Question

If m is the minimum value of K for which the function $f\left( x \right) = x\sqrt {kx - {x^2}}$ is increasing in the interval $\left[ {0,3} \right]$ and M is the maximum value of f in $\left[ {0,3} \right]$ when k=m, then the ordered pair (m, M) is equal to?
$(a){\text{ }}\left( {4,3\sqrt 2 } \right) \\\ (b){\text{ }}\left( {4,3\sqrt 3 } \right) \\\ (c){\text{ }}\left( {3,3\sqrt 3 } \right) \\\ (d){\text{ }}\left( {5,3\sqrt 6 } \right) \\\$

Answer 1

Hint: In this question to find out the minimum value of K, differentiate the given $f\left( x \right)$with respect to x. Then use the concept of the function to be increasing in interval; its first derivative should be greater than 0.

Complete step-by-step answer:
Given function
$f\left( x \right) = x\sqrt {kx - {x^2}}$
To find out the minimum value of k we have to differentiate the function w.r.t. x so we have,
$\dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}x\sqrt {kx - {x^2}}$
Now differentiate it according to product rule of differentiation we have,
$\dfrac{d}{{dx}}f\left( x \right) = \sqrt {kx - {x^2}} \dfrac{d}{{dx}}x + x\dfrac{d}{{dx}}\sqrt {kx - {x^2}}$

Now differentiate it we have,
$\dfrac{d}{{dx}}f\left( x \right) = \sqrt {kx - {x^2}} \left( 1 \right) + x\dfrac{1}{{2\sqrt {kx - {x^2}} }}\left( {\dfrac{d}{{dx}}\left( {kx - {x^2}} \right)} \right)$
$\dfrac{d}{{dx}}f\left( x \right) = \sqrt {kx - {x^2}} \left( 1 \right) + x\dfrac{1}{{2\sqrt {kx - {x^2}} }}\left( {k - 2x} \right)$
Now simplify the above equation we have,
$\dfrac{d}{{dx}}f\left( x \right) = \dfrac{{2\left( {kx - {x^2}} \right) + kx - 2{x^2}}}{{2\sqrt {kx - {x^2}} }} = \dfrac{{3kx - 4{x^2}}}{{2\sqrt {kx - {x^2}} }}$

Now as we know that for increasing function $\dfrac{d}{{dx}}f\left( x \right) \geqslant 0$
Therefore we have,
$\Rightarrow \dfrac{{3kx - 4{x^2}}}{{2\sqrt {kx - {x^2}} }} \geqslant 0$ ........................ (1)
Therefore,
$3kx - 4{x^2} \geqslant 0$
$\Rightarrow 3x\left( {k - \dfrac{4}{3}x} \right) \geqslant 0$
$\Rightarrow k - \dfrac{4}{3}x \geqslant 0$
$\Rightarrow k \geqslant \dfrac{4}{3}x$....................... (2)

Now again from equation (1) for condition to be satisfied the value of
$kx - {x^2} \geqslant 0$
$\Rightarrow {x^2} - kx \leqslant 0$
$\Rightarrow x\left( {x - k} \right) \leqslant 0$
Now it is given that $x \in \left[ {0,3} \right]$
$\Rightarrow x - k \leqslant 0$
Therefore for the minimum value of the function the value of x should be greater than 3.
Therefore $x \geqslant 3$
So the minimum value of x = 3, because $x \in \left[ {0,3} \right]$

Now from equation (2) we have,
$\Rightarrow k \geqslant \dfrac{4}{3} \times 3$
$\Rightarrow k \geqslant 4$
So the minimum value (m) of k is 4.
$\Rightarrow \left( {3 - 4} \right) \leqslant 0$ so it satisfies the condition.
So m = 4.

Now it is given that M is the maximum value of f in the interval [0, 3] when k = m
$\Rightarrow M = 3\sqrt {4 \times 3 - {3^2}} = 3\sqrt 3$
So the value of ordered pair (m, M) = (4, $3\sqrt 3$)
So this is the required answer.
Hence option (B) is correct.

Note: A function is said to be strictly increasing on an interval, if ${x_1} < {x_2}$ implies $f({x_1}) < f({x_2})$. We need not to be confused between strictly increasing and increasing as both are not the same. It can be seen that for an increasing function that the graph rises but can be flat at some places too, however in strictly increasing cases the graph keeps on rising and there are no flat curves.