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Quadratic Equation and Inequalities question

2018 · Shift 0 · Q28
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  5. /2018 · Shift 0 · Q28

Quadratic Equation and Inequalities question

2018 · Shift 0 · Q28

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let S = { x∈x \inx∈ R : x≥x \gex≥ 0 and 2∣x−3∣+x(x−6)+6=02\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 02​x​−3​+x​(x​−6)+6=0}. Then S
  1. A
    contains exactly four elements
  2. B
    is an empty set
  3. C
    contains exactly one element
  4. D
    contains exactly two elements
View written solutionFree

Correct answer: D

Let t=xt=\sqrt{x}t=x​ with the condition x≥0x\ge 0x≥0, so t≥0t\ge 0t≥0.

Then the given equation becomes 2∣t−3∣+t(t−6)+6=0.2|t-3|+t(t-6)+6=0.2∣t−3∣+t(t−6)+6=0. Simplify: 2∣t−3∣+t2−6t+6=0.2|t-3|+t^2-6t+6=0.2∣t−3∣+t2−6t+6=0. So we solve t2−6t+6+2∣t−3∣=0.t^2-6t+6+2|t-3|=0.t2−6t+6+2∣t−3∣=0.

1. Split into cases using the absolute value

Case 1: t≥3t\ge 3t≥3

Then ∣t−3∣=t−3.|t-3|=t-3.∣t−3∣=t−3. So the equation becomes t2−6t+6+2(t−3)=0.t^2-6t+6+2(t-3)=0.t2−6t+6+2(t−3)=0. Simplifying, t2−6t+6+2t−6=0t^2-6t+6+2t-6=0t2−6t+6+2t−6=0 t2−4t=0t^2-4t=0t2−4t=0 t(t−4)=0.t(t-4)=0.t(t−4)=0. Thus, t=0ort=4.t=0 \quad \text{or} \quad t=4.t=0ort=4. But in this case t≥3t\ge 3t≥3, so only t=4t=4t=4 is valid.

Then x=t2=16.x=t^2=16.x=t2=16.

Case 2: t<3t<3t<3

Then ∣t−3∣=3−t.|t-3|=3-t.∣t−3∣=3−t. So the equation becomes t2−6t+6+2(3−t)=0.t^2-6t+6+2(3-t)=0.t2−6t+6+2(3−t)=0. Simplifying, t2−6t+6+6−2t=0t^2-6t+6+6-2t=0t2−6t+6+6−2t=0 t2−8t+12=0.t^2-8t+12=0.t2−8t+12=0. Factorizing, t2−8t+12=(t−2)(t−6)=0.t^2-8t+12=(t-2)(t-6)=0.t2−8t+12=(t−2)(t−6)=0. So t=2ort=6.t=2 \quad \text{or} \quad t=6.t=2ort=6. But in this case t<3t<3t<3, so only t=2t=2t=2 is valid.

Then x=t2=4.x=t^2=4.x=t2=4.

2. Form the set SSS

Thus the solutions are x=4,  16.x=4,\;16.x=4,16. Hence S={4,16}.S=\{4,16\}.S={4,16}. So SSS contains exactly two elements.

3. Check options

  • A: contains exactly four elements — false
  • B: is an empty set — false
  • C: contains exactly one element — false
  • D: contains exactly two elements — true

Therefore, the correct option is D.\boxed{\text{D}}.D​.

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