yup..F(x)..always remained x all thru out the question..
what if we differentiate this function (dy/dx)..... if we do so we r getting x=1 and x=-1.....and then if we find d2y/dx2 then we can see that for x=1 we r getting the maximum value (i.e d2y/dx2
dude its x^2
bilas Sayswhat if we differentiate this function (dy/dx)..... if we do so we r getting x=1 and x=-1.....and then if we find d2y/dx2 then we can see that for x=1 we r getting the maximum value (i.e d2y/dx2
You actually differentiated it with that long formula of differentiation of f(u)/f(v)..and then got the answer..??
what if we differentiate this function (dy/dx)..... if we do so we r getting x=1 and x=-1.....and then if we find d2y/dx2 then we can see that for x=1 we r getting the maximum value (i.e d2y/dx2
If f(x) is a function satisfying f(x) . f(1/x) = f(x) + f(1/x)
and f(4)=65, what will be the value of f(6)?
a) 37
b) 217
c) 64
d) none of these
How to solve this??
and f(4)=65, what will be the value of f(6)?
a) 37
b) 217
c) 64
d) none of these
How to solve this??
At what integral value of x will the function (x2 + 3x + 1)/(x2 - 3x + 1) attain its maximum value??For this type of questions, First divide the expression
a) 3
b) 4
c) -3
d) none of this
it wil be
f(x) = 1 + 6x/(x^2 -3x +1)
Now, differentiate and equate to zero
f'(x) = 0
x(2x-3) = x^2 -3x + 1
=> x^2 = 1
so x = +/-1
Its none of these as per conventional method, but from the options we can choose option 1.
If f(x) is a function satisfying f(x) . f(1/x) = f(x) + f(1/x)
and f(4)=65, what will be the value of f(6)?
a) 37
b) 217
c) 64
d) none of these
How to solve this??
Note that f(x).f(1/x) = f(x) + f(1/x) satisfies for x^n + 1 , where n is an integer.
here, f(4) = 65 = 4^3 + 1
so, f(6) = 6^3 + 1 = 217
If f(x) is a function satisfying f(x) . f(1/x) = f(x) + f(1/x)
and f(4)=65, what will be the value of f(6)?
a) 37
b) 217
c) 64
d) none of these
How to solve this??
lets see this....this function is satisfied by x^n+1....
and we have f(4)=65 which is clearly 4^3+1
now for f(6)..applying the same rule the answer i hope is 217..
how did u derive x^n + 1 ??
bilas Sayshow did u derive x^n + 1 ??
how did u derive x^n + 1 ??
bilas Sayshow did u derive x^n + 1 ??
I dont know y m i posting this..but i got into one more method(you can call it a fluke although)....put f(4)=65 in the given equation and find out f(1/4)..
that is 65/64....that makes it as f(1/4)=65/64.....that is x^3+1/x^3..
now it will be 217/216 for f(1/6)....keep this value in the equation to get f(6) as 217..
altenatively u can directly get the value of f(1/X)...by manipulation in the given equation and then directly proceed and get the answer..
again call it a fluke or backtracking..
Number systems LOD 3 Q-45
there are 25 catalyst tests. each with a DI section having 50 questions out of which 35 questions are exactly unique. find the maximum number of questions prepared.
i solved it like
35*25 = 875
15*12=180 (12 assuming the other 15 Q repeat in another test so 24 and the last test will have repeated questions from this 180)
--------
1055
but the answer is 15 more 1070 can anybody help me out
Number systems LOD 3 Q-45
there are 25 catalyst tests. each with a DI section having 50 questions out of which 35 questions are exactly unique. find the maximum number of questions prepared.
i solved it like
35*25 = 875
15*12=180 (12 assuming the other 15 Q repeat in another test so 24 and the last test will have repeated questions from this 180)
--------
1055
but the answer is 15 more 1070 can anybody help me out
There will be 50 questions in all
35*25 = 875 unique questions.
875 + 15 = 890 is the minimum number of questions.
875 + 12*15 = 1055 is the maximum number of questions.
we can't repeat 13 sets in next 12 sets, but we can repeat 12 sets in next 13 sets, so it should be 12*15
There will be 50 questions in all
35*25 = 875 unique questions.
875 + 15 = 890 is the minimum number of questions.
875 + 12*15 = 1055 is the maximum number of questions.
we can't repeat 13 sets in next 12 sets, but we can repeat 12 sets in next 13 sets, so it should be 12*15
ya even i put it in the same way but the naswer mentioned is 1070 that is why i had a doubt
any1 did the progressions of arun sharma ? have not been able to find solutions to the following series sum type questions 😃 can anybody help me with these by explaining h ow to solve these ? :)
1)The sum of the series represented as
1/(1*5) + 1/(5*9) + 1/(9*13) + 1/(221*225) is
a) 28/221 b)56/221 c) 56/225 d) none of these
2) The sum of the series
1/(sqrt(2) + sqrt(1)) + 1/(sqrt(2) + sqrt(3)) + ........ + 1/(sqrt(120) + sqrt(121)) is :
a)10 b)11 c) 12 d) none of these
3)Find the infinite sum of the series 1/1 + 1/3 + 1/6 + 1/10 + 1/15 ....
a)2 b)2.25 c)3 d)4
4)The sum of the series 1/2 + 1/6 + 1/12 + 1/20 + ....1/156 + 1/182 is
a)12/13 b)13/14 c) 14/13 d) none of these
5) for the above qstn,what is the sum of the series if taken to infinite terms
a)1.1 b)1 c)14/13 d)none of these
any1 did the progressions of arun sharma ? have not been able to find solutions to the following series sum type questions 😃 can anybody help me with these by explaining h ow to solve these ? :)
1)The sum of the series represented as
1/(1*5) + 1/(5*9) + 1/(9*13) + 1/(221*225) is
a) 28/221 b)56/221 c) 56/225 d) none of these
2) The sum of the series
1/(sqrt(2) + sqrt(1)) + 1/(sqrt(2) + sqrt(3)) + ........ + 1/(sqrt(120) + sqrt(121)) is :
a)10 b)11 c) 12 d) none of these
3)Find the infinite sum of the series 1/1 + 1/3 + 1/6 + 1/10 + 1/15 ....
a)2 b)2.25 c)3 d)4
4)The sum of the series 1/2 + 1/6 + 1/12 + 1/20 + ....1/156 + 1/182 is
a)12/13 b)13/14 c) 14/13 d) none of these
5) for the above qstn,what is the sum of the series if taken to infinite terms
a)1.1 b)1 c)14/13 d)none of these
again there are certain LOD2 and LOD3 qstns to which i ve not been able to find solutions :
1)rohit drew a rectangular grid of 529 cells ,arranged in 23 rows and 23 columns and filled each cell with a number.The numbers with which he filled each cell were such that the numbers of each row taken from left to right formed an arithmetic series and the numbers of each column taken from top to bottom formed an arithmetic series.The seventh and the seventeeth numbers of the fifth row were 47 and 63 respectively while the seventh and seventeenth numbers of the fifteenth row were 53 and 77 respectively.What is the sum of all the numbers in the grid ?
a)32798 b)65596 c)52900 d) none of these
2)At burger king - a famous fast food centre on main street in pune,burgers are made only on an automatic burger making machine.The machine continuously makes different sorts of burgers by ading different sorts of fillings on a common bread.The machine makes the burgers at the rate of 1 burger per half a minute.The various fillings are added to the burgers in the following manner.The 1 st,5 th,9 th.... burgers are filled with a chicken patty;the 2nd, 9th,16th ....burgers with vegetable patty;the 1st,5th,9th .....burgers with mushroom patty and the rest with plain cheese and tomato fillings.
the machine makes exactly 660 burgers per day
How many burgers per day are made with the cheese and the tomato as fillings ?
a)424 b)236 c) 237 d) none of these
how many burgers are made with all the three fillings chicken,vegetable and mushroom ?
a)23 b)24 c)25 d)26
3) product of the fourth and the fifth term of an AP is 456.Division of the 9 th term by the 4 th term of the progression gives quotient as 11 and the remainder as 10.Find the first term of the progression .
a)-52 b) -42 c)-56 d)-66 e) -50
4) sum to n terms of the series log m + log m^2/n + log m^3/n^2 + log m^4/n^3 ... is
5) one side of the staircase is to be closed in by rectangular planks from the floor to each step.The wudth of each palnk is 9 inches and their height are successively 6 inches,12 inches,8 inches and so on... There are 24 planks required in total.FInd the area in square feet.
a)112.5 b) 107 c)118.5 d) 105
6) the sum of the squares of the 5 th and the 11 th term of an AP is 3 and the product of the second and the fourteenth term is equal to P.Find the product of the 1 st and 15 th term of the AP
a) (58P - 39)/45 b) (98P + 39)/72 c) (116P - 39)/90 d) ( 98P + 39)/90
any1 did the progressions of arun sharma ? have not been able to find solutions to the following series sum type questions 😃 can anybody help me with these by explaining h ow to solve these ? :)
1)The sum of the series represented as
1/(1*5) + 1/(5*9) + 1/(9*13) + 1/(221*225) is
a) 28/221 b)56/221 c) 56/225 d) none of these
2) The sum of the series
1/(sqrt(2) + sqrt(1)) + 1/(sqrt(2) + sqrt(3)) + ........ + 1/(sqrt(120) + sqrt(121)) is :
a)10 b)11 c) 12 d) none of these
1)1/4
1/4
1/4
1/4
56/225
option (C)
2)1 can be written as (sqrt(2)+sqrt(1))(sqrt(2)+sqrt(1)) and so on for all terms.
so,we get.
sqrt(2)-sqrt(1)+sqrt(3)-sqrt(2)+.....+sqrt(121)-sqrt(120)
sqrt(121)-sqrt(1)
11-1=10
option (A)
any1 did the progressions of arun sharma ? have not been able to find solutions to the following series sum type questions 😃 can anybody help me with these by explaining h ow to solve these ? :)
3)Find the infinite sum of the series 1/1 + 1/3 + 1/6 + 1/10 + 1/15 ....
a)2 b)2.25 c)3 d)4
4)The sum of the series 1/2 + 1/6 + 1/12 + 1/20 + ....1/156 + 1/182 is
a)12/13 b)13/14 c) 14/13 d) none of these
5) for the above qstn,what is the sum of the series if taken to infinite terms
a)1.1 b)1 c)14/13 d)none of these
3)1+1/3+1/6+1/10+....
1+1/(1+2)+1/(1+2+3)+....+1/
2(1/2+1/6+1/12+1/20+....1/n(n+1))
2
2
as 1/(n+1) is 1/infinity=0
answer should be 2.
option (A)
4)
1-1/14
13/14
option (B)
5)for n terms,
1-1/n
here,n->0
so,1
option (B)
any1 did the progressions of arun sharma ? have not been able to find solutions to the following series sum type questions 😃 can anybody help me with these by explaining h ow to solve these ? :)
1)The sum of the series represented as
1/(1*5) + 1/(5*9) + 1/(9*13) + 1/(221*225) is
a) 28/221 b)56/221 c) 56/225 d) none of these
Ans: The above series can be written as:
(1-1/5)+(1/5-1/9)+(1/9-1/13)+.... divided by 4
= 1-1/225 /4
=224/4*225=56/225 .Hence Option C.
2) The sum of the series
1/(sqrt(2) + sqrt(1)) + 1/(sqrt(2) + sqrt(3)) + ........ + 1/(sqrt(120) + sqrt(121)) is :
a)10 b)11 c) 12 d) none of these
Ans: Above 1/(sqrt(2) + sqrt(1)) can be written as sqrt(2) - sqrt(1)
ence the above series becomes
(sqrt(2) - sqrt(1))+(sqrt(3) - sqrt(2))+.....(sqrt(121) - sqrt(120))
=sqrt(121)-sqrt(1)
=11-1=10
Hence Option A
3)Find the infinite sum of the series 1/1 + 1/3 + 1/6 + 1/10 + 1/15 ....
a)2 b)2.25 c)3 d)4
Ans: Above series can be written as 1/1 + 1/3 + 1/6 + 1/10 + 1/15 ....
1/1 + 1/3 + 1/6 + 1/10 + 1/15 ....
Answer is A
See the solution below.
4)The sum of the series 1/2 + 1/6 + 1/12 + 1/20 + ....1/156 + 1/182 is
a)12/13 b)13/14 c) 14/13 d) none of these
Ans:
1-1/2+(1/2-1/3)+(1/3-1/4)+.......(1/13-1/14)
=1-1/14=13/14.
Hence Option B.
5) for the above qstn,what is the sum of the series if taken to infinite terms
a)1.1 b)1 c)14/13 d)none of these
Ans:
If written infinite The above sum becomes 1-1/n
as n=>inf
Answer is 1. Hene Option A.