If mathop {lim }limits_{x to frac{1}{2}} frac | vedclass

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(D) Given $\mathop {\lim }\limits_{x 	o \frac{1}{2}} \frac{{a{x^2} + bx + c}}{{{{\left( {2x - 1} ight)}^2}}} = \frac{1}{2}$.Since the limit exists

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$6$

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(D) Given $\mathop {\lim }\limits_{x 	o \frac{1}{2}} \frac{{a{x^2} + bx + c}}{{{{\left( {2x - 1} ight)}^2}}} = \frac{1}{2}$.Since the limit exists and the denominator approaches $0$,the numerator must also approach $0$ at $x = \frac{1}{2}$.Thus,$a{x^2} + bx + c = k(2x - 1)^2$ for some constant $k$.Comparing the expression,we have $\frac{k(2x - 1)^2}{(2x - 1)^2} = k = \frac{1}{2}$.So,$a{x^2} + bx + c = \frac{1}{2}(2x - 1)^2 = \frac{1}{2}(4x^2 - 4x + 1) = 2x^2 - 2x + \frac{1}{2}$.Comparing coefficients,$a = 2, b = -2, c = \frac{1}{2}$.Now,we evaluate $\mathop {\lim }\limits_{x 	o 2} \frac{{(x - a)(x - b)(x - c)}}{{x - 2}} = \mathop {\lim }\limits_{x 	o 2} \frac{{(x - 2)(x - (-2))(x - \frac{1}{2})}}{{x - 2}}$.$= \mathop {\lim }\limits_{x 	o 2} (x + 2)(x - \frac{1}{2}) = (2 + 2)(2 - \frac{1}{2}) = 4 	imes \frac{3}{2} = 6$.

If $\mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{a{x^2} + bx + c}}{{{{(2x - 1)}^2}}} = \frac{1}{2}$,then $\mathop {\lim }\limits_{x \to 2} \frac{{(x - a)(x - b)(x - c)}}{{x - 2}}$ is

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