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Example- 13

For any real x, show that the value of the expression $$y = 2\left( {k - x} \right)\left( {x + \sqrt {{x^2} + {k^2}} } \right)$$ cannot exceed $$2{k^2}$$

Solution: We have upon rearranging:

\begin{align}&\qquad x + \sqrt {{x^2} + {k^2}} = \frac{y}{{2\left( {k - x} \right)}} \ldots (1)\\ &\Rightarrow \frac{{{k^2}}}{{\sqrt {{x^2} + {k^2}} - x}} = \frac{y}{{2\left( {k - x} \right)}}\end{align}

$\Rightarrow \sqrt {{x^2} + {k^2}} - x = \frac{{2{k^2}\left( {k - y} \right)}}{x} \ldots (2)$

By (1) – (2) and subsequent rearrangement, we can form a quadratic in x:

$4\left( {{k^2} - y} \right){x^2} - 4k\left( {2{k^2} - y} \right)x + \left( {4{k^4} - {y^2}} \right) = 0 \ldots (3)$

Now, we impose the condition that the roots of (3) must be real:

$\begin{array}{l}16{k^2}{\left( {2{k^2} - y} \right)^2} \ge 16\left( {{k^2} - y} \right)\left( {4{k^4} - {y^2}} \right)\\ \Rightarrow {y^2}\left( {y - 2{k^2}} \right) \le 0\\ \Rightarrow y \le 2{k^2}\end{array}$

Therefore, the maximum value of y is $$2{k^2}$$

Example- 14

Find the real values of a for which the equation

${\left( {{{\tan }^2}\theta + 1} \right)^2} + 4a\tan \theta \left( {{{\tan }^2}\theta + 1} \right) + 16{\tan ^2}\theta = 0$

has four distinct real roots in \begin{align}&\left( {0,\;\frac{\pi }{2}} \right)\end{align} .

Solution: Dividing throughout by $${\tan ^2}\theta$$ , we have

${\left( {\tan \theta + \cot \theta } \right)^2} + 4a\left( {\tan \theta + \cot \theta } \right) + 16 = 0$

Using $$\tan \theta + \cot \theta = y$$ ,

${y^2} + 4ay + 16 = 0 \Rightarrow y = \tan \theta + \cot \theta = - 2\left( {a \pm \sqrt {{a^2} - 4} } \right) \ldots (1)$

Now, let $$a \pm \sqrt {{a^2} - 4}$$ be represented as b. Thus,

$y = \tan \theta + \frac{1}{{\tan \theta }} = - 2b \Rightarrow \tan \theta = - b \pm \sqrt {{b^2} - 1} \ldots (2)$

For real roots, we have

From (1) : $${a^2} > 4 \Rightarrow a < - 2\;{\rm{or}}\;a > 2$$

From (2) : $${b^2} > 1 \Rightarrow {\left( {a \pm \sqrt {{a^2} - 4} } \right)^2} > 1$$

$\Rightarrow - \frac{5}{2} < a < \frac{5}{2}$ (Verify!)

Thus, \begin{align}&a \in \left( { - \frac{5}{2}, - 2} \right) \cup \left( {2,\;\frac{5}{2}} \right)\end{align}

However, we have not paid attention to one particular fact: that we need four distinct roots in \begin{align}&\left( {0,\;\frac{\pi }{2}} \right)\end{align} . For that, we need to take only negative values of a, that is, \begin{align}& a \in \left( { - \frac{5}{2}, - 2} \right)\end{align} . The justification of why this should be so is left to the reader as an exercise.

Example- 15

If $${x^2} - 10ax - 11b = 0$$ has roots c and d, and $${x^2} - 10cx - 11d = 0$$ has roots a and b, find$$a + b + c + d$$.

Solution: We have

$\begin{array}{l}a + b = 10c,\;\;c + d = 10a \Rightarrow \left( {a - c} \right) + \left( {b - d} \right) = 10\left( {c - a} \right)\\ \Rightarrow \left( {b - d} \right) = 11\left( {c - a} \right)\end{array}$

Now, since c is a root of $${x^2} - 10ax - 11b = 0$$, we have

 ${c^2} - 10ac - 11b = 0 \ldots (1)$

Similarly

${a^2} - 10ac - 11d = 0 \ldots (2)$

By (1) - (2),

\begin{align}&\qquad \left( {{c^2} - {a^2}} \right) = 11\left( {b - d} \right)\\ &\Rightarrow c + a = \frac{{11\left( {b - d} \right)}}{{c - a}} = 121\\ &\Rightarrow a + b + c + d = 10\left( {a + c} \right) = 1210\end{align}

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