1)

For every pair of continuous function f,g:[0,1] → R such that max { f(x):xε [0,1]}=max {g(x): x ε [0,1]}.

The correct statement(s) is (are)


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

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

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

D) $[f(c)]^{2}=[g(c)]^{2}$ for some $c\epsilon [0,1]$

Answer:

Option A,D

Explanation:

 Plan if a continuous  function has values of opposite sign inside an interval, then it has a root in that  interval

 f,g:[0,1] → R

 We take two cases.

 Let f and g attain their common maximum value at p

   $\Rightarrow$ f(p)=g(p)

 where $p\in [0,1]$

 Let  f and g attain their common maximum value at different points

$\Rightarrow$            f(a)=M   and g(b)= M

$\Rightarrow$      f(a)-g(a)  >0

 and f(b)-g(b) <0

$\Rightarrow$   f(c)-g(c)=0 for some   $c \in [0,1]$ as f and g are continuous functions.

$\Rightarrow$   f(c)-g(c)=0 for some    $c \in [0,1]$ for all cases.              ........(i)

 Option

 (a)  $\Rightarrow$   f2(c)-g2(c)+3[f(c)-g(c)]=0

  which is true from Eq.(i).

 Option (d)  $\Rightarrow$   f2(c)-g2 c)=0 which is true from Eq.(i)

 Now, if we take 

f(x)=1 and g(x)=1,  $\forall x \in [0,1]$

 Option (b) and (c) does not hold.

 Hence, option (a) and (d)  are correct