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Mathematics Question on Quadratic Equations

If [t] denotes the greatest integer ≤ t, then the value of ∫01\int\limits_{0}^{1}[2x−|3x23x^2−5x+2|+1]dx is

A

37+13−46\frac{\sqrt37+\sqrt13-4}{6}

B

37−13−46\frac{\sqrt37-\sqrt13-4}{6}

C

−37−13+46\frac{\sqrt-37-\sqrt13+4}{6}

D

−37+13+46\frac{-\sqrt37+\sqrt13+4}{6}

Answer

37+13−46\frac{\sqrt37+\sqrt13-4}{6}

Explanation

Solution

I=∫01\int\limits_{0}^{1}[2x−|3x23x^2–5x+2|+1]dx

I=\int\limits_{0}^{2/3}$$[\underbrace{ -3x^2+7x-2 }_{I1}]dx+∫2/31\int\limits_{2/3}^{1}[[3x2−3x+2⏟I2][\underbrace{ 3x^2-3x+2 }_{I2}]

I1I_1=∫0t1\int\limits_{0}^{t_1}(−2)dx+∫t11/3\int\limits_{t_1}^{1/3}(−1)dx+∫1/3t2\int\limits_{1/3}^{t_2}0.dx+∫t22/3\int\limits_{t_2}^{2/3}dx
=−t1−t2-t_1-t_2+13\frac{1}{3},

Where, t1=7−376t_1=\frac{7-\sqrt37}{6}, and t2=7−136t_2 = \frac{7-\sqrt13}{6}

I2I_2=∫2/311dx=13\int\limits_{2/3}^{1}1dx=\frac{1}{3}
∴ I=13−t1−t2+13+1I = \frac{1}{3}-t_1-t_2+\frac{1}{3}+1
=53−[7−376+7−136]\frac{5}{3} - [\frac{7-\sqrt37}{6}+\frac{7-\sqrt13}{6}]

=37+13−46=\frac{\sqrt37+\sqrt13-4}{6}
So, the correct option is (A): 37+13−46\frac{\sqrt37+\sqrt13-4}{6}