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For every pair of continuous functions $f, g:[0,1] \rightarrow R$ 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 $c \in[0,1]$
$(B)$ $(f(c))^2+f(c)=(g(c))^2+3 g(c)$ 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 $c \in[0,1]$
$(B,D)$
$(B,C)$
$(A,C)$
$(A,D)$
Solution
Consider
$h(x)=f(x)-g(n) \text { Assume } \quad a < b$
$h(a)=\lambda-g(a)>0 $
$h(b)=f(b)-\lambda<0 $
$\text { else if } a > b h(a) < 0 \text { and } h(b) > 0 .$
By intermediate value theorem $\Rightarrow h(c)=0$ $\quad\quad…….(1)$
$(A)$ $\quad( f ( c ))^2+3 f ( c )=( g ( c ))^2+3 g ( c )$
$( f ( c )- g ( c ))( f ( c )+ g ( c )+3)=0$
So there exist a ' $c$ ' : $f(c)-g(c)$
from (1).
Hence A is correct.
$(D)$ Similarly $( f ( c ))^2=( g ( c ))^2$
$(f(c)-g(c))(f(c)+g(c))=0$
$\Rightarrow \quad( D )$ is correct.
$B$ \& $C$ are wrong as by counter eg
If $f(x)=g(x)=\lambda \neq 0$, then
$B \rightarrow \lambda^2+\lambda=\lambda^2+3 \lambda$ is not possible.
$C \rightarrow \lambda^2+3 \lambda=\lambda^2+\lambda$ is not possible.