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Find for which real value x, the expression $2(k - x)\left[ {x + \sqrt {{x^2} + {k^2}} } \right]$ cannot exceed A.$4{k^2}$ B. $2{k^2}$C. $5{k^2}$D. None of these

Last updated date: 12th Aug 2024
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Hint: Suppose the given expression as y, then form quadratic equation of x. Set the discriminant for greater or equal to zero and solve to obtain the required answer.

Formula used:
The discriminant of a quadratic equation $a{x^2} + bx + c = 0$ is $D = {b^2} - 4ac$ .
For real values of x, $D \ge 0$ .
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$

Complete step by step solution:
Suppose that,
$y = 2(k - x)\left[ {x + \sqrt {{x^2} + {k^2}} } \right]$
$\dfrac{y}{{2(k - x)}} = \left[ {x + \sqrt {{x^2} + {k^2}} } \right]$
$\dfrac{y}{{2(k - x)}} - x = \sqrt {{x^2} + {k^2}}$
Square both sides of the equation,
${\left( {\dfrac{y}{{2(k - x)}} - x} \right)^2} = {\left( {\sqrt {{x^2} + {k^2}} } \right)^2}$
${x^2} - 2.x.\dfrac{y}{{2(k - x)}} + {\left( {\dfrac{y}{{2(k - x)}}} \right)^2} = {x^2} + {k^2}$
Cancel out the term ${x^2}$ from both sides,
$\dfrac{{{y^2}}}{{4{{\left( {k - x} \right)}^2}}} - \dfrac{{xy}}{{k - x}} = {k^2}$
Multiply $4{(k - x)^2}$ to both sides of the equation $\dfrac{{{y^2}}}{{4{{\left( {k - x} \right)}^2}}} - \dfrac{{xy}}{{k - x}} = {k^2}$.
$4{\left( {k - x} \right)^2}\left[ {\dfrac{{{y^2}}}{{4{{\left( {k - x} \right)}^2}}} - \dfrac{{xy}}{{k - x}}} \right] = 4{\left( {k - x} \right)^2} \times {k^2}$
${y^2} - 4xy(k - x) - 4{k^2}{(k - x)^2} = 0$
${y^2} - 4xyk + 4{x^2}y - 4{k^2}\left( {{k^2} - 2kx + {x^2}} \right) = 0$
${y^2} - 4xyk + 4{x^2}y - 4{k^4} + 8{k^3}x - 4{k^2}{x^2} = 0$
$4{x^2}(y - {k^2}) - 4kx(y - 2{k^2}) + {y^2} - 4{k^4} = 0$
This is a quadratic equation of x,
Therefore, the discriminant is greater or equal to zero for real values of x.
$16{k^2}{\left( {y - 2{k^2}} \right)^2} - 16\left( {y - {k^2}} \right)\left( {{y^2} - 4{k^4}} \right) \ge 0$
${k^2}{\left( {y - 2{k^2}} \right)^2} - \left( {y - {k^2}} \right)\left( {{y^2} - 4{k^4}} \right) \ge 0$
${k^2}\left[ {{y^2} - 4y{k^2} + 4{k^4}} \right] - \left[ {{y^3} - 4y{k^4} - {k^2}{y^2} + 4{k^6}} \right] \ge 0$
${k^2}{y^2} - 4y{k^4} + 4{k^6} - {y^3} + 4y{k^4} + {k^2}{y^2} - 4{k^6} \ge 0$
$2{k^2}{y^2} - {y^3} \ge 0$
${y^2}\left[ {2{k^2} - y} \right] \ge 0$
Now, $y \ge 0,y \le 2{k^2}$
$\Rightarrow 0 \le y \le 2{k^2}$

The correct option is B.

Note: Sometime students get confused and set the discriminant as zero and calculate to obtain the values of x, but as the question is two find the value of x for which x cannot exceed the given expression hence, we have to set the discriminant as greater equal to zero.