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Application of Derivatives question

2019 · 10 Jan · Shift 1 · Q40
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  5. /2019 · 10 Jan · Shift 1 · Q40

Application of Derivatives question

2019 · 10 Jan · Shift 1 · Q40

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
The shortest distance between the point (32,0)\left( {{3 \over 2},0} \right)(23​,0) and the curve y =x\sqrt xx​, (x > 0), is -
  1. A
    32{{\sqrt 3 } \over 2}23​​
  2. B
    54{5 \over 4}45​
  3. C
    32{3 \over 2}23​
  4. D
    52{{\sqrt 5 } \over 2}25​​
View written solutionFree

Correct answer: D

  1. Point on the curve

The curve is y=x,x>0.y=\sqrt{x}, \quad x>0.y=x​,x>0. A general point on this curve can be written as (x,x).(x,\sqrt{x}).(x,x​).

We need the shortest distance from the fixed point (32,0)\left(\frac{3}{2},0\right)(23​,0) to this point on the curve.

  1. Distance formula

Let the distance be DDD. Then D2=(x−32)2+(x−0)2.D^2=\left(x-\frac{3}{2}\right)^2+(\sqrt{x}-0)^2.D2=(x−23​)2+(x​−0)2. So, D2=(x−32)2+x.D^2=\left(x-\frac{3}{2}\right)^2+x.D2=(x−23​)2+x.

Expand: D2=x2−3x+94+x=x2−2x+94.D^2=x^2-3x+\frac{9}{4}+x=x^2-2x+\frac{9}{4}.D2=x2−3x+49​+x=x2−2x+49​.

Thus we need to minimize f(x)=x2−2x+94,x>0.f(x)=x^2-2x+\frac{9}{4}, \quad x>0.f(x)=x2−2x+49​,x>0.

  1. Minimize using derivatives

Differentiate: f′(x)=2x−2.f'(x)=2x-2.f′(x)=2x−2. Set it equal to zero: 2x−2=0  ⟹  x=1.2x-2=0 \implies x=1.2x−2=0⟹x=1.

Second derivative: f′′(x)=2>0,f''(x)=2>0,f′′(x)=2>0, so x=1x=1x=1 gives a minimum.

  1. Find the minimum distance

At x=1x=1x=1, D2=1−2+94=54.D^2=1-2+\frac{9}{4}=\frac{5}{4}.D2=1−2+49​=45​. Hence, D=54=52.D=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}.D=45​​=25​​.

  1. Match with options

52\frac{\sqrt{5}}{2}25​​ corresponds to Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So the answer agrees with the stored correct answer.

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