FunctionHard
Question
Let f : (0, ∞) → R and F(x) = 
f(t)dt.
If F(x2) = x2 (1 + x), then f(4) equals
f(t)dt.If F(x2) = x2 (1 + x), then f(4) equals
Options
A.5/4
B.7
C.4
D.2
Solution
Given, F(x) = f(t)dt
By Leibnitz rule, F′(x) = f(x) ......(i)
But F(x2) = x2 (1 + x) = x2 + x3
⇒ F(x) = x + x3/2
⇒ F′(x) = 1 +
⇒ f(x) = 1 +
⇒ f(4) = 1 +
⇒ f(4) = 1 +
× 2 = 4
f(t)dtBy Leibnitz rule, F′(x) = f(x) ......(i)
But F(x2) = x2 (1 + x) = x2 + x3
⇒ F(x) = x + x3/2
⇒ F′(x) = 1 +
⇒ f(x) = 1 +
⇒ f(4) = 1 +
⇒ f(4) = 1 +
× 2 = 4Create a free account to view solution
View Solution FreeMore Function Questions
If f: R → [ - 1, 1], where f (x) = sin , (where [.] denotes the greatest integer function), then...Let f(x) be defined on [-2, 2] and is given by f(x) = and g(x) = f(|x|) + |f(x)|, then correct graph of y = g(x) will be...The domain of f(x) = , is -...The period of f(x) = sin + 2 cos - tan , is -...f(x) and g(x) are two function defined for all values of x. f(x) is an even function and g(x) is periodic function, then...