For every pair of continuous functions `f,g:[0,1]->R` such that `max{f(x):x in [0,1]}= max{g(x):x in[0,1]}` then which are the correct statementsA. `[f(c )]^(2) + 3f(c ) = [g(c )]^(2) + 3g(c )` for some ` c in[0, 1]`B. `[f(c )]^(2)+f(c ) = [g(c )]^(2) + 3g(c )` for some `c in [0, 1]`C. `[f(c )]^(2) + 3f(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]`

For every pair of continuous functions `f,g:[0,1]->R` such that `max{f

1.	For every pair of continuous functions `f,g:[0,1]->R` such that `max{f(x):x in [0,1]}= max{g(x):x in[0,1]}` then which are the correct statementsA. `[f(c )]^(2) + 3f(c ) = [g(c )]^(2) + 3g(c )` for some ` c in[0, 1]`B. `[f(c )]^(2)+f(c ) = [g(c )]^(2) + 3g(c )` for some `c in [0, 1]`C. `[f(c )]^(2) + 3f(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]`
Answer» Correct Answer - A::D PLAN If a continous function has values of opposite sign inside an interval, then it has a root in that interval. `" " f,g : [0, 1] to R` We take two cases. Case I Let f and g attain their common maximum value at p. `rArr" " f(p) = g(p)`, where ` p in [0, 1]` Case II Let f and g attain their common maximum value at different points. ` rArr" " f(a) = M and g(b) = M` ` rArr" " f(a) - g(a) gt 0 and f(b) - g(b) lt 0` ` rArr f(c ) - g(c ) = 0 " for some " c in [0, 1]` as f and g are continuous functions. ` rArr f(c ) - g(c ) = 0 " for some " c in [0, 1]` for all cases. ....(i) Option `(a) rArr f^(2)(c ) - g^(2)(c ) + 3 [f(c ) - g (c ) ] = 0` which is true from Eq. (i) . Option ` (d) rArr f^(2) (c ) - g^(2) (c ) - g^(2) (c ) = 0` which is true from Eq. (i) Now, if we take ` f (x) = 1 and g(x) = 1, AA x in [0, 1]` Options (b) and (c ) does not hold. Hence, options (a) and (d) are correct.