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Application of Derivatives question

2019 · 12 Apr · Shift 1 · Q44
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  5. /2019 · 12 Apr · Shift 1 · Q44

Application of Derivatives question

2019 · 12 Apr · Shift 1 · Q44

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
If m is the minimum value of k for which the function f(x) = x kx−x2\sqrt {kx - {x^2}}kx−x2​ is increasing in the interval [0,3] and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to :
  1. A
    (5,36)\left( {5,3\sqrt 6 } \right)(5,36​)
  2. B
    (4,33)\left( {4,3\sqrt 3 } \right)(4,33​)
  3. C
    (4,32)\left( {4,3\sqrt 2 } \right)(4,32​)
  4. D
    (3,33)\left( {3,3\sqrt 3 } \right)(3,33​)
View written solutionFree

Correct answer: B

  1. Given function

We have

f(x)=xkx−x2=xx(k−x).f(x)=x\sqrt{kx-x^2}=x\sqrt{x(k-x)}.f(x)=xkx−x2​=xx(k−x)​.

We want the minimum value of kkk such that f(x)f(x)f(x) is increasing on [0,3][0,3][0,3].

For the square root to be defined on [0,3][0,3][0,3], we need

kx−x2=x(k−x)≥0∀x∈[0,3].kx-x^2=x(k-x)\ge 0 \quad \forall x\in[0,3].kx−x2=x(k−x)≥0∀x∈[0,3].

Since x≥0x\ge 0x≥0 on this interval, this requires

k−x≥0∀x∈[0,3],k-x\ge 0 \quad \forall x\in[0,3],k−x≥0∀x∈[0,3],

so at least

k≥3.k\ge 3.k≥3.
  1. Differentiate f(x)f(x)f(x)

Let

g(x)=kx−x2.g(x)=kx-x^2.g(x)=kx−x2.

Then

f(x)=xg(x).f(x)=x\sqrt{g(x)}.f(x)=xg(x)​.

Using product rule,

f′(x)=g(x)+x⋅g′(x)2g(x).f'(x)=\sqrt{g(x)}+x\cdot \frac{g'(x)}{2\sqrt{g(x)}}.f′(x)=g(x)​+x⋅2g(x)​g′(x)​.

Now

g′(x)=k−2x.g'(x)=k-2x.g′(x)=k−2x.

So,

f′(x)=kx−x2+x⋅k−2x2kx−x2.f'(x)=\sqrt{kx-x^2}+x\cdot\frac{k-2x}{2\sqrt{kx-x^2}}.f′(x)=kx−x2​+x⋅2kx−x2​k−2x​.

Taking LCM,

f′(x)=2(kx−x2)+x(k−2x)2kx−x2.f'(x)=\frac{2(kx-x^2)+x(k-2x)}{2\sqrt{kx-x^2}}.f′(x)=2kx−x2​2(kx−x2)+x(k−2x)​.

Simplify numerator:

2kx−2x2+kx−2x2=3kx−4x2=x(3k−4x).2kx-2x^2+kx-2x^2=3kx-4x^2=x(3k-4x).2kx−2x2+kx−2x2=3kx−4x2=x(3k−4x).

Hence

f′(x)=x(3k−4x)2kx−x2.f'(x)=\frac{x(3k-4x)}{2\sqrt{kx-x^2}}.f′(x)=2kx−x2​x(3k−4x)​.
  1. Condition for increasing on [0,3][0,3][0,3]

For x∈(0,3)x\in(0,3)x∈(0,3), denominator is positive (when domain is valid), and x>0x>0x>0. Therefore the sign of f′(x)f'(x)f′(x) depends on

3k−4x.3k-4x.3k−4x.

To have f′(x)≥0f'(x)\ge 0f′(x)≥0 for all x∈[0,3]x\in[0,3]x∈[0,3], we need

3k−4x≥0∀x∈[0,3].3k-4x\ge 0 \quad \forall x\in[0,3].3k−4x≥0∀x∈[0,3].

The strongest condition occurs at x=3x=3x=3:

3k−12≥0  ⟹  k≥4.3k-12\ge 0 \implies k\ge 4.3k−12≥0⟹k≥4.

Thus the minimum value is

m=4.m=4.m=4.
  1. Now find maximum value of fff on [0,3][0,3][0,3] when k=m=4k=m=4k=m=4

Substitute k=4k=4k=4:

f(x)=x4x−x2.f(x)=x\sqrt{4x-x^2}.f(x)=x4x−x2​.

Since for k=4k=4k=4, we found f′(x)≥0f'(x)\ge 0f′(x)≥0 on [0,3][0,3][0,3], the function is increasing on [0,3][0,3][0,3]. Therefore its maximum occurs at the right endpoint x=3x=3x=3.

So,

M=f(3)=34⋅3−32=312−9=33.M=f(3)=3\sqrt{4\cdot 3-3^2}=3\sqrt{12-9}=3\sqrt{3}.M=f(3)=34⋅3−32​=312−9​=33​.
  1. Ordered pair

Therefore,

(m,M)=(4,33).(m,M)=(4,3\sqrt{3}).(m,M)=(4,33​).

This matches Option B.

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