For every pair of continuous functions $f, g:[0,1] \rightarrow \mathbb{R}$ such that \[ \max \{f(x): x \in[0,1]\}=\max \{g(x): x \in[0,1]\}, \] the correct statement(s) is(are) :

(A) $(f(c))^{2}+3 f(c)=(g(c))^{2}+3 g(c)$ for some $c \in[0,1]$; (D) $(f(c))^{2}=(g(c))^{2}$ for some $c \in[0,1]$

(B) $(f(c))^{2}+f(c)=(g(c))^{2}+3 g(c)$ for some $c \in[0,1]$

(C) $(f(c))^{2}+3 f(c)=(g(c))^{2}+g(c)$ for some $c \in[0,1]$

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Continuity and Differentiability

For every pair of continuous functions

f, g : [0, 1] \to R

such that

max {f (x) : x \in [0, 1]} = max {g (x) : x \in [0, 1]},

the correct statement(s) is(are) :

(A) $(f (c))^{2} + 3 f (c) = (g (c))^{2} + 3 g (c)$ for some $c \in [0, 1]$ ✓ Correct
(B) $(f (c))^{2} + f (c) = (g (c))^{2} + 3 g (c)$ for some $c \in [0, 1]$
(C) $(f (c))^{2} + 3 f (c) = (g (c))^{2} + g (c)$ for some $c \in [0, 1]$
(D) $(f (c))^{2} = (g (c))^{2}$ for some $c \in [0, 1]$ ✓ Correct

Answer: (A), (D)

Since

f

and

g

are continuous on

[0, 1]

and their maximum values are equal, by the intermediate value theorem applied to

h (x) = f (x) - g (x)

, there exists

c \in [0, 1]

such that

f (c) = g (c)

. When

f (c) = g (c)

, we have

(f (c))^{2} = (g (c))^{2}

, so option (D) is correct. Also,

(f (c))^{2} + 3 f (c) = (g (c))^{2} + 3 g (c)

, so option (A) is correct. Options (B) and (C) can be eliminated by taking

f (x) = g (x) = 1

for all

x

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options