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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 number of solutions of `f(x)=2` is |
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Answer» Correct Answer - B `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)" ` `implies 2^(x) f(y)+4^(y)f(x)=2^(y)f(x)+4^(x)f(y)` `implies (f(x))/(4^(x)-2^(x))=(f(y))/(4^(y)-2^(y))=k` `implies f(x)=k(4^(x)-2^(x))` `f(1)=2` `implies 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 `-1//4` Also `4^(x)-2^(x)=2` `implies (2^(x))^(2)-2^(x)-2=0` `implies (2^(x)-2)(2^(x)+1)=0` `implies2^(x)-2=0` `implies x=1` |
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