JEE Advanced | 2014FunctionHard
Question
For every pair of continuous functions f, g : [0, 1] →R such that
max {f(x) : x ∈ [0, 1]} = max {g(x) : x ∈ [0, 1]}, the correct statement(s) is(are) :
max {f(x) : x ∈ [0, 1]} = max {g(x) : x ∈ [0, 1]}, the correct statement(s) is(are) :
Options
A.(f(c))2 + 3f(c) = (g(c))2 + 3g(c) for some c ∈ [0, 1]
B.(f(c))2 + f(c) = (g(c))2 + 3g(c) for some c ∈ [0, 1]
C.(f(c))2 + 3f(c) = (g(c))2 + g(c) for some c ∈ [0, 1]
D.(f(c))2 = (g(c))2 for some c ∈ [0, 1]
Solution
Let f and g be maximum at c1 and c2 respectively, c1, c2 ∈ (0, 1)
Let h(x) = f(x) - g(x)
Now h(c1) = f(c1) - g(c1) = + ve
and h(c2) = f(c2) - g(c2) = - ve
∴ h(x) = 0 has at least one root in (c1, c2)
∴ f(x) = g(x) for some x = c ∈ (c1, c2)
∴ f(c) = g(c) for some c ∈ (0, 1)
Clearly (A, D) are correct
Let h(x) = f(x) - g(x)
Now h(c1) = f(c1) - g(c1) = + ve
and h(c2) = f(c2) - g(c2) = - ve
∴ h(x) = 0 has at least one root in (c1, c2)
∴ f(x) = g(x) for some x = c ∈ (c1, c2)
∴ f(c) = g(c) for some c ∈ (0, 1)
Clearly (A, D) are correct
Create a free account to view solution
View Solution FreeMore Function Questions
The image of under the mapping f(x) = [x] + (where [ ] denotes greatest integer function)...Let f = x2 + (x ≠ 0), then f(x) equals-...If f : R → R, f(x) = tan x, then pre-image of −1 under f is -...The domain of the function f(x) = is...The range of the function f(x) = ex-e-x, is -...