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Sequences and Series question

2022 · 24 Jun · Shift 2 · Q25
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Sequences and Series question

2022 · 24 Jun · Shift 2 · Q25

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let x, y > 0. If x3y2 = 215, then the least value of 3x + 2y is
  1. A
    30
  2. B
    32
  3. C
    36
  4. D
    40
View written solutionFree

Correct answer: D

  1. Interpret the condition

Given x3y2=215,x>0, y>0x^3y^2 = 2^{15}, \qquad x>0,\ y>0x3y2=215,x>0, y>0 and we need the least value of 3x+2y.3x+2y.3x+2y.

  1. Use AM-GM inequality

Notice that x+x+x+y+y=3x+2y.x+x+x+y+y = 3x+2y.x+x+x+y+y=3x+2y.

By AM-GM, for the five positive numbers x,x,x,y,yx,x,x,y,yx,x,x,y,y, x+x+x+y+y5≥x⋅x⋅x⋅y⋅y5=x3y25.\frac{x+x+x+y+y}{5} \ge \sqrt[5]{x\cdot x\cdot x\cdot y\cdot y} = \sqrt[5]{x^3y^2}.5x+x+x+y+y​≥5x⋅x⋅x⋅y⋅y​=5x3y2​.

So, 3x+2y5≥2155=23=8.\frac{3x+2y}{5} \ge \sqrt[5]{2^{15}} = 2^3 = 8.53x+2y​≥5215​=23=8.

Hence, 3x+2y≥5⋅8=40.3x+2y \ge 5\cdot 8 = 40.3x+2y≥5⋅8=40.

  1. Check when equality occurs

Equality in AM-GM holds when all five numbers are equal: x=x=x=y=y,x=x=x=y=y,x=x=x=y=y, so x=y.x=y.x=y.

Using the constraint, x3y2=x5=215  ⟹  x=23=8.x^3y^2 = x^5 = 2^{15} \implies x=2^3=8.x3y2=x5=215⟹x=23=8. Thus, y=8.y=8.y=8.

Then 3x+2y=3(8)+2(8)=24+16=40.3x+2y = 3(8)+2(8)=24+16=40.3x+2y=3(8)+2(8)=24+16=40.

  1. Conclusion

The least value of 3x+2y3x+2y3x+2y is 40.\boxed{40}.40​.

  1. Option check
  • A: 303030 — incorrect
  • B: 323232 — incorrect
  • C: 363636 — incorrect
  • D: 404040 — correct

So the correct option is D.

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