If f (x) be a continuous function for all real values of x and satisfies x2+x(f(x)−2)+23−3−3f(x)=0∀x∈R then the value of f(3) is

x2+xfx-2+23−3−3fx=0 x2−2x+23−3=(3−x)f(x)⇒f(x)=x2−2x+23−33−xf(3)=limx→3 x2−2x+23−33−x=limx→32x−2−1=23−2−1=2−23=2(1−3)

If f (x) be a continuous function for all real values of x and satisfies x2+x(f(x)−2)+23−3−3f(x)=0∀x∈R then the value of f(3) is

x2+xfx-2+23−3−3fx=0 x2−2x+23−3=(3−x)f(x)⇒f(x)=x2−2x+23−33−xf(3)=limx→3 x2−2x+23−33−x=limx→32x−2−1=23−2−1=2−23=2(1−3)

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If f (x) be a continuous function for all real values of x and satisfies $x^{2} + x (f (x) - 2) + 2 \sqrt{3} - 3 - \sqrt{3} f (x) = 0 \forall x \in R$ then the value of $f (\sqrt{3})$ is

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$2 (\sqrt{3} - 1)$

$2 \sqrt{3} - 1$

$2 (1 - \sqrt{3})$

$\sqrt{3}$

answer is D.

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Detailed Solution

$x^{2} + x (f (x) - 2) + 2 \sqrt{3} - 3 - \sqrt{3} f (x) = 0 \begin{array}{l} x^{2} - 2 x + 2 \sqrt{3} - 3 = (\sqrt{3} - x) f (x) \\ \Rightarrow f (x) = \frac{x^{2} - 2 x + 2 \sqrt{3} - 3}{\sqrt{3} - x} \\ f (\sqrt{3}) = lim_{x \to \sqrt{3}} \frac{x^{2} - 2 x + 2 \sqrt{3} - 3}{\sqrt{3} - x} = lim_{x \to \sqrt{3}} \frac{2 x - 2}{- 1} = \frac{2 \sqrt{3} - 2}{- 1} = 2 - 2 \sqrt{3} = 2 (1 - \sqrt{3}) \end{array}$