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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 Since f(x) and g(x) are continuous on [0,1]. So, they attain their maximum and minimum values in [0,1] Suppose f(x) and g(x) attains their maximum values in [0,1]. Suppose f(x) and g(x) attain their maximum values at `x_(1) and x_(2)` respectively. It is given that `f(x_(1))=g(x_(2))` Let h(x)=f(x)-g(x). Then h(x) is continuous on [0,1] such that `h(x_(1))=f(x_(1))-g(x_(1))ge 0" "[{:(,therefore,f(x_(1))=g(x_(2))ge g(x_(1))),(,therefore,f(x_(1))-g(x_(1))ge 0):}` `and h(x_(2))=f(x_(2))-g(x_(2))le 0" "[{:(,therefore,g(x_(2))=g(x_(1))ge f(x_(2))),(,therefore,f(x_(2))-g(x_(2))ge 0):}` Therefore, there exists `c in (0,1)` such that `h(c)=0 Rightarrow f(c)=g(c)` Clearly, f(c)=g(c) satisfy options (a) and (d). |
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