Question

 

If f (x) = (1 + x)n, then the value of  

           

Solution

Correct option is

2n

 

(0) = 1, f’(x) = n (1 + x)n – 1f’’(x) = n(n – 1) (1 + x)n – 2, …, n(x) =n(n – 1) …1(1 + x)n – n. So f’(0) = nf’’(0) = n (n – 1), …, n(0) = n!. Hence the given expression is equal to  

 nC0 + nC1 + nC2 + … + nCn = 2n.     

SIMILAR QUESTIONS

Q1

The domain of the derivative of the function

           

Q2

If (0) = 0, f’(0) = 2 then the derivative of  at x = 0 is 

Q3

  

If f is differentiable for all x then 

Q4

Let f and g be differentiable function such that f’(x) = 2g(x) and g’(x) = –f(x), and let T(x) = (f (x))2 – (g(x))2. Then T’(x) is equal to

Q5

Let f be a twice differentiable function such that f’’(x) = –f(x) and f’(x) = g(x). If h’(x) = [f(x)]2 + [g(x)]2h(1) = 8 and 

h(0) = 2, then h(2) is equal to

Q6

If y2 = P(x) is a polynomial of degree 3, then    

              is equal to  

Q7

If  then the set of all points where the derivative exist is

Q8

The value of y’’ (1) if x3 – 2x2y2 + 5x + y – 5 = 0 when y(1) = 1, is equal to

Q9

If f(x, then f’(1) equals     

 

Q10

The solution set of f’(x) > g’(x) where f(x) = (1/2)52x + 1 and g(x) = 5x + 4x log 5 is