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Quadratic Equation and Inequalities question

2020 · 8 Jan · Shift 2 · Q26
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  5. /2020 · 8 Jan · Shift 2 · Q26

Quadratic Equation and Inequalities question

2020 · 8 Jan · Shift 2 · Q26

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let S be the set of all real roots of the equation, 3x(3x – 1) + 2 = |3x – 1| + |3x – 2|. Then S :
  1. A
    contains exactly two elements.
  2. B
    is an empty set.
  3. C
    is a singleton.
  4. D
    contains at least four elements.
View written solutionFree

Correct answer: A

We need to solve 3x(3x−1)+2=∣3x−1∣+∣3x−2∣3x(3x-1)+2=|3x-1|+|3x-2|3x(3x−1)+2=∣3x−1∣+∣3x−2∣ and determine the set SSS of all real roots.

1. Simplify the left-hand side

3x(3x−1)+2=9x2−3x+23x(3x-1)+2=9x^2-3x+23x(3x−1)+2=9x2−3x+2

So the equation becomes 9x2−3x+2=∣3x−1∣+∣3x−2∣9x^2-3x+2=|3x-1|+|3x-2|9x2−3x+2=∣3x−1∣+∣3x−2∣

The absolute values change at

\qquad 3x-2=0 \Rightarrow x=\frac23$$ So we solve in three intervals: 1. $x<\frac13$ 2. $\frac13\le x<\frac23$ 3. $x\ge \frac23$ --- ## 2. Case I: $x<\frac13$ Then $$3x-1<0,\qquad 3x-2<0$$ so $$|3x-1|=1-3x,\qquad |3x-2|=2-3x$$ Hence RHS is $$|3x-1|+|3x-2|=(1-3x)+(2-3x)=3-6x$$ Equation becomes $$9x^2-3x+2=3-6x$$ $$9x^2+3x-1=0$$ Solve: $$x=\frac{-3\pm\sqrt{9+36}}{18}=\frac{-3\pm 3\sqrt5}{18}=\frac{-1\pm \sqrt5}{6}$$ Now check the interval $x<\frac13$: - $$\frac{-1+\sqrt5}{6}\approx 0.206<\frac13$$ valid - $$\frac{-1-\sqrt5}{6}<0<\frac13$$ valid So two roots come from this case: $$x=\frac{-1+\sqrt5}{6},\qquad x=\frac{-1-\sqrt5}{6}$$ --- ## 3. Case II: $\frac13\le x<\frac23$ Then $$3x-1\ge 0,\qquad 3x-2<0$$ so $$|3x-1|=3x-1,\qquad |3x-2|=2-3x$$ Thus RHS is $$(3x-1)+(2-3x)=1$$ Equation becomes $$9x^2-3x+2=1$$ $$9x^2-3x+1=0$$ Discriminant: $$\Delta=(-3)^2-4(9)(1)=9-36=-27<0$$ So there is **no real root** in this interval. --- ## 4. Case III: $x\ge \frac23$ Then $$3x-1\ge 0,\qquad 3x-2\ge 0$$ so $$|3x-1|=3x-1,\qquad |3x-2|=3x-2$$ Hence RHS is $$ (3x-1)+(3x-2)=6x-3$$ Equation becomes $$9x^2-3x+2=6x-3$$ $$9x^2-9x+5=0$$ Discriminant: $$\Delta=(-9)^2-4(9)(5)=81-180=-99<0$$ So there is **no real root** in this interval. --- ## 5. Collect all real roots Therefore, $$S=\left\{\frac{-1+\sqrt5}{6},\frac{-1-\sqrt5}{6}\right\}$$ So $S$ contains exactly **two** elements. Hence the correct option is: $$\boxed{\text{A}}$$ --- ## 6. Compare with stored correct answer Stored correct answer is **C** (singleton), but our solution gives **A** (exactly two elements). Therefore, I **disagree** with the stored answer. The equation clearly has two distinct real roots found in the interval $x<\frac13$, and no roots in the other intervals.
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