Let f(x) = x + 1/2. Then,the number of real v | vedclass

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(A) Given,$f(x) = x + \frac{1}{2} = \frac{2x+1}{2}$.$f(2x) = 2x + \frac{1}{2} = \frac{4x+1}{2}$.$f(4x) = 4x + \frac{1}{2} = \frac{8x+1}{2}$.Since $f(x

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(A) Given,$f(x) = x + \frac{1}{2} = \frac{2x+1}{2}$.$f(2x) = 2x + \frac{1}{2} = \frac{4x+1}{2}$.$f(4x) = 4x + \frac{1}{2} = \frac{8x+1}{2}$.Since $f(x), f(2x), f(4x)$ are in $HP$,their reciprocals $\frac{1}{f(x)}, \frac{1}{f(2x)}, \frac{1}{f(4x)}$ are in $AP$.Therefore,$\frac{2}{f(2x)} = \frac{1}{f(x)} + \frac{1}{f(4x)}$.Substituting the values: $\frac{2}{(4x+1)/2} = \frac{2}{2x+1} + \frac{2}{8x+1}$.$\frac{4}{4x+1} = 2 \left( \frac{8x+1 + 2x+1}{(2x+1)(8x+1)} \right)$.$\frac{2}{4x+1} = \frac{10x+2}{(2x+1)(8x+1)}$.$2(2x+1)(8x+1) = (4x+1)(10x+2)$.$2(16x^2 + 10x + 1) = 40x^2 + 18x + 2$.$32x^2 + 20x + 2 = 40x^2 + 18x + 2$.$8x^2 - 2x = 0$.$2x(4x - 1) = 0$.This gives $x = 0$ or $x = 1/4$.If $x = 0$,the terms are $1/2, 1/2, 1/2$,which are not unequal.If $x = 1/4$,the terms are $3/4, 1, 3/2$,which are in $HP$ (since $4/3, 1, 2/3$ are in $AP$ with common difference $-1/3$).Thus,there is only $1$ real value of $x$.

Let $f(x) = x + 1/2$. Then,the number of real values of $x$ for which the three unequal terms $f(x), f(2x), f(4x)$ are in $HP$ is

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