Domain of the function f(x) = {left[ {{{log } | vedclass

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(B) For the function $f(x) = {\left[ {{{\log }_{10}}\left( {\frac{{5x - {x^2}}}{4}} \right)} \right]^{1/2}}$ to be defined,the expression inside the s

Vedclass · Accepted Answer

$1 \le x \le 4$

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(B) For the function $f(x) = {\left[ {{{\log }_{10}}\left( {\frac{{5x - {x^2}}}{4}} \right)} \right]^{1/2}}$ to be defined,the expression inside the square root must be non-negative:${\log _{10}}\left( {\frac{{5x - {x^2}}}{4}} \right) \ge 0$Since the base of the logarithm is $10 > 1$,the inequality remains unchanged when we exponentiate:$\frac{{5x - {x^2}}}{4} \ge {10^0}$$\frac{{5x - {x^2}}}{4} \ge 1$$5x - {x^2} \ge 4$${x^2} - 5x + 4 \le 0$Factoring the quadratic expression:$(x - 1)(x - 4) \le 0$This inequality holds when $x$ lies between the roots $1$ and $4$ inclusive.Therefore,the domain is $[1, 4]$.

Domain of the function $f(x) = {\left[ {{{\log }_{10}}\left( {\frac{{5x - {x^2}}}{4}} \right)} \right]^{1/2}}$ is

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