) is divided by x^{3}x then the remainder is some function of x say g(x). Find the value of g(10)
for getting reminder put x^{3}= x so
x^{135} + x^{125}  x^{115 }+ x^{5} +1 will give
x^{45} + x^{41}*x^{2}  x^{38}*x +x*x^{2} +1
x^{15} + x^{13}*x^{2}*x^{2}  x^{12}*x^{2}*x +x*x^{2} +1
x^{17 }+ x^{3} +1
x^{5}*x^{2} + x + 1
x^{7 }+ x + 1
x^{2}*x + x +1
x^{3} + x +1
x+x+1
2x+1
now put x = 10
Find out for what integral values of n the number 3π is a period of the function :
f(x) = cos nx . sin(5/n)x
Suppose that f is a function such that f(cos x) = cos 17x . Which one of the following functions g has the property that g(sin x) = sin 17x .
(A) g(x) = f(√1x^{2}) (B) g(x) = f(x  π/4) (C) g(x) = √1x^{2 }(D) g(x) = f(x)
all are defined
put π in the first, factorise second and multiply the conjugate of numerator in the third
factorise numerator and denominator
How to inegrate ( In t) dt /(t1)
let lnt = x so dt/t = dx so dt = tdx = e^{x}dx
now ∫(ln t)dt/(t1) = ∫xe^{x}dx/(e^{x}1) = ∫x(e^{x}1+1)dx/(e^{x}1)
now solve
∫√tanx
let √tanx = t
so sec^{2}xdx/2√tanx = dt
so (1+t^{4})dx/2t = dt
so dx = 2tdt/(1+t^{4})
now solve
∫√(x+3)/(x+2) {x>2}
if ƒ:R−{0} > R, 2ƒ(x) − 3ƒ(1⁄x) = x² then ƒ(3)= ?
2ƒ(x) − 3ƒ(1⁄x) = x² (1)
so 2ƒ(1/x) − 3ƒ(x) = 1/x² (2)
multiply (1) to 2 and (2) to 3, then add the two equations, you will get f(x) then calculate f(3)
Sir/Madam,
Suddenly a question struck on my mind: 0×∞= 1 or 0?? as 1⁄0=∞.
it will be 0, assume it by considering that ∞ is a big number
for ex. 0*(1111111111111111111111111) = 0
FIND DOMAIN OF f(x)=log4log3 log2x WHERE 4,3,2 ARE BASES ?
f(x)= log_{4}log_{3}log_{2}x
so 0 < log_{3}log_{2}x < ∞
so 1 < log_{2}x < ∞
so 2 < x < ∞
let us consider the case of ellipse with x and y axes as their axes
eq. is x^{2}/a^{2} + y^{2}/b^{2}= 1
on differentiating we get 2x/a^{2} + 2yy^{'}/b^{2 }= 0 y'is first differential
so yy^{'}/x = b^{2}/a^{2}
again differentiate and get answer.
You should remember that you should differentiate as many times as the number of constants.
for ex. in the case of parabola only first diffrentiation is sufficient.
now complete it for all conics
using the transformation x=r cosθ and y=r sinθ,find the singular solutionof the differential equation x+py=(xy)(p^{2}+1)½ where p=dy/dx
x=rcosθ and y=rsinθ
so dx=rsinθdθ and dy = rcosθdθ
so p = cotθ
so given eq. will become xcotθy = (xy)cosecθ
on putting values of x and y we get
0 = r(cosθsinθ)/sinθ
so tanθ = 1 so θ = nπ+π/4
so x = rcos(nπ+π/4) and y = rsin(nπ+π/4)
show that the family of parabolas y^{2}=4a(x+a) is self orthogonal.
y^{2}= 4a(x+a)
so 2yy^{'} = 4a so a = yy^{'}/2
on putting a we get y^{2}= 4yy^{'}/2(x+yy^{'}/2)
so y^{2} = yy^{'} (2x+yy^{'}) or y = 2xy^{'} + yy^{'2} (1)
now on putting 1/y^{'} in the place of y^{'}
we get y^{2} = y/y^{'}[2xy/y^{'}]
so yy^{'2} = 2xy^{'}  y (2)
similarity of (1) and (2) shows that the given curve is self orthogonal
show that limit /x/ = 0 as x approaches 0.
show that limit /x/ = 0 as x approaches 0.
In a class of 78 students 41 are taking French, 22 are taking German. Of the students taking French or German, 9 are taking both courses. How many students are not enrolled in either course? A. 6
B. 15
C. 24
D. 33
E. 54
Q1. Let f be a twice differentiable function on R.Given that f''(x)>0 for all x element of R.Then which one is true and why?
a.f(x)=0 has exactly two solutions on R.
b.f(x) =0 has a positive solution if f(o)=0 and f'(0)=0
c.f(x)=0 has no positive solution if f(o)=0 and f'(o)>0.
4. f(x)=0 has no positive solution if f(0)=0 and f'(0)<0
If f be decreasing continuous function satisfying f(x+y)=f(x)+f(y) for all x,y belongs to R; f'(0)=1, then
1
∫ f(x)dx is
0
A. 1
B. 1e
C.2e
D. none of these
Solve: dy/dx =[x√(x^21) +y]/(√x^21)
∫√ e^{x}4 dx
integral root of e^{x}4 dx
if f is a real valued differentiable function satisfying If(x)  f(y)I ≤ (xy)^{2} x,yεR and f(0) =0 then f(1) equals
(a) 1
(b) 2
(c) 0
(d) 1
let g(x) be the inverse of the function f(x), and f^{/}(x) = 1/1+x^{3} then g^{/}(x) equals
(1) 1/1+g^{3}
(2) 1/1+f^{3}
(3) 1+ g^{3}
(4) 1+f^{3}
let t_{r} = r/1+r^{2}+ r^{4} then lim n>∞ ∑ t_{r}
lim n>∞ [tanx + 1/2tan(x/2) + 1/2^{2}tan(x/2^{2}) + ............. + 1/2^{n}tan(x/2^{n})]
lim n>∞ cos(x/2)cos(x/4).......cos(x/2^{n})
lt x>0 [(1+sinx)^{cosecx} e +sinx/2]/sin^{2}x
lim X> 0 (1/x^{2}  1/tan^{2}x)
x(√1+y)+y(√1+x)=0,prove that dy/dx=1/(1+x)^{2}
A bank gives an investor double the initial deposit in 5years, interest being simple interest. Then the rate of Interest is:
(A) 20
(B) 18
Sir , i have solved limx>0 (sin^{1}x  tan^{1}x )/x^{3 }. plz have a look on others.
limx>0 (sin^{1}x  tan^{1}x )/x^{3}
use D L Hospital rule or expansion of sin^{1}x and tan^{1}x to solve it
no of solutions of e^{x^2/2}  x^{2 }= 0 are
two intersecting points means two solutions
if f(X) = _{z=1}∑^{n }(x 1/z)(x  1/ 1+z ) then lim x>∞ f(0) is ___________
ans. 1
limx>0 (sin^{1}x  tan^{1}x )/x^{3 }is _________ (without using expansion of sin^{1}x and tan^{1}x )
ans. 1/2
lim x>∞ [x  x^{2} ln(1+1/x) ] is equal to
ans. 1/2
lim x>0^{+ }log_{sinx/2}sinx is equal to
a.1 b.o c. 4
ans. 1
if lim x>0 f(x) exists and is finite and non zero and lim x>∞ [ f(x) + {3f(x) 1/f^{2}(x)} ] =3 then the value of f(x) is equal to
a. 1 b. 1 c. 2
lim x>0 [ sin^{2}(π/(2ax)) ]^{[sec(π/(2bx))]^2}
lim x > (cos(x+1)^{½ }cos(x)^{½} )
lim x»infinity( ((x+1)(x+2)(x+3)(x+4))^¼  x )
multiply with the conjugate in numerator again and again you will get
lim_{x>}∞ ((x+1)(x+2)(x+3)(x+4)  x^{4})/((x+1)(x+2)(x+3)(x+4))^{1/4} + x )*((x+1)(x+2)(x+3)(x+4))^{1/2} + x^{2} )
after solving the numerator get a 3 power expression of x then take 3 power of x common, similarly take x and 2 power of x from the first and second denominator terms
best calculus books for iitjee.
its I. E. Maron
take 1,1,1,1,1,1 to make sum of the value as 37..
clue: use any operations between the numbers.
lim n> ∞ 4^n/n!
limn>∞ 4^{n}/ n!
= lim_{n>}∞ [4/1][4/2][4/3][4/4][4/5][4/6][4/7]................................[4/n]
besides first four bracket, all brackets are real numbers less than 1 so their product will be zero for n tends to infinite
a)integrate (1+x)/lnx from limit 0 to 1
A tarder buys goods at 19% off the list price he wants to get a profit 20% after allowing a discout of 10%. At what % above the list price should he marks the goods.
Answers 
1) 4%
2) 6%
3) 8%
4) none
_{ }∫dx
sin^{5}x^{ }+ cos^{5}x
since first one is a straight line so at every point only one tangent is possible so it is differentiable
but second one which is modulus has a sharp turn at x = 5/2 so two tangents at x = 5/2 are possible so is not differentiable.
DOMAIN OF log (X+4) base of log is 2
A window frame is shaped like a rectangle with an arch forming the top ( ie; a box with a semicircle on the top)
The vertical straight side is Y, the width at the base and at the widest part of the arch (Diam) is X.
I know that the perimeter is 4.
Find X if the area of the frame is at a maximum?
perimeter = 4 so 2Y + X + πX/2 = 4
and area = XY + [π(X/2)^{2}]/2
put Y from 1st equation to 2nd we get A = X[4X{1+(π/2)}]/2 + πX^{2}/8
now calculate dA/dX and then compare it to zero
X will be 8/(4+π)
y=px+a/p
this is a special type of differential equation in which p = dy/dx
its solution will be y = cx+(a/c) here c is a constant
yxp=x+yp
on applying p = dy/dx
(ydxxdy)/dx = (xdx+ydy)/dx
or ydxxdy = xdx+ydy
or dy/dx = (yx)/(y+x)
now it is homogeneous
Importance High!!!!!!!!!!!!!!!!!
Dear Sir / Madam
Myself Rajeev Shrivastava, I am putting a signs series in front of you which based on numbers from 01 to 100. Actually I want to know that what reason of behind Approx equability of signs is
(++), (+), () & (+) at end of the month or year between the below mentioned.
Let's define a kind of mapping:
0↔1
2↔3
4↔5
6↔7
8↔9
(+) means: an Even number
() means: an Odd number
I simply say: for any EO (+) it exists an OE (+) which is paired to.
And that's true since each E is mapped to a O and vice versa (1 to 1 relation where "f(f(x))=x" ) !
E.g.: 29 ↔ 83
There is 25 numbers in the EO (+) group, so 25 in OE (+).
There is 25 numbers in the EE (++) group, so 25 in OO ().
Picking a number at random between 01 and 100 inclusive is choosing equitably in EO, OE, EE or OO group. (25% for each)
So, I really need of your help to solve a puzzle please. I just am trying to make you understand what type of solution I need:
There are some results for your review:
MONTH 
DATE 
DAY 
FIRST_SERIES 
SECOND_SERIES 
FIRST_SERIES 
SECOND_SERIES 

APR 
1 
SUN 
+ 
 
 
+ 
+ 
+ 
APR 
2 
MON 
+ 
+ 
 
+ 
++ 
+ 
APR 
3 
TUE 
+ 
 
+ 
 
+ 
+ 
APR 
4 
WED 
+ 
+ 
+ 
+ 
++ 
++ 
APR 
5 
THU 
 
+ 
 
+ 
+ 
+ 
APR 
6 
FRI 
+ 
+ 
+ 
 
++ 
+ 
APR 
7 
SAT 
+ 
 
 
+ 
+ 
+ 
APR 
8 
SUN 
 
+ 
+ 
 
+ 
+ 
APR 
9 
MON 
 
+ 
 
+ 
+ 
+ 
APR 
10 
TUE 
+ 
+ 
+ 
+ 
++ 
++ 
APR 
11 
WED 
+ 
+ 
 
 
++ 
 
APR 
12 
THU 
 
+ 
 
 
+ 
 
APR 
13 
FRI 
+ 
+ 
+ 
+ 
++ 
++ 
APR 
14 
SAT 
+ 
 
+ 
 
+ 
+ 
APR 
15 
SUN 
+ 
 
+ 
 
+ 
+ 
APR 
16 
MON 
 
 
+ 
+ 
 
++ 
APR 
17 
TUE 
 
 
+ 
+ 
 
++ 
APR 
18 
WED 






APR 
19 
THU 






APR 
20 
FRI 






APR 
21 
SAT 






APR 
22 







APR 
23 







(Complete sheets for the year 2011 and 2012 are attached with this mail)
You can see that there are two times falls in a day of pair make by even (+) and odd () signs, i.e. (++), (+), () & (+) as above said.
The total of pair combination signs become approx equal to each other at the end of the month or year (every year so, I am sending you attachment to go through the year’s result).
By this analysis:
1 I want to know that what is the relation between current falling signs and past fell signs.
2 How can I come to know, what type of combination of sign would be any particular date or day.
3 As you can seen on 22^{nd} and 23^{rd} April’2012 I don’t know what combination of sign is.
4 There is surety that there is some relation between current First fall of combination of signs and past fall or First fall of combination of signs and Second fall of combination of signs.
Please help me to how come to know what type of fall may be for next day by reviewing past or first combination of signs. So reply or elaborate me with an example.
I would be very grateful to you till entire life.
Regards
RAJEEV SHRIVASTAVA
Importance High!!!!!!!!!!!!!!!!!
Dear Sir / Madam
Myself Rajeev Shrivastava, I am putting a signs series in front of you which based on numbers from 01 to 100. Actually I want to know that what reason of behind Approx equability of signs is
(++), (+), () & (+) at end of the month or year between the below mentioned.
Let's define a kind of mapping:
0↔1
2↔3
4↔5
6↔7
8↔9
(+) means: an Even number
() means: an Odd number
I simply say: for any EO (+) it exists an OE (+) which is paired to.
And that's true since each E is mapped to a O and vice versa (1 to 1 relation where "f(f(x))=x" ) !
E.g.: 29 ↔ 83
There is 25 numbers in the EO (+) group, so 25 in OE (+).
There is 25 numbers in the EE (++) group, so 25 in OO ().
Picking a number at random between 01 and 100 inclusive is choosing equitably in EO, OE, EE or OO group. (25% for each)
So, I really need of your help to solve a puzzle please. I just am trying to make you understand what type of solution I need:
There are some results for your review:
MONTH 
DATE 
DAY 
FIRST_SERIES 
SECOND_SERIES 
FIRST_SERIES 
SECOND_SERIES 

APR 
1 
SUN 
+ 
 
 
+ 
+ 
+ 
APR 
2 
MON 
+ 
+ 
 
+ 
++ 
+ 
APR 
3 
TUE 
+ 
 
+ 
 
+ 
+ 
APR 
4 
WED 
+ 
+ 
+ 
+ 
++ 
++ 
APR 
5 
THU 
 
+ 
 
+ 
+ 
+ 
APR 
6 
FRI 
+ 
+ 
+ 
 
++ 
+ 
APR 
7 
SAT 
+ 
 
 
+ 
+ 
+ 
APR 
8 
SUN 
 
+ 
+ 
 
+ 
+ 
APR 
9 
MON 
 
+ 
 
+ 
+ 
+ 
APR 
10 
TUE 
+ 
+ 
+ 
+ 
++ 
++ 
APR 
11 
WED 
+ 
+ 
 
 
++ 
 
APR 
12 
THU 
 
+ 
 
 
+ 
 
APR 
13 
FRI 
+ 
+ 
+ 
+ 
++ 
++ 
APR 
14 
SAT 
+ 
 
+ 
 
+ 
+ 
APR 
15 
SUN 
+ 
 
+ 
 
+ 
+ 
APR 
16 
MON 
 
 
+ 
+ 
 
++ 
APR 
17 
TUE 
 
 
+ 
+ 
 
++ 
APR 
18 
WED 






APR 
19 
THU 






APR 
20 
FRI 






APR 
21 
SAT 






APR 
22 







APR 
23 







(Complete sheets for the year 2011 and 2012 are attached with this mail)
You can see that there are two times falls in a day of pair make by even (+) and odd () signs, i.e. (++), (+), () & (+) as above said.
The total of pair combination signs become approx equal to each other at the end of the month or year (every year so, I am sending you attachment to go through the year’s result).
By this analysis:
1 I want to know that what is the relation between current falling signs and past fell signs.
2 How can I come to know, what type of combination of sign would be any particular date or day.
3 As you can seen on 22^{nd} and 23^{rd} April’2012 I don’t know what combination of sign is.
4 There is surety that there is some relation between current First fall of combination of signs and past fall or First fall of combination of signs and Second fall of combination of signs.
Please help me to how come to know what type of fall may be for next day by reviewing past or first combination of signs. So reply or elaborate me with an example.
I would be very grateful to you till entire life.
Regards
RAJEEV SHRIVASTAVA
If you are making the sequence on the basis of dates then you should take signs accordingly then you can make this sequence for many years
for example if you take 23 june then first sign series will be +, + for even number and  for odd number , now as conversion rule as you gave in your explanation 23 will be converted to 32 so second series will be +
similarly you can make it for any date
for example 29 december first series will be + and its conversion is 38 so next series will be +
if you are thinking something else then explain your question correctly
how to find differential eqn. of all conics whose axes coincide with coordinate axes?? tell me the eqn. of tht. conic
Parabola eq. y^{2} = 4ax ,
on differentiating this eq. with respect to x we get 2y(dy/dx) = 4a
on taking this value of 4a in eq. y^{2}= 4ax we will get the differential eq. of parabola.
similarly for ellipse x^{2}/a^{2} + y^{2}/b^{2} = 1
on differentiaing this eq. twice we will get two additional eq.
by solving these eq. get an eq. free from a and b, this will be the differential eq. of the ellipse.
similarly for hyperbola
∫_{0}π [cot^{1}x]dx
i think the question will be ∫_{0}π [cotx]dx because limits are in angular terms.
by graph given below we get answer as π/2
here in the graph same coloured shaded region are dx*(1) always, because for every similar pair [+k]+[k] = 1 , here k is a real number, so complete area = 1(π/2)
lim x tends to 0 x tan( cos^{1}x)
 what is the integration of cos^{x }2
cos^{x}2 = (cos2)^{x}
so ∫ (cos2)^{x} dx = (cos2) ^{x} / log_{e}cos2
cos^{x}2 = (cos2)^{x}
so ∫ (cos2)^{x} dx = (cos2) ^{x} / log_{e}cos2 + C formula ∫a^{x}dx = a^{x}/log_{e}a
lim ∑ ^{n}_{r=1 }1/n e^{r/n }n tends to infinity.
limn>∞ ∑_{r=1}^{n }e^{r/n}/n
This question is related to definite integration, consider 1 and divide it into n parts, upto nth part total value = r/n is equivalent to x and 1/n is dx
so question will be
_{0}∫^{1} e^{x}dx
now solve it
ABC is an isosceles triangle inscribed in a circle of radius r. If AB=AC and h is the altitude from A to BC. and P and φ denote the perimeter and area of the triangle respectively, then lim n−» 0 φ/p³ is equal to??
first draw a cirle and triange inside it , consider angle B and angle C asα so angle A = 1802α, let centre of the circle is O and D is the point in the line BC where altitude meets in the line BC, given that altitude = h and radius = r
so AB = AC = h cosecα and BC = 2 h cotα so
p = 2 h (cosecα + cotα)
and φ = 1/2 2 h cotα h = h^{2} cotαa
and in triangle OBD angle OBD = 2α  90 so cos(2α  90) = h cotα /r implies that h = 2 r sin^{2}α
now i think limit will be based on h not n
lim _{h−» 0 } φ/p³ = 1/128r
for getting solution put p and φ first then put h in terms of r, you will get an equation based on α and r,
convert lim_{h>0} to limα>0 because when h will be 0, α will also be 0
If f(x) is a monotonically increasing function " x Î R, f "(x) > 0 and f ^{1}(x) exists, then prove that å{f ^{1}(x_{i})/3} < f ^{1}({x_{1}+x_{2}+x_{3}}/3), i=1,2,3
f(x) is monotonically increasing so f '(x)>0 and f '' (x) >0 implies that increment of function increases rapidly with increase in x
These informations provide the following informations about the nature of inverse of f(x)
1) f ^{1} (x) will also be an increasing function but its rate of increase decreases with increasing x
2) for x_{1 }< x_{2} < x_{3} , å{f ^{1}(x_{i})/3} < f ^{1}({x_{1}+x_{2}+x_{3}}/3),
The same result for f(x) will be
å{f (x_{i})/3} > f ({x_{1}+x_{2}+x_{3}}/3),
Evaluate: _{0}ò^{1}1/{ (5+2x2x^{2})(1+e^{(24x)}) } dx
let I = _{0}ò^{1}1/{ (5+2x2x^{2})(1+e^{(24x)}) } dx
on using the property _{0}∫^{a} f(x) dx = _{0}∫^{a}f(ax)dx
we get 2I = _{0}ò^{1}1/(5+2x2x^{2}) dx
now solve this
Let f(x) be a real valued function not identically equal to zero such that f(x+y^{n})=f(x)+(f(y))^{n}; y is real, n is odd and n >1 and f'(0) ³ 0. Find out the value of f '(10) and f(5).
take x = 0 and y = 0 and n = 3
we get f(0) = f(0) + f(0)^{n} or f(0) = 0
now take x = 0, y = 1 and n = 3
so f(1) = f(0) + f(1)^{3}
so get f(1) = 1, other values are not excetable because f(x) be a real valued function not identically equal to zero and f'(0) ³ 0
similarly get the other values
you will get f(5) = 5
so f(x) = x generally then find f ^{' }(x) , i think it will be 1
Evaluate _{0}ò^{x} [x] dx .
integer nearst to x and less than x will be [x]
so _{0}ò^{x} [x] dx = _{0}ò^{1} 0 dx + _{1}ò^{2} 1 dx + _{2}ò^{3} 2 dx + ..................... + _{[x]1}∫^{[x] }[x]1 dx + _{[x]}∫^{x }[x] dx
now solve it
A function f : R® R satisfies f(x+y) = f(x) + f(y) for all x,y Î R and is continuous throughout the domain. If I_{1} + I_{2} + I_{3} + I_{4 }+ I_{5} = 450, where I_{n} = n._{0}ò^{n} f(x) dx. Find f(x).
according to the given condition f(nx) = n f(x)
_{0}ò^{n} f(x) dx
consider x = ny so this integration will become _{0}ò^{1} f(ny) ndy = n^{2 }_{0}ò^{1} f(y) dy = n^{2} _{0}ò^{1} f(x) dx
now by using these conditions I_{1} = 1^{3} _{0}ò^{1} f(x) dx similarly others
put these values and get the answer
Show that _{0}ò^{p} q^{3} ln sin q dq = 3p/2 _{0}ò^{p} q^{2} ln [Ö2 sin q] dq.
let I = _{0}ò^{p} q ln sin q dq
on aplying property of definite integral
I = _{0}ò^{p} (πq) ln sin q dq
so 2 I = _{0}ò^{p} π ln sin q dq
or I = π/2 _{0}ò^{p } ln sin q dq
or I = 2π/2 _{0}ò^{p/2} ln sin q dq this is due to property
similarly solve _{0}ò^{p} q^{2} ln sin q dq and then _{0}ò^{p} q^{3} ln sin q dq
finally solve the right hand side of the equation to prove LHS = RHS
f(x+y) = f(x) + f(y) + 2xy  1 " x,y. f is differentiable and f'(0) = cos a. Prove that f(x) > 0 " x Î R.
f(x+y) = f(x) + f(y) + 2xy  1
so f(0) = 1 obtain this by putting x = 0 and y = 0
now f ^{' }(x+y) = f ^{'} (x) + 2y on differentiating with respect to x
take x = 0, we get f ^{'} (y) = f ^{'}(0) + 2y
or f ' (x) = cosα + 2x
now integrate this equation within the limits 0 to x
we get f(x) = x^{2} + cosα x + 1
its descreminant is negative and coefficient of x^{2} is positive so f(x) > 0
Find area of region bounded by the curve y=[sinx+cosx] between x=0 to x=2p.
[sinx + cosx] = [√2 sin{x+(π/4)}] here [ ] is greatest integer function
its value in different interval are
1 for 0 to π/2
0 for π/2 to 3π/4
1 for 3π/4 to π
2 for π to 3π/2
1 for 3π/2 to 7π/4
0 for 7π/4 to 2π
now solve
Solve: [3Öxy + 4y  7Öy]dx + [4x  7Öxy + 5Öx]dy = 0.
[3Öxy + 4y  7Öy]dx + [4x  7Öxy + 5Öx]dy = 0
=> dy/dx = [3Öxy + 4y  7Öy] / [4x + 7Öxy  5Öx]
=> dy/dx = √y/√x [(3√x + 4√y  7) / (4√x + 7√y  5)]
=> (dy/√y)/(dx/√x) = [(3√x + 4√y  7) / (4√x + 7√y  5)]
now take √x = X + h and √y = Y + k
you will get a homogeneous equation dY/dX = (3X+4Y+3h+4k7)/(7Y4X+7k4h5)
take 3h+4k7 = 0 and 7k4h5 = 0 and solve the equation
[3Öxy + 4y  7Öy]dx + [4x  7Öxy + 5Öx]dy = 0
=> dy/dx = [3Öxy + 4y  7Öy] / [4x + 7Öxy  5Öx]
=> dy/dx = √y/√x [(3√x + 4√y  7) / (4√x + 7√y  5)]
=> (dy/√y)/(dx/√x) = [(3√x + 4√y  7) / (4√x + 7√y  5)]
now take √x = X + h and √y = Y + k
you will get a homogeneous equation dY/dX = (3X+4Y+3h+4k7)/(7Y4X+7k4h5)
take 3h+4k7 = 0 and 7k4h5 = 0 and solve the equation
dY/dX = (3X+4Y)/(7Y4X)
take Y = uX then dY/dX = u + Xdu/dX
after solving put the values of X and Y in terms of x and y,
Let g(x) be a continuous function such that _{0}ò^{1} g(t) dt = 2. Let f(x) = 1/2 _{0}ò^{x} (xt)^{2} g(t) dt then find f '(x) and hence evaluate f "(x).
f(x) = 1/2 _{0}ò^{x} (xt)^{2} g(t) dt
or f(x) = 1/2 _{0}ò^{x }x^{2 }g(t) dt + 1/2 _{0}ò^{x }t^{2 }g(t) dt  _{0}ò^{x }x t g(t) dt
or f(x) = 1/2 x^{2} _{0}ò^{x }^{ }g(t) dt + 1/2 _{0}ò^{x }t^{2 }g(t) dt  x _{0}ò^{x } t g(t) dt
now use newton leibniz formula for middle function and product rule for first and last functions for finding f^{'} (x)
similarly get f^{''} (x)
limit n tends to ∞
then
(x^n)/(n!) equals
let y = lim n>∞ (x^{n}/n!)
or y = x lim n>∞ (1/n) lim n>∞ x^{n1}/(n1)!
or y = 0
limit n tends to ∞
x{ [tan‾¹ (x+1/x+4)]  (π/4)}
I think n is printed by mistake, so considering it as x
the question will be limit x tends to ∞
x{ [tan^{1} (x+1/x+4)]  (π/4)}
consider [tan^{1} (x+1/x+4)]  (π/4) = θ
so [tan^{1} (x+1/x+4)] = (π/4)+θ
so (x+1/x+4) = tan(π/4+θ)
or x = (35tanθ)/2tanθ
so converted question will be limθ>0 (35tanθ)θ/2tanθ
solve
I think n is printed by mistake, so considering it as x
the question will be limit x tends to ∞
x{ [tan^{1} (x+1/x+4)]  (π/4)}
consider [tan^{1} (x+1/x+4)]  (π/4) = θ
so [tan^{1} (x+1/x+4)] = (π/4)+θ
so (x+1/x+4) = tan(π/4+θ)
or x = (35tanθ)/2tanθ
so converted question will be limθ>0 (35tanθ)θ/2tanθ
now we know that limθ>0 tanθ/θ = limθ>0 θ/tanθ = 1
so next line will be limθ>0 (35tanθ)/2 = 3/2
limit n tends to ∞
then
[³√(n²n³) + n ] equals
We have to calculate limit n > ∞ [{³√(n²n³) }+ n ]
Use the formula a^{3}+b^{3} = (a+b)(a^{2}+b^{2}ab) in the format
(a+b) = (a^{3}+b^{3})/(a^{2}+b^{2}ab)
here a = ³√(n²n³) and b = n
on solving we get 1/(1+1+1) = 1/3
if f(X)=xlxl then find f^{1}(x) can you please explain me the meaning and use of sgn
f(x) = xx => f(x) = x^{2} for negative real values of x and f(x) = x^{2} for positive real values of x
so f^{1}(x) = √x for negative real values and = +√x for positive real values of x
which of the following is differntiable at x=0 ?
cos(lxl)+lxl
sin(lxl)lxl
sin(x)+x is differentiable at x = 0
LHD of first = {cos0+h+0+hcos00}/0+h0 = (cosh+h1)/h = (12sin^{2}h+h1)/h = 1 (on taking limits)
now check RHD, its value will be 1
similarly in second both values are 0 so it is differentiable