∫(x−2)dxx4−8x3+18x2−8x+1=f(x)+C (x>2+3)and f(4)=0 then which is/are CORRECT? (C is constant of integration)

x4−8x3+18x2−8x+1 =x4−8x3+16x2+2x2−8x+1 =x2(x−4)+2(x2−4x)+1 =(x2−4x)2+2(x2−4x)+1=(x2−4x+1)2 ∫(x−2)dx(x2−4x+1)=12∫(2x−4)x2−4x+1 =12ln|x2−4x+1|+C

∫(x−2)dxx4−8x3+18x2−8x+1=f(x)+C (x>2+3)and f(4)=0 then which is/are CORRECT? (C is constant of integration)

x4−8x3+18x2−8x+1 =x4−8x3+16x2+2x2−8x+1 =x2(x−4)+2(x2−4x)+1 =(x2−4x)2+2(x2−4x)+1=(x2−4x+1)2 ∫(x−2)dx(x2−4x+1)=12∫(2x−4)x2−4x+1 =12ln|x2−4x+1|+C

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$\int \frac{(x - 2) d x}{\sqrt{x^{4} - 8 x^{3} + 18 x^{2} - 8 x + 1}} = f (x) + C (x > 2 + \sqrt{3})$ and $f (4) = 0$ then which is/are CORRECT? (C is constant of integration)

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Number of solutions of $f (x) = \sqrt{x}$ is 1

Number of solutions of $f (x) = \sqrt{1 - x^{2}}$ is 2

Number of solutions of $f (x) = x^{2}$ is 2

Number of solutions of $f (x) = x^{\frac{1}{3}}$ is 2

answer is D.

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Detailed Solution

$x^{4} - 8 x^{3} + 18 x^{2} - 8 x + 1 = x^{4} - 8 x^{3} + 16 x^{2} + 2 x^{2} - 8 x + 1 = x^{2} (x - 4) + 2 (x^{2} - 4 x) + 1 = {(x^{2} - 4 x)}^{2} + 2 (x^{2} - 4 x) + 1 = {(x^{2} - 4 x + 1)}^{2} \int \frac{(x - 2) d x}{(x^{2} - 4 x + 1)} = \frac{1}{2} \int \frac{(2 x - 4)}{x^{2} - 4 x + 1} = \frac{1}{2} \ln | x^{2} - 4 x + 1 | + C$