Consider a differentiable `f:R to R` for which `f(1)=2 and f(x+y)=2^(x – InterviewSolution

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1.	Consider a differentiable `f:R to R` for which `f(1)=2 and f(x+y)=2^(x)f(y)+4^(y)f(x) AA x , y in R.` The minimum value of `f(x)` isA. `1`B. `-(1)/(2)`C. `-(1)/(4)`D. none of these
Answer» Correct Answer - C `f(x+y)=2^(x)f(y)+4^(y)f(x)" (i)"` Interchanging x and y, we get `f(x+y)=2^(y)f(x)+4^(x)f(y)" (ii)"` `rArr" "2^(x)f(y)+4^(y)f(x)=2^(y)f(x)+4^(x)f(y)` `rArr" "(f(x))/(4^(x)-2^(x))=(f(y))/(4^(y)-2^(y))=k` `rArr" "f(x)=k(4^(x)-2^(x))` `rArr" "f(1)=k(4-2)=2` `rArr" "k=1.` Hence, `f(x)=4^(x)-2^(x).` `f(4)=4^(4)-2^(4)=240` `f(x)=(2^(x))^(2)-2^(x)=(2^(x)-1//2)^(2)-1//4` Thus, f(x) has least value as `-1//4.`

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