If f(x1 + x2 + ... + xn) = f(x1)f(x2)...f(xn) for all x belongs to real; and f(0) = 2 and  f(m) > 0 where m,n belong to natural and m is even and n is odd. 

Evaluate lim ((f(x)-1)/x) where x tends to zero.

2 years ago


Answers : (2)

                                        The answer is 1
2 years ago

Can you kindly give me the solution for the problem?

2 years ago

Post Your Answer

More Questions On Differential Calculus

Ask Experts

Have any Question? Ask Experts
Post Question
Answer ‘n’ Earn
Attractive Gift
To Win!!!
Click Here for details
Please tell how they calculate it sir please see attachment
Hello student, Please find answer to your question By binomial theorm,
Jitender Singh 3 months ago
answer is b
Ans: Hello Student, Please find answer to your question below It is given that exactly one statement is true out of three. Case I: Statement 1 is true which means f(x) = 1 & f(y) not...
Jitender Singh one month ago
i am posting this question from 5 days …..is there anybody who can help me ….….….…...
Ans: Let the resultant of vectors A & B to be C Angle b/w A & C Angle b/w B & C Or you can always say that the resultant will fall toward the vector which have more magnitude.
Jitender Singh 3 months ago
cos a + cos b – cos (a+b) = 3/2 show that a=b= 60 degrees
Let A+B+C=180 cosA+cosB+cosC = cos A+ cos B +cos A+B cosA+cos B- cosC= cos A+ cos B -cos A+B 3/2= 1-4sinA/2 sin B/2 cosC/2 4sinA/2 sin B/2 cosA+B/2=--1/2 A=b=60
Ahsan Barkati 28 days ago
If sinx+siny=root3(cosy-cosx) Prove that sin3x+sin3y=0.
Hello students, please check the solution of your question given below: sinx+siny = root3 ( cosx-cosy) 2sin(x+y/2)cos(x-y/2) = root3 {2sin(x+y/2)sin(y-x/2)} sin(x+y/2)[cos(x-y/2)-root3....
Sunil Raikwar 2 months ago
Dear student, According to the given relation, we would have: If we multiply both sides by a factor of 4, we get: This on simplifying gives us: If the right hand side can be simplified to...
Sumit Majumdar 2 months ago
if a,b,c,d are positive real numbers such that a/3=a+b/4=a+b+c/5=a+b+c+d/6 then a/b+2c+3d
Let a/3 = a + b / 4 = a + b + c / 5 = a + b + c + d/ 6 = k then, a = 3k, b = k, c = k, d = k a/b + 2c + 3d = 3k/k + 2k + 3k = 3k/6k = 1/2
Y RAJYALAKSHMI 2 months ago
hey post raledu kada
kasilaxmi 2 months ago
hello chandu.....
farzana 2 months ago
View all Questions »