JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 21)
If m is the minimum value of k for which the function f(x) = x$$\sqrt {kx - {x^2}} $$ 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 :
$$\left( {5,3\sqrt 6 } \right)$$
$$\left( {4,3\sqrt 3 } \right)$$
$$\left( {4,3\sqrt 2 } \right)$$
$$\left( {3,3\sqrt 3 } \right)$$
Explanation
$$f\left( x \right) = x\sqrt {kx - {x^2}} $$
$$ \Rightarrow $$ $$f'\left( x \right) = \sqrt {kx - {x^2}} + {{(k - 2x)x} \over {2\sqrt {kx - {x^2}} }}$$
$$ \Rightarrow {{2\left( {kx - {x^2}} \right) + kx - 2{x^2}} \over {2\sqrt {kx - {x^2}} }} = {{3kx - 4{x^2}} \over {2\sqrt {kx - {x^2}} }}$$
$$ \Rightarrow {{x(3k - 4x)} \over {2\sqrt {kx - {x^2}} }}$$
Now for increasing function
for f'(x) $$ \ge $$ 0, $$\forall x \in [0,3]$$
$$ \Rightarrow kx - {x^2} \ge 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(3k - 4x) \ge 0,\,\forall x \in \left[ {0,3} \right]\,$$
$$ \Rightarrow x(x - k) \le 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(4x - 3k) \le 0,\,\forall x \in \left[ {0,3} \right]\,$$
k $$ \ge $$ 3 and k $$ \ge $$ 4 $$ \Rightarrow $$ k $$ \ge $$ 4
$$ \Rightarrow $$ m = 4
So f(x) is maximum when k = 4 then $$3\sqrt {4 \times 3 - {3^2}} = 3\sqrt 3 = M$$
$$ \therefore $$ (m, M) = (4, $$3\sqrt 3 $$)
$$ \Rightarrow $$ $$f'\left( x \right) = \sqrt {kx - {x^2}} + {{(k - 2x)x} \over {2\sqrt {kx - {x^2}} }}$$
$$ \Rightarrow {{2\left( {kx - {x^2}} \right) + kx - 2{x^2}} \over {2\sqrt {kx - {x^2}} }} = {{3kx - 4{x^2}} \over {2\sqrt {kx - {x^2}} }}$$
$$ \Rightarrow {{x(3k - 4x)} \over {2\sqrt {kx - {x^2}} }}$$
Now for increasing function
for f'(x) $$ \ge $$ 0, $$\forall x \in [0,3]$$
$$ \Rightarrow kx - {x^2} \ge 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(3k - 4x) \ge 0,\,\forall x \in \left[ {0,3} \right]\,$$
$$ \Rightarrow x(x - k) \le 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(4x - 3k) \le 0,\,\forall x \in \left[ {0,3} \right]\,$$
k $$ \ge $$ 3 and k $$ \ge $$ 4 $$ \Rightarrow $$ k $$ \ge $$ 4
$$ \Rightarrow $$ m = 4
So f(x) is maximum when k = 4 then $$3\sqrt {4 \times 3 - {3^2}} = 3\sqrt 3 = M$$
$$ \therefore $$ (m, M) = (4, $$3\sqrt 3 $$)
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