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QJEEM19Ma93b6: JEE Main 2019 Mathematics - Class 12 JEEM19Ma93b6 Solution: If m is the minimum value of k for which the function f(x) = x…

If m is the minimum value of k for which the function f(x) = x kx−x2\sqrt {kx - {x^2}} 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 :

Multiple Choice Options

  • A (5,36)\left( {5,3\sqrt 6 } \right)
  • B (4,33)\left( {4,3\sqrt 3 } \right)
  • C (4,32)\left( {4,3\sqrt 2 } \right)
  • D (3,33)\left( {3,3\sqrt 3 } \right)

Step-by-Step Verified Solution

f(x)=xkx−x2f\left( x \right) = x\sqrt {kx - {x^2}}

⇒\Rightarrow f′(x)=kx−x2+(k−2x)x2kx−x2f'\left( x \right) = \sqrt {kx - {x^2}} + {{(k - 2x)x} \over {2\sqrt {kx - {x^2}} }}

⇒2(kx−x2)+kx−2x22kx−x2=3kx−4x22kx−x2\Rightarrow {{2\left( {kx - {x^2}} \right) + kx - 2{x^2}} \over {2\sqrt {kx - {x^2}} }} = {{3kx - 4{x^2}} \over {2\sqrt {kx - {x^2}} }}

⇒x(3k−4x)2kx−x2\Rightarrow {{x(3k - 4x)} \over {2\sqrt {kx - {x^2}} }}

Now for increasing function

for f'(x) ≥\ge 0,   ∀x∈[0,3]\forall x \in [0,3]

⇒kx−x2≥0,∀x∈[0,3]  and x(3k−4x)≥0, ∀x∈[0,3] \Rightarrow kx - {x^2} \ge 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(3k - 4x) \ge 0,\,\forall x \in \left[ {0,3} \right]\,

⇒x(x−k)≤0,∀x∈[0,3]  and x(4x−3k)≤0, ∀x∈[0,3] \Rightarrow x(x - k) \le 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(4x - 3k) \le 0,\,\forall x \in \left[ {0,3} \right]\,

k ≥\ge 3 and k ≥\ge 4 ⇒\Rightarrow k ≥\ge 4

⇒\Rightarrow m = 4

So f(x) is maximum when k = 4 then 34×3−32=33=M3\sqrt {4 \times 3 - {3^2}} = 3\sqrt 3 = M

∴\therefore (m, M) = (4, 333\sqrt 3 )

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